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(b) Prove that, for a system in thermal and diffusive equilibrium (with a reser­ voir that can supply both energy and particles),

5.3 Phase Transformations of Pure Substances A phase transformation is a discontinuous change in the properties of a sub­ stance, as its environment is changed only infinitesimally. Familiar examples in­ clude melting ice and boiling water, either of which can be accomplished with only a very small change in temperature. The different forms of the substance-in this case ice, water, and steam-are called phases. Often there is more than one variable that can affect the phase of a substance. For instance, you can condense steam either by lowering the temperature or by raising the pressure. A graph showing the equilibrium phases as a function of temperature and pressure is called a phase diagram. Figure 5.11 shows a qualitative phase diagram for H 2 0, along with some quanti­ tative data on its phase transformations. The diagram is divided into three regions, indicating the conditions under which ice, water, or steam is the most stable phase. It's important to realize, though, that 'metastable' phases can still exist; for in­ stance, liquid water can be 'supercooled' below the freezing point yet remain a liquid for some time. At high pressures there are actually several different phases of ice, with differing crystal structures and other physical properties. The lines on a phase diagram represent conditions under which two different phases can coexist in equilibrium; for instance, ice and water can coexist stably at
5.3
Phase Transformations of Pure Substances T (OC)
Critical point
221
~
eQ)
I-<
::l
rI.l rI.l
Q)
I-<
0.
0.006
-273
0.01 374 Temperature (OC)
-40 -20 0 0.01 25 50 100 150 200 250 300 350 374
Pv (bar)
L (kJjmol)
0.00013 0.00103 0.00611 0.00612 0.0317 0.1234 1.013 4.757 15.54 39.74 85.84 165.2 220.6
51.16 51.13 51.07 45.05 43.99 42.92 40.66 38.09 34.96 30.90 25.30 16.09 0.00
Figure 5.11. Phase diagram for H20 (not to scale). The table gives the vapor pressure and molar latent heat for the solid-gas transformation (first three entries) and the liquid-gas transformation (remaining entries) . Data from Keenan et al. (1978) and Lide (1994).
aOc and 1 atm
(~ 1 bar). The pressure at which a gas can coexist with its solid or liquid phase is called the vapor pressure; thus the vapor pressure of water at room temperature is approximately 0.03 bar. At T = O.Ol°C and P = 0.006 bar, all three phases can coexist; this point is called the triple point. At lower pressures, liquid water cannot exist (in eqUilibrium): ice 'sublimates' directly into vapor. You have probably observed sublimation of 'dry ice,' frozen carbon dioxide. Evidently, the triple point of carbon dioxide lies above atmospheric pressure; in fact it is at 5.2 bars. A qualitative phase diagram for carbon dioxide is shown in Figure 5.12. Another difference between CO 2 and H 2 0 is the slope of the solid-liquid phase boundary. Most substances are like carbon dioxide: Applying more pressure raises the melting temperature. Ice, however, is unusual: Applying
Critical point 73.8 ..--.­
..cd
e ill I-<
;:::l
Gas
rJl
rJl
ill I-<
0.
5.2
-56.6 31 Temperature (0 C)
T (OC)
P v (bar)
-120 -100 -80 -78.6 -60 -56.6 -40 -20 0 20 31
0.0124 0.135 0.889 1.000 4.11 5.18 10.07 19.72 34.85 57.2 73.8
Figure 5.12. Phase diagram for carbon dioxide (not to scale). The table gives the vapor pressure along the solid-gas and liquid-gas equilibrium curves. Data from Lide (1994) and Reynolds (1979).
167
168
Chapter 5
Free Energy and Chemical Thermodynamics
pressure lowers its melting temperature. We will soon see that this is a result of the fact that ice is less dense than water. The liquid-gas phase boundary always has a positive slope: If you have liq­ uid and gas in equilibrium and you raise the temperature, you must apply more pressure to keep the liquid from vaporizing. As the pressure increases, however, the gas becomes more dense, so the difference between liquid and gas grows less. Eventually a point is reached where there is no longer any discontinuous change from liquid to gas. This point is called the critical point, and occurs at 374°C and 221 bars for H 2 0. The critical point of carbon dioxide is more accessible, at 31°C and 74 bars, while that of nitrogen is at only 126 K and 34 bars. Close to the critical point, it's best to hedge and simply call the substance a 'fluid.' There's no critical point on the solid-liquid phase boundary, since the distinction between solids and liquids is a qualitative issue (solids having crystal structure and liquids having randomly arranged molecules), not just a matter of degree. Some materials made of long molecules can, however, form a liquid crystal phase, in which the molecules move around randomly as in a liquid but still tend to be oriented parallel to each other. Helium has the most exotic phase behavior of any element. Figure 5.13 shows the phase diagrams of the two isotopes of helium, the common isotope 4He and the rare isotope 3He. The boiling point of 4He at atmospheric pressure is only 4.2 K, and the critical point is only slightly higher, at 5.2 K and 2.3 bars; for 3He these parameters are somewhat lower still. Helium is the only element that remains a liquid at absolute zero temperature: It will form a solid phase, but only at rather high pressures, about 25 bars for 4He and 30 bars for 3He. The solid-liquid phase boundary for 4He is almost horizontal below 1 K, while for 3He this boundary has a negative slope below 0.3 K. Even more interesting, 4He has two distinct liquid phases: a 'normal' phase called helium I, and a superfiuid ~
c-''
4
ro
o..
i He i
~
3 i He i
C0..
Solid
34
25.3 Helium II (superfluid)
Helium I (normal liquid)
1
Liquid 1
2.2
4.2 5.2 T (K)
3.2 3.3 T (K)
Figure 5.13. Phase diagrams of 4 He (left) and 3He (right). Neither diagram is to scale, but qualitative relations between the diagrams are shown correctly. Not shown are the three different solid phases (crystal structures) of each isotope, or the superfluid phases of 3He below 3 mK.
5.3
Phase Transformations of Pure Substances
phase, below about 2 K, called helium II. The superfluid phase has a number of remarkable properties including zero viscosity and very high thermal conductivity. Helium-3 actually has two distinct superfluid phases, but only at temperatures below 3 mK. Besides temperature and pressure, changing other variables such as composition and magnetic field strength can also cause phase transformations. Figure 5.14 shows phase diagrams for two different magnetic systems. At left is the diagram for a typ­ ical type-I superconductor, such as tin or mercury or lead. The superconducting phase, with zero electrical resistance, exists only when both the temperature and the external magnetic field strength are sufficiently low. At right is the diagram for a ferromagnet such as iron, which has magnetized pointing either up or down, depending on the direction of the applied field. (For simplicity, this dia­ gram assumes that the applied field always points either up or down along a given axis.) When the applied field is zero, phases that are magnetized in both direc­ tions can coexist. As the temperature is raised, however, the magnetization of both phases becomes weaker. Eventually, at the Curie temperature (1043 K for iron), the magnetization disappears completely, so the phase boundary ends at a critical point.* Superconductor
:Q (].)
q:: u
Normal Super­ conducting
t
Ferromagnet Magnetized up
~
Critical point
~ ro S
/ ~--------------~r--------4~ T Magnetized down
Te
T
Figure 5.14. Left: Phase diagram for a typical type-I superconductor. For lead, Te 7.2 K and Be 0.08 T. Right: Phase diagram for a ferromagnet, assuming that the applied field and magnetization are always along a given axis.
*For several decades people have tried to classify phase transformations according to the abruptness of the change. Solid-liquid and liquid-gas transformations are classified as 'first-order,' because S and V, the first derivatives of G, are discontinuous at the phase boundary. Less abrupt transitions (such as critical points and the helium I to helium II transition) used to be classified as 'second-order' and so on, depending on how many successive derivatives you had to take before getting a discontinuous quantity. Because of various problems with this classification scheme, the current fashion is to simply call all the higher-order transitions 'continuous.'
169
170
Chapter 5
Free Energy and Chemical Thermodynamics
Diamonds and Graphite Elemental carbon has two familiar phases, diamond and graphite (both solids, but with different crystal structures). At ordinary pressures the more stable phase is graphite, so diamonds will spontaneously convert to graphite, although this process is extremely slow at room temperature. (At high temperatures the conversion proceeds more rapidly, so if you own any diamonds, be sure not to throw them into the fireplace. *) The fact that graphite is more stable than diamond under standard conditions is reflected in their Gibbs free energies: The Gibbs free energy of a mole of diamond is greater, by 2900 J, than the Gibbs free energy of a mole of graphite. At a given temperature and pressure, the stable phase is always the one with the lower Gibbs free energy, according to the analysis of Section 5.2. But the difference of 2900 J is for standard conditions, 298 K and atmospheric pressure (1 bar). What happens at higher pressures? The pressure dependence of the Gibbs free energy is determined by the volume of the substance,
( ~~)
T,N
(5.41)
= V,
and since a mole of graphite has a greater volume than a mole of diamond, its Gibbs free energy will grow more rapidly as the pressure is raised. Figure 5.15 shows a graph of G vs. P for both substances. If we treat the volumes as constant (neglecting the compressibility of both substances), then each curve is a straight line. The slopes are V = 5.31 X 10- 6 m 3 for graphite and V = 3.42 X 10- 6 m 3 for diamond. As you can see, the two lines intersect at a pressure of about 15 kilobars. Above this very high pressure, diamond should be more stable than graphite. Apparently,
-+---f-------I----+---+-------. P (kbar)
5
10
15
20
Figure 5.15. Molar Gibbs free energies of diamond and graphite as functions of pressure, at room temperature. These straight-line graphs are extrapolated from low pressures, neglecting the changes in volume as pressure increases.
*The temperature required to convert diamond to graphite quickly is actually quite high, about 1500°C. But in the presence of oxygen, either diamond or graphite will easily burn to form carbon dioxide.
5.3
Phase Transformations of Pure Substances
natural diamonds must form at very great depths. Taking rock to be about three times as dense as water, it's easy to estimate that underground pressures normally increase by 3 bars for every 10 meters of depth. So a pressure of 15 kbar requires a depth of about 50 kilometers. The temperature dependence of the Gibbs free energies can be determined in a similar way, using the relation
BG) _-8 (BT P,N -
.
(5.42)
As the temperature is raised the Gibbs free energy of either substance decreases, but this decrease is more rapid for graphite since it has more entropy. Thus, raising the temperature tends to make graphite more stable relative to diamond; the higher the temperature, the more pressure is required before diamond becomes the stable phase. Analyses of this type are extremely useful to geochemists, whose job is to look at rocks and determine the conditions under which they formed. More generally, the Gibbs free energy is the key to attaining a quantitative understanding of phase transformations. Problem 5.24. Go through the arithmetic to verify that diamond becomes more stable than graphite at approximately 15 kbar. Problem 5.25. In working high-pressure geochemistry problems it is usually more convenient to express volumes in units of kJ/kbar. Work out the conversion factor between this unit and m 3. Problem 5.26. How can diamond ever be more stable than graphite, when it has less entropy? Explain how at high pressures the conversion of graphite to diamond can increase the total entropy of the carbon plus its environment. Problem 5.27. Graphite is more compressible than diamond. (a) Taking compressibilities into account, would you expect the transition from graphite to diamond to occur at higher or lower pressure than that pre­ dicted in the text? (b) The isothermal compressibility of graphite is about 3 x 10- 6 bar-l, while that of diamond is more than ten times less and hence negligible in compar­ ison. (Isothermal compressibility is the fractional reduction in volume per unit increase in pressure, as defined in Problem 1.46.) Use this information to make a revised estimate of the pressure at which diamond becomes more stable than graphite (at room temperature). Problem 5.28. Calcium carbonate, CaC03, has two common crystalline forms, calcite and aragonite. Thermodynamic data for these phases can be found at the back of this book. (a) Which is stable at earth's surface, calcite or aragonite? (b) Calculate the pressure (still at room temperature) at which the other phase should become stable.
171
172
Chapter 5
Free Energy and Chemical Thermodynamics
Problem 5.29. Aluminum silicate, AI2Si05, has three different crystalline forms: kyanite, andalusite, and sillimanite. Because each is stable under a different set of temperature-pressure conditions, and all are commonly found in metamorphic rocks, these minerals are important indicators of the geologic history of rock bodies.
(a) Referring to the thermodynamic data at the back of this book, argue that at 298 K the stable phase should be kyanite, regardless of pressure. (b) Now consider what happens at fixed pressure as we vary the temperature. Let I:::!.G be the difference in Gibbs free energies between any two phases, and similarly for I:::!.S. Show that the T dependence of I:::!.G is given by I:::!.G(T2) = I:::!.G(Tt)
(T2 I:::!.S(T) dT.
iT
l
Although the entropy of any given phase will increase significantly as the temperature increases, above room temperature it is often a good approx­ imation to take I:::!.S, the difference in entropies between two phases, to be independent of T. This is because the temperature dependence of S is a function of the heat capacity (as we saw in Chapter 3), and the heat ca­ pacity of a solid at high temperature depends, to a good approximation, only on the number of atoms it contains. (c) Taking I:::!.S to be independent of T, determine the range of temperatures over which kyanite, andalusite, and sillimanite should be stable (at atmo­ spheric pressure). (d) Referring to the room-temperature heat capacities of the three forms of AI2Si05, discuss the accuracy the approximation I:::!.S constant. Problem phases of graphs on would the
5.30. Sketch qualitatively accurate graphs of G vs. T for the three H20 (ice, water, and steam) at atmospheric pressure. Put all three the same set of axes, and label the temperatures O°C and 100°C. How graphs differ at a pressure of 0.001 bar?
Problem 5.31. Sketch qualitatively accurate graphs of G vs. P for the three phases of H20 (ice, water, and steam) at O°C. Put all three graphs on the same set of axes, and label the point corresponding to atmospheric pressure. How would the graphs differ at slightly higher temperatures?
The Clausius-Clapeyron Relation Since entropy determines the temperature dependence of the Gibbs free energy, while volume determines its pressure dependence, the shape of any phase boundary line on a PT diagram is related in a very simple way to the entropies and volumes of the two phases. Let me now derive this relation. For definiteness, I'll discuss the phase boundary between a liquid and a gas, although it could just as well be any other phase boundary. Let's consider some fixed amount of the stuff, say one mole. At the phase boundary neither the liquid phase nor the gas phase is more stable, so their Gibbs free energies must be equal: at phase boundary.
(5.43)
(You can also think of this condition in terms of the chemical potentials: If the liquid and gas are in diffusive equilibrium with each other, then their chemical potentials, i.e., Gibbs free energies per molecule, must be equal.)
5.3
Phase Transformations of Pure Substances
P Figure 5.16. Infinitesimal changes in pressure and temperature, related in such a way as to remain on the phase bound­ ary.
dP
/
~--------+--+------~T
Now increasing the temperature by dT and the pressure by dP, in such a way that the two phases remain in equilibrium (see Figure 5.16). Under this change, the Gibbs free energies must remain equal, so dGl = dG g
to remain on phase boundary.
(5.44)
Therefore, by the thermodynamic identity for G (equation 5.23),
dT+ VgdP.
-BldT+ VldP
(5.45)
(I've omitted the J-L dN terms because I've already assumed that the total amount of stuff is fixed.) Now it's easy to solve for the slope of the phase boundary line, dP/dT: dP (5.46) dT As expected, the slope is determined by the entropies and volumes of the two phases. A large difference in entropy means that a small change in temperature can be very significant in shifting the equilibrium from one phase to the other. This results pressure change is then required to in a steep phase boundary curve, since a compensate the small temperature change. On the other hand, a large difference in volume means that a small change in pressure can be significant after all, making the phase boundary curve shallower. It's often more convenient to write the difference in entropies, Bg Bl , as L/T, where L is the latent heat for converting the material (in whatever quantity we're considering) from liquid to gas. Then equation 5.46 takes the form
dP dT
L
(5.47)
Vg Vl. (Notice that, since both L and ~ V are extensive, their ratio where ~ V is intensive-independent of the amount of material.) This result is known as the Clausius-Clapeyron relation. It applies to the slope of any phase boundary line on a PT diagram, not just to the line separating liquid from gas. As an example, consider again the diamond-graphite system. When a mole of diamond converts to graphite its entropy increases by 3.4 J /K, while its volume increases by 1.9 x 10- 6 m 3 • (Both of these numbers are for room temperature; at
173
174
Chapter 5
Free Energy and Chemical Thermodynamics
higher temperatures the difference in entropy is somewhat greater.) Therefore the slope of the diamond-graphite phase boundary is dP dT
!18 !1V
3.4 J /K 1.9 x 10- 6 m 3
= 1.8 x 106 Pa/K
18 bar/K.
(5.48)
In the previous subsection I showed that at room temperature, diamond is stable at pressures above approximately 15 kbar. Now we see that if the temperature is 100 K higher, we need an additional 1.8 kbar of pressure to make diamond stable. Rapid conversion of graphite to diamond requires still higher temperatures, and correspondingly higher pressures, as shown in the phase diagram in Figure 5.17. The first synthesis of diamond from graphite was accomplished at approximately 1800 K and 60 kbar. Natural diamonds are thought to form at similar pressures but somewhat lower temperatures, at depths of 100-200 km below earth's surface.* 100 80
'i
60
,.J::J
C
Graphite
0. 40 20
o
1000 2000 3000 4000 5000 6000 T (K)
Figure 5.17. The experimen­ tal phase diagram of carbon. The stability region of the gas phase is not visible on this scale; the graphite-liquid-gas triple point is at the bottom of the graphite-liquid phase boundary, at 110 bars pressure. From David A. Young, Phase Dia­ grams of the Elements (U niver­ sity of California Press, Berke­ ley, 1991).
Problem 5.32. The density of ice is 917 kg/m3.
(a) Use the Clausius-Clapeyron relation to explain why the slope of the phase boundary between water and ice is negative.
(b) How much pressure would you have to put on an ice cube to make it melt at -1°C? (c) Approximately how deep under a glacier would you have to be before the weight of the ice above gives the pressure you found in part (b)? (Note that the pressure can be greater at some locations, as where the glacier flows over a protruding rock.) (d) Make a rough estimate of the pressure under the blade of an ice skate, and calculate the melting temperature of ice at this pressure. Some authors have claimed that skaters glide with very little friction because the increased pressure under the blade melts the ice to create a thin layer of water. What do you think of this explanation? *For more on the formation of natural diamonds and the processes that bring them near earth's surface, see Keith G. Cox, 'Kimberlite Pipes,' Scientific American 238, 12(}-132 (April, 1978).
5.3
Phase Transformations of Pure Substances
Problem 5.33. An inventor proposes to make a heat engine using water/ice as the working substance, taking advantage of the fact that water expands as it freezes. A weight to be lifted is placed on top of a piston over a cylinder of water at 1°C. The system is then placed in thermal contact with a low-temperature reservoir at -1°C until the water freezes into ice, lifting the weight. The weight is then removed and the ice is melted by putting it in contact with a high-temperature reservoir at 1°C. The inventor is pleased with this device because it can seemingly perform an unlimited amount of work while absorbing only a finite amount of heat. Explain the flaw in the inventor's reasoning, and use the Clausius-Clapeyron relation to prove that the maximum efficiency of this engine is still given by the Carnot formula, 1 - Tc/Th. Problem 5.34. Below 0.3 K the slope of the 3He solid-liquid phase boundary is negative (see Figure 5.13).
(a) Which phase, solid or liquid, is more dense? Which phase has more entropy (per mole)? Explain your reasoning carefully.
(b) Use the third law of thermodynamics to argue that the slope of the phase boundary must go to zero at T = O. (Note that the 4He solid-liquid phase boundary is essentially horizontal below 1 K.) (c) Suppose that you compress liquid 3He adiabatically until it becomes a solid. If the temperature just before the phase change is 0.1 K, will the temper­ ature after the phase change be higher or lower? Explain your reasoning carefully. Problem 5.35. The Clausius-Clapeyron relation 5.47 is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however , you have to know how both L and ~ V depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ~ V to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions , solve the differential equation explicitly to obtain the following formula for the phase boundary curve: P = (constant) x e- LjRT .
This result is called the vapor pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid. Problem 5.36. Effect of altitude on boiling water.
(a) Use the result of the previous problem and the data in Figure 5.11 to plot a graph of the vapor pressure of water between 50°C and 100°C. How well can you match the data at the two endpoints?
(b) Reading the graph backwards, estimate the boiling temperature of water at each ofthe locations for which you determined the pressure in Problem 1.16. Explain why it takes longer to cook noodles when you're camping in the mountains. (c) Show that the dependence of boiling temperature on altitude is very nearly (though not ' exactly) a linear function, and calculate the slope in degrees Celsius per thousand feet (or in degrees Celsius per kilometer).
175
176
Chapter 5
Free Energy and Chemical Thermodynamics
Problem 5.37. Use the data at the back of this book to calculate the slope of the calcite-aragonite phase boundary (at 298 K). You located one point on this phase boundary in Problem 5.28; use this information to sketch the phase diagram of calcium carbonate. Problem 5.38. In Problems 3.30 and 3.31 you calculated the entropies of diamond and graphite at 500 K. Use these values to predict the slope of the graphite­ diamond phase boundary at 500 K, and compare to Figure 5.17. Why is the slope almost constant at still higher temperatures? Why is the slope zero at T = O? Problem 5.39. Consider again the aluminosilicate system treated in Problem 5.29. Calculate the slopes of all three phase boundaries for this system: kyanite­ andalusite, kyanite-sillimanite, and andalusite-sillimanite. Sketch the phase dia­ gram, and calculate the temperature and pressure of the triple point. Problem 5.40. The methods of this section can also be applied to reactions in which one set of solids converts to another. A geologically important example is the transformation of albite into jadeite + quartz:
Use the data at the back of this book to determine the temperatures and pressures under which a combination of jadeite and quartz is more stable than albite. Sketch the phase diagram of this system. For simplicity, neglect the temperature and pressure dependence of both D.S and D.V.
Problem 5.41. Suppose you have a liquid (say, water) in equilibrium with its gas phase, inside some closed container. You then pump in an inert gas (say, air), thus raising the pressure exerted on the liquid. What happens? (a) For the liquid to remain in diffusive equilibrium with its gas phase, the chemical potentials of each must change by the same amount: dP,l dp,g. Use this fact and equation 5.40 to derive a differential equation for the equilibrium vapor pressure, Pv , as a function of the total pressure P. (Treat the gases as ideal, and assume that none of the inert gas dissolves in the liquid.) (b) Solve the differential equation to obtain e (P-Pv)V/NkT , where the ratio V/ N in the exponent is that of the liquid. (The term Pv (Pv) is just the vapor pressure in the absence of the inert gas.) Thus, the presence of the inert gas leads to a slight increase in the vapor pressure: It causes more of the liquid to evaporate. (c) Calculate the percent increase in vapor pressure when air at atmospheric pressure is added to a system of water and water vapor in equilibrium at 25°C. Argue more generally that the increase in vapor pressure due to the presence of an inert gas will be negligible except under extreme conditions.
5.3
Phase Transformations of Pure Substances
Problem 5.42. Ordinarily, the partial pressure of water vapor in the air is less than the equilibrium vapor pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapor to the equilibrium vapor pressure is called the relative humidity. When the relative humidity is 100%, so that water vapor in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. * The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapor. (a) Use the vapor pressure equation (Problem 5.35) and the data in Figure 5.11 to plot a graph of the vapor pressure of water from O°C to 40°C. Notice that the vapor pressure approximately doubles for every 10° increase in temperature. (b) Suppose that the temperature on a certain summer day is 30°C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%? Problem 5.43. Assume that the air you exhale is at 35°C, with a relative hu­ midity of 90%. This air immediately mixes with environmental air at 5°C and unknown relative humidity; during the mixing, a variety of intermediate tempera­ tures and water vapor percentages temporarily occur. If you are able to 'see your breath' due to the formation of cloud droplets during this mixing, what can you conclude about the relative humidity of your environment? (Refer to the vapor pressure graph drawn in Problem 5.42.) Problem 5.44. Suppose that an unsaturated air mass is rising and cooling at the dry adiabatic lapse rate found in Problem 1.40. If the temperature at ground level is 25°C and the relative humidity there is 50%, at what altitude will this air mass become saturated so that condensation begins and a cloud forms (see Figure 5.18)? (Refer to the vapor pressure graph drawn in Problem 5.42.) Problem 5.45. In Problem 1.40 you calculated the atmospheric temperature gradient required for unsaturated air to spontaneously undergo convection. When a rising air mass becomes saturated, however, the condensing water droplets will give up energy, thus slowing the adiabatic cooling process. (a) Use the first law of thermodynamics to show that, as condensation forms during adiabatic expansion, the temperature of an air mass changes by
2T 2 L dT = ;; P dP - ;; nR dnw, where nw is the number of moles of water vapor present, L is the latent heat of vaporization per mole, and I've assumed I = 715 for air. (b) Assuming that the air is always saturated during this process, the ratio nwln is a known function of temperature and pressure. Carefully express dnwldz in terms of dTldz, dPldz, and the vapor pressure Pv(T). Use the Clausius-Clapeyron relation to eliminate dPvldT. (c) Combine the results of parts (a) and (b) to obtain a formula relating the temperature gradient, dT I dz, to the pressure gradient, dP I dz. Eliminate *This term is widely used, but is unfortunate and misleading. Air is not a sponge that can hold only a certain amount of liquid; even 'saturated' air is mostly empty space. As shown in the previous problem, the density of water vapor that can exist in equilibrium has almost nothing to do with the presence of air.
177
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Chapter 5
Free Energy and Chemical Thermodynamics
Figure 5.18. Cumulus clouds form when rising air expands adiabatically and cools to the dew point (Problem 5.44); the onset of condensation slows the cooling, increasing the tendency ofthe air to rise further (Problem 5.45). These clouds began to form in late morning, in a sky that was clear only an hour before the photo was taken. By mid-afternoon they had developed into thunderstorms. the latter using the 'barometric equation' from Problem 1.16. You should finally obtain Pv L
dT dz
(2M9)
= - '7 I f
1+ PliT
1
~ Pv (~)2
+7
P
RT
'
where M is the mass of a mole of air. The prefactor is just the dry adiabatic lapse rate calculated in Problem 1.40, while the rest of the expression gives the correction due to heating from the condensing water vapor. The whole result is called the wet adiabatic lapse rate; it is the critical temperature gradient above which saturated air will spontaneously convect. (d) Calculate the wet adiabatic lapse rate at atmospheric pressure (1 bar) and 25°C, then at atmospheric pressure and O°C. Explain why the results are different, and discuss their implications. What happens at higher altitudes, where the pressure is lower? Problem 5.46. Everything in this section so far has ignored the boundary be­ tween two phases, as if each molecule were unequivocally part of one phase or the other. In fact, the boundary is a kind of transition zone where molecules are in an environment that differs from both phases. Since the boundary zone is only a few molecules thick, its contribution to the total free energy of a system is very often negligible. One important exception, however, is the first tiny droplets or bub­ bles or grains that form as a material begins to undergo a phase transformation. The formation of these initial specks of a new phase is called nucleation. In this problem we will consider the nucleation of water droplets in a cloud. The surface forming the boundary between any two given phases generally has a fixed thickness, regardless of its area. The additional Gibbs free energy of this surface is therefore directly proportional to its area; the constant of proportionality is called the surface tension, (J: _
(J
=
Gboundary
A
.
5.3
Phase Transformations of Pure Substances
If you have a blob of liquid in equilibrium with its vapor and you wish to stretch it into a shape that has the same volume but more surface area, then (j is the mini­ mum work that you must perform, per unit of additional area, at fixed temperature and pressure. For water at 20°C, (j = 0.073 J 1m2 . (a) Consider a spherical droplet of water containing Nz molecules , surrounded by N - Nz molecules of water vapor. Neglecting surface tension for the moment , write down a formula for the total Gibbs free energy of this system in terms of N , Nz, and the chemical potentials of the liquid and vapor. Rewrite N z in terms of vz, the volume per molecule in the liquid, and r, the radius of the droplet. (b) Now add to your expression for G a term to represent the surface tension, written in terms of rand (j. (c) Sketch a qualitative graph of G vs. r for both signs of J-lg - J-lz, and discuss the implications. For which sign of J-lg - J-lz does there exist a nonzero equilibrium radius? Is this equilibrium stable? (d) Let rc represent the critical equilibrium radius that you discussed quali­ tatively in part (c). Find an expression for rein terms of J-lg - J-lz. Then rewrite the difference of chemical potentials in terms of the relative humid­ ity (see Problem 5.42) , assuming that the vapor behaves as an ideal gas. (The relative humidity is defined in terms of equilibrium of a vapor with a flat surface, or with an infinitely large droplet.) Sketch a graph of the critical radius as a function of the relative humidity, including numbers. Discuss the implications. In particular, explain why it is unlikely that the clouds in our atmosphere would form by spontaneous aggregation of water molecules into droplets. (In fact, cloud droplets form around nuclei of dust particles and other foreign material, when the relative humidity is close to 100%.) Problem 5.47. For a magnetic system held at constant T and 1i (see Prob­ lem 5.17) , the quantity that is minimized is the magnetic analogue of the Gibbs free energy, which obeys the thermodynamic identity
dG m
= -S dT -
J-loM d1i.
Phase diagrams for two magnetic systems are shown in Figure 5.14; the vertical axis on each of these figures is J-lO 1i. (a) Derive an analogue of the Clausius-Clapeyron relation for the slope of a phase boundary in the 1i-T plane. Write your equation in terms of the difference in entropy between the two phases. (b) Discuss the application of your equation to the ferromagnet phase diagram in Figure 5.14. (c) In a type-I superconductor, surface currents flow in such a way as to com­ pletely cancel the magnetic field (B, not 1i) inside. Assuming that M is negligible when the material is in its normal (non-superconducting) state, discuss the application of your equation to the superconductor phase dia­ gram in Figure 5.14. Which phase has the greater entropy? What happens to the difference in entropy between the phases at each end of the phase boundary?
179
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The van der Waals Model To understand phase transformations more deeply, a good approach is to introduce a specific mathematical model. For liquid-gas systems, the most famous model is the van der Waals equation,
aN2) ( P+ V2 (V-Nb) = NkT,
(5.49)
proposed by Johannes van der Waals in 1881. This is a modification of the ideal gas law that takes molecular interactions into account in an approximate way. (Any proposed relation among P, V, and T, like the ideal gas law or the van der Waals equation, is called an equation of state.) The van der Waals equation makes two modifications to the ideal gas law: adding aN 21V 2 to P and subtracting Nb from V. The second modification is easier to understand: A fluid can't be compressed all the way down to zero volume, so we've limited the volume to a minimum value of Nb, at which the pressure goes to infinity. The constant b then represents the minimum volume occupied by a molecule, when it's 'touching' all its neighbors. The first modification, adding aN 2 1V 2 to P, accounts for the short-range attractive forces between molecules when they're not touching (see Figure 5.19). Imagine freezing all the molecules in place, so that the only type of energy present is the negative potential energy due to molecular attraction. If we were to double the density of the system, each molecule would then have twice as many neighbors as before, so the potential energy due to all its interactions with neighbors would double. In other words, the potential energy associated with a single molecule's interactions with all its neighbors is proportional to the density of particles, or to N IV. The total potential energy associated with all molecules' interactions must then be proportional to N 2 IV, since there are N molecules: aN 2 (5.50) total potential energy = - V ' where a is some positive constant of proportionality that depends on the type of molecules. To calculate the pressure, imagine varying the volume slightly while holding the entropy fixed (which isn't a problem if we've frozen all thermal motion); then by the thermodynamic identity, dU = -P dV or P = -(aUlav)s. The contribution to the pressure from just the potential energy is therefore
Pdue to p.e.
=-
d (aN2) dV - V
=-
2 aN V2 .
(5.51)
If we add this negative pressure to the pressure that the fluid would have in the
Figure 5.19. When two molecules come very close together they repel each other strongly. When they are a short distance apart they attract each other.
5.3
Phase Transformations of Pure Substances
absence of attractive forces (namely, NkTI(V - Nb)), we obtain the van der Waals equation, 2 p= NkT aN (5.52) V - Nb - V2 . While the van der Waals equation has the right properties to account for the qualitative behavior of real fluids, I need to emphasize that it is nowhere near exact. In 'deriving' it I've neglected a number of effects, most notably the fact that as a gas becomes more dense it can become inhomogeneous on the microscopic scale: Clusters of molecules can begin to form, violating my assertion that the number of neighbors a molecule has will be directly proportional to N IV. SO throughout this section, please keep in mind that we won't be making any accurate quantitative pre­ dictions. What we're after is qualitative understanding, which can provide a start­ ing point if you later decide to study liquid-gas phase transformations in more depth. The constants a and b will have different values for different substances, and (since the model isn't exact) will even vary somewhat for the same substance under different conditions. For small molecules like N2 and H 2 0, a good value of b is about 6 x 10- 29 m 3 ~ (4 A)3, roughly the cube of the average width of the molecule. The constant a is much more variable, because some types of molecules attract each other much more strongly than others. For N 2, a good value of a is about 4 x 10- 49 J·m 3, or 2.5 eV·A3. If we think of a as being roughly the product of the average interaction energy times the volume over which the interaction can act, then this value is fairly sensible: a small fraction of an electron-volt times a few tens of cubic angstroms. The value of a for H 2 0 is about four times as large, because of the molecule's permanent electric polarization. Helium is at the other extreme, with interactions so weak that its value of a is 40 times less than that of nitrogen. Now let us investigate the consequences of the van der Waals model. A good way to start is by plotting the predicted pressure as a function of volume for a variety of different temperatures (see Figure 5.20). At volumes much greater than Nb the isotherms are concave-up, like those of an ideal gas. At sufficiently high PIPe 2
1
~---;--~--------~------~--------~V/Ve
Figure 5.20. Isotherms (lines of constant temperature) for a van der Waals fluid.
From bottom to top , the lines are for 0.8, 0.9, 1.0, 1.1, and 1.2 times T e , the
temperature at the critical point. The axes are labeled in units of the pressure and
volume at the critical point; in these units the minimum volume (Nb) is 1/3.
181
182
Chapter 5
Free Energy and Chemical Thermodynamics
temperatures, reducing the volume causes the pressure to rise smoothly, eventually approaching infinity as the volume goes to Nb. At lower temperatures, however, the behavior is much more complicated: As V decreases the isotherm rises, falls, and then rises again, seeming to imply that for some states, compressing the fluid can cause its pressure to decrease. Real fluids don't behave like this. But a more careful analysis shows that the van der Waals model doesn't predict this, either. At a given temperature and pressure, the true equilibrium state of a system is determined by its Gibbs free energy. To calculate G for a van der Waals fluid, let's start with the thermodynamic identity for G: dG
=
-SdT + V dP
+ {LdN.
(5.53)
For a fixed amount of material at a given, fixed temperature, this equation reduces to dG V dP. Dividing both sides by dV then gives
(8G) _v(8P)
8V
N,T -
8V
(5.54)
N,T'
The right-hand side can be computed directly from the van der Waals equation (5.52), yielding NkTV 2aN 2 (5.55) (V Nb)2 + ---y-2' To integrate the right-hand side, write the V in the numerator of the first term as (V - Nb) + (Nb), then integrate each of these two pieces separately. The result is
G
-NkTI (V - Nb)
n
+
(NkT)(Nb) _ 2aN V _ Nb V
2
+c
(T)
,
(5.56)
where the integration constant, c(T), can be different for different temperatures but is unimportant for our purposes. This equation allows us to plot the Gibbs free energy for any fixed T. Instead of plotting G as a function of volume, it's more useful to plot G vertically and P horizontally, calculating each as a function of the parameter V. Figure 5.21 shows an example, for the temperature whose isotherm is shown alongside. Al­ though the van der Waals equation associates some pressures with more than one volume, the thermodynamically stable state is that with the lowest Gibbs free en­ ergy; thus the triangular loop in the graph of G (points 2-3-4-5-6) corresponds to unstable states. As the pressure is gradually increased, the system will go straight from point 2 to point 6, with an abrupt decrease in volume: a phase transformation. At point 2 we should call the fluid a gas, because its volume decreases rapidly with increasing pressure. At point 6 we should call the fluid a liquid, because its volume decreases only slightly under a large increase in pressure. At intermediate volumes between these points, the thermodynamically stable state is actually a combination of part gas and part liquid, still at the transition pressure, as indicated by the straight horizontal line on the PV diagram. The curved portion of the isotherm that is cut off by this straight line correctly indicates what the allowed states would be if the fluid were homogeneous; but these homogeneous states are unstable, since
5.3
G
Phase Transformations of Pure Substances
PIPe 7
0.8
3
2
6
0.6 0.4
5
0.2 0.4
0.6
2
1
0.8 PIPe
3 VIVe
Figure 5.21. Gibbs free energy as a function of pressure for a van der Waals fluid
at T = 0.9Te . The corresponding isotherm is shown at right. States in the range
2-3-4-5-6 are unstable.
there is always another state (gas or liquid) at the same pressure with a lower Gibbs free energy. The pressure at the phase transformation is easy enough to determine from the graph of G, but there is a clever method of reading it straight off the PV diagram, without plotting G at all. To derive this method, note that the net change in G as we go around the triangular loop (2-3-4-5-6) is zero:
0=
1.
l oop
dG=
1. (BG) loop
BP
dP= T
1.
VdP.
(5.57)
loop
Written in this last form, the integral can be computed from the PV diagram, though it's easier to turn the diagram sideways (see Figure 5.22). The integral from point 2 to point 3 gives the entire area under this segment, but the integral from point 3 to point 4 cancels out all but the shaded region A. The integral from 4 to 5 gives minus the area under that segment, but then the integral from 5 to 6 adds back all but the shaded region B. Thus the entire integral equals the area
v
Figure 5.22. The same isotherm
as in Figure 5.21, plotted sideways.
Regions A and B have equal areas.
5
~--~--~----~--~--~--~P
183
184
Chapter 5
Free Energy and Chemical Thermodynamics
of A minus the area of B, and if this is to equal zero, we conclude that the two shaded regions must have equal areas. Drawing the straight line so as to enclose equal areas in this way is called the Maxwell construction, after James Clerk Maxwell. Repeating the Maxwell construction for a variety of temperatures yields the results shown in Figure 5.23. For each temperature there is a well-defined pressure, called the vapor pressure, at which the liquid-gas transformation takes place; plotting this pressure vs. temperature gives us a prediction for the entire liquid-gas phase boundary. Meanwhile, the straight segments of the isotherms on the PV diagram fill a region in which the stable state is a combination of gas and liquid, indicated by the shaded area. But what about the high-temperature isotherms, which rise monotonically as V decreases? For these temperatures there is no abrupt transition from low-density states to high-density states: no phase transformation. The phase boundary there­ fore disappears above a certain temperature, called the critical temperature, Tc. The vapor pressure just at T c is called the critical pressure, Pc, while the corre­ sponding volume is called the critical volume, Vc. These values define the critical point, where the properties of the liquid and gas become identical. I find it remarkable that a model as simple as the van der Waals equation predicts all of the important qualitative properties of real fluids: the liquid-gas phase transformation, the general shape of the phase boundary curve, and even the critical point. Unfortunately, the model fails when it comes to numbers. For example, the experimental phase boundary for H 2 0 falls more steeply from the critical point than does the predicted boundary shown above; at T /Tc = 0.8, the measured vapor pressure is only about 0.2Pc , instead of O.4Pc as predicted. More p
P
1.2
1.2
1.0
1.0
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2 1
2
3
4
5
6
7
Critical point

0.2
0.4
0.6
0.8
1.0
-----. V Figure 5.23. Complete phase diagrams predicted by the van der Waals model. The isotherms shown at left are for T jTc ranging from 0.75 to 1.1 in increments of 0.05. In the shaded region the stable state is a combination of gas and liquid. The full vapor pressure curve is shown at right. All axes are labeled in units of the critical values.
5.3
Phase Transformations of Pure Substances
accurate models of the behavior of dense fluids are beyond the scope of this book, * but at least we've taken a first step toward understanding the liquid-gas phase transformation. Problem 5.48. As you can see from Figure 5.20, the critical point is the unique point on the original van der Walls isotherms (before the Maxwell construction) where both the first and second derivatives of P with respect to V (at fixed T) are zero. Use this fact to show that 1 a 8 a and kTc = --. Vc = 3Nb, Pc = 27 b2' 27 b Problem 5.49. Use the result of the previous problem and the approximate values of a and b given in the text to estimate T c , Pc , and VeiN for N2 , H20, and He. (Tabulated values of a and b are often determined by working backward from the measured critical temperature and pressure.) Problem 5.50. The compression factor of a fluid is defined as the ratio PV/NkT; the deviation of this quantity from 1 is a measure of how much the fluid differs from an ideal gas. Calculate the compression factor of a van der Waals fluid at the critical point, and note that the value is independent of a and b. (Ex­ perimental values of compression factors at the critical point are generally lower than the van der Waals prediction, for instance, 0.227 for H20 , 0.274 for C02, 0.305 for He.) Problem 5.51. When plotting graphs and performing numerical calculations, it is convenient to work in terms of reduced variables ,
t T/Tc ,
p
P/Pc,
v
V / Vc .
Rewrite the van der Waals equation in terms of these variables, and notice that the constants a and b disappear.
Problem 5.52. Plot the van der Waals isotherm for T /Tc = 0.95, working in terms of reduced variables. Perform the Maxwell construction (either graphically or numerically) to obtain the vapor pressure. Then plot the Gibbs free energy (in units of NkTc) as a function of pressure for this same temperature and check that this graph predicts the same value for the vapor pressure. Problem 5.53. Repeat the preceding problem for T / Tc
= 0.8 .
Problem 5.54. Calculate the Helmholtz free energy of a van der Waals fluid , up to an undetermined function of temperature as in equation 5.56. Using reduced variables, carefully plot the Helmholtz free energy (in units of NkTc) as a function of volume for T /Tc = 0.8. Identify the two points on the graph corresponding to the liquid and gas at the vapor pressure. (If you haven't worked the preceding problem, just read the appropriate values off Figure 5.23.) Then prove that the Helmholtz free energy of a combination of these two states (part liquid, part gas) can be represented by a straight line connecting these two points on the graph. Explain why the combination is more stable, at a given volume, than the homogeneous state represented by the original curve, and describe how you could have determined the two transition volumes directly from the graph of F. *Chapter 8 introduces an accurate approximation for treating weakly interacting gases, as well as the more general technique of Monte Carlo simulation, which can be applied to
dense fluids .
185
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Chapter 5
Free Energy and Chemical Thermodynamics
Problem 5.55. In this problem you will investigate the behavior of a van der Waals fluid near the critical point. It is easiest to work in terms of reduced variables throughout. (a) Expand the van der Waals equation in a Taylor series in (V - Vc ), keeping terms through order (V - Vc )3. Argue that, for T sufficiently close to T c ) the term quadratic in (V - Vc ) becomes negligible compared to the others and may be dropped. (b) The resulting expression for P(V) is antisymmetric about the point V Vc. Use this fact to find an approximate formula for the vapor pressure as a function of temperature. (You may find it helpful to plot the isotherm.) Evaluate the slope of the phase boundary, dP/ dT, at the critical point. (c) Still working in the same limit, find an expression for the difference in volume between the gas and liquid phases at the vapor pressure. You should find (Vg - Vi) ex: (Tc - Tl, where (3 is known as a critical exponent. Experiments show that (3 has a universal value of about 1/3, but the van der Waals model predicts a larger value. (d) Use the previous result to calculate the predicted latent heat of the trans­ formation as a function of temperature, and sketch this function. (e) The shape of the T = Tc isotherm defines another critical exponent, called 0: (P - Pc) ex: (V - Vc )8. Calculate 0 in the van der Waals modeL (Experi­ mental values of 0 are typically around 4 or 5.) (f) A third critical exponent describes the temperature dependence of the iso­ thermal compressibility,
~ =-~(~~)T' This quantity diverges at the critical point, in proportion to a power of (T - Tc) that in principle could differ depending on whether one approaches the critical point from above or below. Therefore the critical exponents, and ' are defined by the relations ~ex:
(T Tc)-'Y { (Tc - T) -'Y
as T
-l-
Tc from above,
as T
-l-
Tc from below.
I
Calculate ~ on both sides of the critical point in the van der Waals model, and show that, ,'in this model.
5.4 Phase Transformations of Mixtures Phase transformations become a lot more complicated when a system contains two or more types of particles. Consider air, for example, a mixture of approximately 79% nitrogen and 21 % oxygen (neglecting various minor components for simplicity). What happens when you lower the temperature of this mixture, at atmospheric pressure? You might expect that all the oxygen would liquefy at 90.2 K (the boiling point of pure oxygen), leaving a gas of pure nitrogen which would then liquefy at 77.4 K (the boiling point of pure nitrogen). In fact, however, no liquid at all forms until the temperature drops to 81.6 K, when a liquid consisting of 48% oxygen begins to condense. Similar behavior occurs in liquid-solid transitions, such as the crystallization of alloys and igneous rocks. How can we understand this behavior?
5.4
Phase Thansformations of Mixtures
Free Energy of a Mixture As usual, the key is to look at the (Gibbs) free energy,
G
U +PV -TS.
(5.58)
and suppose that Let's consider a system of two types of molecules, A and they are initially separated, sitting side by side at the same temperature and pres­ sure Figure 5.24). Imagine varying the proportions of A and B while holding say at one mole. Let be the free energy the total number of molecules of a mole of pure A, and the free energy of a mole of pure B. For an unmixed combination of part A and part B, the total free energy is just the sum of the separate free energies of the two subsystems:
G
=
(l-x)Gl
+ xGB
(unmixed),
(5.59)
where x is the fraction of B molecules, so that x = 0 for pure A and x = 1 for pure B. A graph of G vs. x for this unmixed system is a straight line, as shown in 5.25. Now suppose that we remove the partition between the two sides and stir the A and B molecules together to form a homogeneous mixture. (I'll use the term mixture only when the substances are mixed at the molecular level. A 'mixture' of salt and pepper does not qualify.) What happens to the free energy? From the definition G = U +PV -TS, we see that G can change because of changes in U, V, and/or S. The energy, U, might increase or decrease, depending on how the forces between dissimilar molecules compare to the forces between identical molecules. The volume, as well, may increase or decrease depending on these forces and on the shapes of the molecules. The entropy, however, will most certainly increase, because there are now many more possible ways to arrange the molecules. As a first approximation, therefore, let us neglect any changes in U and V and assume that the entire change in G comes from the entropy of mixing. As a further simplification, let's also assume that the entropy of can be calculated as in Problem 2.38, so that for one ~Smixing
-R[xlnx + (l-x)ln(l-x)].
(5.60)
A graph of this expression is shown in Figure 5.25. This expression is correct for ideal gases, and also for liquids and solids when the two types of molecules are the same size and have no 'preference' for having like or unlike neighbors. When this A
B
Mixed
Figure 5.24. A collection of two types of molecules, before and after mixing.
187
188
Chapter 5
Free Energy and Chemical Thermodynamics
~Smixing
o
1 Pure B
Pure A
o
1
Pure A
PureB
Figure 5.25. Before mixing, the free energy of a collection of A and B molecules is a linear function of x = NB/(NA + NB). After mixing it is a more complicated function; shown here is the case of an 'ideal' mixture, whose entropy of mixing is shown at right. Although it isn't obvious on this scale, the graphs of both ~Smixing and G (after mixing) have vertical slopes at the endpoints.
expression for the mixing entropy holds and when U and V do not change upon mixing, the free energy of the mixture is
G
=
s+ RT[x lnx + (I-x) In(l-x)]
(l-x)GA+ xG
(ideal mixture).
(5.61)
This function is plotted in Figure 5.25. A mixture having this simple free energy function is called an ideal mixture. Liquid and solid mixtures rarely come close to being ideal, but the ideal case still makes a good starting point for arriving at some qualitative understanding. One important property of expression 5.60 for the entropy of mixing is that its derivative with respect to x goes to infinity at x = 0 and to minus infinity at x = 1. The graph of this expression therefore has a vertical slope at each endpoint. Similarly, expression 5.61 for the Gibbs free energy has an infinite derivative at each endpoint: Adding a tiny amount of impurity to either pure substance lowers the free energy significantly, except when T = 0.* Although the precise formulas written above hold only for ideal solutions, the infinite slope at the endpoints is a general property of the free energy of any mixture. Because a system will spontaneously seek out the state of lowest free energy, this property tells us that equilibrium phases almost always contain impurities. Nonideal mixtures often have the same qualitative properties as ideal mixtures, but not always. The most important exception is when mixing the two substances increases the total energy. This happens in liquids when unlike molecules are less attracted to each other than are like molecules, as with oil and water. The energy change upon mixing is then a concave-down function of x, as shown in Figure 5.26. At T 0 the free energy (G = U + PV T S) is also a concave-down function * Hiding one needle in a stack of pure hay increases the entropy a lot more than does adding a needle to a haystack already containing thousands of needles.
5.4
Phase Transformations of Mixtures
G
o Pure A
1 Pure B
Highest T
o
x----..
1
Figure 5.26. Mixing A and B can often increase the energy of the system; shown
at left is the simple case where the mixing energy is a quadratic function (see
Problem 5.58). Shown at right is the free energy in this case, at four different
temperatures.
(if we any change in V upon mixing). At nonzero T, however, there is a competition in G between the concave-down contribution from the mixing energy and the concave-up contribution from -T times the entropy. At sufficiently high T the entropy contribution always wins and G is everywhere concave-up. But even at very low nonzero T, the entropy contribution still dominates the shape of G near the endpoints x = 0 and x 1. This is because the entropy of mixing has an infinite derivative at the endpoints, while the energy of mixing has only a finite derivative at the endpoints: When there is very little impurity, the mixing energy is simply proportional to the number of impurity molecules. Thus, at small nonzero the free energy function is concave-up near the endpoints and concave-down near the middle, as shown in Figure 5.26. But a concave-down free energy function indicates an unstable mixture. Pick any two points on the graph of G and connect them with a straight line. This line denotes the free energy of an unmixed combination of the two phases represented by the endpoints (just as the straight line in Figure 5.25 denotes the free energy of the unmixed pure phases). Whenever the graph of G is concave-down, you can draw a straight connecting line that lies below the curve, and therefore the unmixed combi­ nation has a lower free energy than the homogeneous mixture. The lowest possible connecting line intersects the curve as a tangent at each end (see Figure 5.27). The tangent points indicate the compositions of the two separated phases, denoted Xa and Xb in the figure. Thus, if the composition of the lies between Xa and Xb, it will spontaneously separate into an A-rich phase of composition Xa and a B-rich phase of composition Xb. We say that the system has a solubility gap, or that the two phases are immiscible. Decreasing the·temperature of this system widens the solubility gap (see Figure 5.26), while increasing the temperature narrows the gap until it disappears when G is everywhere concave-up.
189
190
Chapter 5
Free Energy and Chemical Thermodynamics
G
Figure 5.27. To construct the equilibrium free energy curve, draw the lowest possible straight line across the concave-down sec­ tion, tangent to the curve at both ends. At compositions between the tangent points the mixture will spontaneously sep­ arate into phases whose compo­ sitions are Xa and Xb, in order to lower its free energy.
Homogeneous
a and
o
Xa
If we plot the compositions Xa and Xb at each temperature, we obtain a T vs. x phase diagram like those shown in Figure 5.28. Above the curve the equilibrium state is a single homogeneous mixture, while below the curve the system separates into two phases whose compositions lie on the curve. For the familiar case of oil and water at atmospheric pressure, the critical temperature where complete mixing would occur is far above the boiling point of water. The figure shows a less familiar mixture, water and phenol (C 6 H5 0H), whose critical mixing temperature is 67°C. Solubility gaps occur in solid as well as liquid mixtures. For solids, however, there is often the further complication that the pure-A and pure-B solids have qualitatively different crystal structures. Let us call these structures Q: and (3, respectively. Adding a few B molecules to the Q: phase, or a few A molecules
70
Homogeneous mixture
t
T
Two separated phases
o Pure
1
A
PureB
o Water
1
Phenol
Figure 5.28. Left: Phase diagram for the simple model system whose mixing energy is plotted in Figure 5.26. Right: Experimental data for a real system, water + phenol, that shows qualitatively similar behavior. Adapted with permission from Alan N. Campbell and A. Jean R. Campbell, Journal of the American Chemical Society 59, 2481 (1937). Copyright 1937 American Chemical Society.
5.4
Phase Transformations of Mixtures
to the f3 phase, is always possible, again because of the infinite slope of the entropy of mixing function. But a large amount of impurity will usually stress the crystal lattice significantly, greatly raising is energy. A free energy diagram for such a system therefore looks like Figure 5.29. Again we can draw a straight line tangent to the two concave-up sections, so there is a solubility gap: The stable configu­ ration at intermediate compositions is an unmixed combination of the two phases indicated by the tangent points. For some solids the situation is even more com­ plicated because other crystal structures are stable at intermediate compositions. For example, brass, an alloy of copper and zinc, has five possible crystal structures, each stable within a certain composition range.
Figure 5.29. Free energy graphs
for a mixture of two solids with dif­
ferent crystal structures, a and {3.
Again, the lowest possible straight
connecting line indicates the range
of compositions where an unmixed
combination of a and b phases is
more stable than a homogeneous
mixture.
Problem 5.56. Prove that the entropy of mixing of an ideal mixture has an infinite slope, when plotted vs. x, at x = 0 and x = l. Problem 5.57. Consider an ideal mixture of just 100 molecules, varying in com­ position from pure A to pure B. Use a computer to calculate the mixing entropy as a function of NA, and plot this function (in units of k). Suppose you start with all A and then convert one molecule to type B; by how much does the entropy increase? By how much does the entropy increase when you convert a second molecule, and then a third, from A to B? Discuss. Problem 5.58. In this problem you will model the mixing energy of a mixture in a relatively simple way, in order to relate the existence of a solubility gap to molecular behavior. Consider a mixture of A and B molecules that is ideal in every way but one: The potential energy due to the interaction of neighboring molecules depends upon whether the molecules are like or unlike. Let n be the average number of nearest neighbors of any given molecule (perhaps 6 or 8 or 10). Let Uo be the average potential energy associated with the interaction between neighboring molecules that are the same (A-A or B-B), and let UAB be the potential energy associated with the interaction of a neighboring unlike pair (A-B). There are no interactions beyond the range of the nearest neighbors; the values of Uo and UAB are independent of the amounts of A and B; and the entropy of mixing is the same as for an ideal solution. (a) Show that when the system is unmixed, the total potential energy due to all neighbor-neighbor interactions is ~Nnuo. (Hint: Be sure to count each
191
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Chapter 5
Free Energy and Chemical Thermodynamics neighboring pair only once.)
(b) Find a formula for the total potential energy when the system is mixed, in terms of x, the fraction of B. (c) Subtract the results of parts (a) and (b) to obtain the change in energy upon mixing. Simplify the result as much as possible; you should obtain an expression proportional to x(l-x). Sketch this function vs. x, for both possible signs OfUAB Uo· (d) Show that the slope of the mixing energy function is finite at both end­ points, unlike the slope of the mixing entropy function. (e) For the case U AB > uo, plot a graph of the Gibbs free energy of this system vs. x at several temperatures. Discuss the implications.
(f) Find an expression for the maximum temperature at which this system has a solubility gap.
(g) Make a very rough estimate of
UAB - 'Uo
for a liquid mixture that has a
solubility gap below 100°C. (h) Use a computer to plot the phase diagram (T vs. x) for this system.
Phase Changes of a Miscible Mixture Now let us return to the process described at the beginning of this section, the liquefaction of a mixture of nitrogen and oxygen. Liquid nitrogen and oxygen are completely miscible, so the free energy function of the liquid mixture is everywhere concave-up. The free energy of the gaseous mixture is also everywhere concave-up. By considering the relation between these two functions at various temperatures, we can understand the behavior of this system and sketch its phase diagram. Figure 5.30 shows the free energy functions of a model system that behaves as an ideal mixture in both the gaseous and liquid phases. Think of the components A and B as nitrogen and oxygen, whose behavior should be qualitatively similar. The boiling points of pure A and pure B are denoted TA and T B , respectively. At temperatures greater than TB the stable phase is a gas regardless of composition, so the free energy curve of the gas lies entirely below that of the liquid. As the temperature drops, both free energy functions increase (BC/BT = -S), but that of the gas increases more because the gas has more entropy. At T = TB the curves intersect at x l , where the liquid and gas phases of pure B are in equilibrium. As T decreases further the intersection point moves to the left, until at T = TA the curves intersect at x = O. At still lower temperatures the free energy of the liquid is less than that of the gas at all compositions. At intermediate temperatures, between TA and TB , either the liquid or the gas phase may be more stable, depending on composition. But notice, from the shape of the curves, that you can draw a straight line, tangent to both curves, that lies below both curves. Between the two tangent points, therefore, the stable configuration is an unmixed combination of a gas whose composition is indicated by the left tangent point and a liquid whose composition is indicated by the right tangent point. The straight line denotes the free energy of this unmixed combination. By drawing such a straight line for every temperature between TA and T B , we can construct the T vs. x phase diagram for this system. The mixture is entirely gas in the upper region
5.4
Phase Transformations of Mixtures
G
G
,,
, ,,, I,'
,
,~
I,' ,
.. ,,1
' ' '
,
, ,
71
, ,' ,J.'
, J
, 1' 1.- .......
-------,
'
,',.;' , .. '
I
,
1
T
t
'',
T < TA
' '
-----------------------'
o Pure A
1
PureB
o Pure A
1
PureB
Figure 5.30. The five graphs at left show the liquid and gas free energies of an
ideal mixture at temperatures above, below, at, and between the boiling points
TA and TB. Three graphs at intermediate temperatures are shown at right, along
with the construction of the phase diagram.
of the diagram, entirely liquid in the lower region, and an unmixed combination in the region between the two curves. Figure 5.31 shows the experimental phase diagram for the nitrogen-oxygen sys­ tem. Although this diagram isn't exactly the same as that of the ideal A-B model system, it has all the same qualitative features. From the diagram you can see that if you start with an air-like mixture of 79% nitrogen and 21% oxygen and lower the temperature, it remains a gas until the temperature reaches 81.6 K. At this point a liquid begins to condense. A horizontal line at this temperature intersects the lower curve at x = 0.48, so the liquid is initially 48% oxygen. Because oxygen condenses more easily than nitrogen, the liquid is enriched in oxygen compared to
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Chapter 5
Free Energy and Chemical Thermodynamics 92 90 88
t g h
86 84 82 80 78 76
0 Pure N2
0.2
0.4
0.6 X___00010001o_
0.8
1.0 Pure 02
Figure 5.31. Experimental phase diagram for nitrogen and oxygen at atmospheric pressure. Data from International Critical Tables (volume 3), with endpoints ad­ justed to values in Lide (1994).
the gas. But it is not pure oxygen, because the entropy of mixing gives too much of a thermodynamic advantage to impure phases. As the temperature decreases further, the gas becomes depleted of oxygen and its composition follows the upper curve, down and to the left. IvIeanwhile the composition of the liquid follows the lower curve, also increasing its nitrogen/oxygen ratio. At 79.0 K the liquid com­ position reaches the overall composition of 21 % oxygen, so there can't be any gas left; the last bit of gas to condense contains about 7% oxygen. The liquid-gas transitions of many other mixtures behave similarly. Further­ more, for some mixtures the solid-liquid transition behaves in this way. Examples of such mixtures include copper-nickel, silicon-germanium, and the common miner­ als olivene (varying from Fe2Si04 to :Mg2Si0 4) and plagioclase feldspar (see Prob­ lem 5.64). In all of these systems, the crystal structure of the solid is essentially the same throughout the entire range of composition, so the two pure solids can form approximately ideal mixtures in all proportions. Such a mixture is called a solid solution. Problem 5.59. Suppose you cool a mixture of 50% nitrogen and 50% oxygen until it liquefies. Describe the cooling sequence in detail, including the temperatures and compositions at which liquefaction begins and ends. Problem 5.60. Suppose you start with a liquid mixture of 60% nitrogen and 40% oxygen. Describe what happens as the temperature of this mixture increases. Be sure to give the temperatures and compositions at which boiling begins and ends. Problem 5.61. Suppose you need a tank of oxygen that is 95% pure. Describe a process by which you could obtain such a gas, starting with air. Problem 5.62. Consider a completely miscible two-component system whose overall composition is X, at a temperature where liquid and gas phases coexist. The composition of the gas phase at this temperature is Xu and the composition
5.4
Phase Transformations of Mixtures
of the liquid phase is Xb. Prove the lever rule, which says that the proportion of liquid to gas is (x - Xa)/(Xb - x). Interpret this rule graphically on a phase diagram. Problem 5.63. Everything in this section assumes that the total pressure of the system is fixed. How would you expect the nitrogen-oxygen phase diagram to change if you increase or decrease the pressure? Justify your answer. Problem 5.64. Figure 5.32 shows the phase diagram of plagioclase feldspar, which can be considered a mixture of albite (NaAlSi308) and anorthite (CaAbSi208). (a) Suppose you discover a rock in which each plagioclase crystal varies in composition from center to edge, with the centers of the largest crystals composed of 70% anorthite and the outermost parts of all crystals made of essentially pure albite. Explain in some detail how this variation might arise. What was the composition of the liquid magma from which the rock formed?
(b) Suppose you discover another rock body in which the crystals near the top are albite-rich while the crystals near the bottom are anorthite-rich. Explain how this variation might arise. 1600 1500 1400
6'
~
N 1300 1200 1100 0.0 Albite
0.2
0.6 0.4 x -----.
0.8
1.0 Anorthite
Figure 5.32. The phase diagram of plagioclase feldspar (at atmospheric pressure). From N. L. Bowen, 'The Melting Phenomena of the Plagioclase Feldspars,' American Journal of Science 35, 577-599 (1913).
Problem 5.65. In constructing the phase diagram from the free energy graphs in Figure 5.30, I assumed that both the liquid and the gas are ideal mixtures. Suppose instead that the liquid has a substantial positive mixing energy, so that its free energy curve, while still concave-up, is much flatter. In this case a portion of the curve may still lie above the gas's free energy curve at TA. Draw a qualitatively accurate phase diagram for such a system, showing how you obtained the phase diagram from the free energy graphs. Show that there is a particular composition at which this gas mixture will condense with no change in composition. This special composition is called an azeotrope.
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Problem 5.66. Repeat the previous problem for the opposite case where the liquid has a substantial negative mixing energy, so that its free energy curve dips below the gas's free energy curve at a temperature higher than TB . Construct the phase diagram and show that this system also has an azeotrope. Problem 5.67. In this problem you will derive approximate formulas for the shapes of the phase boundary curves in diagrams such as Figures 5.31 and 5.32, assuming that both phases behave as ideal mixtures. For definiteness, suppose that the phases are liquid and gas. (a) Show that in an ideal mixture of A and B, the chemical potential of species A can be written /-LA
iA + kTln(l -
x),
where /-LA is the chemical potential of pure A (at the same temperature and pressure) and x NBI(NA + N B ). Derive a similar formula for the chemical potential of species B. Note that both formulas can be written for either the liquid phase or the gas phase. (b) At any given temperature T, let Xl and Xg be the compositions of the liquid and gas phases that are in equilibrium with each other. By setting the appropriate chemical potentials equal to each other, show that Xl and Xg obey the equations and Xg
where !J.Go represents the change in G for the pure substance undergoing the phase change at temperature T. (c) Over a limited range of temperatures, we can often assume that the main temperature dependence of !J.Go = !J.Ho T !J.So comes from the ex­ plicit T; both !J.Ho and !J.So are approximately constant. With this sim­ plification, rewrite the results of part (b) entirely in terms of !J.HA, !J.Hi3, TAl and TB (eliminating!J.G and !J.S). Solve for Xl and Xg as functions ofT. (d) Plot your results for the nitrogen-oxygen system. The latent heats of the pure substances are !J.HN2 5570 J/mol and !J.H02 = 6820 J/mol. Com­ pare to the experimental diagram, Figure 5.31. (e) Show that you can account for the shape of Figure 5.32 with suitably chosen !J.Ho values. What are those values?
Phase Changes of a Eutectic System
:NIost two-component solid mixtures do not maintain the same crystal structure over the entire range of composition. The situation shown in Figure 5.29 is more common: two different crystal structures, a and {3, at compositions close to pure A and pure B, with an unmixed combination of a and {3 being stable at intermediate compositions. Let us now consider the solid-liquid transitions of such a system, assuming that A and B are completely miscible in the liquid phase. Again the idea is to look at the free energy functions at various temperatures (see Figure 5.33). For definiteness, suppose that TB, the melting temperature of pure B, is higher than TA, the melting temperature of pure A.
5.4
Phase Transformations of Mixtures
G
G ,
I
' .. ~ ....
--­
I
o Pure A
1
PureE
Figure 5.33. Construction of the phase diagram of a eutectic system from free energy graphs.
At high temperatures the free energy of the liquid will be below that of either solid phase. Then, as the temperature decreases, all three free-energy functions will increase (80/8T = - S), but the free energy of the liquid will increase fastest because it has the most entropy. Below TB the liquid's free energy curve intersects that of the (3 phase, so there is a range of compositions for which the stable config­ uration is an unmixed combination of liquid and (3. As the temperature decreases this range widens and reaches further toward the A side of the diagram. Even­ tually the liquid curve intersects the 0: curve as well and there is an A-rich range of compositions for which the stable phase is an unmixed combination of liquid
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Chapter 5
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Problem 5.66. Repeat the previous problem for the opposite case where the liquid has a substantial negative mixing energy, so that its free energy curve dips below the gas's free energy curve at a temperature higher than TB. Construct the phase diagram and show that this system also has an azeotrope. Problem 5.67. In this problem you will derive approximate formulas for the shapes of the phase boundary curves in diagrams such as Figures 5.31 and 5.32, assuming that both phases behave as ideal mixtures. For definiteness, suppose that the phases are liquid and gas.
(a) Show that in an ideal mixture of A and B, the chemical potential of species A can be written tlA
tlA
+ kTln(1- x),
where tlA is the chemical potential of pure A (at the same temperature and pressure) and x N B I (NA + N B)' Derive a similar formula for the chemical potential of species B. Note that both formulas can be written for either the liquid phase or the gas phase. (b) At any given temperature T, let Xl and Xg be the compositions of the liquid and gas phases that are in equilibrium with each other. By setting the appropriate chemical potentials equal to each other, show that Xl and Xg obey the equations and
Xg
e t:.G~/RT ,
where b.Go represents the change in G for the pure substance undergoing the phase change at temperature T. (c ) Over a limited range of temperatures, we can often assume that the main temperature dependence of b.Go = b.Ho T b.So comes from the ex­ plicit T; both b.Ho and b.So are approximately constant. With this sim­ plification, rewrite the results of part (b) entirely in terms of b.HJ., b.HB, TAl and TB (eliminating b.G and b.S). Solve for Xl and Xg as functions ofT. (d) Plot your results for the nitrogen-oxygen system. The latent heats of the pure substances are b.HN2 5570 J Imol and b.H02 = 6820 J Imol. Com­ pare to the experimental diagram, Figure 5.31.
(e) Show that you can account for the shape of Figure 5.32 with suitably chosen b.Ho values. What are those values?
Phase Changes of a Eutectic System Most two-component solid mixtures do not maintain the same crystal structure over the entire range of composition. The situation shown in Figure 5.29 is more common: two different crystal structures, a and (3, at compositions close to pure A and pure B, with an unmixed combination of a and (3 being stable at intermediate compositions. Let us now consider the solid-liquid transitions of such a system, assuming that A and B are completely miscible in the liquid phase. Again the idea is to look at the free energy functions at various temperatures (see Figure 5.33). For definiteness, suppose that TB, the melting temperature of pure B, is higher than TAl the melting temperature of pure A.
5.4
Phase Transformations of Mixtures
G
t
T
G ,
I
... .1 [
...... ­ ..
­
I
o
1
Pure A
PureB
Figure 5.33. Construction of the phase diagram of a eutectic system from free energy graphs.
At high temperatures the free energy of the liquid will be below that of either solid phase. Then, as the temperature decreases, all three free-energy functions will increase (BG/ BT = - S), but the free energy of the liquid will increase fastest because it has the most entropy. Below TB the liquid's free energy curve intersects that of the (3 phase, so there is a range of compositions for which the stable config­ uration is an unmixed combination of liquid and (3. As the temperature decreases this range widens and reaches further toward the A side of the diagram. Even­ tually the liquid curve intersects the a curve as well and there is an A-rich range of compositions for which the stable phase is an unmixed combination of liquid
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Chapter 5
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and 0:. As T decreases further this range reaches toward the B side of the diagram until finally it intersects the liquid + f3 range at the eutectic point. At still lower temperatures the stable configuration is an unmixed combination of the 0: and f3 solids; the free energy of the liquid is higher than that of this combination. The eutectic point defines a special composition at which the melting temper­ ature is as low as possible, lower than that of either pure substance. A liquid near the eutectic composition remains stable at low temperatures because it has more mixing entropy than the unmixed combination of solids would have. (A solid mixture would have about as much mixing entropy, but is forbidden by the large positive mixing energy that results from stressing the crystal lattice.) A good example of a eutectic mixture is the tin-lead solder used in electrical circuits. Figure 5.34 shows the phase diagram of the tin-lead system. Common electrical solder is very close to the eutectic composition of 38% lead by weight (or 26% by number of atoms). Using this composition has several advantages: the melting temperature is as low as possible (183°C); the solder freezes suddenly rather than gradually; and the cooled metal is relatively strong, with small crystals of the two phases uniformly alternating at the microscopic scale. Many other mixtures behave in a similar way. Most pure liquid crystals freeze at inconveniently high temperatures, so eutectic mixtures are often used to obtain liquid crystals for use at room temperature. A less exotic mixture is water + table salt (NaCI), which can remain a liquid at temperatures as low as -21°C, at the eutectic composition of 23% NaCI by weight.* Another familiar example is the coolant used in automobile engines, a mixture of water and ethylene glycol (HOCH 2 CH 2 0H). Pure water freezes at O°C, and pure ethylene glycol at -13°C,
10
Atomic percent lead 20 40 50 60 70 80 90 30
300
G
~
200
N
100 0 Sn
10
20
30 40 50 60 70 Weight percent lead
80
90
100 Pb
Figure 5.34. Phase diagram for mixtures of tin and lead. From Thaddeus B.
Massalski, ed., Binary Alloy Phase Diagrams, second edition (ASM International,
Materials Park, OR, 1990).
*The water
+
NaCl phase diagram is shown in Zemansky and Dittman (1997).
5.4
Phase Transformations of Mixtures
so neither would remain a liquid on winter nights in a cold climate. Fortunately, a 50-50 mixture (by volume) of these two liquids does not begin to freeze until the temperature reaches -31°C. The eutectic point is lower still, at -49°C and a composition of 56% ethylene glycol by volume. * Although the phase diagram of a eutectic system may seem complicated enough, many two-component systems are further complicated by the existence of other crystal structures of intermediate compositions; Problems 5.71 and 5.72 explore some of the possibilities. Then there are three-component systems, for which the composition axis of the phase diagram is actually a plane (usually represented by a triangle). You can find hundreds of intricate phase diagrams in books on metall urgy, ceramics, and petrology. All can be understood qualitatively in terms of free energy graphs as we have done here. Because this is an introductory text, though, let us move on and explore the properties of some simple mixtures more quantitatively. Problem 5.68. Plumber's solder is composed of 67% lead and 33% tin by weight. Describe what happens to this mixture as it cools, and explain why this composi­ tion might be more suitable than the eutectic composition for joining pipes. Problem 5.69. What happens when you spread salt crystals over an icy sidewalk? Why is this procedure rarely used in very cold climates? Problem 5.70. What happens when you add salt to the ice bath in an ice cream maker? How is it possible for the temperature to spontaneously drop below O°C? Explain in as much detail as you can. Problem 5.71. Figure 5.35 (left) shows the free energy curves at one particu­ lar temperature for a two-component system that has three possible solid phases (crystal structures), one of essentially pure A, one of essentially pure B, and one of intermediate composition. Draw tangent lines to determine which phases are present at which values of x. To determine qualitatively what happens at other temperatures, you can simply shift the liquid free energy curve up or down (since the entropy of the liquid is larger than that of any solid). Do so, and construct
G
G
.
..
.
,
,, ,, ,
Liquid,,' ,,
Liquid ,..
- . _-------_ . -­
------- --- --
o
x-------'
1
o
1
Figure 5.35. Free energy diagrams for Problems 5.71 and 5.72. *For the full phase diagram see J. Bevan Ott, J. Rex Goates, and John D. Lamb, Journal of Chemical Thermodynamics 4, 123-126 (1972).
199
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Chapter 5
Free Energy and Chemical Thermodynamics
a qualitative phase diagram for this system. You should find two eutectic points. Examples of systems with this behavior include water + ethylene glycol and tin + magnesium. Problem 5.72. Repeat the previous problem for the diagram in Figure 5.35 (right), which has an important qualitative difference. In this phase diagram, you should find that {3 and liquid are in equilibrium only at temperatures below the point where the liquid is in equilibrium with infinitesimal amounts of 0: and {3. This point is called a peritectic point. Examples of systems with this behavior include water + NaCI and leucite + quartz.
5.5 Dilute Solutions A solution is the same thing as a mixture, except that we think of one component (the solvent) as being primary and the other components (the solutes) as being secondary. A solution is called dilute if the solute molecules are much less abundant than the solvent molecules (see Figure 5.36), so that each solute molecule is 'always' surrounded by solvent molecules and 'never' interacts directly with other solute molecules. In many ways the solute in a dilute solution behaves like an ideal gas. We can therefore predict many of the properties of a dilute solution (including its boiling and freezing points) quantitatively.
Figure 5.36. A dilute solution, in which the solute is much less abundant than the solvent.
Solvent and Solute Chemical Potentials To predict the properties of a dilute solution interacting with its environment, we'll need to know something about the chemical potentials of the solvent and solutes. The chemical potential, J.1A, of species A is related to the Gibbs free energy by J.1A = 8G/8NA, so what we need is a formula for the Gibbs free energy of a dilute solution in terms of the numbers of solvent and solute molecules. Coming up with the correct formula for G is a bit tricky but very worthwhile: Once we have this formula, a host of applications become possible. Suppose we start with a pure solvent of A molecules. Then the Gibbs free energy is just NA times the chemical potential:
G = N AJ.1o(T, P)
(pure solvent),
(5.62)
where J.10 is the chemical potential of the pure solvent, a function of temperature and pressure.
5.5
Dilute Solutions
Now imagine that we add a single B molecule, holding the temperature and pressure fixed. Under this operation the (Gibbs) free energy changes by de
= dU + P dV - T dS.
(5.63)
The important thing about dU and P dV is that neither depends on N A : Both depend only on how the B molecule interacts with its immediate neighbors, re­ gardless of how many other A molecules are present. For the T dS term, however, the situation is more complicated. Part of dS is independent of N A , but another part comes from our freedom in choosing where to put the B molecule. The num­ ber of choices is proportional to NA, so this operation increases the multiplicity by a factor proportional to NA, and therefore dS, the increase in entropy, includes a term k InNA: (5.64) dS = kIn NA + (terms independent of NA)' The change in the free energy can therefore be written de
= f(T, P) - kTlnNA
(adding one B molecule),
(5.65)
where f(T, P) is a function of temperature and pressure but not of NA. Next imagine that we add two B molecules to the pure solvent. For this oper­ ation we can almost apply the preceding argument twice and conclude that de = 2f(T, P) - 2kT In NA
(wrong).
(5.66)
The problem is that there is a further change in entropy resulting from the fact that the two B molecules are identical. Interchanging these two molecules does not result in a distinct state, so we need to divide the multiplicity by 2, or subtract kIn 2 from the entropy. With this modification, de
= 2f(T,P) - 2kTlnNA + kTln2
(adding two B molecules).
(5.67)
The generalization to many B molecules is now straightforward. In the free energy we get NB times f(T, P) and NB times -kTlnNA. To account for the inter­ changeability of the B molecules, we also get a term kT In N B! ~ kTN B (In N B-1). Adding all these terms to the free energy of the pure solvent, we finally obtain (5.68) This expression is valid in the limit NB « N A , that is, when the solution is dilute. For a nondilute solution the situation would be much more complicated because the B molecules would also interact with each other. If a solution contains more than one solute, all the terms in equation 5.68 except the first get repeated, with NB replaced by Nc, ND, and so on. The solvent and solute chemical potentials follow immediately from equation 5.68:
J.1-A =
( ae)
NBkT
= J.1-o(T,P) - - N ;
aN
A
T ,P ,NB
(5.69)
A
(5.70)
201
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Chapter 5
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As we would expect, adding more solute reduces the chemical potential of A and increases the chemical potential of B. Note also that both of these quantities, being intensive, depend only on the ratio NB/NA) not on the absolute number of solvent or solute molecules. It is conventional to rewrite equation 5.70 in terms of the molality* of the solution, which is defined as the number of moles of solute per kilogram of solvent: molality = m
moles of solute kilograms of solvent'
(5.71)
The molality is a constant times the ratio NB/NA , and the constant can be absorbed into the function f(T, P) to give a new function called fLO(T, P). The solute chemical potential can then be written (5.72) where mB is the molality of the solute (in moles per kilogram) and flO is the chemical L Values of flO can be obtained potential under the 'standard' condition mB from tables of Gibbs free energies, so equation 5.72 relates the tabulated value to the value at any other molality (so long as the solution is dilute). Problem 5.73. If expression 5.68 is correct, it must be extensive: Increasing both NA and NB by a common factor while holding all intensive variables fixed should increase G by the same factor. Show that expression 5.68 has this property. Show that it would not have this property had we not added the term proportional to InNB !. Problem 5.74. Check that equations 5.69 and 5.70 satisfy the identity G NAJ-LA + NBJ-LB (equation 5.37). Problem 5.75. Compare solution to expression 5.61 what circumstances should under these circumstances,
expression 5.68 for the Gibbs free energy of a dilute for the Gibbs free energy of an ideal mixture. Under these two expressions agree? Show that they do agree and identify the function f(T, P) in this case.
Osmotic Pressure As a first application of equation 5.69, consider a solution that is separated from some pure solvent by a membrane that allows only solvent molecules, not solute molecules, to pass through (see Figure 5.37). One example of such a semiperme­ able membrane is the membrane surrounding any plant or animal cell, which is permeable to water and other very small molecules but not to larger molecules or charged ions. Other semipermeable membranes are used in industry, for instance, in the desalination of seawater. According to equation 5.69, the chemical potential of the solvent in the solution is less than that of the pure solvent, at a given temperature and pressure. Particles *Molality is not the same as molarity, the number of moles of solute per liter of solution. For dilute solutions in water, however, the two are almost identicaL
5.5
Dilute Solutions
Semipermeable membrane Figure 5.31. When a solution is separated by a semipermeable membrane from pure solvent at the same temperature and pres­ sure, solvent will spontaneously flow into the solution.
•
tend to flow toward lower chemical potential, so in this situation the solvent mole­ cules will spontaneously flow from the pure solvent into the solution. This flow of molecules is called osmosis. That osmosis should happen is hardly surprising: Sol­ vent molecules are constantly bombarding the membrane on both sides, but more frequently on the side where the solvent is more concentrated, so naturally they hit the holes and pass through from that side more often. If you want to prevent osmosis from happening, you can do so by applying some additional pressure to the solution (see Figure 5.38). How much pressure is required? Well, when the pressure is just right to stop the osmotic flow, the chemical potential of the solvent must be the same on both sides of the membrane. Using equation 5.69, this condition is (5.73) where PI is the pressure on the side with pure solvent and P2 is the pressure on the side of the solution. Assuming that these two pressures are not too different, we can approximate (5.74) and plug this expression into equation 5.73 to obtain (P2
-
Pd 8J-lO = NBkT. 8P NA
(5.75)
To evaluate the derivative 8J-loI8P, recall that the chemical potential of a pure substance is just the Gibbs free energy per particle, GIN. Since 8GI8P = V (at
Figure 5.38. To prevent osmosis, P2 must exceed PI by an amount called the osmotic pressure.
203
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Chapter 5
Free Energy and Chemical Thermodynamics
fixed T and N), our derivative is
OJ.Lo OP
V
(5.76)
N'
the volume per molecule of the pure solvent. But since the solution is dilute, its volume per solvent molecule is essentially the same. Let's therefore take V in equation 5.76 to be the volume of the solution, and N to be the number of solvent molecules in the solution, that is, N A . Then equation 5.75 becomes (5.77) or simply
nBRT V
(5.78)
(where nB/V is the number of moles of solute per unit volume). This pressure difference is called the osmotic pressure. It is the excess pressure required on the side of the solution to prevent osmosis. Equation 5.78 for the osmotic pressure of a dilute solution is called van't Hoff's formula, after Jacobus Hendricus van't Hoff. It says that the osmotic pressure is exactly the same as the pressure of an ideal gas of the same concentration as the solute. In fact, it's tempting to think of the osmotic pressure as being exerted entirely by the solute, once we have balanced the pressure of the solvent on both sides. This interpretation is bad physics, but I still use it as a mnemonic aid to remember the formula. As an example, consider the solution of ions, sugars, amino acids, and other molecules inside a biological cell. In a typical cell there are about 200 water mole­ cules for each molecule of something else, so this solution is reasonably dilute. Since a mole of water has a mass of 18 g and a volume of 18 cm3 , the number of moles of solute per unit volume is
(
1 200 )
(
1 mol 18 cm3 )
(
100 cm 1m )
3
3
= 278 mol/m .
(5.79)
If you put a cell into pure water, it will absorb water by osmosis until the pres­ sure inside exceeds the pressure outside by the osmotic pressure, which at room temperature is
(278 moljm 3 )(8.3 J/mol·K)(300 K) = 6.9
X
105 N/m2,
(5.80)
or about 7 atm. An animal cell membrane subjected to this much pressure will burst, but plant cells have rigid walls that can withstand such a pressure.
5.5
Dilute Solutions
Problem 5.76. Seawater has a salinity of 3.5%, meaning that if you boil away a kilogram of seawater, when you're finished you'll have 35 g of solids (mostly NaCI) left in the pot. When dissolved, sodium chloride dissociates into separate Na+ and CI- ions. (a) Calculate the osmotic pressure difference between seawater and fresh water. Assume for simplicity that all the dissolved salts in seawater are N aCl. (b) If you apply a pressure difference greater than the osmotic pressure to a solution separated from pure solvent by a semipermeable membrane, you get reverse osmosis: a flow of solvent out of the solution. This process can be used to desalinate seawater. Calculate the minimum work required to desalinate one liter of seawater. Discuss some reasons why the actual work required would be greater than the minimum. Problem 5.77. Osmotic pressure measurements can be used to determine the molecular weights of large molecules such as proteins. For a solution of large molecules to qualify as 'dilute,' its molar concentration must be very low and hence the osmotic pressure can be too small to measure accurately. For this reason, the usual procedure is to measure the osmotic pressure at a variety of concentrations, then extrapolate the results to the limit of zero concentration. Here are some data* for the protein hemoglobin dissolved in water at 3°C: ~h
Concentration (grams/liter)
(cm)
5.6 16.6 32.5 43.4 54.0
2.0 6.5 12.8 17.6 22.6
The quantity ~h is the equilibrium difference in fluid level between the solution and the pure solvent, as shown in Figure 5.39. From these measurements, determine the approximate molecular weight of hemoglobin (in grams per mole).
Figure 5.39. An experimental arrange­ ment for measuring osmotic pressure. Solvent flows across the membrane from left to right until the difference in fluid level, ~h, is just enough to supply the osmotic pressure.
*From H. B. Bull, An Introduction to Physical Biochemistry, second edition (F. A. Davis, Philadelphia, 1971), p. 182. The measurements were made by H. Gutfreund.
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Problem 5.78. Because osmotic pressures can be quite large, you may wonder whether the approximation made in equation 5.74 is valid in practice: Is /-to re­ ally a linear function of P to the required accuracy? Answer this question by discussing whether the derivative of this function changes significantly, over the relevant pressure range, in realistic examples.
Boiling and Freezing Points In Section 5.4 we saw how impurities can shift the boiling and freezing points of a substance. For dilute solutions, we are now in a position to compute this shift quantitatively. Consider first the case of a dilute solution at its boiling point , when it is in equilibrium with its gas phase (see Figure 5.40). Assume for simplicity that the solute does not evaporate at all-this is an excellent approximation for salt in water, for instance. Then the gas contains no solute, so we need only consider the equilibrium condition for the solvent:
J.LA ,liq(T, P) = J.LA,gas(T, P).
(5.81)
Using equation 5.69 to rewrite the left-hand side, this condition becomes
J.Lo(T, P) -
NBkT NA = J.Lgas(T, P),
(5.82)
where J.Lo is the chemical potential of the pure solvent. Now, as in the osmotic pressure derivation above, the procedure is to expand each J.L function about the nearby point where the pure solvent would be in equi­ librium. Because J.L depends on both temperature and pressure, we can hold either fixed while allowing the other to vary. Let us first vary the pressure. Let Po be the vapor pressure of the pure solvent at temperature T, so that
J.Lo(T, Po) = J.Lgas(T, Po).
(5.83)
In terms of the chemical potentials at Po, equation 5.82 becomes
J.Lo(T, Po)
8J.Lo NBkT Po) 8P - ~ = J.Lgas(T, Po)
+ (P -
+ (P -
8J.Lgas
Po) 8P .
(5.84)
The first term on each side cancels by equation 5.83, and each 8J.L/8P is the volume . . . . . '
. ,
.
.
Figure 5.40. The presence of a solute reduces the tendency of a solvent to evaporate.
5.5
Dilute Solutions
.
(5.85)
per particle for that phase, so
(V)
(P - Po) N
NBkT . - NA
(V)
= (P - Po) N
hq
gas
The volume per particle in the gas phase is just kT I Po, while the volume per particle in the liquid is negligible in comparison. This equation therefore reduces to or
(5.86)
The vapor pressure is reduced by a fraction equal to the ratio of the numbers of solute and solvent molecules. This result is known as Raoult's law. At the molecular level, the reduction in vapor pressure happens because the addition of solute reduces the number of solvent molecules at the surface of the liquid-hence they escape into the vapor less frequently. Alternatively, we could hold the pressure fixed and solve for the shift in tem­ perature needed to maintain equilibrium in the presence of the solute. Let To be the boiling point of the pure solvent at pressure P, so that {Lo (To , P)
=
{Lgas (To , P).
(5.87)
In terms of the chemical potentials at To, equation 5.82 becomes {Lo(To, P )
8{Lo NBkT To ) 8T NA
+ (T -
= {Lgas
( ) ( ) 8{Lgas To, P + T - To 8T'
(5.88)
Again the first term on each side cancels. Each 8{L18T is just minus the entropy per particle for that phase (because 8C I aT = - 8), so
(8)
-(T - To) N
liq
NBkT - NA
=
-(T - To)
(8) N
.
(5.89)
gas
It's simplest to set the N under each 8 equal to NA, remembering that each 8 now refers to NA molecules of solvent. The difference in entropy between the gas and the liquid is LITo, where L is the latent heat of vaporization. Therefore the temperature shift is
6
nB RT
(5.90)
L
where I've approximated T ~ To on the right-hand side. As an example let's compute the boiling temperature of seawater. A convenient quantity to consider is one kilogram; then L is 2260 kJ. A kilogram of seawater contains 35 g of dissolved salts, mostly N aCl. The average atomic mass of N a and CI is about 29, so 35 g of salt dissolves into 35/29 = 1.2 moles of ions. Therefore the boiling temperature is shifted (relative to fresh water) by
T _
T,
o
= (1.2 mol)(8.3 J/mol·K)(373 K)2 = 2260 kJ
K 0.6
.
(5.91)
207
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Chapter 5
Free Energy and Chemical Thermodynamics
To compute the shift in the vapor pressure at a given temperature, we need to know that a kilogram of water contains 1000/18 = 56 moles of water molecules. Hence, by Raoult's law, fJ.P 1.2 mol = -0.022. (5.92) 56 mol Po Both of these effects are quite small: Seawater evaporates almost as easily as fresh water. Ironically, the shift in boiling temperature becomes large only for a nondilute solution, when the formulas of this section become inaccurate (though they can still give a rough estimate). A formula essentially identical to equation 5.90 gives the shift in the freezing temperature of a dilute solution. Because the proof is so similar, I'll let you do it (see Problem 5.81). I'll also let you think about why the freezing temperature is reduced rather than increased. For water and most other solvents the shift in freezing temperature is somewhat larger than the shift in boiling temperature, due to the smaller value of L. Together with osmotic pressure, the shifts in the vapor pressure, boiling temper­ ature, and freezing temperature are all known as colligative properties of dilute solutions. All of these effects depend only on the amount of solute, not on what the solute is. Problem 5.79. Most pasta recipes instruct you to add a teaspoon of salt to a pot of boiling water. Does this have a significant effect on the boiling temperature? Justify your answer with a rough numerical estimate. Problem 5.80. Use the Clausius-Clapeyron relation to derive equation 5.90 di­ rectly from Raoult's law. Be sure to explain the logic carefully. Problem 5.81. Derive a formula, similar to equation 5.90, for the shift in the freezing temperature of a dilute solution. Assume that the solid phase is pure solvent, no solute. You should find that the shift is negative: The freezing temper­ ature of a solution is less than that of the pure solvent. Explain in general terms why the shift should be negative. Problem 5.82. Use the result of the previous problem to calculate the freezing temperature of seawater.
5.6 Chemical Equilibrium One interesting fact about chemical reactions is that they hardly ever go to com­ pletion. Consider, for example, the dissociation of water into H+ and OH- ions: (5.93)
Under ordinary conditions, this reaction tends strongly to go to the left; an ordinary glass of water at equilibrium contains about 500 million water molecules for every pair of H+ and OH- ions. Naively, we tend to think of the water molecule as being 'more stable' than the ions. But this can't be the whole story-otherwise there would be no ions in a glass full of water, when in fact there are quadrillions of them.
5.6
Chemical Equilibrium
One way to understand why there are always some 'unstable' ions even at equi­ librium is to visualize the collisions at the molecular level. At room temperature, the water molecules are constantly colliding with each other at rather high speed. Every once in a while, one of these collisions is violent enough to break a molecule apart into two ions. The ions then tend to become separated, and do not recombine until they chance to meet new partners in the very dilute solution. Eventually an equilibrium is reached between the breaking apart and recombining, both of which occur rather rarely. At a more abstract level, we can think about equilibrium in terms of the Gibbs free energy. At room temperature and atmospheric pressure, the concentration of each species at equilibrium is determined by the condition that the total Gibbs free energy, (5.94) G = U - TS + PV, be minimized. We might expect that the minimum would occur when there are only water molecules, with no ions present. Indeed, it is true that a glass of water has much less Gibbs free energy than a glass full of H+ and OH- ions, simply because it has so much less energy. However, breaking just a few molecules apart into ions can lower the Gibbs free energy still further, because the entropy increases substantially. At higher temperature this entropy will contribute more to G, so there will be more ions. It is instructive to plot the Gibbs free energy as a function of the extent of the reaction, which in this case is the fraction x of the water molecules that are split into ions. If every water molecule is intact then x = 0, while if every molecule were dissociated then x would equal 1. If we were to keep the dissociated ions separate from the intact water, then a graph of G vs. x would be a straight line with a large positive slope (see Figure 5.41). When we let the ions mix with the molecules, however, the entropy of mixing introduces an additional concave-upward term in G. As discussed in Section 5.4, the derivative of this term is infinite at the endpoints x = 0 and x = 1. Therefore, no matter how great the energy difference between the reactants and the products, the equilibrium point-the minimum of G-will lie G Without mixing
Figure 5.41. If reactants and products remained sep­ arate, the free energy would be a linear function of the extent of the reaction. With mixing, however, G has a minimum somewhere be­ tween x = 0 and x = 1.
>',
'
With mixing
o
1
Reactants
Products
Equilibrium
209
210
Chapter 5
Free Energy and Chemical Thermodynamics
at least a little bit inward from the lower of the two endpoints. (In practice it may be closer than one part in Avogadro's number; in such cases the reaction effectively does go to completion.) We can characterize the equilibrium point by the condition that the slope of the Gibbs free energy graph is zero. This means that if one more H2 0 molecule dissociates, G is unchanged:
(5.95) In the last expression I've used the thermodynamic identity for G, plus the as­ sumption that the temperature and pressure are fixed. The sum runs over all three species: H 2 0, H+, and OH-. But the changes in the three N/s are not indepen­ dent: An increase of one H+ is always accompanied by an increase of one OH- and a decrease of one H2 0. One set of possible changes is
dNoH - = 1.
(5.96)
Plugging these numbers into equation 5.95 yields
(5.97) This relation among the chemical potentials must be satisfied at equilibrium. Since each chemical potential is a function of the concentration of that species (a higher concentration implying a higher chemical potential), this condition determines the various concentrations at equilibrium. Before generalizing this result to an arbitrary chemical reaction, let's consider another example, the combination of nitrogen and hydrogen to form ammonia:
(5.98) Again, the reaction is at equilibrium when is
I-£i
dNi = O. One possible set of dN's
(5.99) resulting in the equilibrium condition
(5.100) By now you can probably see the pattern: The equilibrium condition is always the same as the reaction equation itself, but with the names of the chemical species replaced by their chemical potentials and ~ replaced by =. To write this rule as a formula we need some notation. Let Xi represent the chemical name of the ith species involved in a reaction, and let Vi represent the stoichiometric coefficient of this species in the reaction equation, that is, the number of i molecules that participate each time the reaction happens. (For instance, v H2 = 3 in the previous example.) An arbitrary reaction equation then looks like this:
(5.101)
5.6
Chemical Equilibrium
In the corresponding equilibrium condition, we simply replace each species name with its chemical potential:
(5.102) The next step in understanding chemical equilibrium is to write each chemical potential /-Li in terms of the concentration of that species; then one can solve for the equilibrium concentrations. I could try to explain how to do this in general, but because gases, solutes, solvents, and pure substances must all be treated differently, I think it's easier (and more interesting) to demonstrate the procedure through the four worked examples that make up the rest of this section.
Problem 5.83. Write down the equilibrium condition for each of the following reactions:
(a) 2H
~
H2
(b) 2CO + 02 ~ 2C02 (c) CH4 + 202 ~ 2H20 + C02 (d) H2S04 ~ 2H+ + SO~-
(e) 2p
+ 2n
~ 4He
Nitrogen Fixation First consider the gaseous reaction 5.98, in which N2 and H2 combine to form ammonia (NH3). This reaction is called nitrogen 'fixation' because it puts the nitrogen into a form that can be used by plants to synthesize amino acids and other important molecules. The equilibrium condition for this reaction is written in equation 5.100. If we assume that each species is an ideal gas, we can use equation 5.40 for each chemical potential to obtain
Here each /-L0 represents the chemical potential of that species in its 'standard state,' when its partial pressure is po. Normally we take po to be 1 bar. Gathering all the /-L0 's on the right and all the logarithms on the left gives
Now if we multiply through by Avogadro's number, what's on the right is the 'standard' Gibbs free energy of the reaction, written t::.Go. This quantity is the hypothetical change in G when one mole of pure N2 reacts with three moles of pure H2 to form two moles of pure ammonia, all at 1 bar. The important thing about t::.Go is that you can often look it up in reference tables. Meanwhile, we can combine the logarithms on the left into one big logarithm; thus,
(5.105)
211
212
Chapter 5
Free Energy and Chemical Thermodynamics
or with a bit more rearranging,
(5.106) .LJ~u.U>'~V.l.l 5.106 is our final result. On the left-hand side are the equilibrium partial pressures of the three gases, raised to the powers of their stoichiometric co­ ettJlCl(mts, with reactants in the denominator and products in the numerator. There are also enough powers of the reference pressure po to make the whole expres­ sion dimensionless. The quantity on the right-hand side is called the equilibrium constant, K: (5.107)
It is a function of temperature (both through !:lGo and the explicit T) but not of the amounts of the gases that are present. Often we compute K once and for all a given T) and then simply write
p,2
NH3
(po)2 K.
(5.108)
This equation is called the law of mass action (don't ask me why). Even if you don't know the value of K, equation 5.108 tells you quite a bit about this reaction. If the gases are initially in equilibrium and you add more nitrogen or hydrogen, some of what you add will have to react to form ammonia in order to maintain equilibrium. If you add more ammonia, some will have to convert to nitrogen and hydrogen. If you double the partial pressure of both the hydrogen and the nitrogen, the partial pressure of the ammonia must quadruple in order to maintain equilibrium. Increasing the total pressure therefore favors the production of more ammonia. One way to remember how a system in equilibrium will respond to these kinds of changes, at least qualitatively, is Le Chatelier's principle: When you disturb a system in equilibrium, it will respond in a way that partially offsets the disturbance. So for instance, when you increase the total pressure, more nitrogen and hydrogen will react to form ammonia, decreasing the total number of molecules and thus reducing the total pressure. To be more quantitative, we need a numerical value for the equilibrium con­ stant K. Sometimes you can find tabulated values of constants, but more often you need to compute them from !:lGo values using equation 5.107. For the production of two moles of ammonia at 298 K, standard tables give the value !:lGo -32.9 kJ. The equilibrium constant at room temperature is therefore K =
. ( +32,900 J ) exp (8.31 JjK)(298 K)
5.9
X
105 ,
(5.109)
so this reaction tends strongly to the right, favoring the production of ammonia from nitrogen and hydrogen.
5.6
Chemical Equilibrium
At higher temperatures, K becomes much smaller (see Problem 5.86), so you might think that industrial production of ammonia would be carried out at rel­ atively low temperature. However, the equilibrium condition tells us absolutely nothing about the rate of the reaction. It turns out that, unless a good catalyst is present, this reaction proceeds negligibly slowly at temperatures below about 700°C. Certain bacteria do contain excellent catalysts (enzymes) that can fix nitro­ gen at room temperature. For industrial production, though, the best known cata­ lyst still requires a temperature of about 500°C to achieve an acceptable production rate. At this temperature the equilibrium constant is only 6.9 x 10- 5 , so very high pressures are needed to produce a reasonable amount of ammonia. The industrial nitrogen-fixation process used today, employing an iron-molybdenum catalyst, tem­ peratures around 500°C, and total pressures around 400 atm, was developed in the early 20th century by the German chemist Fritz Haber. This process has revolu­ tionized the production of chemical fertilizers, and also, unfortunately, facilitated the manufacture of explosives. Problem 5.84. A mixture of one part nitrogen and three parts hydrogen is heated, in the presence of a suitable catalyst, to a temperature of 500°C. What fraction of the nitrogen (atom for atom) is converted to ammonia, if the final total pressure is 400 atm? Pretend for simplicity that the gases behave ideally despite the very high pressure. The equilibrium constant at 500°C is 6.9 X 10- 5 . (Hint: You'll have to solve a quadratic equation.) Problem 5.85. Derive the van't Hoff equation, dInK
6.Ho RT2 '
which gives the dependence of the equilibrium constant on temperature. * Here 6.Ho is the enthalpy change of the reaction, for pure substances in their standard states (1 bar pressure for gases). Notice that if 6.H o is positive (loosely speaking, if the reaction requires the absorption of heat), then higher temperature makes the reaction tend more to the right, as you might expect. Often you can neglect the temperature dependence of 6.Ho; solve the equation in this case to obtain
Problem 5.86. Use the result of the previous problem to estimate the equilib­ rium constant of the reaction N2 + 3H2 +-+ 2NH3 at 500°C, using only the room­ temperature data at the back of this book. Compare your result to the actual value of K at 500° C quoted in the text.
*Van't Hoff's equation is not to be confused with van't Hoff's formula for osmotic pressure. Same person, different physical principle.
213
214
Chapter 5
Free Energy and Chemical Thermodynamics
Dissociation of Water
s(
As a second example of chemical equilibrium, consider again the dissociation of water into H+ and OH- ions, discussed briefly at the beginning of this section: (5.110)
s~
ir
P h
At equilibrium the chemical potentials of the three species satisfy (5.111) Assuming that the solution is dilute (a very good approximation under normal conditions), the chemical potentials are given by equations 5.69 (for H 2 0) and 5.72 (for H+ and OH-). Furthermore, the deviation of J.lH 2 0 from its value for pure water is negligible. The equilibrium condition is therefore (5.112) where each J.l 0 represents the chemical potential of the substance in its 'standard state'~pure liquid for H2 0 and a concentration of one mole per kilogram solvent for the ions. The m's are molalities, understood to be measured in units of one mole solute per kilogram of solvent. As in the previous example, the next step is to gather the J.l 0 's on the right, the logarithms on the left, and multiply through by Avogadro's number:
(5.113) where l::!.Go is the standard change in G for the reaction, again a value that can be looked up in tables. A bit of rearrangement gives (5.114) the equilibrium condition for the ion molalities. Before plugging in numbers, it's worthwhile to pause and compare this result to the equilibrium condition in the previous example, equation 5.106. In both cases the right-hand side is called the equilibrium constant, (5.115) and is given by the same exponential function of the standard change in the Gibbs free energy. But the 'standard' states are now completely different: pure liquid for the solvent and 1 molal for the solutes instead of 1 bar partial pressure for the gases of the previous example. Correspondingly, the left-hand side of equation 5.114 involves molalities instead of partial pressures (but still raised to the powers of their stoichiometric coefficients, in this case both equal to 1). 1-1ost significantly, the amount or concentration of water does not appear at all on the left-hand side of equation 5.114. This is because in a dilute solution there is always plenty of
a is
Ii IT
1 d
1:
t a v
5.6
Chemical Equilibrium
solvent available for the reaction, no matter how much has already reacted. (The same would be true of a pure liquid or solid that reacts only at its surface.) A final difference between ideal gas reactions and reactions in solution is that in the latter case, the equilibrium constant can in principle depend on the total pressure. In practice, however, this dependence is usually negligible except at very high (e.g., geological) pressures (see Problem 5.88). The value of !)'Go for the dissociation of one mole of water at room temperature and atmospheric pressure is 79.9 kJ, so the equilibrium constant for this reaction is 79,900 -14 (5.116) K = exp ( (8.31 JjK)(298 K) = 1.0 x 10 .
J)
If all the H+ and OH- ions come from dissociation of water molecules, then they must be equally abundant, so in this case
m H + = m OH -- = 1.0 x 10- 7 .
(5.117)
The 7 in this result is called the pH of pure water. More generally, the pH is defined as minus the base-10 logarithm of the molality of H+ ions: (5.118) If other substances are dissolved in water, the pH can shift significantly. When the pH is less than 7 (indicating a higher H+ concentration) we say the solution is acidic, while when the pH is greater than 7 (indicating a lower H+ concentration) we say the solution is basic. Problem 5.87. Sulfuric acid, H2S04, readily dissociates into H+ and HSOi ions: H2S04
H+
+ HSOi.
The hydrogen sulfate ion, in turn, can dissociate again:
HSO 4
f----t
H+
+ SO~- .
The equilibrium constants for these reactions, in aqueous solutions at 298 K, are approximately 102 and 10-1.9, respectively. (For dissociation of acids it is usually more convenient to look up K than t:::.Go. By the way, the negative base-l0 log­ arithm of K for such a reaction is called pK, in analogy to pH. So for the first reaction pK -2, while for the second reaction pK = 1.9.) (a) Argue that the first reaction tends so strongly to the right that we might as well consider it to have gone to completion, in any solution that could pos­ sibly be considered dilute. At what pH values would a significant fraction of the sulfuric acid not be dissociated? (b) In industrialized regions where lots of coal is burned, the concentration of sulfate in rainwater is typically 5 x 10- 5 mol/kg. The sulfate can take any of the chemical forms mentioned above. Show that, at this concentration, the second reaction will also have gone essentially to completion, so all the sulfate is in the form of SO~-. What is the pH of this rainwater?
(c) Explain why you can neglect dissociation of water into H+ and OH- in answering the previous question. (d) At what pH would dissolved sulfate be equally distributed between HSO 4 and SO~-?
215
216
Chapter 5
Free Energy and Chemical Thermodynamics
Problem 5.88. Express a(~G O )/ap in terms of the volumes of solutions of reactants and products, for a chemical reaction of dilute solutes. Plug in some reasonable numbers , to show that a pressure increase of 1 atm has only a negligible effect on the equilibrium constant.
Oxygen Dissolving in Water When oxygen (0 2 ) gas dissolves in water (see Figure 5.42), there is no chemical reaction per se, but we can still apply the techniques of this section to find out how much O 2 will dissolve. The 'reaction' equation and tabulated ~G o value are ~Go
= 16.4 kJ,
(5.119)
where g is for 'gas' and aq is for 'aqueous' (i.e., dissolved in water). The ~Go value is for one mole of oxygen at 1 bar pressure dissolving in 1 kg of water (to give a solution with molality 1), all at 298 K. When the dissolved oxygen is in equilibrium with the oxygen in the adjacent gas, their chemical potentials must be equal: J-Lgas
=
(5.120)
J-Lsolute ·
Using equation 5.40 for J-Lgas and equation 5.72 for J-Lsolute, we can write both chem­ ical potentials in terms of standard-state values and the respective concentrations: J-L~as
+ kT In(PI PO) =
J-L~olute
+ kT In m.
(5.121)
Here P is the partial pressure of O 2 in the gas, po is the standard pressure of 1 bar, and m is the molality of the dissolved oxygen in moles per kilogram of water. Once again, the procedure is to gather the J-L° 'S on the right and the logarithms on the left, then multiply through by Avogadro's number: (5.122)
or equivalently, ~ _
PIPo - e
.. ..
...
...
. . .
..
...
. . . .
-f:,.Go/RT
.
(5.123)
.. ..
Figure 5.42. The dissolution of a gas in a liq­ uid, such as oxygen in water , can be treated as a chemical reaction with its own equilibrium con­ stant.
5.6
Chemical Equilibrium
Equation 5.123 says that the ratio of the amount of dissolved oxygen to the amount in the adjacent gas is a constant, at any given temperature and total pressure. This result is known as Henry's law. As in the previous example, the dependence of !:::'Go on the total pressure is usually negligible unless the pressure is very The constant on the right-hand side of the equation is sometimes called a 'Henry's law constant,' but one often finds these constants tabulated in very different ways-as reciprocals and/or in terms of mole fraction rather than molality. For oxygen in water at room temperature the right-hand side of equation 5.123 is 1 16,400 J ) (5.124) exp ( - (8.31 J/K)(298 K) = 0.00133 750' meaning that if the partial pressure of oxygen is 1 bar, about 1/750 of a mole of oxygen will dissolve in each kilogram of water. In our atmosphere at sea level the partial pressure of oxygen is only about 1/5 as much, and the amount of dissolved oxygen in water is proportionally less. Still, each liter of water contains the equiv­ alent of about 7 cm3 of pure oxygen (if it were a gas at atmospheric pressure), enough for fish to breathe. Problem 5.89. The standard enthalpy change upon dissolving one mole of oxygen at 25°C is -11.7 kJ. Use this number and the van't Hoff equation (Problem 5.85) to calculate the equilibrium (Henry's law) constant for oxygen in water at O°C and at 100°C. Discuss the results briefly. Problem 5.90. When solid quartz 'dissolves' in water, it combines with water molecules in the reaction
(a) Use this data in the back of this book to compute the amount of silica dissolved in water in equilibrium with solid quartz, at 25°C. (b) Use the van't Hoff equation (Problem 5.85) to compute the amount of silica dissolved in water in equilibrium with solid quartz at 100°C.
Problem 5.91. When carbon dioxide 'dissolves' in water, essentially all of it reacts to form carbonic acid, H2C03:
The carbonic acid can then dissociate into H+ and bicarbonate ions,
(The table at the back of this book gives thermodynamic data for both of these reactions.) Consider a body of otherwise pure water (or perhaps a raindrop) that is in equilibrium with the atmosphere near sea level, where the partial pressure of carbon dioxide is 3.4 x bar (or 340 parts per million). Calculate the molality of carbonic acid and of bicarbonate ions in the water, and determine the pH of the solution. Note that even 'natural' precipitation is somewhat acidic.
217
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Chapter 5
Free Energy and Chemical Thermodynamics
Ionization of Hydrogen As a final example of chemical equilibrium, let's consider the ionization of atomic hydrogen into a proton and an electron, H
+----+
p + e,
(5.125)
an important reaction in stars such as our sun. This reaction is so simple that we can compute the equilibrium constant from first principles, without looking up anything in a table. Following the same steps as in the previous examples, we can write the equilib­ rium condition for the partial pressures as
k Tl n (PHPO) PpP e
=
° fLp° + fLe° - fLH'
(5.126)
where each flo is the chemical potential of that species at 1 bar pressure. Under most conditions we can treat all three species as structureess monatomic g,ases~ for which we derived an explicit formula for fL in Section 3.5:
fL
= -kTI
[V (27rmkT)3/2] = -kTl [kT (27rmkT)3/2] h2 n P h2 .
n N
(5.127)
(Here m is the mass of the particle, not molality.) The only subtlety is that this formula includes only kinetic energy, taking the energy zero-point to be the state where all the particles are at rest. In computing the difference of the flo,S we also need to include the ionization energy, I = 13.6 eV, that you need to put in to convert H to p + e even if there is no kinetic energy. I'll put this in by subtracting I from fLH: (5.128) For p and e, the formulas for flo are identical but with different masses and without the final - I. ~w~)it ~~.~~ ~ i ~') ~.'O.)~'0 ~'L~ ~~~~ '0.).~'W.~~~~ )'. ~.).,,~ ~') ~ m'i.~ everything except the - I in the chemical potentials of these two specIes cancels. Dividing through by -kT, we're left with
PHPO) In ( PpPe
=
In[kT (27rmekT)3/2] h2 po
kIT'
(5.129)
A bit of algebra then yields the following result:
Pp
kT (27rmekT)3/2 e-I/kT.
PH
Pe
(5.130)
h2
This formula is called the Saha equation. It gives the ratio of the amount of protons) to the amount of un-ionized hydrogen as a ionized hydrogen (that
5.6
Chemical Equilibrium
function of temperature and the concentration of electrons. (Note that the combi­ nation Pe/kT is the same as Ne/V, the number of electrons per unit volume.) At the surface of the sun the temperature is about 5800 K, so the exponential factor is only e- 11kT 1.5 x 10- 12 . Meanwhile the electron concentration is roughly 19 3 2 x 10 m- ; the Saha equation thus predicts a ratio of (5.131) Even at the surface of the sun, less than one hydrogen atom in 10,000 is ionized. Problem 5.92. Suppose you have a box of atomic hydrogen, initially at room temperature and atmospheric pressure. You then raise the temperature, keeping the volume fixed. (a) Find an expression for the fraction of the hydrogen that is ionized as a function of temperature. (You'll have to solve a quadratic equation.) Check that your expression has the expected behavior at very low and very high temperatures. (b) At what temperature is exactly half of the hydrogen ionized? (c) Would raising the initial pressure cause the temperature you found in part (b) to increase or decrease? Explain. (d) Plot the expression you found in part (a) as a function of the dimension­ less variable t = kT/ I. Choose the range of t values to clearly show the interesting part of the graph.
Tbermodynamics bas sometbing to say about everytbing but does not tell us everytbing about anytbing. -Martin Goldstein and lnge F. Goldstein, Tbe Refrigerator and tbe Universe. Copy­ right 1993 by the President and Fellows of Harvard College. Reprinted by permission of Harvard University Press.
219
6
Boltzmann Statistics
Most of this book so far has dealt with the second law of thermodynamics: its origin in the statistical behavior of large numbers of particles, and its applications in physics, chemistry, earth science, and engineering. However, the second law by itself usually doesn't tell us all we would like to know. In the last two chapters especially, we have often had to rely on experimental measurements (of enthalpies, entropies, and so on) before we could extract any predictions from the second law. This approach to thermodynamics can be extremely powerful, provided that the needed measurements can be made to the required precision. Ideally, though, we would like to be able to calculate all thermodynamic quan­ tities from first principles, starting from microscopic models of various systems of interest. In this book we have already worked with three important microscopic models: the two-state paramagnet, the Einstein solid, and the monatomic ideal gas. For each of these models we were able to write down an explicit combinatoric formula for the multiplicity, 0, and from there go on to compute the entropy, tem­ perature, and other thermodynamic properties. In this chapter and the next two, we will study a number of more complicated models, representing a much greater variety of physical systems. For these more complicated models the direct combi­ natoric approach used in Chapters 2 and 3 would be too difficult mathematically. We therefore need to develop some new theoretical tools.
6.1 The Boltzmann Factor In this section I will introduce the most powerful tool in all of statistical mechanics: an amazingly simple formula for the probability of finding a system in any particular microstate, when that system is in thermal equilibrium with a 'reservoir' at a specified temperature (see Figure 6.1). The system can be almost anything, but for definiteness, let's say it's a single atom. The microstates of the system then correspond to the various energy levels of 220
6.1
I
The Boltzmann Factor
I 'Reservoir' Energy = UR Temperature = T
~
~
'System' Energy = E
liMil
Figure 6.1. A 'system' in thermal contact with a much larger 'reservoir' at some well-defined temperature.
the atom, although for a given energy level there is often more than one independent state. For instance, a hydrogen atom has only one ground state (neglecting spin), with energy -13.6 eV. But it has four independent states with energy -3.4 eV, nine states with energy -1.5 eV, and so on (see Figure 6.2). Each of these independent states counts as a separate microstate. When an energy level corresponds to more than one independent state, we say that level is degenerate. (For a more precise definition of degeneracy, and a more thorough discussion of the hydrogen atom, see Appendix A.) If our atom were completely isolated from the rest of the universe, then its energy would be fixed, and all microstates with that energy would be equally probable. Now, however, we're interested in the situation where the atom is not isolated, but instead is exchanging energy with lots of other atoms, which form a large 'reservoir' with a fixed temperature. In this case the atom could conceivably be found in any of its microstates, but some will be more likely than others, depending on their energies. (Microstates with the same energy will still have the same probability.) Energy
-1.5 eV I
-3.4 eV
,
--'~-,~ 82
__ 81
/-~,
,--,~ I
-13.6 eV
,
,
I
/
Figure 6.2. Energy level diagram for a hydrogen atom, showing the three lowest energy levels. There are four independent states with energy -3.4 eV, and nine independent states with energy -1. 5 e V.
221
222
Chapter 6
Boltzmann Statistics
Since the probability of finding the atom in any particular microstate depends on how many other microstates there are, I'll simplify the problem, at first, by looking only at the ratio of probabilities for two particular microstates of interest (such as those circled in Figure 6.2). Let me call these states Sl and S2, their energies E(Sl) and E(S2), and their probabilities P(Sl) and P(S2)' How can I find a formula for the ratio of these probabilities? Let's go all the way back to the fundamental assumption of statistical mechanics: For an isolated system, all accessible microstates are equally probable. Our atom is not an isolated system, but the atom and the reservoir together do make an isolated system, and we are equally likely to find this combined system in any of its accessible microstates. Now we don't care what the state of the reservoir is; we just want to know what state the atom is in. But if the atom is in state Sl, then the reservoir will have some very large number of accessible states, all equally probable. I'll call this number nR(Sl): the multiplicity of the reservoir when the atom is in state Sl. Similarly, I'll use nR (S2) to denote the multiplicity of the reservoir when the atom is in state S2. These two multiplicities will generally be different, because when the atom is in a lower-energy state, more energy is left for the reservoir. Suppose, for instance, that state Sl has the lower energy, so that nR(SJ) > nR (S2). As a specific example, let's say nR (Sl) = 100 and nR (S2) = 50 (though in a true reservoir the multiplicities would be much larger). Now fundamentally, all microstates of the combined system are equally probable. Therefore since there are twice as many states of the combined system when the atom is in state Sl than when it is in state S2, the former state must be twice as probable as the latter. More generally, the probability of finding the atom in any particular state is directly proportional to the number of microstates that are accessible to the reservoir. Thus the ratio of probabilities for any two states is
(6.1)
Now we just have to get this expression into a more convenient form, using some math and a bit of thermodynamics. First I'll rewrite each n in terms of entropy, using the definition S = k In n: (6.2) The exponent now contains the change in the entropy of the reservoir, when the atom undergoes a transition from state 1 to state 2. This change will be tiny, since the atom is so small compared to the reservoir. We can therefore invoke the thermodynamic identity:
(6.3)
The right-hand side involves the changes in the reservoir's energy, volume, and number of particles. But anything gained by the reservoir is lost by the atom, so
6.1
The Boltzmann Factor
we can write each of these changes as minus the change in the same quantity for the atom. I'm going to throwaway the P dV and J-L dN terms, but for different reasons. The quantity P dVR is usually nonzero, but much smaller than dUR and therefore negligible. For instance, when an atom goes into an excited state, its effective volume might increase by a cubic angstrom, so at atmospheric pressure, the term P dV is of order 10- 25 J. This is a million times less than the typical change in the atom's energy of a few electron-volts. Meanwhile, dN really is zero, at least when the small system is a single atom, and also in the other cases that we'll consider in this chapter. (In the following chapter I'll put the dN term back, in order to deal with other types of systems.) So the difference of entropies in equation 6.2 can be rewritten
where E is the energy of the atom. Plugging this expression back into equation 6.2, we obtain
P(82) = P(8d
e-[E(S2)-E(sdl/ kT
=
E e- (S2)/kT e-E(Sl)/kT·
(6.5)
The ratio of probabilities is equal to a ratio of simple exponential factors, each of which is a function of the energy of the corresponding microstate and the temper­ ature of the reservoir. Each of these exponential factors is called a Boltzmann factor: Boltzmann factor = e-E(s)/kT. (6.6) It would be nice if we could just say that the probability of each state is equal to the corresponding Boltzmann factor. Unfortunately, this isn't true. To arrive at the correct statement, let's manipulate equation 6.5 to get everything involving 81 on one side and everything involving 82 on the other side:
(6.7) Notice that the left side of this equation is independent of 81; therefore the right side must be as well. Similarly, since the right side is independent of 82, so is the left side. But if both sides are independent of both 81 and 82, then in fact both sides must be equal to a constant, the same for all states. The constant is called liZ; it is the constant of proportionality that converts a Boltzmann factor into a probability. In conclusion, for any state 8,
P(8) = ~ e-E(s)/kT
Z
.
(6.8)
This is the most useful formula in all of statistical mechanics. Memorize it. * *Equation 6.8 is sometimes called the Boltzmann distribution or the canonical distribution.
223
224
Chapter 6
Boltzmann Statistics
To interpret equation 6.8, let's suppose for a moment that the ground state energy of our atom is Eo = 0, while the excited states all have positive energies. Then the probability of the ground state is liZ, and all other states have smaller probabilities. States with energies much less than kT have probabilities only slightly less than liZ, while states with energies much greater than kT have negligible probabilities, suppressed by the exponential Boltzmann factor. Figure 6.3 shows a bar graph of the probabilities for the states of a hypothetical system. But what if the ground state energy is not zero? Physically, we should expect that shifting all the energies by a constant has no effect on the probabilities, and indeed, the probabilities are unchanged. It's true that all the Boltzmann factors get multiplied by an additional factor of e- E o / kT , but we'll see in a moment that Z gets multiplied by this same factor, so it cancels out in equation 6.8. Thus, the ground state still has the highest probability, and the remaining states have probabilities that are either a little less or a lot less, depending on how their energies, as measured from the ground state, compare to kT .
-U--~~------t-~JL_~Jl_JtL=;:::::J:rrr:::::::-:~ E( s)
kT
2kT
3kT
Figure 6.3. Bar graph of the relative probabilities of the states of a hypothetical system. The horizontal axis is energy. The smooth curve represents the Boltzmann distribution, equation 6.8, for one particular t emperature. At lower temperatures it would fall off more suddenly, while at higher temperatures it would fall off more gradually. Problem 6.1. Consider a system of two Einstein solids, where the first 'solid' contains just a single oscillator, while the second solid contains 100 oscillators. The total number of energy units in the combined system is fixed at 500. Use a computer to make a table of the multiplicity of the combined system, for each possible value of the energy of the first solid from 0 units to 20. Make a graph of the total multiplicity vs. the energy of the first solid, and discuss, in some detail, whether the shape of the graph is what you would expect. Also plot the logarithm of the total multiplicity, and discuss the shape of this graph. Problem 6.2. Prove that the probability of finding an atom in any particular energy level is P(E) = (ljZ)e- F / kT , where F = E - TS and the 'entropy' of a level is k times the logarithm of the number of degenerate states for that level.
6.1
The Boltzmann Factor
The Partition Function By now you're probably wondering how to actually calculate Z. The trick is to remember that the total probability of finding the atom in some state or other must be 1: 1 = LP(s) = L ~ e-E(s)/kT = ~ Le-E(s)/kT. (6.9) s s s Solving for Z therefore gives Z = L
e-E(s)/kT
= sum of all
Boltzmann factors.
(6.10)
s
This sum isn't always easy to carry out, since there may be an infinite number of states s and you may not have a simple formula for their energies. But the terms in the sum smaller and smaller as the energies Es get larger, so often you can just compute the first several terms numerically, neglecting states with energies much greater than kT. The quantity Z is called the partition function, and turns out to be far more useful than I would have suspected. It is a 'constant' in that it does not depend on any particular state s, but it does depend on temperature. To interpret it further, suppose once again that the ground state has energy zero. Then the Boltzmann factor for the ground state is 1, and the rest of the Boltzmann factors are less than 1, by a little or a lot, in proportion to the probabilities of the associated states. Thus the partition function essentially counts how many states are accessible to the atom, weighting each one in proportion to its probability. At very low temperature, Z ~ I, since all the excited states have very small Boltzmann factors. At high temperature, Z will be much larger. And if we shift all the energies by a constant Eo, the whole partition function just gets multiplied by an uninteresting factor of e- Eo / kT , which cancels when we compute probabilities. Problem 6.3. Consider a hypothetical atom that has just two states: a ground state with energy zero and an excited state with energy 2 eV. Draw a graph of the partition function for this system as a function of temperature, and evaluate the partition function numerically at T 300 K, 3000 K, 30,000 K, and 300,000 K. Problem 6.4. Estimate the partition function for the hypothetical system rep­ resented in Figure 6.3. Then estimate the probability of this system being in its ground state. Problem 6.5. Imagine a particle that can be in only three states, with energies -0.05 eV, 0, and 0.05 eV. This particle is in equilibrium with a reservoir at 300 K. (a) Calculate the partition function for this particle. (b) Calculate the probability for this particle to be in each of the three states. (c ) Because the zero point for measuring energies is arbitrary, we could just as well say that the energies of the three states are 0, +0.05 eV, and +0.10 eV, respectively. Repeat parts (a) and (b) using these numbers. Explain what changes and what doesn't.
225
226
Chapter 6
Boltzmann Statistics
Thermal Excitation of Atoms As a simple application of Boltzmann factors, let's consider a hydrogen atom in the atmosphere of the sun, where the temperature is about 5800 K. (We'll see in the following chapter how this temperature can be measured from earth.) I'd like to compare the probability of finding the atom in one of its first excited states (82) to the probability of finding it in the ground state (81). The ratio of probabilities is the ratio of Boltzmann factors, so
(6.11)
The difference in energy is 10.2 eV, while kT is (8.62 x 10- 5 eV /K)(5800 K) = 0.50 eV. So the ratio of probabilities is approximately e- 20 .4 = 1.4 x 10- 9 . For every billion atoms in the ground state, roughly 1.4 (on average) will be in anyone of the first excited states. Since there are four such excited states, all with the same energy, the total number of atoms in these states will be four times as large, about 5.6 (for every billion in the ground state). Atoms in the atmosphere of the sun can absorb sunlight on its way toward earth, but only at wavelengths that can induce transitions of the atoms into higher excited states. A hydrogen atom in its first excited state can absorb wavelengths in the Balmer series: 656 nm, 486 nm, 434 nm, and so on. These wavelengths are therefore missing, in part, from the sunlight we receive. If you put a narrow beam of sunlight through a good diffraction grating, you can see dark lines at the missing wavelengths (see Figure 6.4). There are also other prominent dark lines, created by other types of atoms in the solar atmosphere: iron, magnesium, sodium, calcium, and so on. The weird thing is, all these other wavelengths are absorbed by atoms H Ca H Fe H
I 11(
I I
Ca Fe H
(
I
Mg
H
I
I
16 Cyg A 5800 K 'Y UMa 9500 K
380
400
420 440 Wavelength (nm)
460
480
Figure 6.4. Photographs of the spectra of two stars. The upper spectrum is of a sunlike star (in the constellation Cygnus) with a surface temperature of about 5800 K; notice that the hydrogen absorption lines are clearly visible among a number of lines from other elements. The lower spectrum is of a hotter star (in Ursa Major ) the Big Dipper)) with a surface temperature of 9500 K. At this temperature a much larger fraction of the hydrogen atoms are in their first excited states, so the hydrogen lines are much more prominent than any others. Reproduced with permission from Helmut A. Abt et al., An Atlas of Low-Dispersion Grating Stellar Spectra (Kitt Peak National Obgervatory, Tucson, AZ, 1968).
6.1
The Boltzmann Factor
(or ions) that start out either in their ground states or in very low-energy excited states (less than 3 eV above the ground state). The Balmer lines, by contrast, come only from the very rare hydrogen atoms that are excited more than 10 eV above the ground state. (A hydrogen atom in its ground state does not absorb any visible wavelengths.) Since the Balmer lines are quite prominent among the others, we can only conclude that hydrogen atoms are much more abundant in the sun's atmosphere than any of these other types. * Problem 6.6. Estimate the probability that a hydrogen atom at room temper­ ature is in one of its first excited states (relative to the probability of being in the ground state). Don't forget to take degeneracy into account. Then repeat the calculation for a hydrogen atom in the atmosphere of the star I UMa, whose surface temperature is approximately 9500 K. Problem 6.7. Each of the hydrogen atom states shown in Figure 6.2 is actually twofold degenerate, because the electron can be in two independent spin states, both with essentially the same energy. Repeat the calculation given in the text for the relative probability of being in a first excited state, taking spin degeneracy into account. Show that the results are unaffected. Problem 6.S. The energy required to ionize a hydrogen atom is 13.6 eV, so you might expect that the number of ionized hydrogen atoms in the sun's atmosphere would be even less than the number in the first excited state. Yet at the end of Chapter 5 I showed that the fraction of ionized hydrogen is much larger, nearly one atom in 10,000. Explain why this result is not a contradiction, and why it would be incorrect to try to calculate the fraction of ionized hydrogen using the methods of this section. Problem 6.9. In the numerical example in the text, I calculated only the ratio of the probabilities of a hydrogen atom being in two different states. At such a low temperature the absolute probability of being in a first excited state is essentially the same as the relative probability compared to the ground state. Proving this rigorously, however, is a bit problematic, because a hydrogen atom has infinitely many states. (a) Estimate the partition function for a hydrogen atom at 5800 K, by adding the Boltzmann factors for all the states shown explicitly in Figure 6.2. (For simplicity you may wish to take the ground state energy to be zero, and shift the other energies accordingly.) (b) Show that if all bound states are included in the sum, then the partition function of a hydrogen atom is infinite, at any nonzero temperature. (See Appendix A for the full energy level structure of a hydrogen atom.)
(c) When a hydrogen atom is in energy level n, the approximate radius of the electron wavefunction is aon 2 , where ao is the Bohr radius, about 5 x 10- 11 m. Going back to equation 6.3, argue that the P dV term is not negligible for the very high-n states, and therefore that the result of part (a), not that of part (b), gives the physically relevant partition function for this problem. Discuss. *The recipe of the stars was first worked out by Cecilia Payne in 1924. The story is beautifully told by Philip and Phylis Morrison in The Ring of Truth (Random House, New York, 1987).
227
228
Chapter 6
Boltzmann Statistics
Problem 6.10. A water molecule can vibrate in various ways, but the easiest type of vibration to excite is the 'flexing' mode in which the hydrogen atoms move toward and away from each other but the HO bonds do not stretch. The oscillations of this mode are approximately harmonic, with a frequency of 4.8 x 10 13 Hz. As and so for any quantum harmonic oscillator, the energy levels are ~hf, ~hf, on. None of these levels are degenerate.
(a) Calculate the probability of a water molecule being in its ground state and in each of the first two excited states, assuming that it is in equi­ librium with a reservoir (say the atmosphere) at 300 K. (Hint: Calculate Z by adding up the first few Boltzmann factors, until the rest are negligible.) (b) Repeat the calculation for a water molecule in equilibrium with a reservoir at 700 K (perhaps in a steam turbine). Problem 6.11. A lithium nucleus has four independent spin orientations, conven­ -3/2, -1/2, 1/2,3/2. In a magnetic tionally labeled by the quantum number m field B, the of these four states are E = mJLB, where the constant JL is In the Purcell-Pound experiment described in Section 3.3, the 1.03 X 10- 7 maximum field strength was 0.63 T and the temperature was 300 K. Calculate the probability of a lithium nucleus being in each of its four spin states under these conditions. Then show if the field is suddenly reversed, the probabilities of the four states obey the Boltzmann distribution for T -300 K. Problem 6.12. Cold interstellar molecular clouds often contain the molecule cyanogen (CN), whose first rotational excited states have an energy of 4.7x 10- 4 eV (above the ground state). There are actually three such excited states, all with the same energy. In 1941, studies of the absorption spectrum of that passes through these molecular clouds showed that for every ten CN molecules that are in the ground state, approximately three others are in the three first excited states (that is, an average of one in each of these states). To account for this data, astronomers that the molecules might be in thermal equilibrium with some 'reservoir' with a well-defined temperature. What is that temperature?* Problem 6.13. At very high temperatures (as in the very early universe), the proton and the neutron can be of as two different states of the same particle, called the 'nucleon.' (The reactions that convert a proton to a neutron or vice versa the absorption of an electron or a positron or a neutrino, but all of these particles tend to be very abundant at sufficiently high temperatures.) Since the neutron's mass is higher than the proton's by 2.3 x its energy is higher by this amount times c 2 . Suppose, then, that at some very early time, the nucleons were in thermal equilibrium with the rest of the universe at 1011 K. What fraction of the nucleons at that time were protons, and what fraction were neutrons? Problem 6.14. Use Boltzmann factors to derive the exponential formula for the density of an isothermal atmosphere, already derived in Problems 1.16 and 3.37. (Hint: Let the system be a single air molecule, let 81 be a state with the molecule at sea level, and let 82 be a state with the molecule at height
*For a review of these measurements and calculations, see Patrick Thaddeus, Annual Reviews of Astronomy and Astrophysics 10, 305-334 (1972).
6.2
A verage Values
6.2 Average Values In the previous section we saw how to calculate the probability that a system is in any particular one of its microstates s, given that it is in equilibrium with a reservior at temperature T: (6.12) where f3 is an abbreviation for l/kT. The exponential factor is called the Boltz­ mann factor, while Z is the partition function ,
Z =
2.:.: e-f3 E (s) ,
(6.13)
that is, the sum of the Boltzmann factors for all possible states. Suppose, though, that we're not interested in knowing all the probabilities of all the various states our system could be in- suppose we just want to know the average value of some property of the system, such as its energy. Is there an easy way to compute this average, and if so, how? Let me give a simple example. Suppose my system is an atom that has just three possible states: The ground state with energy 0 eV, a state with energy 4 eV, and a state with energy 7 eV. Actually, though, I have five such atoms, and at the moment, two of them are in the ground state, two are in the 4-eV state, and one is in the 7-eV state (see Figure 6.5). What is the average energy of all my atoms? Just add 'em up and divide by 5:
E
= (0 eV) . 2 + (4 eV) . 2 + (7 eV) . 1 = 3 eV.
(6.14)
5
But there's another way to think about this computation. Instead of computing the numerator first and then dividing by 5, we can group the 1/5 with the factors of 2, 2, and 1, which represent the numbers of atoms in each state:
--
2 E = (0 eV) . -
+ (4 eV) . -2 + (7 eV) . -1 = 3 eV.
(6.15)
555
In this expression, the energy of each state is multiplied by the probability of that state occurring (in any particular atom chosen from among my sample of 5); those probabilities are just 2/5, 2/5, and 1/5, respectively.
7eV Figure 6.5. Five hypothetical atoms distributed among three different states.
t
•
4eV
••
o
••
229
230
Chapter 6
Boltzmann Statistics
It's not hard to generalize this example into a formula. If I have a large sample of N atoms, and N (s) is the number of atoms in any particular state s, then the average value of the energy is
E =
=.!::..~N~~ = LE(s) N};) s
LE(s)P(s),
(6.16)
s
where P(s) is the probability of finding an atom in state s. So the average energy is just the sum of all the energies, weighted by their probabilities. In the statistical mechanical systems that we're considering, each probability is given by equation 6.12, so (6.17) Notice that the sum is similar to the partition function (6.13), but with an extra factor of E( s) in each term. * The average value of any other variable of interest can be computed in exactly the same way. If the variable is called X, and has the value X (s) in state s, then
LX(s)P(s) =
1
z
LX(s)
s
(6.18)
s
One nice feature of average values is that they are additive; for example, the average total energy of two objects is the sum of their individual average nn,''t'n'oe> This means that if you have a collection of many identical, independent particles, you can compute their total (average) energy from the average energy of just one, simply by multiplying by how many there are:
U=NE.
(6.19)
(Now you see why I've been using the symbol E for the energy of the atom; I've system that contains it.) So reserved U for the total energy of the much when I divided things into an 'atom' and a 'reservoir' in the previous section, it was partly just a trick. Even if you want to know the total energy of the whole system, you can often find it by concentrating first on one particle in the system, treating the rest as the reservoir. Once you know the average value of the quantity of interest for your particle, just multiply by N to the total. Technically, the U in equation 6.19 is merely the average energy of the entire system is in thermal contact with other objects, then the system. If even.this instantaneous value of U will fluctuate away from the average. However, if N is large, these fluctuations will almost always be negligible. Problem 6.17 shows how to calculate the size of typical fluctuations. *In this chapter, the set of systems that we average over will be a hypothetical set to states according to the Boltzmann probability distribution. whose members are This hypothetical set of systems is often called a canonical ensemble. In Chapters 2 and 3 we instead worked with isolated systems, where all allowed states had the same probability; a set of hypothetical systems with that (trivial) probability distribution is called a microcanonical ensemble.
6.2
Average Values
231
Problem 6.15. Suppose you have 10 atoms of weberium: 4 with energy 0 eV, 3 with energy 1 eV, 2 with energy 4 eV, and 1 with energy 6 eV.
(a) Compute the average energy of all your atoms, by adding up all their energies and dividing by 10. (b) Compute the probability that one of your atoms chosen at random would have energy E, for each of the four values of E that occur. (c ) Compute the average energy again, using the formula E
=
E(s)P(s).
Problem 6.16. Prove that, for any system in equilibrium with a reservoir at temperature T, the average value of the energy is
E=
1 [)Z Z [)f3
[)
[)f3 In Z,
where f3 l/kT. These formulas can be extremely useful when you have an explicit formula for the partition function. Problem 6.17. The most common measure of the fluctuations of a set of numbers away from the average is the standard deviation, defined as follows.
(a) For each atom in the five-atom toy model of Figure 6.5, compute the devi­ ation of the energy from the average energy, that is, Ei for i 1 to 5. Call these deviations llEi. (b) Compute the average of the squares of the five deviations, that is, Then compute the square root of this quantity, which is the root-mean­ square (rms) deviation, or standard deviation. Call this number a E. Does a E give a reasonable measure of how far the individual values tend to stray from the average? (c ) Prove in general that that is, the standard deviation squared is the average of the squares minus the square of the average. This formula usually the easier way of computing a standard deviation.
(d) Check the preceding formula for the five-atom toy model of Figure 6.5. Problem 6.18. Prove that, for any system in equilibrium with a reservoir at temperature T, the average value of E2 is
Then use this result and the results of the previous two problems to derive a formula for aE in terms of the heat capacity, C [)E/[)T. You should find
Problem 6.19. Apply the result of Problem 6.18 to obtain a formula for the standard deviation of the energy of a system of N identical harmonic oscillators (such as in an Einstein solid), in the high-temperature limit. Divide by the average energy to obtain a measure of the fractional fluctuation in energy. Evaluate this fraction numerically for N 1, 10 4 , and 1020 . Discuss the results briefly.
232
Chapter 6
Boltzmann Statistics
Paramagnetism As a first application of these tools, I'd like to rederive some of our earlier results (see Section 3.3) for the ideal two-state paramagnet. Recall that each elementary dipole in an ideal two-state paramagnet has just two possible states: an 'up' state with energy - J-LB , and a 'down' state with energy +J-LB. (Here B is the strength of the externally applied magnetic field , while the component of the dipole's magnetic moment in the direction of the field is ±J-L.) The partition function for a single dipole is therefore (6.20) The probability of finding the dipole in the 'up' state is e+/3 p,B
Pi
e+/3p,B
= - Z - = -2-co-s-h(-:-f3-J-L-B--:-) ,
(6.21)
while the probability of finding it in the 'down' state is e-/3p,B
e- /3 p,B
P! = - Z - = -2-co-s-h(-:-f3-J-L-B--:-r
(6.22)
You can easily check that these two probabilities add up to 1. The average energy of our dipole is
(6.23)
If we have a collection of N such dipoles, the total energy is
(6.24) in agreement with equation 3.31. In Section 3.3, however, we had to work much harder to derive this result: We started with the exact combinatoric formula for the multiplicity, then applied Stirling's approximation to simplify the entropy, then took a derivative ~nd did lots of algebra to finally get U as a function of T. Here all we needed was Boltzmann factors. According to the result of Problem 6.16, we can also compute the average energy by differentiating Z with respect to f3, then multi plying by -1/ Z : -
E
18Z
=
-2 8f3·
(6.25)
Let's check this formula for the two-state paramagnet: (6.26)
6.2
Average Values
Yep, it works. Finally, we can compute the average value of a dipole's magnetic moment along the direction of B:
(6.27) Thus the total magnetization of the sample is
(6.28) in agreement with equation 3.32. Problem 6.20. This problem concerns a collection of N identical harmonic os­ cillators (perhaps an Einstein solid or the internal vibrations of gas molecules) at temperature T. As in Section 2.2, the allowed energies of each oscillator are 0, hj , 2hj, and so on.
(a) Prove by long division that 1 2 3 --=1+x+x +x 1-x
+ ..
For what values of x does this series have a finite sum?
(b) Evaluate the partition function for a single harmonic oscillator. Use the result of part (a) to simplify your answer as much as possible. (c) Use formula 6.25 to find an expression for the average energy of a single oscillator at temperature T. Simplify your answer as much as possible. (d) What is the total energy of the system of N oscillators at temperature T? Your result should agree with what you found in Problem 3.25. (e) If you haven't already done so in Problem 3.25, compute the heat capacity of this system and check that it has the expected limits as T ---+ and T ---+ 00.
°
Problem 6.21. In the real world , most oscillators are not perfectly harmonic. For a quantum oscillator, this means that the spacing between energy levels is not exactly uniform. The vibrational levels of an H2 molecule, for example, are more accurately described by the approximate formula
n
= 0,
1, 2, .. ,
where E is the spacing between the two lowest levels. Thus, the levels get closer together with increasing energy. (This formula is reasonably accurate only up to about n = 15; for slightly higher n it would say that En decreases with increasing n. In fact, the molecule dissociates and there are no more discrete levels beyond n ;:::;; 15.) Use a computer to calculate the partition function, average energy, and heat capacity of a system with this set of energy levels. Include all levels through n = 15, but check to see how the results change when you include fewer levels. Plot the heat capacity as a function of kT/ Eo Compare to the case of a perfectly harmonic oscillator with evenly spaced levels, and also to the vibrational portion of the graph in Figure 1.13.
233
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Chapter 6
Boltzmann Statistics
Problem 6.22. In most paramagnetic materials, the individual magnetic particles have more than two independent states (orientations). The number of independent states depends on the particle's angular momentum 'quantum number' j, which must be a multiple of 1/2. For j 1/2 there are just two independent states, as discussed in the text above and in Section 3.3. More generally, the allowed values of the z component of a particle's magnetic moment are
where Jp, is a constant, equal to the difference in J-tz between one state and the next. (When the particle's angular momentum comes entirely from electron spins, Jp, equals twice the Bohr magneton. When orbital angular momentum also con­ tributes, Jp, is somewhat different but comparable in magnitude. For an atomic nucleus, Jp, is roughly a thousand times smaller.) Thus the number of states is 2j + 1. In the presence of a magnetic field B pointing in the z direction, the particle's magnetic energy (neglecting interactions between dipoles) is -J-tzB. (a) Prove the following identity for the sum of a finite geometric series: 1+x
+ x 2 + .. + xn
-x
(Hint: Either prove this formula by induction on n, or write the series as a difference between two infinite series and use the result of Problem 6.20(a).) (b) Show that the partition function of a single magnetic particle is
z= --~--::---=--where b = j3Jp,B. (c ) Show that the total magnetization of a system of N such particles is
M
NJp,
[(j + ~) coth[b(j + ~)] -
~ coth ~],
where cothx is the hyperbolic cotangent, equal to cosh x/ sinhx. Plot the quantity M/NJp, vs. b, for a few different values of j.
(d) Show that the magnetization has the expected behavior as T O. (e) Show that the magnetization is proportional to l/T (Curie's law) in the limit T -+ 00. (Hint: First show that coth x ~ + ~ when x « 1.) (f) Show that for j 1/2, the result of part (c) reduces to the formula derived in the text for a two-state paramagnet.
±
Rotation of Diatomic Molecules Now let's consider a more intricate application of Boltzmann factors and average values: the rotational motion of a diatomic molecule (assumed to be isolated, as in a low-density gas). Rotational energies are quantized. (For details, see Appendix A.) For a diatomic molecule like 00 or HOI, the allowed rotational energies are E(j) = j(j + l)E,
(6.29)
where j can be 0, 1, 2, etc., and E is a constant that is inversely proportional to the molecule's moment of inertia. The number of degenerate states for level j is 2j + 1,
6.2
Average Values
Energy 12E
--------------j
Figure 6.6. Energy level dia­ gram for the rotational states of a diatomic molecule.
----------j
= 3
= 2
------j=1
o
--j=O
as shown in Figure 6.6. (I'm assuming, for now, that the two atoms making up the molecule are of different types. For molecules made of identical atoms, like H2 or N2 , there is a subtlety that I'll deal with later.) Given this energy level structure, we can write the partition function as a sum over j: 00
Zrot
00
= 2:(2j + l)e- E (j)/kT = 2:(2j + l)e- j (j+l)€/kT. j=O
(6.30)
j=O
Figure 6.7 shows a pictorial representation of this sum as the area under a bar graph. Unfortunately, there is no way to evaluate the sum exactly in closed form. But it's not hard to evaluate the sum numerically, for any particular temperature. Even better, in most cases of interest we can approximate the sum as an integral that yields a very simple result. Let's look at some numbers. The constant E, which sets the energy scale for rotational excitations, is never more than a small fraction of an electron-volt. For a CO molecule, for instance, E = 0.00024 eV, so that Elk = 2.8 K. Ordinarily we are interested only in temperatures much higher than Elk, so the quantity kT/E will be much greater than 1. In this case the number of terms that contribute significantly
kT/E = 30 kT/E = 3
o
o
2
4
6
8
10
12
14
Figure 6.7. Bar-graph representations of the partition sum 6.30, for two different temperatures. At high temperatures the sum can be approximated as the area under a smooth curve.
235
236
Chapter 6
Boltzmann Statistics
to the partition function will be quite large, so we can, to a good approximation, replace the bar graph in Figure 6.7 with the smooth curve. The partition function is then approximately the area under this curve: Zrot
~ (X! (2j + l)e- j (j+l)E j kT dj
Jo
= kT
(when kT ~ E).
(6.31)
€
(To evaluate the integral, make the substitution x j(j + l)€/kT.) This result should be accurate in the high-temperature limit where Zrot » 1. As expected, the partition function increases with increasing temperature. For CO at room temperature, Zrot is slightly greater than 100 (see Problem 6.23). Still working in the high-temperature approximation, we can calculate the av­ erage rotational energy of a molecule using the magical formula 6.25: 1 8Z
Z
=
-({3 ) ~ ~ € 8{3 {3€
=~ {3
kT
(when kT» E).
(6.32)
This is just the prediction of the equipartition theorem, since a diatomic molecule has two rotational degrees of freedom. Differentiating with respect to T gives the contribution of this energy to the heat capacity, simply k (for each molecule), again in agreement with the equipartition theorem. At low temperature, however, the third law tells us that the heat capacity must go to zero; and indeed it does, as you can confirm from the exact expression 6.30 Problem 6.26). So much for diatomic molecules made of distinguishable atoms. Now, what about the case of identical atoms, such as the important molecules N2 and 0 2? The subtlety here is that turning the molecule by 180 0 does not change its spatial configuration, so the molecule actually has only half as many states as it other­ wise would. In the high-temperature limit, when Z » 1, we can account for this symmetry by inserting a factor of 1/2 into the partition function: (identical atoms, kT
» €).
(6.33)
The factor of 1/2 cancels out of the average energy (equation 6.32), so it has no effect on the heat capacity. At lower temperatures, however, things become more complicated: One must figure out exactly which terms should be omitted from the partition function (equation 6.30). At ordinary pressures, all diatomic gases except hydrogen will liquefy long before such low temperatures are reached. The behavior of hydrogen at low temperature is the subject of Problem 6.30. Problem 6.23. For a CO molecule, the constant E is approximately 0.00024 eV. (This number is measured using microwave spectroscopy, that is, by measuring the microwave frequencies needed to excite the molecules into higher rotational states. ) Calculate the rotational partition function for a CO molecule at room temperature (300 K), first using the exact formula 6.30 and then the approximate formula 6.31. Problem 6.24. For an 02 molecule, the constant E is approximately 0.00018 eV. Estimate the rotational partition function for an 02 molecule at room temperature.
6.2
Average Values
Problem 6.25. The analysis of this section applies also to linear polyatomic mole­ cules, for which no rotation about the axis of symmetry is possible. An example is C02, with E = 0.000049 e V. Estimate the rotational partition function for a C02 molecule at room temperature. (Note that the arrangement of the atoms is OCO, and the two oxygen atoms are identical.) Problem 6.26. In the low-temperature limit (kT « E), each term in the ro­ tational partition function (equation 6.30) is much smaller than the one before. Since the first term is independent of T , cut off the sum after the second term and compute the average energy and the heat capacity in this approximation. Keep only the largest T-dependent term at each stage of the calculation. Is your re­ sult consistent with the third law of thermodynamics? Sketch the behavior of the heat capacity at all temperatures, interpolating between the high-temperature and low-temperature expressions. Problem 6.27. Use a computer to sum the exact rotational partition function (equation 6.30) numerically, and plot the result as a function of kT/ E. Keep enough terms in the sum to be confident that the series has converged. Show that the approximation in equation 6.31 is a bit low, and estimate by how much. Explain the discrepancy by referring to Figure 6.7. Problem 6.28. Use a computer to sum the rotational partition function (equation 6.30) algebraically, keeping terms through j = 6. Then calculate the average energy and the heat capacity. Plot the heat capacity for values of kT/ E ranging from 0 to 3. Have you kept enough terms in Z to give accurate results within this temperature range? Problem 6.29. Although an ordinary H2 molecule consists of two identical atoms, this is not the case for the molecule HD, with one atom of deuterium (Le. , heavy hydrogen, 2H). Because of its small moment of inertia, the HD molecule has a relatively large value of E: 0.0057 eV. At approximately what temperature would you expect the rotational heat capacity of a gas of HD molecules to 'freeze out,' that is, to fall significantly below the constant value predicted by the equipartition theorem? Problem 6.30. In this problem you will investigate the behavior of ordinary hydrogen, H2, at low temperatures. The constant E is 0.0076 e V. As noted in the text, only half of the terms in the rotational partition function, equation 6.30, contribute for any given molecule. More precisely, the set of allowed j values is determined by the spin configuration of the two atomic nuclei. There are four independent spin configurations, classified as a single 'singlet' state and three 'triplet' states. The time required for a molecule to convert between the singlet and triplet configurations is ordinarily quite long, so the properties of the two types of molecules can be studied independently. The singlet molecules are known as parahydrogen while the triplet molecules are known as orthohydrogen.
(a) For parahydrogen, only the rotational states with even values of j are allowed.* Use a computer (as in Problem 6.28) to calculate the rotational *For those who have studied quantum mechanics, here's why: Even-j wavefunctions are symmetric (unchanged) under the operation of replacing r with -r, which is equivalent to interchanging the two nuclei; odd-j wavefunctions are antisymmetric under this operation. The two hydrogen nuclei (protons) are fermions , so their overall wavefunction must be antisymmetric under interchange. The singlet state (T 1 - 1j) is already antisymmetric in
237
238
Chapter 6
(b) (c)
(d)
(e)
Boltzmann Statistics partition function, average energy, and heat capacity of a parahydrogen molecule. Plot the heat capacity as a function of kT/ E. * For orthohydrogen, only the rotational states with odd values of j are allowed. Repeat part (a) for orthohydrogen. At high temperature, where the number of accessible even-j states is es­ sentially the same as the number of accessible odd-j states, a sample of hydrogen gas will ordinarily consist of a mixture of 1/4 parahydrogen and 3/4 orthohydrogen. A mixture with these proportions is called normal hydrogen. Suppose that normal hydrogen is cooled to low temperature without allowing the spin configurations of the molecules to change. Plot the rotational heat capacity of this mixture as a function of temperature. At what temperature does the rotational heat capacity fall to half its high­ temperature value (Le., to k/2 per molecule)? Suppose now that some hydrogen is cooled in the presence of a catalyst that allows the nuclear spins to frequently change alignment. In this case all terms in the original partition function are allowed, but the odd-j terms should be counted three times each because of the nuclear spin degener­ acy. Calculate the rotational partition function, average energy, and heat capacity of this system, and plot the heat capacity as a function of kT/ E. A deuterium molecule, D2, has nine independent nuclear spin configura­ tions, of which six are 'symmetric' and three are 'antisymmetric.' The rule for nomenclature is that the variety with more independent states gets called 'ortho-,' while the other gets called 'para-.' For orthodeuterium onlyeven-j rotational states are allowed, while for paradeuterium only odd­ j states are allowed. t Suppose, then, that a sample of D2 gas, consisting of a normal equilibrium mixture of 2/3 ortho and 1/3 para, is cooled without allowing the nuclear spin configurations to change. Calculate and plot the rotational heat capacity of this system as a function of temperature.:j:
6.3 The Equipartition Theorem I've been invoking the equipartition theorem throughout this book, and we've ver­ ified that it is true in a number of particular cases, but so far I haven't shown you an actual proof. The proof is quite easy, if you use Boltzmann factors. The equipartition theorem doesn't apply to all systems. It applies only to systems whose energy is in the form of quadratic 'degrees of freedom,' of the form
E(q)
=
cq2,
(6.34)
where c is a constant coefficient and q is any coordinate or momentum variable, like x, or Px, or Lx (angular momentum). I'm going to treat just this single degree spin, so its spatial wavefunction must be symmetric, while the triplet states (ii, 11, and i 1 + 1i) are symmetric in spin, so their spatial wavefunctions must be antisymmetric. *For a molecule such as 02 with spin-O nuclei, this graph is the whole story; the only nuclear spin configuration is a singlet and only the even-j states are allowed. tDeuterium nuclei are bosons, so the overall wavefunction must be symmetric under interchange. +For a good discussion of hydrogen at low temperature, with references to experiments, see Gopal (1966).
6.3 '
, , , '
, , , '
, , ''
, , '
'
, , , '
-----+l I+­ llq
The Equipartition Theorem , , '
, , , , '.- q
Figure 6.S. To count states over a continuous variable q, pretend that they're discretely spaced, separated by llq.
of freedom as my 'system,' assume that it's in equilibrium with a reservoir at temperature T, and calculate its average energy, E. I'll analyze this system in classical mechanics, where each value of q corresponds to a separate, independent state. To count the states, I'll pretend that they're discretely spaced, separated by small intervals f:1q, as shown in Figure 6.8. As long as f:1q is extremely small, we expect it to cancel out of the final result for E. The partition function for this system is
(6.35) q
q
To evaluate the sum, I'll multiply by f:1q inside the sum and divide by f:1q outside the sum: q2 z=~ e-{3c f:1q. (6.36) q q
L
Now the sum can be interpreted as the area under a bar graph whose height is determined by the Boltzmann factor (see Figure 6.9). Since f:1q is very small, we can approximate the bar graph by the smooth curve, changing the sum into an integral:
1
00
Z
= -1
f:1q
e-{3c q 2
dq.
(6.37)
-00
Before trying to evaluate the integral, let's change variables to x dq = dx / v7JC. Then 00 1 1 _x2 z= ~ I7.i':. e dx. uq v (3c - 0 0
1
Boltzmann factor,
=
v7JC q, so that (6.38)
e-{3 cq 2
Figure 6.9. The partition function is the area under a bar graph whose height is the Boltzmann factor, e-{3 cq 2. To calculate this area, we pretend that the bar graph is a smooth curve.
239
240
Chapter 6
Boltzmann Statistics
The integral over x is now just some number, whose value isn't very important as far as the physics is concerned. However, the integral is rather interesting mathemati­ x2 cally. The function e- is called a Gaussian, and unfortunately its antiderivative cannot be written in terms of elementary functions. But there is a clever trick (described in Appendix B) for evaluating the definite integral from -00 to 00, and the result is simply Vii. So our final result for the partition function is Z
= ~ ~ = C/3-1/2 !1q V/3c '
(6.39)
where C is just an abbreviation for v;TC1 !1q. Once you have an explicit formula for the partition function, it's easy to calculate the average energy, using the magical formula 6.25: E
18Z
Z 8/3
__1~ ~C/3-1/2
8/3
1
(6.40)
= ~kT.
This is just the equipartition theorem. Notice that the constants c, !1q, and fo have all canceled out. The most important fact about this proof is that it does not carryover to quantum-mechanical systems. You can sort of see this from Figure 6.9: If the number of distinct states that have significant probabilities is too small, then the smooth Gaussian curve will not be a good approximation to the bar graph. And indeed, as we've seen in the case of an Einstein solid, the equipartition theorem is true only in the high-temperature limit, where many distinct states contribute and therefore the spacing between the states is unimportant. In general, the equipar­ tition theorem applies only when the spacing between energy levels is much less than kT. Problem 6.31. Consider a classical of freedom' that is linear rather than quadratic: E = clql for some constant c. (An example would be the kinetic energy of a highly relativistic particle in one dimension, written in terms of its momentum.) Repeat the derivation of the equipartition theorem for this system, and show that the average energy is E kT. Problem 6.32. Consider a classical particle moving in a one-dimensional potential well u(x), as shown in Figure 6.10. The particle is in thermal equilibrium with a reservoir at temperature so the probabilities of its various states are determined by Boltzmann statistics. (a) Show that the average position of the particle is given by x = -'----:::--;-;:-­ where each integral is over the entire x axis.
6.3
The Equipartition Theorem
u(x) Figure 6.10. A one-dimensional po­
tential well. The higher the temper­
ature, the farther the particle will
stray from the equilibrium point.
xo
x
(b) If the temperature is reasonably low (but still high enough for classical me­ chanics to apply), the particle will spend most of its time near the bottom of the potential welL In that case we can expand u(x) in a Taylor series about the equilibrium point Xo: u(x)
u(xo)
+ (x -
+~
xo) dul Xo
+ ~(x
XO)3
~~ Ixo + ..
Show that the linear term must be zero, and that truncating the series after the quadratic term results in the trivial prediction x Xo.
(c) If we keep the cubic term in the Taylor series as well, the integrals in the formula for x become difficult. To simplify them, assume that the cubic term is small, so its exponential can be expanded in a Taylor series (leaving the quadratic term in the exponent). Keeping only the smallest temperature-dependent term, show that in this limit x differs from Xo by a term proportional to kT. Express the coefficient of this term in terms of the coefficients of the Taylor series for u(x).
(d) The interaction of noble gas atoms can be modeled using the Lennard­ Jones potential,
u(x)
XO )12 = uo [( -;;
Sketch this function, and show that the minimum of the potential well is at x = Xo, with depth uo· For argon, Xo = 3.9 A and Uo = O.OlD eV. Ex­ pand the Lennard-Jones potential in a Taylor series about the equilibrium point, and use the result of part (c) to predict the linear thermal expansion coefficient (see Problem 1.8) of a noble gas crystal in terms of uo. Evalu­ ate the result numerically for argon, and compare to the measured value Q: = 0.0007 (at 80 K).
241
242
Chapter 6
Boltzmann Statistics
6.4 The Maxwell Speed Distribution For our next application of Boltzmann factors, I'd like to take a detailed look at the motion of molecules in an ideal gas. We already know (from the equipartition theorem) that the root-mean-square speed of the molecules is given by the formula (6.41) But this is just a sort of average. Some of the molecules will be moving faster than this, others slower. In practice, we might want to know exactly how many molecules are moving at any given speed. Equivalently, let's ask what is the probability of some particular molecule moving at a given speed. Technically, the probability that a molecule is moving at any given speed v is zero. Since speed can vary continuously, there are infinitely many possible speeds, and therefore each of them has infinitesimal probability (which is essentially the same as zero). However, some speeds are less probable than others, and we can still represent the relative probabilities of various speeds by a graph, which turns out to look like Figure 6.11. The most probable speed is where the graph is the highest, and other speeds are less probable, in proportion to the height of the graph. Furthermore, if we normalize the graph (that is, adjust the vertical scale) in the right way, it has a more precise interpretation: The area under the graph between any two speeds VI and V2 equals the probability that the molecule's speed is between VI and V2: Probability(vI'
V2)
=
l
V2
V(v) dv,
(6.42)
VI
where V(v) is the height of the graph. If the interval between VI and V2 is infinites­ imal, then V(v) doesn't change significantly within the interval and we can write simply Probability(v .. v+dv) = V(v) dv. (6.43)
D(v)
~~--------------+---~----~~----~v
Figure 6.11. A graph of the relative probabilities for a gas molecule to have various speeds. More precisely, the vertical scale is defined so that the area under the graph within any interval equals the probability of the molecule having a speed in that interval.
6.4
The Maxwell Speed Distribution
The function V( v) is called a distribution function. Its actual value at any point isn't very meaningful by itself. Instead, V( v) is a function whose purpose in life is to be integrated. To turn V( v) into a probability you must integrate over some interval of v's (or, if the interval is small, just multiply by the width of the interval). The function V( v) itself doesn't even have the right units (namely, none) for a probability; instead, it has units of l/v, or (m/s)-I. Now that we know how to interpret the answer, I'd like to derive a formula for the function V(v). The most important ingredient in the derivation is the Boltzmann factor. But another important element is the fact that space is three dimensional, which implies that for any given speed, there are many possible velocity vectors. In fact, we can write the function V( v) schematically as
V( v) ex (probabi~itY of a :nol~cule) x ( number ~f vectors if ) . havIng velocIty v correspondIng to speed v
(6.44)
There's also a constant of proportionality, which we'll worry about later. The first factor in equation 6.44 is just the Boltzmann factor. Each velocity vector corresponds to a distinct molecular state, and the probability of a molecule being in any given state s is proportional to the Boltzmann factor e-E(s)/kT. In this case the energy is just the translational kinetic energy, ~mv2 (where v = i1i), so probability of a mOlecule) -mv2/2kT (6.45) ( ex e . having velocity if I've neglected any variables besides velocity that might affect the state of the mole­ cule, such as its position in space or its internal motion. This simplification is valid for an ideal gas, where the translational motion is independent of all other variables. Equation 6.45 says that the most likely velocity vector for a molecule in an ideal gas is zero. Given what we know about Boltzmann factors, this result should hardly be surprising: Low-energy states are always more probable than high-energy states, for any system at finite (positive) temperature. However, the most likely velocity vector does not correspond to the most likely speed, because for some speeds there are more distinct velocity vectors than for others. So let us turn to the second factor in equation 6.44. To evaluate this factor, imagine a three-dimensional 'velocity space' in which each point represents a ve­ locity vector (see Figure 6.12). The set of velocity vectors corresponding to any given speed v lives on the surface of a sphere with radius v. The larger v is, the bigger the sphere, and the more possible velocity vectors there are. So I claim that the second factor in equation 6.44 is the surface area of the sphere in velocity space: number of vectors if ) 4 2 . ex rrv . ( correspondIng to speed v
(6.46)
Putting this 'degeneracy' factor together with the Boltzmann factor (6.45), we obtain V( v) = C . 4rrv 2 e- mv2 /2kT, (6.47)
243
244
Chapter 6
Boltzmann Statistics
vz
Figure 6.12. In 'velocity space' each point represents a possible velocity vector. The set of all vectors for a given speed v lies on the surface of a sphere with radius v.
Vy
Vx
where C is a constant of proportionality. To determine C, note that the total probability of finding the molecule at some speed must equal 1: 1
fo= D(v) dv
Changing variables to x
4rrC fooo v 2 e -=v' /2kT dv.
(6.48)
vVm/2kT puts this integral into the form 2kT)3/2 1 = 47rC ( - m
foCXl x 2
dx.
(6.49)
0 2
Like the pure Gaussian , the function x 2 e- x cannot be anti-differentiated in terms of elementary functions. Again, however, there are tricks (explained in Appendix B) for evaluating the definite integral from 0 to 00; in this case the answer is v'iF/4. The 4 cancels, leaving us with C = (m/27rkT)3/2. Our final result for the distribution function V(v) is therefore V(v)
(
~
27rkT )
3/2
47rv 2 e -mv 2 /2kT.
(6.50)
This result is called the Maxwell distribution (after James Clerk 11axwell) for the speeds of molecules in an ideal gas. It's a complicated formula, but I hope you won't find it hard to remember the important parts: the Boltzmann factor involving the translational kinetic energy, and the geometrical factor of the surface area of the sphere in velocity space. Figure 6.13 shows another plot of the Maxwell distribution. At very small v, the Boltzmann factor is approximately 1 so the curve is a parabola; in particular, the distribution goes to zero at v O. This result does not contradict the fact that zero is the most likely velocity vector, because now we're talking about speeds, and there are simply too few velocity vectors corresponding to very small speeds. IvIeanwhile, the Maxwell distribution also goes to zero at very high speeds (much greater than VkT/rn), because of the exponential fall-off of the Boltzmann factor. In between v 0 and v 00, the Maxwell distribution rises and falls. By setting the derivative of equation 6.50 equal to zero, you can show that the maximum value V2kT/rn. As you would expect, the peak shifts to the of V(v) occurs at V max
6.4
The Maxwell Speed Distribution
D{v)
Dies exponentially
/ ~~--------~~~--------------~~--~v
V
Vmax
Vrms
Figure 6.13. The Maxwell speed distribution falls off as v ~ 0 and as v ~ 00. The average speed is slightly larger than the most likely speed, while the rms speed is a bit larger still.
right as the temperature is increased. The most likely speed is not the same as the rms speed; referring to equation 6.41, we see that the rms speed is the greater of the two, by about 22%. The average speed is different still; to compute it, add up all the possible speeds, weighted by their probabilities:
v=
L
(6.51)
vV(v)dv.
all v
In this equation I'm imagining the various speeds to be discretely spaced, separated by dv. If you turn the sum into an integral and evaluate it, you get V=
J8kT, 7rm
(6.52)
which lies in between V max and v rms . As an example, consider the nitrogen molecules in air at room temperature. You can easily calculate the most probable speed, which turns out to be 422 mls at 300 K. But some of the molecules are moving much faster than this, while others are moving much slower. What is the probability that a particular molecule is moving faster than 1000 m/s? First let's make a graphical estimate. The speed 1000 mls exceeds V max by a factor of 1000 mls = 2.37. (6.53) 422 mls Looking at Figure 6.13, you can see that at this point the Maxwell distribution is rapidly dying out but not yet dead. The area under the graph beyond 2.37vmax looks like only one or two percent of the total area under the graph. Quantitatively, the probability is given by the integral of the Maxwell distribu­ tion from 1000 mls to infinity: Probability (v > 1000 m/s)
= 47r
3/2100 v (27rmkT ) m/s 1000
2
e- mv2 /2kT dv.
(6.54)
245
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Chapter 6
Boltzmann Statistics
With this nontrivial lower limit, the integral cannot be carried out analytically; the best option is to do it numerically, by calculator or computer. You could go ahead and plug in numbers at this point, instructing the computer to work in units of m/s. But it's much cleaner to first change variables to x = vJm/2kT = vlv max ) just as in equation 6.49. The integral then becomes (6.55) where the lower limit is the value of x when v = 1000 mis, that is, Xmin = (1000 m/s)/(422 m/s) 2.37. Now it's easy to type the integral into a com­ puter. I did so and got an answer of 0.0105 for the probability. Only about 1% of the nitrogen molecules are moving faster than 1000 m/s. Problem 6.33. Calculate the most probable speed, average speed, and rms speed for oxygen (02) molecules at room temperature. Problem 6.34. Carefully plot the Maxwell speed distribution for nitrogen mole­ cules at T 300 K and at T 600 K. Plot both graphs on the same axes, and label the axes with numbers. Problem 6.35. Verify from the Maxwell speed distribution that the most likely speed of a molecule is y'2kT /m. Problem 6.36. Fill in the steps between equations 6.51 and 6.52, to determine the average speed of the molecules in an ideal gas. Problem 6.37. Use the Maxwell distribution to calculate the average value of v 2 for the molecules in an ideal gas. Check that your answer agrees with equation 6.41. Problem 6.38. At room temperature, what fraction of the nitrogen molecules in the air are moving at less than 300 m/s? Problem 6.39. A particle near earth's surface traveling faster than about 11 km/s has enough kinetic energy to completely escape from the earth, despite earth's gravitational pull. Molecules in the upper atmosphere that are moving faster than this will therefore escape if they do not suffer any collisions on the way out. (a) The temperature of earth's upper atmosphere is actually quite high, around 1000 K. Calculate the probability of a nitrogen molecule at this temperature moving faster than 11 km/s, and comment on the result. (b) Repeat the calculation for a hydrogen molecule (H2) and for a helium atom, and discuss the implications. (c) Escape speed from the moon's surface is only about 2.4 km/s. Explain why the moon has no atmosphere. Problem 6.40. You might wonder why all the molecules in a gas in thermal equilibrium don't have exactly the same speed. After all, when two molecules collide, doesn't the faster one always lose energy and the slower one energy? And if so, wouldn't repeated collisions eventually bring all the molecules to some common speed? Describe an example of a billiard-ball collision in which this is not the case: The faster ball gains energy and the slower ball loses energy. Include numbers, and be sure that your collision conserves both energy and momentum.
6.5
Partition Functions and Free Energy
Problem 6.41. Imagine a world in which space is two-dimensional, but the laws of physics are otherwise the same. Derive the speed distribution formula for an ideal gas of nonrelativistic particles in this fictitious world, and sketch this distribution. Carefully explain the similarities and differences between the two-dimensional and three-dimensional cases. What is the most likely velocity vector? What is the most likely speed?
6.5 Partition Functions and Free Energy For an isolated system with fixed energy U, the most fundamental statistical quan­ tity is the multiplicity, f2(U) - the number of available microstates. The logarithm of the multiplicity gives the entropy, which tends to increase. For a system in equilibrium with a reservoir at temperature T (see Figure 6.14), the quantity most analogous to f2 is the partition function, Z(T). Like f2(U), the partition function is more or less equal to the number of microstates available to the system (but at fixed temperature, not fixed energy). We might therefore expect its logarithm to be a quantity that tends to increase under these conditions. But we already know a quantity that tends to decrease under these conditions: the Helmholtz free energy, F. The quantity that tends to increase would be -F, or, if we want a dimensionless quantity, - F/ kT. Taking a giant intuitive leap, we might therefore guess the formula
F = -kTlnZ
or
(6.56)
Indeed, this formula turns out to be true. Let me prove it. First recall the definition of F:
F
=U -T5.
(6.57)
Also, recall the partial derivative relation
(aF) aT
= -5 V,N
(6.58) .
T fixed
U fixed
S
=
kInO
F
= - kTlnZ
Figure 6.14. For an isolated system (left), S tends to increase. For a system at constant temperature (right), F tends to decrease. Like S, F can be written as the logarithm of a statistical quantity, in this case Z.
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Chapter 6
Boltzmann Statistics
Solving equation 6.57 for S and plugging into equation 6.58
F-U
(6.59)
T
This is a differential equation for the function F(T), for any fixed V and N. To prove equation 6.56, I'll show that the quantity -kTln Z obeys the same differential equation, with the same 'initial' condition at T = O. Let me define the symbol F to stand for the quantity -kT In Z. Holding V and N fixed, I want to find a formula for the derivative of this quantity with respect to T:
a (-kTlnZ) = -klnZ -
a
kT aT InZ.
(6.60)
In the second term I'll use the chain rule to rewrite the derivative in terms of (3 = l/kT: a(3 a -1 1 az U -lnZ (6.61) In Z = kT2 Z a(3 = kT2' aT
a
(Here I'm using U instead of for the system's average energy, because these ideas are most useful when applied to fairly large systems.) Plugging this result back into equation 6.60, we obtain
-klnZ
(6.62)
that is, F obeys exactly the same differential equation as F. A first-order differential equation has an infinite family of solutions, correspond­ ing to different 'initial' conditions. So to complete the proof that F = F, I need to show that they're the same for at least one particular value of T, say T O. At T = 0, the original F is simply equal to U, the energy of the system when it is at zero temperature. This energy must be the lowest possible energy, Uo, since the Boltzmann factors e-U(s)/kT for all excited states will be infinitely suppressed in comparison to the ground state. the partition function at T = 0 is simply e- Uo / kT , again since all other Boltzmann factors are infinitely suppressed in comparison. Therefore
F(O)
-kTlnZ(O)
Uo
F(O),
(6.63)
completing the proof that F F for all T. The usefulness of the formula F -kTln Z is that from F we can compute the entropy, pressure, and chemical potential, the partial-derivative formulas
s = - (~~) V,N ,
p
(6.64)
In this way we can compute all the thermodynamic properties of a system, once we know its partition function. In Section 6.7 I'll apply this technique to analyze an ideal gas.
6.6
Partition Functions for Composite Systems
Problem 6.42. In Problem 6.20 you computed the partition function for a quan­ tum harmonic oscillator: Zh .o. = l/(l-e-,BE), where E = hf is the spacing between energy levels. (a) Find an expression for the Helmholtz free energy of a system of N harmonic oscillators.
(b) Find an expression for the entropy of this system as a function of temper­ ature. (Don't worry, the result is fairly complicated.)
Problem 6.43. Some advanced textbooks define entropy by the formula
S
= -k L 1'(s) In 1'(s),
where the sum runs over all microstates accessible to the system and 1'(s) is the probability of the system being in microstate s.
(a) For an isolated system, 1'(s) = l/D for all accessible states s. Show that in this case the preceding formula reduces to our familiar definition of entropy.
(b) For a system in thermal equilibrium with a reservoir at temperature T , 1'(s) = e-E(s)/kT /Z. Show that in this case as well, the preceding formula agrees with what we already know about entropy.
6.6 Partition Functions for Composite Systenl.s Before trying to write down the partition function for an ideal gas, it is useful to ask in general how the partition function for a system of several particles is related to the partition function for each individual particle. For instance, consider a system of just two particles, 1 and 2. If these particles do not interact with each other, so their total energy is simply E1 + E 2 , then (6.65) s
where the sum runs over all states 8 for the composite system. If, in addition, the two particles are distinguishable (either by their fixed positions or by some intrinsic properties), then the set of states for the composite system is equivalent to the set of all possible pairs of states, (81, 82), for the two particles individually. In this case, (6.66) where 81 represents the state of particle 1 and 82 represents the state of particle 2. The first Boltzmann factor can be moved outside the sum over 82. This sum, now of just the second Boltzmann factor, is simply the partition function for particle 2 alone, Z2' This partition function is independent of 81, and can therefore be taken out of the remaining sum. Finally, the sum over 81 gives simply Z1, leaving us with (noninteracting, distinguishable particles).
(6.67)
If the particles are indistinguishable, however, the step going from equation 6.65 to equation 6.66 is not valid. The problem is exactly the same as the one
249
250
Chapter 6
Boltzmann Statistics
1·--...
Figure 6.15. Interchanging the states of two indistinguishable particles leaves the system in the same state as before.
we encountered when computing the multiplicity of an ideal gas in Section 2.5: Putting particle 1 in state A and particle 2 in state B is the same thing as putting particle 2 in state A and particle 1 in state B Figure 6.15). Equation 6.66 therefore counts nearly every state twice, and a more accurate formula would be 1
Ztotal
= '2Z1Z2
(noninteracting, indistinguishable particles).
(6.68)
This formula still isn't precisely correct, because there are some terms in the double sum of equation 6.66 in which both particles are in the same state, that is, SI S2. These terms have not been double-counted, so we shouldn't divide their number by 2. But for ideal gases and many other familiar the density is low enough that the chances of both particles being in the same state are negligible. The terms with SI = S2 are therefore only a tiny fraction of all the terms in equation 6.66, and it doesn't much matter whether we count them correctly or not. * The generalization of equations 6.67 and 6.68 to systems of more than two particles is straightforward. If the particles are distinguishable, the total partition function is the product of all the individual partition functions: (noninteracting, distinguishable systems).
(6.69)
This equation also applies to the total partition function of a single particle that can store energy in several ways; for instance, could be the partition function for its motion in the x direction, Z2 for its motion in the y direction, and so on. For a not-too-dense system of N indistinguishable particles, the general formula is (noninteracting, indistinguishable particles), (6.70) Ztotal where ZI is the partition function for anyone of the particles individually. The number of ways of interchanging N particles with each other is N!, hence the prefactor. When we deal with multiparticle systems, one point of terminology can be confusing. It is important to distinguish the 'state' of an individual particle from *The following chapter deals with very dense systems for which this issue is important. Until then, don't worry about it.
6.7
Ideal Gas Revisited
the 'state' of the entire system. Unfortunately, I don't know of a good concise way to distinguish between these two concepts. When the context is ambiguous, I'll write single-particle state or system state, as appropriate. In the preceding discussion, 8 is the system state while 81 and 82 are single-particle states. In general, to specify the system state, you must specify the single-particle states of all the particles in the system. Problem 6.44. Consider a large system of N indistinguishable, noninteracting molecules (perhaps in an ideal gas or a dilute solution). Find an expression for the Helmholtz free energy of this system, in terms of Z 1, the partition function for a single molecule. (Use Stirling's approximation to eliminate the NL) Then use your result to find the chemical potential, again in terms of Z1.
6.7 Ideal Gas Revisited The Partition Function We now have all the tools needed to calculate the partition function, and hence all the other thermal quantities, of an ideal gas. An ideal gas , as before, means one in which the molecules are usually far enough apart that we can neglect any energy due to forces between them. If the gas contains N molecules (all identical), then its partition function has the form (6.71 ) where Z1 is the partition function for one individual molecule. To calculate Z1, we must add up the Boltzmann factors for all possible mi­ crostates of a single molecule. Each Boltzmann factor has the form (6.72) where E tr is the molecule's translational kinetic energy and Eint is its internal energy (rotational, vibrational, or whatever), for the state 8. The sum over all single-particle states can be written as a double sum over translational states and internal states, allowing us to factor the partition function as in the previous section. The result is simply (6.73) Z1 = ZtrZint, where Ztr
e-Etr/kT
= translational states
and
Z int=
'e-Eint/kT. ~
(6.74)
internal states
The internal partition functions for rotational and vibrational states are treated in Section 6.2. For a given rotational and vibrational state, a molecule can also have various electronic states, in which its electrons are in different independent wave­ functions. For most molecules at ordinary temperatures, electronic excited states have negligible Boltzmann factors, due to their rather high energies. The electronic
251
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Chapter 6
Boltzmann Statistics
ground state, however, can sometimes be degenerate. An oxygen molecule, for ex­ ample, has a threefold-degenerate ground state, which contributes a factor of 3 to its internal partition function. Now let us put aside the internal partition function and concentrate on the translational part, Ztr. To compute Ztr, we need to add up the Boltzmann factors for all possible translational states of a molecule. One way to enumerate these states is to count all the possible position and momentum vectors for a molecule, slipping in a factor of 11h 3 to account for quantum mechanics as in Section 2.5. Instead, however, I'd now like to use the more rigorous method of counting all the independent definite-energy wavefunctions, just as we've been doing with internal states of atoms and molecules. I'll start with the case of a molecule confined to a one-dimensional box, then generalize to three dimensions. A few of the definite-energy wavefunctions for a molecule in a one-dimensional box are shown in Figure 6.16. Because the molecule is confined to the box, its wavefunction must go to zero at each end, and therefore the allowed standing-wave patterns are limited to wavelengths of An = 2L
n
n
= 1, 2, .. ,
(6.75)
where L is the length of the box and n is the number of 'bumps.' Each of these standing waves can be thought of as a superposition of left- and right-moving trav­ eling waves with equal and opposite momenta; the magnitude of the momentum is given by the de Broglie relation p hiA, that is, h Pn = An
hn 2L
(6.76)
Finally, the relation between energy and momentum for a nonrelativistic particle is E = p2/2m, where m is its mass. So the allowed energies for a molecule in a one-dimensional box are (6.77)
2£
Al = 2£
Figure 6.16. The three lowest-energy wavefunctions for a particle confined to a one-dimensional box.
6.7
Ideal Gas Revisited
Knowing the energies, we can immediately write down the translational partition function for this molecule (still in one dimension):
=2:
=
2: e-h2n2/8mL2kT.
(6.78)
n
n
Unless Land/or T is extremely small, the energy levels are extremely close together, so we may as well approximate the sum as an integral:
ft
--­ =
V27rmkT L h
L
2
2
(6.79)
where tQ is defined as the reciprocal of the square root in the previous expression: h tQ --;::.
(6.80)
I like to call tQ the quantum length; aside from the factor of 7r, it is the de Broglie wavelength of a particle of mass m whose kinetic energy is kT. For a nitrogen molecule at room temperature, the quantum length works out to 1.9 X 10- 11 m. The ratio L / tQ is therefore quite large for any realistic box, meaning that many translational states are available to the molecule under these conditions: roughly the number of de Broglie wavelengths that would fit inside the box. So much for a molecule moving in one dimension. For a molecule moving in three dimensions, the total kinetic energy is 2
=
Px
2m
2
Py
+ 2m + 2m'
(6.81)
where each momentum component can take on infinitely many different values according to formula 6.76. Since the n's for the three momentum components can be chosen independently, we can again factor the partition function, into a piece for each of the three dimensions:
v,
vQ
(6.82)
where V is the total volume of the box and
t3
_
Q -
(
vQ
is the quantum volume,
h J27rmkT )
3
(6.83)
The quantum volume is just the cube of the quantum length, so it's very small for a molecule at room temperature. The translational partition function is essentially the number of de Broglie-wavelength cubes that would fit inside the entire volume under ordinary conditions. of the box, and is again quite
253
254
Chapter 6
Boltzmann Statistics
Combining this result with equation 6.73, we obtain for the single-particle par­ tition function V (6.84) Zint, vQ
where Zint is a sum over all relevant internal states. The partition function for the entire gas of N molecules is then
~ (VZint)N
Z
N!
(6.85)
vQ
For future reference, the logarithm of the partition function is
InZ
N[ln V
+ In
InN -lnvQ
+ 1J.
(6.86)
Predictions At this point we can compute all of the thermal properties of an ideal gas. Let's start with the total (average) energy, using the formula derived in Problem 6.16:
U=
18Z
Z 8/3
8
The quantities in equation 6.86 that depend on
8 1 8vQ U=-N-InZ.t+ N - - -
8/3
III
VQ
8/3
(6.87)
8/3 In Z. /3 are Zint
31 N E·IIIt + N· -2/3
and vQ, so 3
= U t + -NkT 2 . III
(6.88)
Here Eint is the average internal energy of a molecule. The average translational kinetic energy is ~kT, as we already knew from the equipartition theorem. Taking another derivative gives the heat capacity, 8U Cv = 8T
(6.89)
For a diatomic gas, the internal contribution to the heat capacity comes from rotation and vibration. As shown in Section 6.2, each of these contributions adds approximately N k to the heat capacity at sufficiently high temperatures, but goes to zero at lower temperatures. The translational contribution could also freeze out in theory, but only at temperatures so low that I!Q is of order L, so replacing the sum by an integral in equation 6.79 becomes invalid. We have now explained all the features in the graph of Cv for hydrogen shown in Figure 1.13. To compute the remaining thermal properties of an ideal gas, we need the Helmholtz free energy,
F
-kTlnZ
= -NkT[ln V + InZint -InN -lnvQ + 1J -NkT[ln V InN -In vQ + 1J + F int ,
(6.90)
6.7
F,
where F int is the internal contribution to expression it's easy to compute the pressure,
p __ (aF)
-
av
T,N
namely
Ideal Gas Revisited
-NkTln Zint.
From this
NkT
---v-.
(6.91)
I'll let you work out the entropy and chemical potential. The results are
s = - (aF) aT aT V,N = Nk[ln(~) NVQ + ~l 2
(6.92)
and f.L
=
(aF) = _kTln(VZint). aN T,v NVQ
(6.93)
If we neglect the internal contributions, both of these quantities reduce to our earlier results for a monatomic ideal gas. Problem 6.45. Derive equations 6.92 and 6.93 for the entropy and chemical potential of an ideal gas. Problem 6.46. Equations 6.92 and 6.93 for the entropy and chemical potential involve the logarithm of the quantity V Zint!N vQ. Is this logarithm normally positive or negative? Plug in some numbers for an ordinary gas and discuss. Problem 6.47. Estimate the temperature at which the translational motion of a nitrogen molecule would freeze out, in a box of width 1 cm. Problem 6.48. For a diatomic gas near room temperature, the internal parti­ tion function is simply the rotational partition function computed in Section 6.2, multiplied by the degeneracy Ze of the electronic ground state.
(a) Show that the entropy in this case is
Calculate the entropy of a mole of oxygen (Ze = 3) at room temperature and atmospheric pressure, and compare to the measured value in the table at the back of this book. * (b) Calculate the chemical potential of oxygen in earth's atmosphere near sea level, at room temperature. Express the answer in electron-volts. Problem 6.49. For a mole of nitrogen (N2) gas at room temperature and atmo­ spheric pressure, compute the following: U, H, F, G, S, and p,. (The electronic ground state of nitrogen is not degenerate.) Problem 6.50. Show explicitly from the results of this section that G an ideal gas.
= N p, for
*See Rock (1983) or Gopal (1966) for a discussion of the comparison of theoretical and experimental entropies.
255
256
Chapter 6
Boltzmann Statistics
Problem 6.51. In this section we computed the single-particle translational par­ tition function, Ztr, by summing over all definite-energy wavefunctions. An al­ ternative approach, however, is to sum over all possible position and momentum vectors, as we did in Section 2.5. Because position and momentum are continuous variables, the sums are really and we need to slip in a factor of 1/ h 3 to get a unitless number that actually counts the independent wavefunctions. Thus, we might guess the formula
Z
tr
h13
Jd r dpe 3
3
-Etr/kT
,
where the integral sign actually represents six integrals, three over the po­ sition components (denoted d 3 r) and three over the momentum components (de­ noted d3 p). The of integration includes all momentum vectors, but only those position vectors that lie within a box of volume V. By evaluating the inte­ grals explicitly, show that this expression yields the same result for the translational partition function as that obtained in the text. (The only time this formula would not be valid would be when the box is so small that we could not justify converting the sum in equation 6.78 to an integral.) Problem 6.52. Consider an ideal gas of highly relativistic particles (such as photons or fast-moving electrons), whose energy-momentum relation is E pc instead of E = p2/ 2m . Assume that these particles live in a one-dimensional universe. By following the same logic as above, derive a formula for the particle partition function, , for one particle in this gas. Problem 6.53. The dissociation of molecular hydrogen into atomic hydrogen,
H2
+---+
2H,
can be treated as an ideal gas reaction using the techniques of Section 5.6. The equilibrium constant K for this reaction is defined as
K=
where pO is a reference pressure conventionally taken to be 1 bar, and the other P's are the partial pressures of the two at equilibrium. Now, using the methods of Boltzmann statistics developed in this chapter, you are ready to calculate K from first principles. Do so. That is, derive a formula for K in terms of more basic quantities such as the energy needed to dissociate one molecule (see Problem 1.53) and the internal partition function for molecular hydrogen. This internal partition function is a product of rotational and vibrational contributions, which you can estimate using the methods and data in Section 6.2. (An H2 molecule doesn't have any electronic spin degeneracy, but an H atom does-the electron can be in two different spin states. electronic excited states, which are important only at very high temperatures. The degeneracy due to nuclear alignments cancels, but include it if you wish.) Calculate K numerically at T 300 K, 1000 K, 3000 K, and 6000 K. Discuss the implications, working out a couple of numerical examples to show when hydrogen is mostly dissociated and when it is not.
7
Quantum Statistics
7.1 The Gibbs Factor In deriving the Boltzmann factor in Section 6.1, I allowed the small system and the reservoir to exchange energy, but not particles. Often, however, it is useful to consider a system that can exchange particles with its environment (see Figure 7.1). Let me now modify the previous derivation to allow for this possibility. As in Section 6.1, we can write the ratio of probabilities for two different mi­ crostates as OR(S2) --(7.1) OR(Sl) The exponent now contains the change in the entropy of the reservoir as the system goes from state 1 to state 2. This is an infinitesimal change from the reservoir's viewpoint, so we can invoke the thermodynamic identity:
(7.2)
I --+.
'Reservoir'
UR,NR
I
.~
T,j.l
'System' E,N
I I
J Figure 7.1. A system in thermal and diffusive contact with a much larger reser­ voir, whose temperature and chemical potential are effectively constant. 257
258
Chapter 7
Quantum Statistics
Since any energy, volume, or particles gained by the reservoir must be lost by the system, each of the changes on the right-hand side can be written as minus the same change for the system. As in Section 6.1, I'll throwaway the P dV term; this term is often zero, or at least very small compared to the others. This time, however, I'll keep the /-l dN term. Then the change in entropy can be written
(7.3) On the right-hand side both E and N refer to the small system, hence the overall minus sign. Plugging this expression into equation 7.1 gives
(7.4)
As before, the ratio of probabilities is a ratio of simple exponential factors, each of which is a function of the temperature of the reservoir and the energy of the corresponding microstate. Now, however, the factor depends also on the number of particles in the system for state s. This new exponential factor is called a Gibbs factor: Gibbs factor = e-[E(s)-p,N(s)l/kT. (7.5) If we want an absolute probability instead of a ratio of probabilities, again we have to slip a constant of proportionality in front of the exponential:
P(s) = ~e-[E(S)-P,N(S)l/kT.
(7.6)
The quantity Z is called the grand partition function* or the Gibbs sum. By requiring that the sum of the probabilities of all states equal 1, you can easily show that
(7.7)
s
where the sum runs over all possible states (including all possible values of N). If more than one type of particle can be present in the system, then the /-l dN term in equation 7.2 becomes a sum over species of /-li dNi , and each subsequent equation is modified in a similar way. For instance, if there are two types of particles, the Gibbs factor becomes (two species).
(7.8)
*In analogy with the terms 'microcanonical' and 'canonical' used to describe the methods of Chapters 2-3 and 6, the approach used here is called grand canonical. A hypothetical set of systems with probabilities assigned according to equation 7.6 is called a grand canonical ensemble.
7.1
The Gibbs Factor
An Example: Carbon Monoxide Poisoning A good example of a system to illustrate the use of Gibbs factors is an adsorp­ tion site on a hemoglobin molecule, which carries oxygen in the blood. A single hemoglobin molecule has four adsorption sites, each consisting of an Fe 2 + ion sur­ rounded by various other molecules. Each site can carry one O 2 molecule. For simplicity I'll take the system to be just one of the four sites, and pretend that it is completely independent of the other three. * Then if oxygen is the only molecule that can occupy the site, the system has just two possible states: unoccupied and occupied (see Figure 7.2). I'll take the energies of these two states to be 0 and f, with f = -0.7 eV.t The grand partition function for this single-site system has just two terms:
Z = 1 + e-(E-p,)/kT.
(7.9)
The chemical potential J.L is relatively high in the lungs, where oxygen is abundant, but is much lower in the cells where the oxygen is used. Let's consider the situation near the lungs. There the blood is in approximate diffusive equilibrium with the atmosphere, an ideal gas in which the partial pressure of oxygen is about 0.2 atm. The chemical potential can therefore be calculated from equation 6.93: J.L
= -kTln ( VZint) NVQ
~ -0.6 eV
(7.10)
at body temperature, 310 K. Plugging in these numbers gives for the second Gibbs factor e -(E-p,) / kT ~ e(O.l eV)/kT ~ 40. (7.11) The probability of any given site being occupied is therefore . P( occupIed by O 2 )
=
40 --41+ 0
= 98%.
(7.12)
4
Fe2 +
E=O
E
= -0.7 eV
. ~;
E = -0.85 eV
Figure 7.2. A single heme site can be unoccupied, occupied by oxygen, or occu­ pied by carbon monoxide. (The energy values are only approximate.) *The assumption of independent sites is quite accurate for myoglobin, a related protein that binds oxygen in muscles, which has only one adsorption site per molecule. A more accurate model of hemoglobin is presented in Problem 7.2. tBiochemists never express energies in electron-volts. In fact, they rarely talk about individual bond energies at all (perhaps because these energies can vary so much under different conditions). I've chosen the E values in this section to yield results that are in rough agreement with experimental measurements.
259
260
Chapter 7
Quantum Statistics
Suppose, however, that there is also some carbon monoxide present, which also be adsorbed into the heme site. Now there are three states available to the unoccupied, occupied by O 2 , and occupied by CO. The grand partition tUllctlOn
where t' is the negative energy of a bound CO molecule and J-l' is the Ch(~mllCal potential of CO in the environment. On the one hand, CO will never be as abundallt as oxygen. If it is 100 times less abundant, then its chemical potential is roughly kTln 100 = 0.12 eV, so J-l' is roughly -0.72 eV. On the other hand, more tightly bound to the site than oxygen, with t' ~ -0.85 eV. Plugging in numbers gives for the third Gibbs factor e-(E'-p,')/kT ~ e(O.13eV)/kT ~
120.
The probability of the site being occupied by an oxygen molecule therefore to P(occupied by O 2 )
=
1+
4~0+ 120 = 25%.
Problem 7.1. Near the cells where oxygen is used, its chemical potential is sig­ nificantly lower than near the lungs. Even though there is no gaseous oxygen these cells, it is customary to express the abundance of oxygen in terms of the par­ tial pressure of gaseous oxygen that would be in equilibrium with the blood. Using the independent-site model just presented, with only oxygen present, calculate and plot the fraction of occupied heme sites as a function of the partial pressure oxygen. This curve is called the Langmuir adsorption isotherm ('isotherm' because it's for a fixed temperature). Experiments show that adsorption by follows the shape of this curve quite accurately. Problem 7.2. In a real hemoglobin molecule, the tendency of oxygen to bind a heme site increases as the other three heme sites become occupied. To model this effect in a simple way, imagine that a hemoglobin molecule has just two sites, either or both of which can be occupied. This system has four possible states (with only oxygen present). Take the energy of the unoccupied state to be zero, the energies of the two singly occupied states to be -0.55 eV, and the energy the doubly occupied state to be -1.3 eV (so the change in energy upon binding the second oxygen is -0.75 eV). As in the previous problem, calculate and plot the fraction of occupied sites as a function of the effective partial pressure of oxygen. Compare to the graph from the previous problem (for independent sites). Can you think of why this behavior is preferable for the function of hemoglobin? Problem 7.3. Consider a system consisting of a single hydrogen atom/ion, which has two possible states: unoccupied (i.e., no electron present) and occupied (i.e., one electron present, in the ground state). Calculate the ratio of the probabilities of these two states, to obtain the Saha equation, already derived in Section 5.6. Treat the electrons as a monatomic ideal gas, for the purpose of determining J-l. Neglect the fact that an electron has two independent spin states.
7.1
The Gibbs Factor
Problem 7.4. Repeat the previous problem, taking into account the two inde­ pendent spin states of the electron. Now the system has two 'occupied' states, one with the electron in each spin configuration. However, the chemical potential of the electron gas is also slightly different. Show that the ratio of probabilities is the same as before: The degeneracy cancels out of the Saha equation. Problem 7.5. Consider a system consisting of a impurity atom/ion in a semiconductor. Suppose that the impurity atom has one 'extra' electron compared to the neighboring atoms, as would a phosphorus atom occupying a lattice site in a silicon crystal. The extra electron is then easily leaving behind a positively charged ion. The ionized electron is called a conduction electron, because it is free to move through the material; the impurity atom is called a donor, because it can 'donate' a conduction electron. This system is analogous to the hydrogen atom considered in the previous two problems except that the ionization energy is much mainly due to the of the ionic charge by the dielectric behavior of the medium. (a) Write down a formula for the probability of a single donor atom ionized. Do not the fact that the if present, can have two independent spin states. Express your formula in terms of the temperature, the ionization energy I, and the chemical potential of the 'gas' of ionized electrons. (b) Assuming that the conduction electrons behave like an ordinary ideal gas (with two spin states per particle), write their chemical potential in terms of the number of conduction electrons per unit volume, Nc/V. (c) Now assume that every conduction electron comes from an ionized donor atom. In this case the number of conduction electrons is equal to the number of donors that are ionized. Use this condition to derive a quadratic equation for Nc in terms of the number of donor atoms (Nd ), eliminating j.t. Solve for Nc the quadratic formula. It's helpful to introduce some abbreviations for dimensionless Try x = Nc/Nd, t kT/ I, and so on.) (d) For phosphorus in silicon, the ionization energy is 0.044 e V. Suppose that there are 10 17 P atoms per cubic centimeter. Using these numbers, calcu­ late and plot the fraction of ionized donors as a function of temperature. Discuss the results. Problem 7.6. Show that when a system is in thermal and diffusive equilibrium with a reservoir, the average number of particles in the system is
N = kT8Z
Z
where the partial derivative is taken at fixed temperature and volume. Show also that the mean square number of particles is
Use these results to show that the standard deviation of N is (IN
=
in analogy with Problem 6.18. Finally, apply this formula to an ideal gas, to obtain a simple expression for (J N in terms of N. Discuss your result briefly.
261
262
Chapter 7
Quantum Statistics
Problem 7.7. In Section 6.5 I derived the useful relation F -kTln Z between the Helmholtz free energy and the ordinary partition function. Use an analogous argument to prove that .p
-kTlnZ,
where Z is the grand partition function and .p is the grand free energy introduced in Problem 5.23.
7.2 Bosons and Fermions The most important application of Gibbs factors is to quantum statistics, the study of dense systems in which two or more identical particles have a reasonable chance of wanting to occupy the same single-particle state. In this situation, my derivation (in Section 6.6) of the partition function for a system of N indistinguish­ able, noninteracting particles, (7.16) breaks down. The problem is that the counting factor of N!, the number of ways of interchanging the particles among their various states, is correct only if the particles are always in different states. (In this section I'll use the word 'state' to mean a single-particle state. For the state of the system as a whole I'll always say 'system state.') To better understand this issue, let's consider a very simple example: a system containing two noninteracting particles, either of which can occupy any of five states (see 7.3). Imagine that all five of these states have energy zero, so every Boltzmann factor equals 1 (and therefore Z is the same as n). If the two particles are distinguishable, then each has five available states and the total number of system states is Z = 5 x 5 25. If the two particles are 2 indistinguishable, equation 7.16 would predict Z 5 /2 = 12.5, and this can't be right, since Z must (for this system) be an integer. So let's count the system states more carefully. Since the particles are indis­ tinguishable, all that matters is the number of particles in any given state. I can state by a sequence of five integers, each representing therefore represent any the number of particles in a particular state. For instance, 01100 would represent the state in which the second and third states each contain one particle,
.
~
DDDDD
Figure 7.3. A simple model of five single-particle states, with two particles that can occupy these states.
7.2
Bosons and Fermions
while the rest contain none. Here, then, are all the allowed system states: 11000 10100 10010 10001 01100
01010 01001 00110 00101 00011
20000 02000 00200 00020 00002
(If you pretend that the states are harmonic oscillators and the particles are energy units, you can count the system states in the same way as for an Einstein solid.) There are 15 system states in all, of which 10 have the two particles in different states while 5 have the two particles in the same state. Each of the first 10 system states would actually be two different system states if the particles were distinguish­ able, since then they could be placed in either order. These 20 system states, plus the last 5 listed above, make the 25 counted in the previous paragraph. The factor of l/N! in equation 7.16 correctly cuts the 20 down to 10, but also incorrectly cuts out half of the last five states. Here I'm implicitly assuming that two identical particles can occupy the same state. It turns out that some types of particles can do this while others can't. Par­ ticles that can share a state with another of the same species are called bosons, * and include photons, pions, helium-4 atoms, and a variety of others. The number of identical bosoils in a given state is unlimited. Experiments show, however, that many types of particles cannot share a state with another particle of the same type- not because they physically repel each other, but due to a quirk of quantum mechanics that I won't try to explain here (see Appendix A for some further dis­ cussion of this point). These particles are called fermions, t and include electrons, protons, neutrons, neutrinos, helium-3 atoms, and others. If the particles in the preceding example are identical fermions, then the five system states in the final column of the table are not allowed, so Z is only 10, not 15. (In formula 7.16, a system state with two particles in the same state is counted as half a system state, so this formula interpolates between the correct result for fermions and the correct result for bosons.) The rule that two identical fermions cannot occupy the same state is called the Pauli exclusion principle. You can tell which particles are bosons and which are fermions by looking at their spins. Particles with integer spin (0, 1, 2, etc., in units of h/27r) are bosons, while particles with half-integer spin (1/2, 3/2, etc.) are fermions. This rule is not the definition of a boson or fermion, however; it is a nontrivial fact of nature, a deep consequence of the theories of relativity and quantum mechanics (as first derived by Wolfgang Pauli). * After Satyendra Nath Bose, who in 1924 introduced the method of treating a photon gas presented in Section 7.4. The generalization to other bosons was provided by Einstein shortly thereafter. t After Enrico Fermi, who in 1926 worked out the basic implications of the exclusion principle for statistical mechanics. Paul A. M. Dirac independently did the same thing, in the same year.
263
264
Chapter 7
Quantum Statistics
In many situations, however, it just doesn 't matter whether the particles in a fluid are bosons or fermions. When the number of available single-particle states is much greater than the number of particles, (7.17) the chance of any two particles wanting to occupy the same state is negligible. More precisely, only a tiny fraction of all system states have a significant number of states doubly occupied. For an ideal gas, the single-particle partition function is Zl = V Zint!VQ , where Zint is some reasonably small number and vQ is the quantum volume, V
Q -
f3 Q -
h ( J27rmkT
)3
'
(7.18)
roughly the cube of the average de Broglie wavelength. The condition (7.17) for the formula Z = zf /N! to apply then translates to V
N »vQ ,
(7.19)
which says that the average distance between particles must be much greater than the average de Broglie wavelength. For the air we breathe, the average distance between molecules is about 3 nm while the average de Broglie wavelength is less than 0.02 nm, so this condition is definitely satisfied. Notice, by the way, that this condition depends not only on the density of the system, but also on the temperature and the mass of the particles, both through vQ. It's hard to visualize what actually happens in a gas when condition 7.17 breaks down and multiple particles start trying to get into the same state. Figure 7.4, though imperfect, is about the best I can do. Picture each particle as being smeared out in a quantum wavefunction filling a volume equal to vQ. (This is equivalent to putting the particles into wavefunctions that are as localized in space as possible. To squeeze them into narrower wavefunctions we would have to introduce uncer­
•
•
•
•
• •
•
•
• •
•
•
•
•
• •
•
• •
Normal gas, VIN» vQ
• Quantum gas, VIN ~ vQ
Figure 7.4. In a normal gas, the space between particles is much greater than the typical size of a particle's wavefunction. When the wavefunctions begin to 'touch' and overlap, wecall it a quantum gas.
7.2
Bosons and Fermions
tainties in momentum that are large compared to the average momentum h/RQ' thus increasing the energy and temperature of the system.) In a normal gas, the effective volume thus occupied by all the particles will be much less than the volume of the container. (Often the quantum volume is less than the physical volume of a molecule.) But if the gas is sufficiently dense or vQ is sufficiently large, then the wavefunctions will start trying to touch and overlap. At this point it starts to matter whether the particles are fermions or bosons; either way, the behavior will be much different from that of a normal gas. There are plenty of systems that violate condition 7.17, either because they are very dense (like a neutron star), or very cold (like liquid helium), or composed of very light particles (like the electrons in a metal or the photons in a hot oven). The rest of this chapter is devoted to the study of these fascinating systems. Problem 7.8. Suppose you have a 'box' in which each particle may occupy any of 10 single-particle states. For simplicity, assume that each of these states has energy zero. (a) What is the partition function of this system if the box contains only one particle? (b) What is the partition function of this system if the box contains two dis­ tinguishable particles? (c) What is the partition function if the box contains two identical bosons?
(d) What is the partition function if the box contains two identical fermions? (e) What would be the partition function of this system according to equa­ tion 7.16?
(f) What is the probability of finding both particles in the same single-particle state, for the three cases of distinguishable particles, identical bosons, and identical fermions? Problem 7.9. Compute the quantum volume for an N2 molecule at room tem­ perature, and argue that a gas of such molecules at atmospheric pressure can be treated using Boltzmann statistics. At about what temperature would quantum statistics become relevant for this system (keeping the density constant and pre­ tending that the gas does not liquefy)? Problem 7.10. Consider a system of five particles, inside a container where the allowed energy levels are nondegenerate and evenly spaced. For instance, the particles could be trapped in a one-dimensional harmonic oscillator potential. In this problem you will consider the allowed states for this system, depending on whether the particles are identical fermions, identical bosons, or distinguishable particles. (a) Describe the ground state of this system, for each of these three cases. (b) Suppose that the system has one unit of energy (above the ground state). Describe the allowed states of the system, for each of the three cases. How many possible system states are there in each case? (c) Repeat part (b) for two units of energy and for three units of energy.
(d) Suppose that the temperature of this system is low, so that the total energy is low (though not necessarily zero). In what way will the behavior of the bosonic system differ from that of the system of distinguishable particles? Discuss.
265
266
Chapter 7
Quantum Statistics
The Distribution Functions When a system violates the condition Z 1 » N, so that we cannot treat it using the methods of Chapter 6, we can use Gibbs factors instead. The idea is to first consider a 'system' consisting of one single-particle state, rather than a particle itself. Thus the system will consist of a particular spatial wavefunction (and, for particles with spin, a particular spin orientation). This idea seems strange at first, because we normally work with wavefunctions of definite energy, and each of these wavefunctions shares its space with all the other wavefunctions. The 'system' and the 'reservoir' therefore occupy the same physical space, as in Figure 7.5. Fortunately, the mathematics that went into the derivation of the Gibbs factor couldn't care less whether the system is spatially distinct from the reservoir, so all those formulas still apply to a single-particle-state system. So let's concentrate on just one single-particle state of a system (say, a particle in a box), whose energy when occupied by a single particle is E. When the state is unoccupied, its energy is 0; if it can be occupied by n particles, then the energy will be nE. The probability of the state being occupied by n particles is
P(n) =
2. e -(m-J.£n)/kT = 2. e - n (t:-JL)/kT z
z
'
(7.20)
where Z is the grand partition function, that is, the sum of the Gibbs factors for all possible n. If the particles in question are fermions, then n can only be 0 or 1, so the grand partition function is (fermions) .
(7.21)
From this we can compute the probability of the state being occupied or unoccupied, as a function of E, p' and T. We can also compute the average number of particles in the state, also called the occupancy of the state: e-(t:-JL)/kT
n=
L nP(n) = o· P(O) + 1 . P(1) = 1 + n
e-(t:-JL)/kT
(7.22)
1 e(t:-JL)/kT
+1
(fermions) .
System Figure 7.5. To treat a quantum gas using Gibbs factors, we consider a 'system' consisting of one single-particle state (or wavefunction). The 'reservoir' consists of all the other possible single-particle states.
7.2
Bosons and Fermions
This important formula is called the Fermi-Dirac distribution; I'll call it 'fiFO: 1
(7.23)
The Fermi-Dirac distribution goes to zero when € » ft, and goes to 1 when € «: ft. Thus, states with energy much less than ft tend to be occupied, while states with energy much greater than ft tend to be unoccupied. A state with energy exactly equal to ft has a 50% chance of being occupied, while the width of the fall-off from 1 to 0 is a few times kT. A graph of the Fermi-Dirac distribution vs. € for three different temperatures is shown in Figure 7.6. If instead the particles in question are bosons, then n can be any nonnegative integer, so the grand partition function is
1 + e-(€-/L)/kT + (e-(€-/L)/kT)2 1
+ ..
(7.24)
(bosons).
(Since the Gibbs factors cannot keep growing without limit, ft must be less than € and therefore the series must converge.) Meanwhile, the average number of particles in the state is fi
= L nP(n)
o· P(O) + 1· P(l) + 2· P(2) + ..
(7.25)
n
To evaluate this sum let's abbreviate x (€ - ft)/kT. Then (7.26)
1 >,
u
!=: cd
0. ~
u u 0
II Q
r..
I~
0
€
JL
Figure 7.6. The Fermi-Dirac distribution goes to 1 for very low-energy states and to zero for very high-energy states. It equals 1/2 for a state with energy JL, falling off suddenly for low T and gradually for high T. (Although JL is fixed on this graph, in the next section we'll see that JL normally varies with temperature.)
267
268
Chapter 7
Quantum Statistics
You can easily check that this formula works for fermions. For bosons, we have
(7.27)
1 e(~-/-L)/kT -
1
(bosons).
This important formula is called the Bose-Einstein distribution; I'll call it nBE:
nBE --
1
(7.28)
~~~----
e(~-/-L)/kT -
1·
Like the Fermi-Dirac distribution, the Bose-Einstein distribution goes to zero when f »j.L. Unlike the Fermi-Dirac distribution, however, it goes to infinity as f approaches j.L from above (see Figure 7.7). It would be negative if f could be less than j.L, but we've already seen that this cannot happen. To better understand the Fermi-Dirac and Bose-Einstein distributions, it's use­ ful to ask what n would be for particles obeying Boltzmann statistics. In this case, the probability of any single particle being in a certain state of energy f is (Boltzmann) ,
(7.29)
so if there are N independent particles in total, the average number in this state is _
_
nBoltzmann -
_ N
NP(s) - Zl e
-~/kT
(7.30)
.
But according to the result of Problem 6.44, the chemical potential for such a system is j.L = -kTln(ZdN). Therefore the average occupancy can be written _ nBoltzmann -
e
/-L/kT
e
-~/kT
_
- e
-(~-/-L)/kT
.
(7.31)
When f is sufficiently greater than j.L, so that this exponential is very small, we can neglect the 1 in the denominator of either the Fermi-Dirac distribution (7.23) or the Bose-Einstein distribution (7.28), and both reduce to the Boltzmann distribu­ tion (7.31). The equality of the three distribution functions in this limit is shown in Figure 7.7. The precise condition for the three distributions to agree is that the exponent (f - j.L) / kT be much greater than 1. If we take the lowest-energy state to have f ~ 0, then this condition will be met for all states whenever j.L « -kT, that is, when Zl » N. This is the same condition that we arrived at through different reasoning at the beginning of this section. We now know how to compute the average number of particles occupying a single-particle state, whether the particles are fermions or bosons, in terms of the en­ ergy of the state, the temperature, and the chemical potential. To apply these ideas to any particular system, we still need to know what the energies of all the states are. This is a problem in quantum mechanics, and can be extremely difficult in many cases. In this book we'll deal mostly with particles in a 'box,' where the quantum­ mechanical wavefunctions -are simple sine waves and the corresponding energies can
7.2
Bosons and Fermions
~----+-----+-----+-----+=~~~~~~E
Figure 7.7. Comparison of the Fermi-Dirac, Bose-Einstein, and Boltzmann distri­ butions, all for the same value of J1. When (E - J1) / kT » 1, the three distributions become equal.
be determined straightforwardly. The particles could be electrons in a metal, neu­ trons in a neutron star, atoms in a fluid at very low temperature, photons inside a hot oven, or even 'phonons,' the quantized units of vibrational energy in a solid. For any of these applications, before we can apply the Fermi-Dirac or Bose­ Einstein distribution, we 'll also have to figure out what the chemical potential is. In a few cases this is quite easy, but in other applications it will require considerable work. As we'll see, fL is usually determined indirectly by the total number of particles in the system. Problem 7.11. For a system of fermions at room temperature, compute the probability of a single-particle state being occupied if its energy is (a) 1 eV less than J1 (b) 0.01 eV less than J1 (c ) equal to J1 (d) 0.01 eV greater than J1 (e ) 1 e V greater than J1 Problem 7.12. Consider two single-particle states, A and B , in a system of fermions, where EA = J1 - x and EB = J1 + x ; that is, level A lies below J1 by the same amount that level B lies above J1. Prove that the probability of level B being occupied is the same as the probability of level A being unoccupied. In other words, the Fermi-Dirac distribution is 'symmetrical' about the point where E = J1. Problem 7.13. For a system of bosons at room temperature, compute the average occupancy of a single-particle state and the probability of the state containing 0, 1, 2, or 3 bosons, if the energy of the state is (a) 0.001 eV greater than J1 (b) 0.01 e V greater than J1 (c) 0.1 eV greater than J1 (d ) 1 e V greater than J1
269
270
Chapter 7
Quantum Statistics
Problem 7.14. For a system of particles at room temperature, how large must E fL be before the Fermi-Dirac, Bose-Einstein, and Boltzmann distributions agree within 1%7 Is this condition ever violated for the gases in our atmosphere? Ex­ plain. Problem 7.15. For a system obeying Boltzmann statistics, we know what fL is from Chapter 6. Suppose, though, that you knew the distribution function (equation 7.31) but didn't know 1-£. You could still determine fL by requiring that the total number of particles, summed over all single-particle states, equal N. Carry out this calculation, to rederive the formula fL = -kTln(ZdN). (This is normally how 1-£ is determined in quantum statistics, although the math is usually more difficult.) Problem 7.16. Consider an isolated system of N identical fermions, inside a con­ tainer where the allowed energy levels are nondegenerate and evenly spaced. * For instance, the fermions could be trapped in a one-dimensional harmonic oscillator potentiaL For simplicity, neglect the fact that fermions can have multiple spin orientations (or assume that they are all forced to have the same spin orientation). Then each energy level is either occupied or unoccupied, and any allowed system state can be represented by a column of dots, with a filled dot representing an oc­ cupied level and a hollow dot representing an unoccupied level. The lowest-energy system state has all levels below a certain point occupied, and all levels above that point unoccupied. Let 'fJ be the spacing between energy levels, and let q be the number of energy units (each of size 'fJ) in excess of the ground-state energy. Assume that q < N. Figure 7.8 shows all system states up to q 3.
(a) Draw dot diagrams, as in the figure, for all allowed system states with q
4, q
5, and q = 6.
(b) According to the fundamental assumption, all allowed system states with a given value of q are equally probable. Compute the probability of each energy level being occupied, for q = 6. Draw a graph of this probability as a function of the energy of the leveL (c) In the thermodynamic limit where q is large, the probability of a level being occupied should be given by the Fermi-Dirac distribution. Even though 6 is not a large number, estimate the values of fL and T that you would have to plug into the Fermi-Dirac distribution to best fit the graph you drew in part (b). q=O
t
» bO
I-<
Q.l
~
I=il
0 0 0 0
••• •
q
1
0 0 0
• •• • 0
q
2
0 0
0 0 0
• •• • 0 0
•• •• 0
q=3 0
• ••• 0 0 0
0 0
• • •• 0
0
0 0 0
•• • • 0
Figure 7.8. A representation of the system states of a fermionic sys­ tern with evenly spaced, nondegen­ erate energy levels. A filled dot rep­ resents an occupied single-particle state, while a hollow dot represents an unoccupied single-particle state.
*This problem and Problem 7.27 are based on an article by J. Arnaud et aL, American Journal of Physics 67, 215 (1999).
7.3
Degenerate Fermi Gases
(d) Calculate the entropy of this system for each value of q from 0 to 6, and draw a graph of entropy vs. energy. Make a rough estimate of the slope of this graph near q = 6, to obtain another estimate of the temperature of this system at that point. Check that it is in rough agreement with your answer to part (c). Problem 7.17. In analogy with the previous problem, consider a system of iden­ tical spin-O bosons trapped in a region where the energy levels are evenly spaced. Assume that N is a large number, and again let q be the number of energy units. (a) Draw diagrams representing all allowed system states from q = 0 up to q = 6. Instead of using dots as in the previous problem, use numbers to indicate the number of bosons occupying each level.
(b) Compute the occupancy of each energy level, for q = 6. Draw a graph of the occupancy as a function of the energy of the level. (c) Estimate the values of J1, and T that you would have to plug into the Bose­ Einstein distribution to best fit the graph of part (b). (d) As in part (d) of the previous problem, draw a graph of entropy vs. energy and estimate the temperature at q = 6 from this graph. Problem 7.18. Imagine that there exists a third type of particle, which can share a single-particle state with one other particle of the same type but no more. Thus the number of these particles in any state can be 0, 1, or 2. Derive the distribution function for the average occupancy of a state by particles of this type, and plot the occupancy as a function of the state's energy, for several different temperatures.
7.3 Degenerate Fermi Gases As a first application of quantum statistics and the Fermi-Dirac distribution, I'd like to consider a 'gas' of fermions at very low temperature. The fermions could be helium-3 atoms, or protons and neutrons in an atomic nucleus, or electrons in a white dwarf star, or neutrons in a neutron star. The most familiar example, though, is the conduction electrons inside a chunk of metal. In this section I'll say 'electrons' to be specific, even though the results apply to other types of fermions as well. By 'very low temperature,' I do not necessarily mean low compared to room temperature. What I mean is that the condition for Boltzmann statistics to apply to an ideal gas, V/N » vQ, is badly violated, so that in fact V/N «vQ. For an electron at room temperature, the quantum volume is
VQ
)3 = (4.3 nm)3. h =( v21rmkT
(7.32)
But in a typical metal there is about one conduction electron per atom, so the volume per conduction electron is roughly the volume of an atom, (0.2 nm)3. Thus, the temperature is much too low for Boltzmann statistics to apply. Instead, we are in the opposite limit, where for many purposes we can pretend that T = O. Let us therefore first consider the properties of an electron gas at T = 0, and later ask what happens at small nonzero temperatures.
271
272
Chapter 7
Quantum Statistics
Zero Temperature
At T 0 the Fermi-Dirac distribution becomes a step function (see Figure 7.9). All single-particle states with energy less than J.L are occupied, while all states with energy greater than J.L are unoccupied. In this context J.L is also called the Fermi energy, denoted fF:
(7.33)
When a gas of fermions is so cold that nearly all states below fF are occupied while nearly all states above fF are unoccupied, it is said to be degenerate. (This use of the word is completely unrelated to its other use to describe a set of quantum states that have the same energy.) The value of fF is determined by the total number of electrons present. Imagine an empty box, to which you add electrons one at a time, with no excess energy. Each electron goes into the lowest available state, until the last electron goes into a state with energy just below fF' To add one more electron you would have to give it an energy essentially equal to fF J.L; in this context, the equation J.L (aU/aN)s,v makes perfect physical sense, since dU J.L when dN = 1 (and S is fixed at zero when all the electrons are packed into the lowest-energy states). In order to calculate fF' as well as other interesting quantities such as the total energy and the pressure of the electron gas, I'll make the approximation that the electrons are free particles, subject to no forces whatsoever except that they are confined inside a box of volume V = L3. For the conduction electrons in a metal, this approximation is not especially accurate. Although it is reasonable to neglect long-range electrostatic forces in any electrically neutral material, each conduction electron still feels attractive forces from nearby ions in the crystal lattice, and I'm neglecting these forces. * The definite-energy wavefunctions of a free electron inside a box are just sine waves, exactly as for the gas molecules treated in Section 6.7. For a one-dimensional
1~------------------~
O~I--------------------~------------------~~ Figure 7.9. At T = 0, the Fermi-Dirac distribution equals 1 for all states with < p, and equals 0 for all states with t > p,.
t
*Problems 7.33 and 7.34 treat some of the effects of the crystal lattice on the conduction electrons. For much more detail, see a solid state physics textbook such as Kittel (1996) or Ashcroft and Mermin (1976).
7.3
Degenerate Fermi Gases
box the allowed wavelengths and momenta are (as before) h
Pn
hn
= An = 2L'
(7.34)
where n is any positive integer. In a three-dimensional box these equations apply separately to the x, y, and z directions, so hny Py = 2L'
hnx Px = 2L '
(7.35)
where (nx, ny, n z ) is a triplet of positive integers. The allowed energies are therefore
(7.36) To visualize the set of allowed states, I like to draw a picture of 'n-space,' the three-dimensional space whose axes are n x , ny, and n z (see Figure 7.10). Each allowed n vector corresponds to a point in this space with positive integer coor­ dinates; the set of all allowed states forms a huge lattice filling the first octant of n-space. Each lattice point actually represents two states, since for each spatial wavefunction there are two independent spin orientations. In n-space, the energy of any state is proportional to the square of the distance from the origin, n; + n~ + n;. So as we add electrons to the box, they settle into states starting at the origin and gradually working outward. By the time we're done, the total number of occupied states is so huge that the occupied region of n-space is essentially an eighth of a sphere. (The roughness of the edges is insignificant, compared to the enormous size of the entire sphere.) I'll call the radius of this sphere n max . It's now quite easy to relate the total number of electrons, N, to the chemical potential or Fermi energy, J-L = EF. On one hand, EF is the energy of a state that sits just on the surface of the sphere in n-space, so
(7.37)
Figure 7.10. Each triplet of integers (nx, ny, n z ) represents a pair of definite-energy elec­ tron states (one with each spin orientation). The set of all inde­ pendent states fills the positive octant of n-space.
_ nx
273
274
Chapter 7
Quantum Statistics
On the other hand, the total volume of the eighth-sphere in n-space equals the number of lattice points enclosed, since the separation between lattice points is 1 in all three directions. Therefore the total number of occupied states is twice this volume (because of the two spin orientations):
. 1 4 N = 2 x (volume of eIghth-sphere) = 2· 8 . 37rn~ax =
3
(7.38)
Combining these two equations gives the Fermi energy as a function of N and the volume V L3 of the box: C = h (7.39) F 8m 7rV
2(3N)2/3
Notice that this quantity is intensive, since it depends only on the number density of electrons, N IV. For a larger container with correspondingly more electrons, €F comes out the same. Although I have derived this result only for electrons in a cube-shaped box, it actually applies to macroscopic containers (or chunks of metal) of any shape. The Fermi energy is the highest energy of all the electrons. On average, they'll have somewhat less energy, a little more than half CF' To be more precise, we have to do an integral, to find the total energy of all the electrons; the average is just the total divided by N. To calculate the total energy of all the electrons, I'll add up the energies of the electrons in all occupied states. This entails a triple sum over n x , ny, and n z :
!!!
2
c(fi) dnx dny dn z .
(7.40)
The factor of 2 is for the two spin orientations for each fi. I'm allowed to change the sum into an integral because the number of terms is so huge, it might as well be a continuous function. To evaluate the triple integral I'll use spherical coordinates, as illustrated in Figure 7.11. Note that the volume element dnx dny dnz becomes
d~dB
~
~
,,/:
nsinOd¢
/'n /
/ /
----~~------_.--------~ny
Figure 7.11. In spherical coordinates (n, B, ¢), the infinitesimal volume element is (dn)(n dB)(n sin B71¢).
7.3
Degenerate Fermi Gases
n 2 sin {} dn d{} d¢. The total energy of all the electrons is therefore 2J
The angular integrals give leaves us with
U
{7r /2
{nmax
U
dn Jo
o 7r /2,
{nmax 2 = 7r J E(n) n dn
2
(7.41 )
d¢n sin{}c(n).
one-eighth the surface area of a unit sphere. This
(nmax
7r h 2
8mL2
o
{7r /2 d{} Jo
J
o
4
_
n dn - ---::.~
3
= 5NEF '
(7.42)
The average energy of the electrons is therefore 3/5 the Fermi energy. If you plug in some numbers, you'll find that the Fermi energy for conduction electrons in a typical metal is a few electron-volts. This is huge compared to the average thermal energy of a particle at room temperature, roughly kT ~ 1/40 eV. In fact, comparing the Fermi energy to the average thermal energy is essentially the same as comparing the quantum volume to the average volume per particle, as I did at the beginning of this section:
V
is the same as
N «vQ
(7.43)
When this condition is met, the approximation T ~ 0 is fairly accurate for many purposes, and the gas is said to be degenerate. The temperature that a Fermi gas would have to have in order for kT to equal EF is called the Fermi temperature: EF/k. This temperature is purely hypothetical for electrons in a metal, since before it is reached. metals liquefy and evaporate -(BU/BV)S,N, which you can derive from the thermo­ Using the formula P dynamic identity or straight from classical mechanics, we can calculate the pressure of a degenerate electron gas: P =
-~ [~N~ (3N)2/3 V _2/3] BV 5
8m
7r
2NEF 5V
(7.44)
This quantity is called the degeneracy pressure. It is positive because when you compress a degenerate electron gas, the wavelengths of all the wavefunctions are reduced, hence the energies of all the wavefunctions increase. Degeneracy pressure is what keeps matter from collapsing under the huge electrostatic forces that try to pull electrons and protons together. Please note that degeneracy pressure has absolutely nothing to do with electrostatic repulsion between the electrons (which we've completely ignored); it arises purely by virtue of the exclusion principle. Numerically, the degeneracy pressure comes out to a few billion N/m 2 for a typical metal. But this number is not directly measurable-it is canceled by the electrostatic forces that hold the electrons inside the metal in the first place. A more measurable quantity is the bulk modulus, that is, the change in pressure when the material is compressed, divided by the fractional change in volume: B
_V(BP) BV T
IOU 9 V
(7.45)
275
276
Chapter 7
Quantum Statistics
This quantity is also quite in 8I units, but it is not completely canceled by the electrostatic the formula actually agrees with experiment, within a factor of 3 or so, for most metals. Problem 7.19. Each atom in a chunk of copper contributes one conduction electron. Look up the density and atomic mass of copper, and calculate the Fermi energy, the Fermi temperature, the degeneracy pressure, and the contribution of the degeneracy pressure to the bulk modulus. Is room temperature sufficiently low to treat this system as a degenerate electron gas? Problem 7.20. At the center of the sun, the temperature is approximately 10 7 K and the concentration of electrons is approximately 1032 per cubic meter. Would it be (approximately) valid to treat these electrons as a 'classical' ideal gas (using Boltzmann statistics), or as a degenerate Fermi gas (with T :::::: 0), or neither? Problem 7.21. An atomic nucleus can be crudely modeled as a gas of nucleons with a number density of 0.18 (where 1 fm m). Because nucleons come in two different types (protons and neutrons), each with spin 1/2, each spatial wavefunction can hold four nucleons. Calculate the Fermi energy of this system, in MeV. Also calculate the Fermi temperature, and comment on the result. Problem 7.22. Consider a degenerate electron gas in which essentially all of the electrons are highly relativistic (E » mc2 ), so that their energies are E pc (where p is the magnitude of the momentum vector).
(a) Modify the derivation given above to show that for a relativistic electron gas at zero temperature, the chemical potential (or Fermi energy) is given by J1 = hc(3N/87fV)1/3. (b) Find a formula for the total energy of this system in terms of Nand J1.
Problem 7.23. A white dwarf star (see Figure 7.12) is essentially a degenerate electron gas, with a bunch of nuclei mixed in to balance the charge and to provide the gravitational attraction that holds the star together. In this problem you will derive a relation between the mass and the radius of a white dwarf star, modeling the star as a uniform-density sphere. White dwarf stars tend to be extremely hot by our standards; nevertheless, it is an excellent approximation in this problem to set T O. (a) Use dimensional analysis to argue that the gravitational potential energy of a uniform-density sphere (mass M, radius R) must equal Ugrav
=
GA1 2
-(constant)~,
where (constant) is some numerical constant. Be sure to explain the minus The constant turns out to equal 3/5; you can derive it by calculating the (negative) work needed to assemble the shell by shell, from the inside out. (b) Assuming that the star contains one proton and one neutron for each elec­ tron, and that the electrons are nonrelativistic, show that the total (kinetic) energy of the electrons equals
=
(0.0086)
7.3
Figure 7.12. The double star system Sir­
ius A and B. Sirius A (greatly overexposed in the photo) is the brightest star in our night sky. Its companion, Sirius B, is hot­
ter but very faint, indicating that it must be extremely small-a white dwarf. From the orbital motion of the pair we know that Sirius B has about the same mass as our sun. (UCO/Lick Observatory photo.)
(c)
(d) (e)
(f)
(g)
Degenerate Fermi Gases
The numerical factor can be expressed exactly in terms of 7r and cube roots and such, but it's not worth it. The equilibrium radius of the white dwarf is that which minimizes the total energy U grav + Ukinetic. Sketch the total energy as a function of R, and find a formula for the equilibrium radius in terms of the mass. As the mass increases, does the radius increase or decrease? Does this make sense? Evaluate the equilibrium radius for M = 2 X 1030 kg, the mass of the sun. Also evaluate the density. How does the density compare to that of water? Calculate the Fermi energy and the Fermi temperature, for the case con­ sidered in part (d). Discuss whether the approximation T = 0 is valid. Suppose instead that the electrons in the white dwarf star are highly rel­ ativistic. Using the result of the previous problem, show that the total kinetic energy of the electrons is now proportional to 1/ R instead of 1/ R2. Argue that there is no stable equilibrium radius for such a star. The transition from the nonrelativistic regime to the ultrarelativistic regime occurs approximately where the average kinetic energy of an electron is equal to its rest energy, mc 2 . Is the nonrelativistic approximation valid for a one-solar-mass white dwarf? Above what mass would you expect a white dwarf to become relativistic and hence unstable?
Problem 7.24. A star that is too heavy to stabilize as a white dwarf can collapse further to form a neutron star: a star made entirely of neutrons, supported against gravitational collapse by degenerate neutron pressure. Repeat the steps of the previous problem for a neutron star, to determine the following: the mass­ radius relation; the radius, density, Fermi energy, and Fermi temperature of a one-solar-mass neutron star; and the critical mass above which a neutron star becomes relativistic and hence unstable to further collapse.
Small Nonzero Temperatures One property of a Fermi gas that we cannot calculate using the approximation T = 0 is the heat capacity, since this is a measure of how the energy of the system depends on T. Let us therefore consider what happens when the temperature is very small but nonzero. Before doing any careful calculations, I'll explain what happens qualitatively and try to give some plausibility arguments. At temperature T, all particles typically acquire a thermal energy of roughly kT. However, in a degenerate electron gas, most of the electrons cannot acquire such a small amount of energy, because all the states that they might jump into are already occupied (recall the shape of the Fermi-Dirac distribution, Figure 7.6).
277
278
Chapter 7
Quantum Statistics
The only electrons that can acquire some thermal energy are those that are already within about kT of the Fermi energy~these can jump up into unoccupied states above EF' (The spaces they leave behind allow some, but not many, of the lower­ lying electrons to also gain energy.) Notice that the number of electrons that can be affected by the increase in T is proportional to T. This number must also be proportional to N, because it is an extensive quantity. Thus, the additional energy that a degenerate electron gas acquires when its temperature is raised from zero to T is doubly proportional to T: additional energy ex (number of affected electrons) x (energy acquired by each) ex (NkT) x (kT) ex N(kT)2.
(7.46)
We can guess the constant of proportionality using dimensional analysis. The quan­ tity N(kT)2 has units of (energy)2, so to get something with units of (energy)l, we need to divide by some constant with units of energy. The only such constant available is EFl so the additional energy must be N(kT)2/ EF , times some constant of order 1. In a few pages we'll see that this constant is 1r2 /4, so the total energy of a degenerate Fermi gas for T « EF / k is
u
3
1r2
(kT)2
5
4
EF
-NEF+- N - - .
(7.47)
From this result we can easily calculate the heat capacity: Cv
(7.48)
Notice that the heat capacity goes to zero at T = 0, as required by the_third __ law of therI!1Qdynarnics.:. The approach to zero is linear in and this prediction agrees welr with experiments on metals at low temperatures. (Above a few kelvins, lattice vibrations also contribute significantly to the heat capacity of a metal.) The numerical coefficient of 1r 2 /2 usually agrees with experiment to within 50% or better, but there are exceptions. Problem 7.25. Use the results of this section to estimate the contribution of conduction electrons to the heat capacity of one mole of copper at room temper­ ature. How does this contribution compare to that of lattice vibrations, assuming that these are not frozen out? (The electronic contribution has been measured at low temperatures, and turns out to be about 40% more than predicted by the free electron model used Problem 7.26. In this problem you will model helium-3 as a noninteracting Fermi gas. Although liquefies at low temperatures, the liquid has an unusually low density and behaves in many ways like a gas because the forces between the atoms are so weak. Helium-3 atoms are spin-l/2 because of the unpaired neutron in the nucleus. (a) Pretending that liquid 3 He is a noninteracting Fermi gas, calculate the Fermi energy and the Fermi temperature. The molar volume (at low pres­ sures) is 37 cm 3 .
7.3
DegeIler.ate Fermi Gases
(b) Calculate the heat capacity for T «TF , and compare to the eXlpeI'lmlental result Cv (2.8 K-1)NkT (in the low-temperature limit). (Don't expect agreement. )
(c) The entropy of solid 3He below 1 K is almost entirely due to its multiplicity of nuclear spin alignments. Sketch a graph S vs. T for liquid and solid 3He at low temperature, and estimate the temperature at which the liquid and solid have the same entropy. Discuss the shape of the solid-liquid boundary shown in Figure 5.13.
Problem 7.27. The argument given above for why Cv ex: T does not on the details of the energy levels available to the fermions, so it should also apply to the model considered in Problem 7.16: a gas of fermions trapped in such a way that the energy levels are evenly spaced and nondegenerate. (a) Show that, in this model, the number of possible system states for a value of q is equal to the number of distinct ways of writing q as a sum of (For example, there are three system states for q 3, cOlTe~;pond.ing to the sums 3, 2 + 1, and 1 + 1 + 1. Note that 2 + 1 and 1 + 2 are not counted separately.) This combinatorial function is called the number of unrestricted partitions of q, denoted p(q). For eX'ij,mlpl~3,
p(3)
3.
(b) By emlmeratin.g the partitions explicitly, compute p(7) and p(8). (c) Make a table of p(q) for values of q up to 100, by either looking up the values in a mathematical reference book, or using a software that can compute or writing your own program to compute them. From this compute the entropy, temperature, and heat capacity of this system, the same methods as in Section 3.3. Plot the heat capacity as a function of temperature, and note that it is approximately linear. (d) Ramanujan and Hardy (two famous mathematicians) have shown that when q is the number of unrestricted partitions of q is ap­ proximately by
p(q)
~
e
nV2 q/3 ;7)
.
4v3q
Check the accuracy of this formula for q = 10 and for q = 100. Working in this approximation, calculate the entropy, temperature, and heat CalJac:Ity of this system. the heat capacity as a series in decreasing powers of kT fry, that this ratio is large and keeping the two largest terms. Compare to the numerical results you obtained in part (c). Why is the heat capacity of this system independent of N, unlike that of the three­ dimensional box of fermions discussed in the text?
The Density of States To better visualize-and quantify-the behavior of a Fermi gas at small nonzero temperatures, I need to introduce a new concept. Let's go back to the energy integral (7.42), and change variables from n to the electron energy E:
E
=
---'1'1-
n
VE,
dn = J8m£2 h2
1 dE.
(7.49)
279
280
Chapter 7
Quantum Statistics
With this substitution, you can show that the energy integral for a Fermi gas at zero temperature becomes
u= lFE[H8~~2t2Jf]dE
(T
=
0).
(7.50)
The quantity in square brackets has a nice interpretation: It is the number of single­ particle states per unit energy. To compute the total energy of the system we carry out a sum over all energies of the energy in question times the number of states with that energy. The number of single-particle states per unit energy is called the density of states. The symbol for it is g( €), and it can be written in various ways:
g(€)=
7r(8m)3/ 2 3N 2h 3 VyE=~VE.
(7.51)
2€F
The second expression is compact and handy, but perhaps rather confusing since it seems to imply that g( €) depends on N , when in fact the N dependence is canceled by €F. I like the first expression better , since it shows explicitly that g(€) is proportional to V and independent of N. But either way, the most important point is that g( €), for a three-dimensional box of free particles, is proportional to J€. A graph of the function is a parabola opening to the right, as shown in Figure 7.13. If you want to know how many states there are between two energies €l and €2, you just integrate this function over the desired range. The density of states is a function whose purpose in life is to be integrated. The density-of-states idea can be applied to lots of other systems besides this one. Equation 7.51 and Figure 7.13 are for the specific case of a gas of 'free' electrons, confined inside a fixed volume but not subject to any other forces. In more realistic models of metals we would want to take into account the attraction of the electrons toward the positive ions of the crystal lattice. Then the wavefunctions and their energies would be quite different, and therefore g( €) would be a much more complicated function. The nice thing is that determining g is purely a problem of quantum mechanics, having nothing to do with thermal effects or temperature. And
. E
Figure 7.13. Density of states for a system of noninteracting, nonrelativistic particles in a three-dimensional box. The number of states within any energy interval is the area under the graph. For a Fermi gas at T = 0, all states with E < EF are occupied 'Yhile all states with E > EF are unoccupied.
7.3
Degenerate Fermi Gases
once you know 9 for some system, you can then forget about quantum mechanics and concentrate on the thermal physics. For an electron gas at zero temperature, we can get the total number of electrons by just integrating the density of states up to the Fermi energy:
(T = 0).
(7.52)
(For a free electron gas this is the same as equation 7.50 for the energy, but without the extra factor of Eo) But what if T is nonzero? Then we need to multiply g(E) by the probability of a state with that energy being occupied, that is, by the Fermi­ Dirac distribution function. Also we need to integrate all the way up to infinity, since any state could conceivably be occupied: N =
00 g(E) nFD(E) dE = 100 g(E) ( e 1 o
0
E-p,
1 )/kT
+ 1 dE
(any T).
And to get the total energy of all the electrons, just slip in an
roo
roo
1
U = io Eg(E)nFD(E) dE = io Eg(E) e(E-p,)/kT
+ 1 dE
(7.53)
E:
(any T).
(7.54)
Figure 7.14 shows a graph of the integrand of the N-integral (7.53), for a free electron gas at nonzero T. Instead of falling immediately to zero at E = EF, the number of electrons per unit energy now drops more gradually, over a width of a few times kT. The chemical potential, f.L, is the point where the probability of a state being occupied is exactly 1/2, and it's important to note that this point is no longer the same as it was at zero temperature: f.L(T)
'I EF
except when T
=
o.
(7.55)
~----~--------------~--~ ~~--~~~------~t J-t tF
Figure 7.14. At nonzero T, the number of fermions per unit energy is given by the density of states times the Fermi-Dirac distribution. Because increasing the temperature does not change the total number of fermions, the two lightly shaded areas must be equal. Since g(t) is greater above tF than below, this means that the chemical potential decreases as T increases. This graph is drawn for T /TF = 0.1; at this temperature J-t is abol:1t 1% less than tF.
281
282
Chapter 7
Quantum Statistics
Why not? Recall from Problem 7.12 that the Fermi-Dirac distribution function is symmetrical about E IL: The probability of a state above IL being occupied is the same as the probability of a state the same amount below IL being unoccupied. Now suppose that IL were to remain constant as T increases from zero. Then since the density of states is greater to the right of IL than to the left, the number of electrons we would be adding at E > IL would be greater than the number we are losing from E < IL. In other words, we could increase the total number of electrons just by raising the temperature! To prevent such nonsense, the chemical potential has to decrease slightly, thus lowering all of the probabilities by a small amount. The precise formula for IL(T) is determined implicitly by the integral for N, equation 7.53. If we could carry out this integral, we could take the resulting formula and solve it for IL(T) (since N is a fixed constant). Then we could plug our value of IL(T) into the energy integral (7.54), and try to carry out that integral to find U(T) (and hence the heat capacity). The bad news is that these integrals cannot be evaluated exactly, even for the simple case of a free electron gas. The good news is that they can be evaluated approximately, in the limit kT « EF' ill this limit the answer for the energy integral is what I wrote in equation 7.47. Problem 7.28. Consider a free Fermi gas in two dimensions, confined to a square area A = L2. (a) Find the Fermi energy (in terms of N and A), and show that the average energy of the particles is EF /2. (b) Derive a formula for the density of states. You should find that it is a constant, independent of E. (c) Explain how the chemical potential of this system should behave as a func­ tion of temperature, both when kT « EF and when T is much higher. (d) Because g(E) is a constant for this system, it is possible to carry out the integral 7.53 for the number of particles analytically. Do so, and solve for /J as a function of N. Show that the resulting formula has the expected qualitative behavior. (e) Show that in the high-temperature limit, kT »EF, the chemical potential of this system is the same as that of an ordinary ideal gas.
The Sommerfeld Expansion After talking about the integrals 7.53 and 7.54 for so long, it's about time I explained how to evaluate them, to find the chemical potential and total energy of a free electron gas. The method for doing this in the limit kT « EF is due to Arnold Sommerfeld, and is therefore called the Sommerfeld expansion. None of the steps are particularly difficult, but taken as a whole the calculation is rather tricky and intricate. Hang on. I'll start with the integral for N:
1
00
N =
g(E) 'po (E) dE = go
1
00
El/2 'po (E) dE.
(7.56)
(In the second expression I've introduced the abbreviation go for the constant that multiplies JE in equation 7.51 for the density of states.) Although this integral
7.3
Degenerate Fermi Gases
runs over all positive E, the most interesting region is near E = J-L, where nFO(E) falls off steeply (for T « EF)' SO the first trick is to isolate this region, by integrating by parts: 2 3/22 3/2 ( dnF o ) (7.57) N = '3g0E nFO(E) 0 + '3 go Jo E -~ dE.
1=
{=
The boundary term vanishes at both limits, leaving us with an integral that is much nicer, because d nFO / dE is negligible everywhere except in a narrow region around E = J-L (see Figure 7.15). Explicitly, we can compute (7.58) where x
= (E - J-L)/kT. Thus the integral that we need to evaluate is
_ _2 N - 3 go
1= 0
_1 ex kT (ex + 1) 2
3/2 _ E dE -
_2
3 go
1=
-1-£ /
ex
kT (ex
3/2
+ 1) 2 E
dx,
(7.59)
where in the last expression I've changed the integration variable to x. Because the integrand in this expression dies out exponentially when IE - J-LI » kT, we can now make two approximations. First, we can extend the lower limit on the integral down to -00; this makes things look more symmetrical, and it's harmless because the integrand is utterly negligible at negative E values anyway. Second, we can expand the function E3 / 2 in a Taylor series about the point E = J-L, and keep only the first few terms:
(7.60)
With these approximations our integral becomes
(7.61) Now, with only integer powers of x appearing, the integrals can actually be per­ formed, term by term.
Figure 7.15. The derivative of the Fermi-Dirac distribution is negligible everywhere except within a few kT of J.L.
€
283
284
Chapter 7
Quantum Statistics
The first term is easy:
j_(x +
eX
e
1
)2 dx =
j=
dnFD _= - a dE = nFD ( - 00) -
nFD ( 00) = 1 - 0 = 1.
(7.62)
E
The second term is also easy, since the integrand is an odd function of x:
x
{= __x_e__ dx = {=
J_= (ex + 1)2
x
J_= (ex + 1)(1 + e- x )
dx = 0 .
(7.63)
The third integral is the hard one. It can be evaluated analytically, as shown in Appendix B:
{=
7r 2
x 2 eX
(7.64)
J_= (ex + 1)2 dx = 3·
You can also look it up in tables, or evaluate it numerically. Assembling the pieces of equation 7.61, we obtain for the number of electrons
N = -2 gof.L 3/2 3
_ (~)3/2 - N EF
2
1 2 -1/2 . -7r + .. + -go (kT) f.L
4
3
2
(kT)2 + N 8 EF3/2 f.L 1/2 7r
+
..
(7.65)
.
(In the second line I've plugged in go = 3N/2E~2, from equation 7.51.) Canceling the N's, we now see that f.L/EF is approximately equal to 1, with a correction proportional to (kT/EF)2 (which we assume to be very small). Since the correction term is already quite small, we can approximate f.L ~ EF in that term, then solve for f.L/ EF to obtain 2 kT 2 f.L [ 7r ]2/3 EF = 1 - 8 ( ~) + . . (7.66) 2 = 1 _ 7r (kT)2 + .. . 12 EF As predicted, the chemical potential gradually decreases as T is raised. The behav­ ior of f.L over a wide range of temperatures is shown in Figure 7.16. The integral (7.54) for the total energy can be evaluated using exactly the same series of tricks. I'll leave it for you to do in Problem 7.28; the result is
3 f.L5/2 37r 2 (kT)2 U=-N-+-N--+···. 5 EF 3/2 8 EF
(7.67)
Finally you can plug in formula 7.66 for f.L and do just a bit more algebra to obtain (7.68) as I claimed in equation 7.47.
7.3
Degenerate Fermi Gases
Figure 7.16. Chemical potential of a noninteracting, nonrelativistic Fermi gas in a three-dimensional box, calculated numerically as described in Problem 7.31. At low temperatures /l is given approximately by equation 7.66, while at high temperatures /l becomes negative and approaches the form for an ordinary gas obeying Boltzmann statistics.
Now admittedly, that was a lot of work just to get a factor of 7r 2 /4 (since we had already guessed the rest by dimensional analysis). But I've presented this calculation in detail not so much because the answer is important, as because the methods are so typical of what professional physicists (and many other scientists and engineers) often do. Very few real-world problems can be solved exactly, so it 's crucial for a scientist to learn when and how to make approximations. And more often than not, it's only after doing the hard calculation that one develops enough intuition to see how to guess most of the answer. Problem 7.29. Carry out the Sommerfeld expansion for the energy integral (7.54), to obtain equation 7.67. Then plug in the expansion for /l to obtain the final answer, equation 7.68. Problem 7.30. The Sommerfeld expansion is an expansion in powers of kT/EF, which is assumed to be small. In this section I kept all terms through order (kT/EF)2 , omitting higher-order terms. Show at each relevant step that the term proportional to T3 is zero, so that the next nonvanishing terms in the expansions for /l and U are proportional to T4. (If you enjoy such things, you might try evaluating the T4 terms, possibly with the aid of a computer algebra program.) Problem 7.31. In Problem 7.28 you found the density of states and the chemical potential for a two-dimensional Fermi gas. Calculate the heat capacity of this gas in the limit kT « EF' Also show that the heat capacity has the expected behavior when kT » EF' Sketch the heat capacity as a function of temperature. Problem 7.32. Although the integrals (7.53 and 7.54) for Nand U cannot be carried out analytically for all T , it's not difficult to evaluate them numerically using a computer. This calculation has little relevance for electrons in metals (for which the limit kT « EF is always sufficient), but it is needed for liquid 3He and for astrophysical systems like the electrons at the center of the sun. (a) As a warm-up exercise, evaluate the N integral (7.53) for the case kT = EF and /l = 0, and check tnat your answer is consistent with the graph shown
285
286
Chapter 7
Quantum Statistics above. (Hint: As always when solving a problem on a computer, it's best to first put everything in terms of dimensionless variables. So let t = kT/cF, C = /-l/cF, and x = c/kT. Rewrite everything in terms of these variables, and then put it on the computer.)
(b) The next step is to vary /-l, holding T fixed, until the integral works out to the desired value, N. Do this for values of kT/cF ranging from 0.1 up to 2, and plot the results to reproduce Figure 7.16. (It's probably not a good idea to try to use numerical methods when kT /cF is much smaller than 0.1, since you can start getting overflow errors from exponentiating large numbers. But this is the region where we've already solved the problem analytically.)
(c) Plug your calculated values of /-l into the energy integral (7.54), and evaluate that integral numerically to obtain the energy as a function of temperature for kT up to 2cF' Plot the results , and evaluate the slope to obtain the heat capacity. Check that the heat capacity has the expected behavior at both low and high temperatures.
Problem 1.33. When the attractive forces of the ions in a crystal are taken into account, the allowed electron energies are no longer given by the simple formula 7.36; instead, the allowed energies are grouped into bands, separated by gaps where there are no allowed energies. In a conductor the Fermi energy lies within one of the bands; in this section we have treated the electrons in this band as 'free' particles confined to a fixed volume. In an insulator, on the other hand, the Fermi energy lies within a gap, so that at T = 0 the band below the gap is completely occupied while the band above the gap is unoccupied. Because there are no empty states close in energy to those that are occupied, the electrons are 'stuck in place' and the material does not conduct electricity. A semiconductor is an insulator in which the gap is narrow enough for a few electrons to jump across it at room temperature. Figure 7.17 shows the density of states in the vicinity of the Fermi energy for an idealized semiconductor, and defines some terminology and notation to be used in this problem.
(a) As a first approximation, let us model the density of states near the bottom of the conduction band using the same function as for a free Fermi gas, with an appropriate zero-point: g(c) = gO~, where go is the same constant as in equation 7.51. Let us also model the density of states near the top
g(C)
j-.---
Gap
., I Conduction band
~----~------------+-------~---------+----------------~C
Figure 1.11. The periodic potential of a crystal lattice results in a density­ of-states function consisting of 'bands' (with many states) and 'gaps' (with no states). For an insulator or a semiconductor, the Fermi energy lies in the middle of a gap so that at T = 0, the 'valence band' is completely full while the-'conduction band' is completely empty.
7.3
Degenerate Fermi Gases
of the valence band as a mirror image of this function. Explain why, in this approximation, the chemical potential must always lie precisely in the middle of the gap, regardless of temperature. (b) Normally the width of the gap is much greater than kT. Working in this limit, derive an expression for the number of conduction electrons per unit volume, in terms of the temperature and the width of the gap. (c) For silicon near room temperature, the gap between the valence and con­ duction bands is approximately 1.11 eV. Roughly how many conduction electrons are there in a cubic centimeter of silicon at room temperature? How does this compare to the number of conduction electrons in a similar amount of copper?
(d) Explain why a semiconductor conducts electricity much better at higher temperatures. Back up your explanation with some numbers. (Ordinary conductors like copper, on the other hand, conduct better at low temper­ atures.) (e) Very roughly, how wide would the gap between the valence and conduction bands have to be in order to consider a material an insulator rather than a semiconductor?
Problem 7.34. In a real semiconductor, the density of states at the bottom of the conduction band will differ from the model used in the previous problem by a numerical factor, which can be small or large depending on the material. Let us therefore write for the conduction band g(E) = gOeVE - Ee, where gOe is a new normalization constant that differs from go by some fudge factor. Similarly, write g( E) at the top of the valence band in terms of a new normalization constant gOv' (a) Explain why, if gov =I- gOe, the chemical potential will now vary with tem­ perature. When will it increa.',e, and when will it decrease? (b) Write down an expression for the number of conduction electrons, in terms of T, f.,t, Ee, and gOe. Simplify this expression as much as possible, assuming Ee f.,t» kT. (c) An empty state in the valence band is called a hole. In analogy to part (b), write down an expression for the number of holes, and simplify it in the limit f.,t - Ev » kT.
(d) Combine the results of parts (b) and (c) to find an expression for the chemical potential as a function of temperature. (e) For silicon, goelgo = 1.09 and gov / gO silicon at room temperature.
=
0.44. * Calculate the shift in
f.,t
for
Problem 7.35. The previous two problems dealt with pure semiconductors, also called intrinsic semiconductors. Useful semiconductor devices are instead made from doped semiconductors, which contain substantial numbers of impurity atoms. One example of a doped semiconductor was treated in Problem 7.5. Let us now consider that system again. (Note that in Problem 7.5 we measured all energies relative to the bottom of the conduction band, Ee. We also neglected the distinction between gO and gOei this simplification happens to be ok for conduction electrons in silicon.) *These values can be calculated from the 'effective masses' of electrons and holes. See, for example, S. M. Sze, Physics of Semiconductor Devices, second edition (Wiley, New York, 1981).
287
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Quantum Statistics
(a) Calculate and plot the chemical as a function of temperature, for silicon doped with 10 17 phosphorus atoms per cm 3 (as in Problem 7.5). Continue to assume that the conduction electrons can be treated as an ordinary ideal gas. (b) Discuss whether it is legitimate to assume for this system that the con­ duction electrons can be treated as an ordinary ideal gas, as opposed to a Fermi gas. Give some numerical eXl1.mpl€:s. (c) Estimate the temperature at which the number of valence electrons ex­ cited to the conduction band would become comparable to the number of conduction electrons from donor impurities. Which source of conduction electrons is more important at room temperature? Problem 7.36. Most spin-l/2 fermions, including electrons and helium-3 atoms, have nonzero moments. A gas of such particles is therefore paramagnetic. Consider, for a gas of free electrons, confined inside a three-dimensional In the box. The z component of the magnetic moment of each electron is presence of a field B pointing in the direction, each 'up' state acquires an additional energy of -/hBB, while each 'down' state acquires an additional energy of +/hBB. (a) Explain why you would expect the magnetization of a degenerate electron gas to be substantially less than that of the electronic paramagnets studied in Chapters 3 and 6, for a given number of particles at a given field str·engtJl. (b) Write down a formula for the density of states of this system in the presence of a magnetic field B, and interpret your formula graphically. (c) The magnetization of this system is /hB(NT -Nt), where NT and Nt are the numbers of electrons with up and down magnetic moments, rp;;:nl',.ti,vp!v Find a formula for the magnetization of this system at T 0, in terms of N, /hB, B, and the Fermi energy. (d) Find the first temperature-dependent correction to your answer to part (c), You may assume that /hBB « kT; this implies in the limit T « that the presence of the magnetic field has effect on the chemical potential/h. avoid confusing /hB with /h, I suggest using an abbreviation such as 8 for the quantity /hB B. )
7.4 Blackbody Radiation As a next application of quantum statistics, I'd like to consider the electromagnetic radiation inside some 'box' (like an oven or kiln) at a given temperature. First let me discuss what we would expect of such a system in classical (Le., non-quantum) physics.
The Ultraviolet Catastrophe In classical physics, we treat electromagnetic radiation as a continuous 'field' that permeates all space. Inside a box, we can think of this field as a combination of various standing-wave patterns, as shown in Figure 7.18. Each standing-wave pattern behaves as a harmonic oscillator with frequency f = c/ A. Like a me­ chanical oscillator, each electromagnetic standing wave has two degrees of freedom,
7.4
Blackbody Radiation
E=kT Total energy
= kT . 00
E=kT
E=kT
Figure 7.18. We can analyze the electromagnetic field in a box as a superposition of standing-wave modes of various wavelengths. Each mode is a harmonic oscil­ lator with some well-defined frequency. Classically, each oscillator should have an average energy of kT. Since the total number of modes is infinite, so is the total energy in the box.
with an average thermal energy of 2 . ~kT. Since the total number of oscilla­ tors in the electromagnetic field is infinite, the total thermal energy should also be infinite. Experimentally, though, you're not blasted with an infinite amount of electromagnetic radiation every time you open the oven door to check the cookies. This disagreement between classical theory and experiment is called the ultravio­ let catastrophe (because the infinite energy would come mostly from very short wavelengths) . The Planck Distribution The solution to the ultraviolet catastrophe comes from quantum mechanics. (His­ torically, the ultraviolet catastrophe led to the birth of quantum mechanics.) In quantum mechanics, a harmonic oscillator can't have just any amount of energy; its allowed energy levels are
En = 0, hj, 2hj, ..
(7.69)
(As usual I'm measuring all energies relative to the ground-state energy. See Ap­ pendix A for more discussion of this point.) The partition function for a single oscillator is therefore Z = 1 + e­ J3hj + e- 2J3hj + .. (7.70) 1 - 1 - e- J3hj '
and the average energy is -
E =
1 8Z
--Z 8(3 =
hj ehflkT -
l'
(7.71)
If we think of the energy as coming in 'units' of hj, then the average number of units of energy in the oscillator is
1
riPl
= e hjjkT -1 .
(7.72)
289
290
Chapter 7
Quantum Statistics
This formula is called the Planck distribution (after Max Planck). According to the Planck distribution, short-wavelength modes of the electro­ magnetic field, with hf » kT, are exponentially suppressed: They are 'frozen out,' and might as well not exist. Thus the total number of electromagnetic os­ cillators that effectively contribute to the energy inside the box is finite, and the ultraviolet catastrophe does not occur. Notice that this solution requires that the oscillator energies be quantized: It is the size of the energy units, compared to kT, that provides the exponential suppression factor.
Photons 'U nits' of energy in the electromagnetic field can also be thought of as particles, called photons. They are bosons, so the number of them in any 'mode' or wave pattern of the field ought to be given by the Bose-Einstein distribution: riBE
=
1 e(€-j-t)/kT _
(7.73)
1.
Here € is the energy of each particle in the mode, that is, equation 7.72 therefore requires
€
for photons.
h f. Comparison with (7.74)
But why should this be true? I'll give you two reasons, both based on the fact that photons can be created or destroyed in any quantity; their total number is not conserved. First consider the Helmholtz free energy, which must attain the minimum pos­ sible value at equilibrium with T and V held fixed. In a system of photons, the number N of particles is not constrained, but rather takes whatever value will minimize F. If N then changes infinitesimally, F should be unchanged:
aF) _ (aN
0
TV , ­
(at equilibrium).
(7.75)
But this partial derivative is precisely equal to the chemical potential. A second argument makes use of the condition for chemical equilibrium derived in Section 5.6. Consider a typical reaction in which a photon (,) is created or absor bed by an electron: e f----+ e + ,. (7.76) As we saw in Section 5.6, the equilibrium condition for such a reaction is the same as the reaction equation, with the name of each species replaced by its chemical potential. In this case, J.Le
= J.Le +J.L,
(at equilibrium).
(7.77)
In other words, the chemical potential for photons is zero. By either argument, the chemical potential for a 'gas' of photons inside a box at fixed temperature is zero, so the Bose-Einstein distribution reduces to the Planck distribution, as required.
7.4
Blackbody Radiation
Summing over Modes The Planck distribution tells us how many photons are in any single 'mode' (or 'single-particle state') of the electromagnetic field. Next we might want to know the total number of photons inside the box, and also the total energy of all the photons. To compute either one, we have to sum over all possible states, just as we did for electrons. I'll compute the total energy, and let you compute the total number of photons in Problem 7.44. Let's start in one dimension, with a 'box' of length L. The allowed wavelengths and momenta are the same for photons as for any other particles:
A = 2L., n
p
hn 2L'
(7.78)
(Here n is a positive integer that labels which mode we're talking about, not to be confused with ri pl , the average number of photons in a given mode.) Photons, however, are ultrarelativistic particles, so their energies are given by E
= pe
hen 2L
(7.79)
instead of E p2/2m. (You can also derive this result straight from the Einstein relation E hf between a photon's energy and its frequency. For light, f = e/ A, so E he/A hen/2L.) In three dimensions, momentum becomes a vector, with each component given by h/2L times some The energy is e times the magnitude of the momentum vector: he hen (7.80) 2L 2L' where in the last expression I'm using n for the magnitude of the n vector, as in Section 7.3. Now the average energy in any particular mode is equal to E times the occupancy of that mode, and the occupancy is given by the Planck distribution. To get the total energy in all modes, we sum over n x , ny, and n z . We also need to slip in a factor of 2, since each wave shape can hold photons with two independent polarizations. So the total energy is
U=2
hen L
1
--:-----
1
(7.81)
As in Section 7.3, we can convert the sums to integrals and carry out the integration in spherical coordinates (see 7.11). This time, however, the upper limit on the integration over n is infinity:
rOO {7r/2 {7r/2 hen 1 2 U = } 0 dn } 0 de } 0 d¢ n sin e L --:---;-:-::-:-:-=--
(7.82)
Again the angular integrals give 11'/2, the surface area of an eighth of a unit sphere.
291
292
Chapter 7
Quantum Statistics
The Planck Spectrum The integral over n looks a little nicer if we change variables to the photon energy, hcn/2L. We then get an overall factor of L3 = V, so the total energy per unit volume is E
(7.83) Here the integrand has a nice interpretation: It is the energy density per unit photon energy, or the spectrum of the photons: 811'
(hC)3
(7.84)
1
This function, first derived by Planck, gives the relative intensity of the radiation as a function of photon energy (or as a function of frequency, if you change variables the energy per unit again to f = E/h). If you integrate U(E) from El to E2, you volume within that range of photon energies. To actually evaluate the integral over E, it's convenient to change variables again, to x = E/kT. Then equation 7.83 becomes
u
(7.85)
V
The integrand is still proportional to the Planck spectrum; this function is plotted in Figure 7.19. The spectrum peaks at x = 2.82, or E = 2.82kT. Not surpris­ ingly, higher temperatures tend to give higher photon energies. (This fact is called 1.4 1.2 1.0 x3
0.8
eX -1
0.6 0.4 0.2 2
4 x
6 €/kT
8
10
12
Figure 7.19. The Planck spectrum, plotted in terms of the dimensionless variable x = €/kT hi/kT. The area under any portion of this graph, multiplied by 87r(kT)4 /(hc)3, equals the energy density of electromagnetic radiation within the corresponding frequency (or photon energy) range; see equation 7.85.
7.4
Blackbody Radiation
Wien's law.) You can measure the temperature inside an oven (or more likely, a kiln) by letting a bit of the radiation out and looking at its color. For instance, a typical clay-firing temperature of 1500 K gives a spectrum that peaks at E = 0.36 eV, in the near infrared. (Visible-light photons have higher energies, in the range of about 2-3 eV.) Problem 7.37. Prove that the peak of the Planck spectrum is at x
= 2.82.
Problem 7.38. It's not obvious from Figure 7.19 how the Planck spectrum changes as a function of temperature. To examine the temperature dependence, make a quantitative plot of the function u( E) for T = 3000 K and T = 6000 K (both on the same graph). Label the horizontal axis in electron-volts. Problem 7.39. Change variables in equation 7.83 to A = he/E, and thus derive a formula for the photon spectrum as a function of wavelength. Plot this spectrum, and find a numerical formula for the wavelength where the spectrum peaks, in terms of he/kT. Explain why the peak does not occur at he/(2.82kT). Problem 7.40. Starting from equation 7.83 , derive a formula for the density of states of a photon gas (or any other gas of ultrarelativistic particles having two polarization states). Sketch this function. Problem 7.41. Consider any two internal states, Sl and S2, of an atom. Let S2 be the higher-energy state, so that E(S2) - E(sd = E for some positive constant E. If the atom is currently in state S2, then there is a certain probability per unit time for it to spontaneously decay down to state Sl, emitting a photon with energy E. This probability per unit time is called the Einstein A coefficient:
A
= probability of spontaneous decay per unit time.
On the other hand, if the atom is currently in state Sl and we shine light on it with frequency f = E/ h, then there is a chance that it will absorb a photon, jumping into state S2 . The probability for this to occur is proportional not only to the amount of time elapsed but also to the intensity of the light, or more precisely, the energy density of the light per unit frequency, u(f). (This is the function which, when integrated over any frequency interval, gives the energy per unit volume within that frequency interval. For our atomic transition, all that matters is the value of u(f) at f = E/ h.) The probability of absorbing a photon, per unit time per unit intensity, is called the Einstein B coefficient:
B = probability of absorption per unit time u(f) . Finally, it is also possible for the atom to make a stimulated transition from S2 down to Sl, again with a probability that is proportional to the intensity of light at frequency f. (Stimulated emission is the fundamental mechanism of the laser: Light Amplification by Stimulated Emission of Radiation.) Thus we define a third coefficient, B', that is analogous to B:
B
'
=
probability of stimulated emission per unit time
u(f)
.
As Einstein showed in 1917, knowing anyone of these three coefficients is as good as knowing them all.
293
294
Chapter .7
Quantum Statistics
(a) Imagine a collection of many of these atoms, such that NI of them are in state SI and N2 are in state S2. Write down a formula for dNI/dt in terms of A, B, B', NI, and u(J).
(b) Einstein's trick is to imagine that these atoms are bathed in thermal ra­ diation, so that u(J) is the Planck spectral function. At equilibrium, Nl and N2 should be constant in time, with their ratio given by a simple Boltzmann factor. Show, then, that the coefficients must be related by B' =B
and
Total Energy Enough about the spectrum-what about the total electromagnetic energy inside the box? Equation 7.85 is essentially the final answer, except for the integral over x, which is just some dimensionless number. From Figure 7.19 you can estimate that this number is about 6.5; a beautiful but very tricky calculation (see Appendix B) gives it exactly as 1T.4 /15. Therefore the total energy density, summing over all frequencies, is
u
V
(7.86)
The most important feature of this result is its dependence on the fourth power of the temperature. If you double the temperature of your oven, the amount of electromagnetic energy inside increases by a factor of 24 16. Numerically, the total electromagnetic energy inside a typical oven is quite small. At cookie-baking temperature, 375°F or about 460 K, the energy per unit volume comes out to 3.5 X 10- 5 J /m 3 • This is tiny compared to the thermal energy of the air inside the oven. Formula 7.86 may look complicated, but you could have guessed the answer, aside from the numerical coefficient, by dimensional analysis. The average energy per photon must be something of order kT, so the total energy must be proportional to NkT, where N is the total number of photons. Since N is extensive, it must be proportional to the volume V of the container; thus the total energy must be of the form (7.87) u (constant) . VkT where £ is something with units of length. (If you want, you can pretend that each photon occupies a volume of £3 .) But the only relevant length in the problem is the typical de Broglie wavelength of the photons, A = h / p = he/ E ex he/ kT. Plugging this in for £ yields equation 7.86, aside from the factor of 8IT 5 /15. Problem 7.42. Consider the electromagnetic radiation inside a kiln, with a vol­ ume of 1 m 3 and a temperature of 1500 K.
(a) What is the total energy of this radiation? (b) Sketch the spectrum of the radiation as a function of photon energy. (c ) What fraction of all the energy is in the visible portion of the spectrum, with wavelengths between 400 nm and 700 nm?
7.4
Blackbody Radiation
Problem 7.43. At the surface of the sun, the temperature is approximately 5800 K. (a) How much energy is contained in the electromagnetic radiation filling a cubic meter of space at the sun's surface? (b) Sketch the spectrum of this radiation as a function of photon energy. Mark the region of the spectrum that corresponds to visible wavelengths, between 400 nm and 700 nm. (c) What fraction of the energy is in the visible portion of the spectrum? (Hint: Do the integral numerically.)
Entropy of a Photon Gas Besides the total energy of a photon gas, we might want to know a number of other quantities, for instance, the total number of photons present or the total entropy. These two quantities turn out to be equal, up to a constant factor. Let me now compute the entropy. The easiest way to compute the entropy is from the heat capacity. For a box of thermal photons with volume V, Cv =
(au)
3
(7.88)
aT v = 4aT ,
where a is an abbreviation for 81T 5 k 4 V/15(hc)3. This expression is good all the way down to absolute zero, so we can integrate it to find the absolute entropy. Introducing the symbol T' for the integration variable,
dT' = 4a {T(T')2 dT' = ~aT3 io 3
=
5
321T V (kT)3 k. 45
hc
(7.89)
The total number of photons is given by the same formula, with a different numerical coefficient, and without the final k (see Problem 7.44).
The Cosmic Background Radiation The grandest example of a photon gas is the radiation that fills the entire observ­ able universe, with an almost perfect thermal spectrum at a temperature of 2.73 K. Interpreting this temperature is a bit tricky, however: There is no longer any mech­ anism to keep the photons in thermal equilibrium with each other or with anything else; the radiation is instead thought to be left over from a time when the universe was filled with ionized gas that interacted strongly with electromagnetic radiation. At that time, the temperature was more like 3000 K; since then the universe has expanded a thousandfold in all directions, and the photon wavelengths have been stretched out accordingly (Doppler-shifted, if you care to think of it this way), preserving the shape of the spectrum but shifting the effective temperature down to 2.73 K. The photons making up the cosmic background radiation have rather low en­ ergies: The spectrum peaks at € = 2.82kT = 6.6 x 10- 4 eV. This corresponds to
295
296
Chapter 7
Quantum Statistics
wavelengths of about a millimeter, in the far infrared. These wavelengths don't penetrate our atmosphere, but the long-wavelength tail of the spectrum, in the microwave region of a few centimeters, can be detected without much difficulty. It was discovered accidentally by radio astronomers in 1965. Figure 7.20 shows a more recent set of measurements over a wide range of wavelengths, made from above earth's atmosphere by the Cosmic Background Explorer satellite. According to formula 7.86, the total energy in the cosmic background radiation is only 0.26 MeV 1m3 . This is to be contrasted with the average energy density of ordinary matter, which on cosmic scales is of the order of a proton per cubic meter or 1000 MeV 1m3 . (Ironically, the density of the exotic background radiation is known to three significant figures, while the average density of ordinary matter is uncertain by nearly a factor of 10.) On the other hand, the entropy of the background radiation is much greater than that of ordinary matter: According to equation 7.89, every cubic meter of space contains a photon entropy of (2.89 x 109 )k, nearly three billion 'units' of entropy. The entropy of ordinary matter is not easy to calculate but if we pretend that this matter is an ordinary ideal gas we can estimate that its entropy is N k times some small number, in other words, only a few k per cubic meter.
1.6 ,--.,
... I
00
---S ---
1.4 1.2
M
'':l
It')
N
I
1.0 0.8
0
,...; '--'
0.6 0.4 0.2 0
1
2
f
3 (lOll s-l)
4
5
6
Figure 7.20. Spectrum of the cosmic background radiation, as measured by the Cosmic Background Explorer satellite. Plotted vertically is the density per unit frequency, in SI units. Note that a frequency of 3 x lOll corresponds to a of A = c/ f = 1.0 mm. Each square represents a measured data point. The point-by-point uncertainties are too small to show up on this scale; the size of the squares instead represents a liberal estimate of the uncertainty due to sYf'ltema,tlc effects. The solid curve is the theoretical Planck spectrum, with the temperature adjusted to 2.735 K to the best fit. From J. C. Mather et aL, Astrophysical Journal Letters 354, L37 (1990); adapted courtesy of NASA/GSFC and the COBE Science Working Group. Subsequent measurements from this ex­ periment and others now give a best-fit temperature of 2.728 ± 0.002 K.
7.4
Blackbody Radiation
Problem 7.44. Number of photons in a photon gas.
(a) Show that the number of photons in equilibrium in a box of volume V at
temperature T is
kT)3 N = 81rV ( he
(X)
io
x2
ex-! dx.
The integral cannot be done analytically; either look it up in a table or evaluate it numerically.
(b) How does this result compare to the formula derived in the text for the
entropy of a photon (What is the entropy per photon, in terms of k?)
(c) Calculate the number of photons per cubic meter at the following temper­
atures: 300 K; 1500 K (a typical kiln); 2.73 K (the cosmic background
radiation).
Problem 7.45. Use the formula P = -(8U/8V)S,N to show that the pressure of a photon gas is 1/3 times the energy density (UIV). Compute the pressure exerted by the radiation inside a kiln at 1500 K, and compare to the ordinary gas pressure exerted by the air. Then compute the pressure of the radiation at the center of the sun, where the temperature is 15 million K. Compare to the gas pressure of the ionized hydrogen, whose density is approximately 10 5 kg/m 3 . Problem 7.46. Sometimes it is useful to know the free energy of a photon gas.
(a) Calculate the (Helmholtz) free energy directly from the definition F =
U
TS. (Express the answer in terms of T and V.)
(b) Check the formula S
-(8F/8T)v for this system.
(c) Differentiate F with respect to V to obtain the pressure of a photon gas.
Check that your result agrees with that of the previous problem.
= -kT In Z
separately to each mode (that is, each effective oscillator), then sum over
all modes. Carry out this calculation, to obtain
(d) A more interesting way to calculate F is to apply the formula F
F
(kT)4 81rV (he)3
roo
io
2
x In(l
Integrate by parts, and check that your answer agrees with part (a). Problem 7.47. In the text I claimed that the universe was filled with ionized gas until its temperature cooled to about 3000 K. To see why, assume that the universe contains only photons and hydrogen atoms, with a constant ratio of 109 photons per hydrogen atom. Calculate and plot the fraction of atoms that were ionized as a function of temperature, for temperatures between 0 and 6000 K. How does the result change if the ratio of photons to atoms is 10 8 , or 101O? (Hint: Write everything in terms of dimensionless variables such as t kT/ I, where I is the ionization energy of hydrogen. ) Problem 7.48. In addition to the cosmic background radiation of photons, the universe is thought to be permeated with a background radiation of neutrinos (v) and antineutrinos (D), currently at an effective temperature of 1.95 K. There are three species of neutrinos, each of which has an antiparticle, with only one allowed polarization state for each particle or antiparticle. For parts (a) through (c) below, assume that all three species are exactly massless.
297
298
Chapter 7
Quantum Statistics
(a) It is reasonable to assume that for each species, the concentration of neu­ trinos equals the concentration of antineutrinos, so that their chemical potentials are equal: /-Lv = /-LiJ. Furthermore, neutrinos and antineutrinos can be produced and annihilated in pairs by the reaction v
+ D .... 2'(
(where '( is a photon). Assuming that this reaction is at equilibrium (as it would have been in the very early universe), prove that /-L = 0 for both the neutrinos and the antineutrinos.
(b) If neutrinos are massless, they must be highly relativistic. They are also fermions: They obey the exclusion principle. Use these facts to derive a formula for the total energy density (energy per unit volume) of the neutrino-antineutrino background radiation. (Hint: There are very few differences between this 'neutrino gas' and a photon gas. Antiparticles still have positive energy, so to include the antineutrinos all you need is a factor of 2. To account for the three species, just multiply by 3.) To evaluate the final integral, first change to a dimensionless variable and then use a computer or look it up in a table or consult Appendix B.
(c) Derive a formula for the number of neutrinos per unit volume in the neutrino background radiation. Evaluate your result numerically for the present neutrino temperature of 1.95 K.
(d) It is possible that neutrinos have very small, but nonzero, masses. This wouldn't have affected the production of neutrinos in the early universe, when mc2 would have been negligible compared to typical thermal ener­ gies. But today, the total mass of all the background neutrinos could be significant. Suppose, then, that just one of the three species of neutrinos (and the corresponding antineutrino) has a nonzero mass m. What would mc2 have to be (in eV), in order for the total mass of neutrinos in the universe to be comparable to the total mass of ordinary matter?
Problem 7.49. For a brief time in the early universe, the temperature was hot enough to produce large numbers of electron-positron pairs. These pairs then constituted a third type of 'background radiation,' in addition to the photons and neutrinos (see Figure 7.21). Like neutrinos, electrons and positrons are fermions. Unlike neutrinos, electrons and positrons are known to be massive (each with the same mass), and each has two independent polarization states. During the time period of interest the densities of electrons and positrons were approximately equal, so it is a good approximation to set the chemical potentials equal to zero as in
Figure 7.21. When the temperature was greater than the electron mass times c 2 /k, the universe was filled with three types of radiation: electrons and positrons (solid arrows); neutri­ nos (dashed); and photons (wavy). Bathed in this radiation were a few protons and neutrons, roughly one for every billion radiation particles.
7.4
Blackbody Radiation
the previous problem. Recall from special relativity that the energy of a massive particle is E = J(pc)2 + (mc 2)2.
(a) Show that the energy density of electrons and positrons at temperature T is given by
Q= V
161r(kT)4 (T)
(hc)3 u ,
where
(b) Show that u(T) goes to zero when kT « mc2, and explain why this is a reasonable result. (c) Evaluate u(T) in the limit kT » mc2 , and compare to the result of the previous problem for the neutrino radiation. (d) Use a computer to calculate and plot u(T) at intermediate temperatures. (e) Use the method of Problem 7.46, part (d), to show that the free energy density of the electron-positron radiation is F V
where J(T)
=
10
00
_ 161r(kT)4 J(T) (hc)3 ' 2
2
x 2 ln(1 + e -Jx +(mc /kT)2) dx.
Evaluate J(T) in both limits, and use a computer to calculate and plot J(T) at intermediate temperatures.
(f) Write the entropy of the electron-positron radiation in terms of the func­ tions u(T) and J(T). Evaluate the entropy explicitly in the high-T limit. Problem 1.50. The results of the previous problem can be used to explain why the current temperature of the cosmic neutrino background (Problem 7.48) is 1.95 K rather than 2.73 K. Originally the temperatures of the photons and the neutrinos would have been equal, but as the universe expanded and cooled, the interactions of neutrinos with other particles soon became negligibly weak. Shortly thereafter, the temperature dropped to the point where kT/ c2 was no longer much greater than the electron mass. As the electrons and positrons disappeared during the next few minutes, they 'heated' the photon radiation but not the neutrino radiation. (a) Imagine that the universe has some finite total volume V, but that V is increasing with time. Write down a formula for the total entropy of the electrons, positrons, and photons as a function of V and T, using the aux­ illiary functions u(T) and J(T) introduced in the previous problem. Argue that this total entropy would have been conserved in the early universe, assuming that no other species of particles interacted with these. (b) The entropy of the neutrino radiation would have been separately conserved during this time period, because the neutrinos were unable to interact with anything. Use this fact to show that the neutrino temperature Tv and the photon temperature T are related by
1
+ u(T) + J(T) = constant
299
300
Chapter 7
Quantum Statistics as the universe expands and cools. Evaluate the constant by assuming that T = Tv when the temperatures are very high.
(c) Calculate the ratio T lTv in the limit of low temperature, to confirm that the present neutrino temperature should be 1.95 K . (d) Use a computer to plot the ratio TITv as a function of T, for kTlmc 2 ranging from 0 to 3. *
Photons Escaping through a Hole So far in this section I have analyzed the gas of photons inside an oven or any other box in thermal equilibrium. Eventually, though, we'd like to understand the photons emitted by a hot object. To begin, let's ask what happens if you start with a photon gas in a box, then poke a hole in the box to let some photons out (see Figure 7.22). All photons travel at the same speed (in vacuum), regardless of their wave­ lengths. So low-energy photons will escape through the hole with the same prob­ ability as high-energy photons, and thus the spectrum of the photons coming out will look the same as the spectrum of the photons inside. What's harder to figure out is the total amount of radiation that escapes; the calculation doesn't involve much physics, but the geometry is rather tricky. The photons that escape now, during a time interval dt, were once pointed at the hole from somewhere within a hemispherical shell, as shown in Figure 7.23. The radius R of the shell depends on how long ago we're looking, while the thickness of the shell is edt. I'll use spherical coordinates to label various points on the shell, as shown. The angle () ranges from 0, at the left end of the shell, to 7r/2, at the extreme edges on the right. There's also an azimuthal angle cp, not shown, which ranges from 0 to 27r as you go from the top edge of the shell into the page, down
Figure 7.22. When you open a hole in a container filled with radiation (here a kiln), the spectrum of the light that escapes is the same as the spectrum of the light inside. The total amount of energy that escapes is proportional to the size of the hole and to the amount of time that passes.
*Now that you've finished this problem, you'll find it relatively easy to work out the dynamics of the early universe, to determine when all this happened. The basic idea is to assume that the universe is expanding at 'escape velocity.' Everything you need to know is in Weinberg (1977).
7.4
Blackbody Radiation
1 A
T~
/ Figure 7.23. The photons that escape now were once somewhere within a hemi­ spherical shell inside the box. From a given point in this shell, the probability of escape depends on the distance from the hole and the angle ().
to the bottom, out of the page, and back to the top. Now consider the shaded chunk of the shell shown Figure 7.23. Its volume is volume of chunk = (Rd()) x (Rsin()d¢) x (edt).
(7.90)
(The depth of the chunk, perpendicular to the page, is R sin () d¢, since R sin () is the radius of a ring of constant () swept out as ¢ ranges from 0 to 27r.) The energy density of the photons within this chunk is given by equation 7.86: U V
87r 5 (kT)4 15 (he)3'
(7.91)
In what follows I'll simply call this quantity UIV; the total energy in the chunk is thus energy in chunk =
~ edt R2 sin () d() d¢.
(7.92)
But not all the energy in this chunk of space will escape through the hole, because most of the photons are pointed in the wrong direction. The probability of a photon being pointed in the right direction is equal to the apparent area of the hole, as viewed from the chunk, divided by the total area of an imaginary sphere of radius R centered on the chunk: probability of escape
A cos () 47rR2 .
(7.93)
Here A is the area of the hole, and A cos () is its foreshortened area, as seen from the chunk. The amount of energy that escapes from this chunk is therefore . Acos() U . energy escapmg from chunk = ~ V edt sm () d() d¢.
(7.94)
301
302
Chapter 7
Quantum Statistics
To find the total energy that escapes through the hole in the time interval dt, we just integrate over e and ¢:
r d¢ io{11' 2
total energy escaping = io
27r
=
11'
/2
de
A cos 47r
eU
v edt sine
A U 111'/2 vedt cos e sine de 47r 0
(7.95)
AU 4 V edt.
The amount of energy that escapes is naturally proportional to the area A of the hole, and also to the duration dt of the time intervaL If we divide by these quantities we get the power emitted per unit area: eU 4
power per unit area
(7.96)
Aside from the factor of 1/4, you could have guessed this result using dimensional analysis: To turn energy/volume into power/area, you have to multiply by some­ thing with units of distance/time, and the only relevant speed in the problem is the speed of light. Plugging in formula 7.91 for the energy density inside the box, we obtain the more explicit result power per unit area where
(j
(7.97)
is known as the Stefan-Boltzmann constant, 5.67 x 10
-8
W m 2 K4 .
(7.98)
(This number isn't hard to memorize: Just think '5-6-7-8,' and don't forget the minus sign.) The dependence of the power radiated on the fourth power of the temperature is known as Stefan's law, and was discovered empirically in 1879.
Radiation from Other Objects Although I derived Stefan's law for photons emitted from a hole in a box, it also applies to photons emitted by any nonreflecting ('black') surface at temperature T. Such radiation is therefore called blackbody radiation. The proof that a black object emits photons exactly as does a hole in a box is amazingly simple. Suppose you have a hole in a box, on one hand, and a black object, on the other hand, both at the same temperature, facing each other as in Figure 7.24. Each object emits photons, some of which are absorbed by the other. If the objects are the same size, each will absorb the same fraction of the other's radiation. Now suppose that the blackbody does not emit the same amount of power as the hole; perhaps
7.4
Blackbody Radiation
~
/
'-' ~
~
~
~
~
~
~
Box of photons
Blackbody
Figure 7.24. A thought experiment to demonstrate that a perfectly black surface emits radiation identical to that emitted by a hole in a box of thermal photons.
it emits somewhat less. Then more energy will flow from the hole to the blackbody than from the blackbody to the hole, and the blackbody will gradually get hotter. Oops! This process would violate the second law of thermodynamics. And if the blackbody emits more radiation than the hole, then the blackbody gradually cools off while the box with the hole gets hotter; again, this can't happen. So the total power emitted by the blackbody, per unit area at any given temper­ ature, must be the same as that emitted by the hole. But we can say more. Imagine inserting a filter, which allows only a certain range of wavelengths to pass through, between the hole and the blackbody. Again, if one object emits more radiation at these wavelengths than the other, its temperature will decrease while the other's temperature increases, in violation of the second law. Thus the entire spectrum of radiation emitted by the blackbody must be the same as for the hole. If an object is not black, so that it reflects some photons instead of absorbing them, things get a bit more complicated. Let's say that out of every three photons (at some given wavelength) that hit the object, it reflects one back and absorbs the other two. Now, in order to remain in thermal equilibrium with the hole, it only needs to emit two photons, which join the reflected photon on its way back. More generally, if e is the fraction of photons absorbed (at some given wavelength), then e is also the fraction emitted, in comparison to a perfect blackbody. This number e is called the emissivity of the material. It equals 1 for a perfect blackbody, and equals 0 for a perfectly reflective surface. Thus, a good reflector is a poor emitter, and vice versa. Generally the emissivity depends upon the wavelength of the light, so the spectrum of radiation emitted will differ from a perfect blackbody spectrum. If we use a weighted average of e over all relevant wavelengths, then the total power radiated by an object can be written power = creAT 4 ,
(7.99)
where A is the object's surface area. Problem 7.51. The tungsten filament of an incandescent light bulb has a tem­ perature of approximately 3000 K. The emissivity of tungsten is approximately 1/3, and you may assume that it is independent of wavelength.
303
304
Chapter 7
Quantum Statistics
(a) If the bulb gives off a total of 100 watts, what is the surface area of its filament in square millimeters? (b) At what value of the photon energy does the peak in the bulb's spectrum occur? What is the wavelength corresponding to this photon energy? (c) Sketch (or use a computer to plot) the spectrum of light given off by the filament. Indicate the region on the graph that corresponds to visible wave­ lengths, between 400 and 700 nm. (d) Calculate the fraction of the bulb's energy that comes out as visible light. (Do the integral numerically on a calculator or computer.) Check your result qualitatively from the graph of part (c). (e) To increase the efficiency of an incandescent bulb, would you want to raise or lower the temperature? (Some incandescent bulbs do attain slightly higher efficiency by using a different temperature.)
(f) Estimate the maximum possible efficiency (i.e., fraction of energy in the visible spectrum) of an incandescent bulb, and the corresponding filament temperature. Neglect the fact that tungsten melts at 3695 K.
Problem 7.52. (a) Estimate (roughly) the total power radiated by your body, neglecting any energy that is returned by your clothes and environment. (Whatever the color of your skin, its emissivity at infrared wavelengths is quite close to 1; almost any nonmetal is a near-perfect blackbody at these wavelengths.) (b) Compare the total energy radiated by your body in one day (expressed in kilocalories) to the energy in the food you eat. Why is there such a large discrepancy? (c) The sun has a mass of 2 x 10 30 kg and radiates energy at a rate of 3.9 x 1026 watts. Which puts out more power per units mass-the sun or your body?
Problem 7.53. A black hole is a blackbody if ever there was one, so it should emit blackbody radiation, called Hawking radiation. A black hole of mass M has a total energy of M c2 , a surface area of 161rG 2 M2 / c4 ) and a temperature of hc3 /161r 2kGM (as shown in Problem 3.7). (a) Estimate the typical wavelength of the Hawking radiation emitted by a one-solar-mass (2 x 1030 kg) black hole. Compare your answer to the size of the black hole. (b) Calculate the total power radiated by a one-solar-mass black hole. (c) Imagine a black hole in empty space, where it emits radiation but absorbs nothing. As it loses energy, its mass must decrease; one could say it 'evap­ orates.' Derive a differential equation for the mass as a function of time, and solve this equation to obtain an expression for the lifetime of a black hole in terms of its initial mass. (d) Calculate the lifetime of a one-solar-mass black hole, and compare to the estimated age of the known universe (1010 years). (e) Suppose that a black hole that was created early in the history of the universe finishes evaporating today. What was its initial mass? In what part of the electromagnetic spectrum would most of its radiation have been emitted?
7.4
Blackbody Radiation
The Sun and the Earth From the amount of solar radiation received by the earth (1370 W 1m 2 , known as the solar constant) and the earth's distance from the sun (150 million kilometers), it's pretty easy to calculate the sun's total energy output or luminosity: 3.9 x 10 26 watts. The sun's radius is a little over 100 times the earth's: 7.0 x 108 m; so its surface area is 6.1 x 1018 m 2 . From this information, assuming an emissivity of 1 (which is not terribly accurate but good enough for our purposes), we can calculate the sun's surface temperature: T =
cum~~sity)1/4
5800 K.
(7.100)
Knowing the temperature, we can predict that the spectrum of sunlight should peak at a photon energy of c = 2.82 kT
1.41 eV,
(7.101)
which corresponds to a wavelength of 880 nm, in the near infrared. This is a testable prediction, and it agrees with experiment: The sun's spectrum is approximately by the Planck formula, with a peak at this energy. Since the peak is so close to the red end of the visible spectrum, much of the sun's energy is emitted as visible light. (If you've learned elsewhere that the sun's spectrum peaks in the middle of the visible spectrum at about 500 nm, and you're worried about the discrepancy, go back and work Problem 7.39.) A tiny fraction of the sun's radiation is absorbed by the earth, warming the earth's surface to a temperature suitable for life. But the earth doesn't just keep getting hotter and hotter; it also emits radiation into space, at the same rate, on average. This balance between absorption and emission gives us a way to estimate the earth's equilibrium surface temperature. As a first crude estimate, let's pretend that the earth is a perfect blackbody at all wavelengths. Then the power absorbed is the solar constant times the earth's cross-sectional area as viewed from the sun, . The power emitted, meanwhile, is by Stefan's law, with A being the full surface area of the earth, 47r R2, and T being the effective average surface Setting the power absorbed equal to the power emitted gives (solar constant) . 7r R2
47r R2 a-T 4 (7.102)
> T
=
= 279 K.
This is extremely close to the measured average temperature of 288 K (15°C). However, the earth is not a perfect blackbody. About 30% of the sunlight striking the earth is reflected directly back into space, mostly by clouds. Taking reflection into account brings the earth's predicted average temperature down to a frigid 255 K.
305
306
Chapter 7
Quantum Statistics
Since a poor absorber is also a poor emitter, you might think we could bring the earth's predicted temperature back up by taking the imperfect emissivity into account on the right-hand side of equation 7.102. Unfortunately, this doesn't work. There's no particular reason why the earth's emissivity should be the same for the infrared light emitted as for the visible light absorbed, and in fact, the earth's surface (like almost any nonmetal) is a very efficient emitter at infrared wavelengths. But there's another mechanism that saves us: Water vapor and carbon dioxide in earth's atmosphere make the atmosphere mostly opaque at wavelengths above a few microns, so if you look at the earth from space with an eye sensitive to infrared light, what you see is mostly the atmosphere, not the surface. The equilibrium temperature of 255 K applies (roughly) to the atmosphere, while the surface below is heated both by the incoming sunlight and by the atmospheric 'blanket.' If we model the atmosphere as a single layer that is transparent to visible light but opaque to infrared, we get the situation shown in Figure 7.25. Equilibrium requires that the energy of the incident sunlight (minus what is reflected) be equal to the energy emitted upward by the atmosphere, which in turn is equal to the energy radiated downward by the atmosphere. Therefore the earth's surface receives twice as much energy (in this simplified model) as it would from sunlight alone. According to equation 7.102, this mechanism raises the surface temperature by a factor of 21/4, to 303 K. This is a bit high, but then, the atmosphere isn't just a single perfectly opaque layer. By the way, this mechanism is called the greenhouse effect, even though most greenhouses depend primarily on a different mechanism (namely, limiting convective cooling). Sunlight
Atmosphere
1111111111111111'-11__11111111111111111 Ground Figure 7.25. Earth's atmosphere is mostly transparent to incoming sunlight, but opaque to the infrared light radiated upward by earth's surface. If we model the atmosphere as a single layer, then equilibrium requires that earth's surface receive as much energy from the atmosphere as from the sun. Problem 7.54. The sun is the only star whose size we can easily measure directly; astronomers therefore estimate the sizes of other stars using Stefan's law. (a) The spectrum of Sirius A, plotted as a function of energy, peaks at a photon energy of 2.4 eV, while Sirius A is approximately 24 times as luminous as the sun. How does the radius of Sirius A compare to the sun's radius? (b) Sirius B, the companion of Sirius A (see Figure 7.12), is only 3% as luminous as the sun. Its spectrum, plotted as a function of energy, peaks at about 7 eV. Howdoes its radius compare to that of the sun?
7.5
Debye Theory of Solids
(c) The spectrum of the star Betelgeuse, plotted as a function of energy, peaks at a photon energy of 0.8 eV, while Betelgeuse is approximately 10,000 times as luminous as the sun. How does the radius of Betelgeuse compare to the sun's radius? Why is Betelgeuse called a 'red supergiant'? Problem 7.55. Suppose that the concentration of infrared-absorbing gases in earth's atmosphere were to double, effectively creating a second 'blanket' to warm the surface. Estimate the equilibrium surface temperature of the earth that would result from this catastrophe. (Hint: First show that the lower atmospheric blanket is warmer than the upper one by a factor of 21/4. The surface is warmer than the lower blanket by a smaller factor.) Problem 7.56. The planet Venus is different from the earth in several respects. First, it is only 70% as far from the sun. Second, its thick clouds reflect 77% of all incident sunlight. Finally, its atmosphere is much more opaque to infrared light.
(a) Calculate the solar constant at the location of Venus, and estimate what the average surface temperature of Venus would be if it had no atmosphere and did not reflect any sunlight.
(b) Estimate the surface temperature again, taking the reflectivity of the clouds into account. (c) The opaqueness of Venus's atmosphere at infrared wavelengths is roughly 70 times that of earth's atmosphere. You can therefore model the atmosphere of Venus as 70 successive 'blankets' of the type considered in the text, with each blanket at a different equilibrium temperature. Use this model to estimate the surface temperature of Venus. (Hint: The temperature of the top layer is what you found in part (b). The next layer down is warmer by a factor of 21/ 4 . The next layer down is warmer by a smaller factor. Keep working your way down until you see the pattern.)
7.5 Debye Theory of Solids In Section 2.2 I introduc:ed the Einstein model of a solid crystal, in which each atom is treated as an independent three-dimensional harmonic oscillator. In Prob­ lem 3.25, you used this model to derive a prediction for the heat capacity, (Einstein model),
(7.103)
where N is the number of atoms and E = hi is the universal size of the units of energy for the identical oscillators. When kT » E, the heat capacity approaches a constant value, 3Nk, in agreement with the equipartition theorem. Below kT ~ E, the heat capacity falls off, approaching zero as the temperature goes to zero. This prediction agrees with experiment to a first approximation, but not in detail. In particular, equation 7.103 predicts that the heat capacity goes to zero exponen­ tially in the limit T ---t 0, whereas experiments show that the true low-temperature behavior is cubic: C v ex: T3. The problem with the Einstein model is that the atoms in a crystal do not vibrate independently of each other. If you wiggle one atom, its neighbors will also start to wiggle, in a complicated way that depends on the frequency of oscillation.
307
308
Chapter 7
Quantum Statistics
There are low-frequency modes of oscillation in which large groups of atoms are all moving together, and also high-frequency modes in which atoms are moving opposite to their neighbors. The units of energy come in different sizes, proportional to the frequencies of the modes of vibration. Even at very low temperatures, when the high-frequency modes are frozen out, a few low-frequency modes are still active. This is the reason why the heat capacity goes to zero less dramatically than the Einstein model predicts. In many ways, the modes of oscillation of a solid crystal are similar to the modes of oscillation of the electromagnetic field in vacuum. This similarity suggests that we try to adapt our recent treatment of electromagnetic radiation to the mechanical oscillations of the crystal. Mechanical oscillations are also called sound waves, and behave very much like light waves. There are a few differences, however: • Sound waves travel much slower than light waves, at a speed that depends on the stiffness and density of the material. I'll call this speed cs , and treat it as a constant, neglecting the fact that it can depend on wavelength and direction. • Whereas light waves must be transversely polarized, sound waves can also be longitudinally polarized. (In seismology, transversely polarized waves are called shear waves, or S-waves, while longitudinally polarized waves are called pres­ sure waves, or P-waves.) So instead of two polarizations we have three. For simplicity, I'll pretend that all three polarizations have the same speed. • Whereas light waves can have arbitrarily short wavelengths, sound waves in solids cannot have wavelengths shorter than twice the atomic spacing. The first two differences are easy to take into account. The third will require some thought. Aside from these three differences, sound waves behave almost identically to light waves. Each mode of oscillation has a set of equally spaced energy levels, with the unit of energy equal to f
= hf
(7.104)
In the last expression, L is the length of the crystal and n = Iiii is the magnitude of the vector in n-space specifying the shape of the wave. When this mode is in equilibrium at temperature T, the number of units of energy it contains, on average, is given by the Planck distribution:
-1
(7.105)
(This n is not to be confused with the n in the previous equation.) As with elec­ tromagnetic waves, we can think of these units of energy as particles obeying Bose­ Einstein statistics with It O. This time the 'particles' are called phonons. To calculate the total thermal energy of the crystal, we add up the energies of all allowed modes: (7.106)
7.5
Debye
of Solids
The factor of 3 counts the three polarization states for each ii. The next will be to convert the sum to an integral. But first we'd better worry about what values of ii are being summed over. If these were electromagnetic oscillations, there would be an infinite number of allowed modes and each sum would go to infinity. But in a crystal, the atomic spacing puts a strict lower limit on the wavelength. Consider a lattice of atoms in just one dimension (see Figure 7.26). Each mode of oscillation has its own distinct shape, with the number of 'bumps' equal to n. Because each bump must contain at least one atom, n cannot exceed the number of atoms in a row. If the three­ dimensional crystal is a perfect cube, then the number of atoms along any direction is ifN, so each sum in equation 7.106 should go from 1 to ifN. In other words, we're summing over a cube in n-space. If the crystal itself is not a perfect then neither is the corresponding volume of n-space. Still, however, the sum will run over a region in n-space whose total volume is N. Now comes the tricky approximation. Summing (or integrating) over a cube or some other complicated region of n-space is no fun, because the function we're summing depends on n X1 ny, and n z in a very complicated way (an exponential of a square root). On the other hand, the function depends on the magnitude of ii in a simpler way, and it doesn't depend on the angle in n-space at all. So Peter got the clever idea to pretend that the relevant region of n-space is a sphere, or rather, an eighth of a sphere. To preserve the total number of degrees of freedom, he chose a sphere whose total volume is N. You can easily show that the radius of the sphere has to be (7.107)
n=
ifN
n=3
n=2
n=l Figure 7.26. Modes of oscillation of a row of atoms in a crystal. If the crystal is a cube, then the number of atoms any row is ifN. This is also the total number of modes along this direction, because each 'bump' in the wave form must contain at least one atom.
309
310
Chapter 7
Quantum Statistics
Figure 7.27. The sum in equation 7.106 is technically over a cube in n-space whose width is ifN. As an approximation, we instead sum over an eighth-sphere with the same total volume.
Figure 7.27 shows the cube in n-space, and the sphere that approximates it. Remarkably, Debye's approximation is exact in both the high-temperature and low-temperature limits. At high temperature, all that matters is the total number of modes, that is, the total number of degrees of freedom; this number is preserved by choosing the sphere to have the correct volume. At low temperature, modes with large ii are frozen out anyway, so we can count them however we like. At intermediate temperatures, we'll get results that are not exact, but they'll still be surprisingly good. When we make Debye's approximation, and convert the sums to integrals in spherical coordinates, equation 7.106 becomes
u = 3 [n
rr /2 dB J[rr / 2d¢ n 2 sin B eE/k; - 1 .
m ax
dn
Jo
Jo
o
(7.108)
The angular integrals give 7r/2 (yet again), leaving us with
U= -37r 2
i
nmax
0
-hc s 2L
n
3
ehcsn/2LkT -
1
dn.
(7.109)
This integral cannot be done analytically, but it's at least a little cleaner if we change to the dimensionless variable x
=
hcsn 2LkT'
(7.110)
The upper limit on the integral will then be
Xmax
=
hcsn max 2LkT
=
hc s (6N)1 /3 2kT 7rV
TD
= T'
(7.111)
where the last equality defines the Debye temperature, TD-essentially an ab­ breviation for all the constants. Making the variable change and collecting all the constants is now straightforward. When the smoke clears, we obtain _ 9NkT4 U3
-
TD
i
0
TD T /
x3
--dx. eX - 1
(7.112)
7.5
Debye Theory of Solids
At this point you can do the integral on a computer if you like, for any desired temperature. Without a computer, though, we can still check the low-temperature and high-temperature limits. When T :» TD , the upper limit of the integral is much less than 1, so x is always very small and we can approximate eX ~ 1 + x in the denominator. The 1 cancels, leaving the x to cancel one power of x in the numerator. The integral then gives simply ~ (TD/T)3, leading to the final result
u= 3NkT
when T» T D ,
(7.113)
in agreement with the equipartition theorem (and the Einstein model). The heat capacity in this limit is just Cv 3Nk. When T « TD, the upper limit on the integral is so large that by the time we get to it, the integrand is dead (due to the eX in the denominator). So we might as well replace the upper limit by infinity-the extra modes we're adding don't contribute anyway. In this approximation, the integral is the same as the one we did for the photon gas (equation 7.85), and evaluates to Jr4/15. So the total energy is when T« TD.
(7.114)
To get the heat capacity, differentiate with respect to T:
C v = 12Jr 5
4(~)3
Nk
TD
when T«
(7.115)
The prediction Cv ex: T3 agrees beautifully with low-temperature experiments on almost any solid material. For though, there is also a linear contribution to the heat capacity from the conduction electrons, as described in Section 7.3. The total heat capacity at low temperature is therefore
C where 'Y
= 'YT + 12Jr
4
Nk T3
(metal, T«
(7.116)
Jr2 Nk2 /2fF in the free electron model. Figure 7.28 shows plots of C /T
8
Figure 7.28. Low-temperature measurements of the heat capac­ ities (per mole) of copper, sil­ ver, and gold. Adapted with per­ mission from William S. Corak et al., Physical Review 98, 1699 (1955).
6
4 2
o
2
4
6
8 10 12 T2 (K2)
14
16
18
311
312
Chapter 7
Quantum Statistics
vs. T2 for three familiar metals. The linearity of the data confirms the Debye theory of lattice vibrations, while the intercepts give us the experimental values of!. At intermediate temperatures, you have to do a numerical integral to get the total thermal energy in the crystaL If what you really want is the heat capacity, it's best to differentiate equation 7.109 analytically, then change variables to x. The result is T x4 eX C v = 9Nk ( -;:r;- - - dx. (7.117) 10 a
)31TDIT
A computer-generated plot of this function is shown in Figure 7.29. For comparison, the Einstein model prediction, equation 7.103, is also plotted, with the constant € chosen to make the curves agree at relatively high temperatures. As you can see, the two curves still differ significantly at low temperatures. Figure 1.14 shows further comparisons of experimental data to the prediction of the Debye model. The Debye temperature of any particular substance can be predicted from the speed of sound in that substance, using equation 7.111. Usually, however, one obtains a better fit to the data by choosing To so that the measured heat capacity best fits the theoretical prediction. Typical values of To range from 88 K for lead (which is soft and dense) to 1860 K for diamond (which is stiff and light). Since the heat capacity reaches 95% of its maximum value at T the Debye temperature gives you a rough idea of when you can away with just using the equipartition theorem. When you can't, Debye's formula usually gives a good, but not great, estimate of the heat capacity over the full range of temperatures. To do better, we'd have to do a lot more work, taking into account the fact that the speed of a phonon depends on its wavelength, polarization, and direction of travel with respect to the crystal axes. That kind of analysis belongs in a book on solid state physics. 1.0 0.8 0.6
Cv 3Nk
- - - - Debye model . - . - . - . _. Einstein model
0.4 0.2
~~~~~~~~~~·······~~~~···············~~~··~~··~~T/To
0.2
0.4
0.6
0.8
1.0
Figure 7.29. The Debye prediction for the heat capacity of a solid, with the prediction of the Einstein model plotted for comparison. The constant € in the Einstein model ha.'3 been chosen to obtain the best agreement with the Debye model at high temperatures. Note that the Einstein curve is much flatter than the Debye curve at low temperatures.
7.5
Debye Theory of Solids
Problem 7.57. Fill in the steps to derive equations 7.112 and 7.117. Problem 7.58. The speed of sound in copper is 3560 Use this value to calculate its theoretical Debye temperature. Then determine the experimental Debye temperature from Figure 7.28, and compare. Problem 7.59. Explain in some detail why the three graphs in Figure 7.28 all intercept the vertical axis in about the same place, whereas their slopes differ considerably. Problem 7.60. Sketch the heat capacity of copper as a function of temperature from 0 to 5 K, showing the contributions of lattice vibrations and conduction electrons separately. At what temperature are these two contributions equal? Problem 7.61. The heat capacity of liquid below 0.6 K is proportional to T 3 , with the measured value Cv/Nk = (T/4.67 K)3. This behavior suggests that the dominant excitations at low temperature are long-wavelength phonons. The only important difference between phonons in a liquid and phonons in a solid is that a liquid cannot transmit transversely polarized waves-sound waves must be longitudinaL The speed of sound in liquid 4He is 238 mis, and the density is 0.145 g/cm 3 . From these numbers, calculate the phonon contribution to the heat capacity of 4He in the low-temperature limit, and compare to the measured value. Problem 7.62. Evaluate the integrand in equation 7.112 as a power series in x, keeping terms through x4. Then carry out the integral to find a more accurate ex­ pression for the energy in the high-temperature limit. Differentiate this expression to obtain the heat capacity, and use the result to estimate the percent deviation of C v from 3Nk at T = and T = Problem 7.63. Consider a two-dimensional solid, such as a stretched drumhead or a layer of mica or graphite. Find an expression (in terms of an integral) for the thermal energy of a square chunk of this material of area A = L 2 , and evaluate the result approximately for very low and very high temperatures. Also find an expression for the heat capacity, and use a computer or a calculator to plot the heat capacity as a function of temperature. Assume that the material can only vibrate perpendicular to its own plane, i.e., that there is only one 'polarization.' Problem 7.64. A ferromagnet is a material (like iron) that magnetizes sponta­ neously, even in the absence of an externally applied magnetic field. This happens because each elementary dipole has a strong tendency to align parallel to its neigh­ bors. At T 0 the magnetization of a ferromagnet has the maximum possible value, with all dipoles perfectly lined up; if there are N atoms, the total magneti­ zation is typically rv2fLBN, where fLB is the Bohr magneton. At somewhat higher temperatures, the excitations take the form of spin waves, which can be visualized classically as shown in Figure 7.30. Like sound waves, spin waves are quantized: Each wave mode can have only integer multiples of a basic energy unit. In analogy with phonons, we think of the energy units as particles, called magnons. Each magnon reduces the total spin of the system by one unit of h/27r, and therefore reduces the magnetization by rv2fLB. However, whereas the frequency of a sound wave is inversely proportional to its wavelength, the frequency of a spin wave is proportional to the square of 1/ A (in the limit of long wavelengths). Therefore, since E hi and p h/ A for any 'particle,' the energy of a magnon is proportional
313
314
Chapter 7
Quantum Statistics
Ground
state:
11111 11 1 1 11
Spin wave: V'
v··
Wavelength Figure 7.30. In the ground state of a ferromagnet, all the elementary dipoles point in the same direction. The lowest-energy excitations above the ground state are spin waves, in which the dipoles precess in a conical motion. A long-wavelength spin wave carries very little energy, because the difference in direction between neighboring dipoles is very small. to the square of its momentum. In analogy with the energy-momentum relation for an ordinary nonrelativistic particle, we can write E = p2 /2m*, where m* is a constant related to the spin-spin interaction energy and the atomic spacing. For iron, m* turns out to equal 1.24 x 10 29 kg, about 14 times the mass of an electron. Another difference between magnons and phonons is that each magnon (or spin wave mode) has only one possible polarization. (a) Show that at low temperatures, the number of magnons per unit volume in a three-dimensional ferromagnet is given by
Nm _ (2m*kT)3/21°O VI - 27r h2 dx. V o e X -1 Evaluate the integral numerically. (b) Use the result of part (a) to find an expression for the fractional reduction in magnetization, (M(O) - M(T»/M(O). Write your answer in the form (T/TO)3/2, and estimate the constant To for iron.
(c) Calculate the heat capacity due to magnetic excitations in a ferromagnet at low temperature. You should find Cv / Nk = (T /Td 3/ 2 , where Tl differs from To only by a numerical constant. Estimate Tl for iron, and compare the magnon and phonon contributions to the heat capacity. (The Debye temperature of iron is 470 K.)
(d) Consider a two-dimensional array of magnetic dipoles at low tempera­ ture. Assume that each elementary dipole can still point in any (three­ dimensional) direction, so spin waves are still possible. Show that the integral for the total number of magnons diverges in this case. (This re­ sult is an indication that there can be no spontaneous magnetization in such a two-dimensional system. However, in Section 8.2 we will consider a different two-dimensional model in which magnetization does occur.)
7.6
Bose-Einstein Condensation
7.6 Bose-Einstein Condensation The previous two sections treated bosons (photons and phonons) that can be cre­ ated in arbitrary numbers-whose total number is determined by the condition of thermal equilibrium. But what about more 'ordinary' bosons, such as atoms with integer spin, whose number is fixed from the outset? I've saved this case for last because it is more difficult. In order to apply the Bose-Einstein distribution we'll have to determine the chemical potential, which (rather than being fixed at zero) is now a nontrivial function of the density and temperature. Determining /-L will require some careful analysis, but is worth the trouble: We'll find that it behaves in a most peculiar way, indicating that a gas of bosons will abruptly 'condense' into the ground state as the temperature goes below a certain critical value. It's simplest to first consider the limit T ---+ O. At zero temperature, all the atoms will be in the lowest-energy available state, and since arbitrarily many bosons are allowed in any given state, this means that every atom will be in the ground state. (Here again, when I say simply 'state' I mean a single-particle state.) For atoms confined to a box of volume V = L 3 , the energy of the ground state is
h2
EO
2
2
2
3h
2
(7.118)
= 8mL2 (1 + 1 + 1 ) = 8mL2'
which works out to a very small energy provided that L is macroscopic. At any temperature, the average number of atoms in this state, which I'll call No, is given by the Bose-Einstein distribution:
(7.119) When T is sufficiently low, No will be quite large. In this case, the denominator of this expression must be very small, which implies that the exponential is very close to 1, which implies that the exponent, (EO - /-L)/kT, is very small. We can therefore expand the exponential in a Taylor series and keep only the first two terms, to obtain No
1 = --------------1+
(EO -
/-L) / kT - 1
kT EO -
/-L
(when No
»
1).
(7.120)
The chemical potential /-L, therefore, must be equal to EO at T = 0, and just a tiny bit less than EO when T is nonzero but still sufficiently small that nearly all of the atoms are in the ground state. The remaining question is this: How low must the temperature be, in order for No to be large? The general condition that determines /-L is that the sum of the Bose-Einstein distribution over all states must add up to the total number of atoms, N:
N=' ~
1
e(Es-J-t)/kT - 1 .
(7.121)
all s
In principle, we could keep guessing values of /-L until this sum works out correctly
315
316
Chapter 7
Quantum Statistics
(and repeat the process for each value of T). In practice, it's usually easier to convert the sum to an integral: N =
10' '9 (E) -,.----,---,1-_-1 dE.
(7.122)
This approximation should be valid when kT » EO, so that the number of terms that contribute significantly to the sum is large. The function g(E) is the density of states: the number of single-particle states per unit energy. For spin-zero bosons confined in a box of volume V, this function is the same as what we used for electrons in Section 7.3 (equation 7.51) but divided by 2 because now there is only one spin orientation: 2 (21Tm)3/2 (7.123)
g(E) =
vJE.
ft h2
Figure 7.31 shows graphs of the density of states, the Bose-Einstein distribution (drawn for f-L slightly less than zero), and the product of the two, which is the distribution of particles as a function of energy. Unfortunately, the integral 7.122 cannot be performed analytically. Therefore we must guess values of f-L until we find one that works, doing the integral numer­ 0, which should ically each time. The most interesting (and easiest) guess is f-L work (to a good approximation) at temperatures that are low enough for No to be large. Plugging in f-L 0 and changing variables to x E/kT gives ~ (21Tm ft h 2
N
=
)3/2 V roo 10
2 (21TmkT)3/2 V h2
vic dE e€/kT -
1 (7.124)
roo vxdx.
10
eX - 1
The integral over x is equal to 2.315; combining this number with the factor of 2/ ft yields the formula N
2.612 (
21TmkT)3/2 h2 V.
(7.125)
This result is obviously wrong: Everything on the right-hand side is independent of states
Bose-Einstein distribution
Particle distribution
x
+------------------€
h-----~~~~-----€
kT
+-----+----~4-----~€
kT
Figure 7.31. The distribution of bosons as a function of energy is the product of two functions, the den&ity of states and the Bose-Einstein distribution.
7.6
Bose-Einstein Condensation
of temperature except T, so it says that the number of atoms depends on the temperature, which is absurd. In fact, there can be only one particular temperature for which equation 7.125 is correct; I'll call this temperature Tc: N
= 2.612 (
27rmkTc h2
)3/2 V,
or
kTc
h2 ) (N)2/3 V .
= 0.527 ( 27rm
(7.126)
But what's wrong with equation 7.125 when T f=. T c? At temperatures higher than T c , the chemical potential must be significantly less than zero; from equation 7.122 you can see that a negative value of J.L will yield a result for N that is smaller than the right-hand side of equation 7.125, as desired. At temperatures lower than T c , on the other hand, the solution to the paradox is more subtle; in this case, replacing the discrete sum 7.121 with the integral 7.122 is invalid. Look carefully at the integrand in equation 7.124. As E goes to zero, the density of states (proportional to y'E) goes to zero while the Bose-Einstein distribution blows up (in proportion to 1/ E). Although the product is an integrable function, it is not at all clear that this infinite spike at E = 0 correctly represents the sum 7.121 over the actual discretely spaced states. In fact , we have already seen in equation 7.120 that the number of atoms in the ground state can be enormous when J.L ~ 0, and this enormous number is not included in our integral. On the other hand, the integral should correctly represent the number of particles in the vast majority of the states, away from the spike, where E » EO. If we imagine cutting off the integral at a lower limit that is somewhat greater than EO but much less than kT, we'll still obtain approximately the same answer,
Nexcited
= 2.612 (
27rmkT)3/2 V h2
(when T < T c).
(7.127)
This is then the number of atoms in excited states, not including the ground state. (Whether this expression correctly accounts for the few lowest excited states, just above the ground state in energy, is not completely clear. If we assume that the difference between N and the preceding expression for Nexcited is sufficiently large, then it follows that J.L must be much closer to the ground-state energy than to the energy of the first excited state, and therefore that no excited state contains anywhere near as many atoms as the ground state. However, there will be a narrow range of temperatures, just below T c , where this condition is not met. When the total number of atoms is not particularly large, this range of temperatures might not even be so narrow. These issues are explored in Problem 7.66.) So the bottom line is this: At temperatures higher than T c , the chemical poten­ tial is negative and essentially all of the atoms are in excited states. At temperatures lower than T c , the chemical potential is very close to zero and the number of atoms in excited states is given by equation 7.127; this formula can be rewritten more simply as T )3 / 2 (7.128) Nexcited = ( Tc N
317
318
Chapter 7
Quantum Statistics
The rest of the atoms must be in the ground state, so No
=
N
Nexcited
(7.129)
Figure 7.32 shows a graph of No and Nexcited as functions of temperature; Fig­ ure 7.33 shows the temperature dependence of the chemical potential. The abrupt accumulation of atoms in the ground state at temperatures below Tc is called Bose-Einstein condensation. The transition temperature Tc is called the condensation temperature, while the ground-state atoms themselves are called the condensate. Notice from equation 7.126 that the condensation tem­ perature is (aside from the factor of 2.612) precisely the temperature at which the quantum volume (vQ (h 2/27rmkT)3/2) equals the average volume per particle (V/ N). In other words, if we imagine the atoms being in wavefunctions that are as localized in space as possible (as in Figure 7.4), then condensation begins to occur
N
4-=-------------------~----------------------~T
Figure 7.32. Number of atoms in the ground state (No) and in excited states, for an ideal Bose gas in a three-dimensional box. Below Tc the number of atoms in excited states is proportional to T3/2.
-0.4
-0.8+ Figure 7.33. Chemical potential of an ideal Bose gas in a three-dimensional box. Below the condensation temperature, J.L differs from zero by an amount that is too small to show on this scale. Above the condensation temperature J.L be­ comes negative; the values plotted here were calculated numerically as described in Problem 7.69.
7.6
Bose-Einstein Condensation
just as the wavefunctions to overlap significantly. (The condensate atoms themselves have wavefunctions that occupy the entire container, which I won't try to draw.) Numerically, the condensation temperature tUrns out to be very small in all realistic experimental situations. However, it's not as low as we might have guessed. If you put a single particle into a box of volume V, it's reasonably likely to be found in the ground state only when kT is of order EO or smaller (so that the excited states, which have energies of 2Eo and are significantly less probable). However, if you put a large number of identical bosons into the same box, you can get most of them into the ground state at temperatures only somewhat less than T e , which is much higher: From equations 7.118 and 7.126 we see that kTe is greater than EO by a factor of order N 2 / 3 . The hierarchy of energy scales-(EO - f.L) « EO « kTe-is depicted schematically in 7.34.
Single-particle states
//
1111111111111111111111111111111111111111111111
III
€
kTe
Figure 7.34. Schematic representation of the energy scales involved in Bose­ Einstein condensation. The short vertical lines mark the energies of various single­ particle states. (Aside from growing closer together (on average) with increasing energy, the locations of these lines are not quantitatively accurate.) The conden­ sation temperature (times k) is many times larger than the spacing between the lowest energy levels, while the chemical potential, when T < T e , is only a tiny amount below the ground-state energy.
Real-World Examples
Bose-Einstein condensation of a gas of weakly interacting atoms was first achieved in 1995, using rubidium-87. * In this experiment, roughly 104 atoms were confined (using the laser cooling and trapping technique described in Section 4.4) in a volume of order 10- 15 . A large fraction of the atoms were observed to condense into the ground state at a temperature of about 10- 7 K, a hundred times than the temperature at which a single isolated atom would have a good chance of being in the ground state. 7.35 shows the velocity distribution of the atoms in this experiment, at temperatures above, just below, and far below the condensation temperature. As of 1999, Bose-Einstein condensation has also been achieved with dilute gases of atomic sodium, lithium, and hydrogen. * For a beautiful description of this experiment see Carl E. Wieman, 'The Richtmyer Memorial Lecture: Bose-Einstein Condensation in an Ultracold Gas,' American Journal of Physics 64, 847-855 (1996). ­
319
320
Chapter 7
T
Quantum Statistics
= 200 nK
T = 100 nK
T-;::;:,O
Figure 7.35. Evidence for Bose-Einstein condensation of rubidium-87 atoms. These images were made by turning off the magnetic field that confined the atoms, letting the gas expand for a moment, and then shining light on the expanded cloud to map its distribution. Thus, the positions of the atoms in these images give a measure of their velocities just before the field was turned off. Above the conden­ sation temperature (left) , the velocity distribution is broad and isotropic, in accord with the Maxwell-Boltzmann distribution. Below the condensation temperature (center), a substantial fraction of the atoms fall into a small, elongated region in velocity space. These atoms make up the condensate; the elongation occurs because the trap is narrower in the vertical direction, causing the ground-state wavefunction to be narrower in position space and thus wider in velocity space. At the lowest temperatures achieved (right) , essentially all of the atoms are in the ground-state wavefunction. From Carl E. Wieman, American Journal of Physics 64, 854 (1996).
Bose-Einstein condensation also occurs in systems where particle interactions are significant, so that the quantitative treatment of this section is not very accu­ rate. The most famous example is liquid helium-4, which forms a superfluid phase, with essentially zero viscosity, at temperatures below 2.17 K (see Figure 5.13). More precisely, the liquid below this temperature is a mixture of normal and su­ perfluid components, with the superfluid becoming more predominant as the tem­ perature decreases. This behavior suggests that the superfluid component is a Bose-Einstein condensate; indeed, a naive calculation, ignoring interatomic forces, predicts a condensation temperature only slightly greater than the observed value (see Problem 7.68). Unfortunately, the superfluid property itself cannot be under­ stood without accounting for interactions between the helium atoms. If the superfluid component of helium-4 is a Bose-Einstein condensate, then you would think that helium-3, which is a fermion , would have no such phase. And indeed, it has no superfluid transition anywhere near 2 K. Below 3 millikelvin, however, 3He turns out to have not one but two distinct superfluid phases.* How *These phases were discovered in the early 1970s. To achieve such low temperatures the experimenters used a helium dilution refrigerator (see Section 4.4) in combination with the cooling technique described in Problem 5.34.
7.6
Bose-Einstein Condensation
is this possible for a system of fermions? It turns out that the 'particles' that condense are actually pairs of 3He atoms, held together by the interaction of their nuclear magnetic moments with the surrounding atoms. * A pair of fermions has integer spin and is therefore a boson. An analogous phenomenon occurs in a su­ perconductor, where pairs of electrons are held together through interactions with the vibrating lattice of ions. At low temperature these pairs 'condense' into a superconducting state, yet another example of Bose-Einstein condensation. t
Why Does it Happen? Now that I've shown you that Bose-Einstein condensation does happen, let me return to the question of why it happens. The derivation above was based entirely on the Bose-Einstein distribution function-a powerful tool, but not terribly intuitive. It's not hard, though, to gain some understanding of this phenomenon using more elementary methods. Suppose that, instead of a collection of identical bosons, we have a collection of N distinguishable particles all confined inside a box. (Perhaps they're all painted different colors or something.) Then, if the particles don't interact with each other, we can treat each one of them as a separate system using Boltzmann statistics. At temperature T, a given particle has a decent chance of occupying any single­ particle state whose energy is of order kT, and the number of such states will be quite large under any realistic conditions. (This number is essentially equal to the single-particle partition function, Zl.) The probability of the particle being in the ground state is therefore very small, namely llZl. Since this conclusion applies separately to each one of the N distinguishable particles, only a tiny fraction of the particles will be found in the ground state. There is no Bose-Einstein condensation. It's useful to analyze this same situation from a different perspective, treating the entire system all at once, rather than one particle at a time. From this view­ point, each system state has its own probability and its own Boltzmann factor. The system state with all the particles in the ground state has a Boltzmann factor of 1 (taking the ground-state energy to be zero for simplicity), while a system state with total energy U has a Boltzmann factor of e- UjkT . According to the conclusion of the previous paragraph, the dominant system states are those for which nearly all of the particles are in excited states with energies of order kT; the total system energy is therefore U rv NkT, so the Boltzmann factor of a typical system state is something like e- NkT jkT = e- N . This is a very small number! How can it be that the system prefers these states, rather than condensing into the ground state with its much larger Boltzmann factor? The answer is that while any particular system state with energy of order NkT is highly improbable, the number of such states is so huge that taken together they *For an overview of the physics of both isotopes of liquid helium, see Wilks and Betts (1987). tFor review articles on Bose-Einstein condensation in a variety of systems, see A. Griffin, D. W. Snoke, and S. Stringari, eds., Bose-Einstein Condensation (Cambridge University Press, Cambridge, 1995).
321
322
Chapter 7
Quantum Statistics Identical bosons
Distinguishable particles
Ivoel0
o '--------' D~ ~
Excited states (E ' kT)
....- Ground state (E = 0)
---Jiooo­
Figure 7.36. When most particles are in excited states, the Boltzmann factor for the entire system is always very small (of order e- N). For distinguishable particles, the number of arrangements among these states is so large that system states of this type are still very probable. For identical bosons, however, the number of arrangements is much smaller.
are quite probable after all (see Figure 7.36). The number of ways of arranging N distinguishable particles among Zl single-particle states is zf, which overwhelms the Boltzmann factor e- N provided that Zl » 1. Now let's return to the case of identical bosons. Here again, if essentially all the particles are in single-particle states with energies of order kT, then the system state has a Boltzmann factor of order . But now, the number of such system states is much smaller. This number is essentially the number of ways of arranging N indistinguishable particles among Zl single-particle states, which is mathematically the same as the number of ways of arranging N units of energy among Zl oscillators in an Einstein solid: number of ) ( system states
I'V
(N+Zl-1) N
I'V
{(eZ1/N)N (eN/Z1)Zl
when Zl
::» Nj
when Zl« N.
(7.130)
When the number of available single-particle states is much larger than the number of bosons, the combinatoric factor is again large enough to overwhelm the Boltz­ mann factor , so system states with essentially all the bosons in excited states will again predominate. On the other hand, when the number of available single­ particle states is much smaller than the number of bosons, the combinatoric factor is not large enough to compensate for the Boltzmann factor, so these system states, even all taken together, will be exponentially improbable. (This last conclusion is not quite clear from looking at the formulas, but here is a simple numerical exam­ 100 and Zl = 25, a system state with all the bosons in excited ple: When N states has a Boltzmann factor of order e- lOO 4 x 10- 44 , while the number of such system states is only G~6) 3 x 1025 .) In general, the combinatoric factor will be sufficiently large to get about one boson, on average, into each available excited state. Any remaining bosons condense into the ground state, because of the way the Boltzmann factor favors system states with lower energy. So the explanation of Bose-Einstein condensation lies in the combinatorics of counting arrangements of identical particles: Since the number of distinct ways of arranging identical ~articles among the excited states is relatively small, the ground
7.6
Bose-Einstein Condensation
state becomes much more favored than if the particles were distinguishable. You may still be wondering, though, how we know that bosons of a given species are truly identical and must therefore be counted in this way. Or alternatively, how do we know that the fundamental assumption, which gives all distinct states (of the system plus its environment) the same statistical weight, applies to systems of identical bosons? These questions have good theoretical answers, but the an­ swers require an understanding of quantum mechanics that is beyond the scope of this book. Even then, the answers are not completely airtight-there is still the possibility that some undiscovered type of interaction may be able to distinguish supposedly identical bosons from each other, causing a Bose-Einstein condensate to spontaneously evaporate. So far, the experimental fact is that such interactions do not seem to exist. Let us therefore invoke Occam's Razor and conclude, if only tentatively, that bosons of a given species are truly indistinguishable; as David Griffiths has said,* even God cannot tell them apart. Problem 7.65. Evaluate the integral in equation 7.124 numerically, to confirm the value quoted in the text. Problem 7.66. Consider a collection of 10,000 atoms of rubidium-87, confined inside a box of volume (10- 5 m)3. (a) Calculate EO, the energy of the ground state. £.ix:pr1ess your answer in both joules and electron-volts.)
(b) Calculate the condensation temperature, and compare kTe to
EO.
(c) Suppose that T 0.9Te. How many atoms are in the ground state? How close is the chemical potential to the ground-state energy? How many atoms are in each of the (threefold-degenerate) first excited states?
(d) Repeat parts (b) and (c) for the case of 10 6 atoms, confined to the same volume. Discuss the conditions under which the number of atoms in the ground state will be much greater than the number in the first excited state. Problem 7.67. In the first achievement of Bose-Einstein condensation with atomic hydrogen, t a gas of approximately 2 x 1010 atoms was trapped and cooled until its peak density was 1.8 x 10 14 atoms/cm 3 . Calculate the condensation tem­ perature for this system, and compare to the measured value of 50 J.lK. Problem 7.68. Calculate the condensation temperature for liquid helium-4, pre­ Tt:>1'1r11na that the liquid is a gas of noninteracting atoms. Compare to the observed temperature of the superfluid transition, 2.17 K. (The density of liquid helium-4 is 0.145 g/cm 3 .) Problem 7.69. If you have a computer system that can do numerical integrals, it's not particularly difficult to evaluate J.l for T > Te. (a) As usual when solving a problem on a computer, it's best to start by putting everything in terms of dimensionless variables. So define t = T ITc)
* Introduction to Quantum Mechanics (Prentice-Hall, Englewood Cliffs, NJ, 1995), page 179. tDale G. Fried et al., Physical. Review Letters 81, 3811 (1998).
323
324
Chapter 7
Quantum Statistics c = J-t/kTc , and x = E/kTc. Express the integral that defines J-t, equation 7.122, in terms of these variables. You should obtain the equation 2.315
=
1
.jXdx
= o eX ( _ c )/t -1 .
c when T = 2Tc is approx­ imately -0.8. Plug in these values and check that the equation above is approximately satisfied.
(b) According to Figure 7.33, the correct value of
(c ) Now vary J-t, holding T fixed, to find the precise value of J-t for T = 2Tc . Repeat for values of T /Tc ranging from 1.2 up to 3.0, in increments of 0.2. Plot a graph of J-t as a function of temperature.
Problem 7.70. Figure 7.37 shows the heat capacity of a Bose gas as a function of temperature. In this problem you will calculate the shape of this unusual graph.
(a) Write down an expression for the total energy of a gas of N bosons confined to a volume V, in terms of an integral (analogous to equation 7.122).
(b) For T < Tc you can set J-t = O. Evaluate the integral numerically in this case, then differentiate the result with respect to T to obtain the heat capacity. Compare to Figure 7.37.
(c) Explain why the heat capacity must approach ~Nk in the high-T limit. (d) For T > Tc you can evaluate the integral using the values of J-t calculated in Problem 7.69. Do this to obtain the energy as a function of temperature, then numerically differentiate the result to obtain the heat capacity. Plot the heat capacity, and check that your graph agrees with Figure 7.37. 2.0
1.5 Cv Nk
1.0
0.5
0.5
1.0
1.5
2.0
2.5
3.0
T/Tc
Figure 7.37. Heat capacity of an ideal Bose gas in a three-dimensional box.
Problem 7.71. Starting from the formula for Cv derived in Problem 7.70(b), calculate the entropy, Helmholtz free energy, and pressure of a Bose gas for T < Tc. Notice that the pressure is independent of volume; how can this be the case?
7.6
Bose-Einstein Condensation
Problem 7.72. For a gas of particles confined inside a two-dimensional box, the density of states is constant, independent of E (see Problem 7.28). Investigate the behavior of a gas of noninteracting bosons in a two-dimensional box. You should find that the chemical potential remains less than zero as long as T is significantly greater than zero, and hence that there is no abrupt condensation of particles into the ground state. Explain how you know that this is the case, and describe what does happen to this system as the temperature decreases. What property must g( E) have in order for there to be an abrupt Bose­ Einstein condensation? Problem 7.73. Consider a gas of N identical spin-O bosons confined by an isotropic three-dimensional harmonic oscillator potential. (In the rubidium ex­ periment discussed above, the confining potential was actually harmonic, though not isotropic.) The energy levels in this potential are E nhl, where n is any nonnegative integer and I is the classical oscillation frequency. The degeneracy of level n is (n + l)(n + 2)/2. (a) Find a formula for the density of states, g(E), for an atom confined by this potential. (You may assume n » 1.)
(b) Find a formula for the condensation temperature of this system, in terms of the oscillation
I.
(c) This potential confines particles inside a volume of roughly the cube of the oscillation amplitude. The oscillation amplitude, in turn, can be estimated by the particle's total energy kT) equal to the potential energy of the 'spring.' Making these and neglecting all factors of 2 and 7r and so on, show that your answer to part (b) is roughly equivalent to the formula derived in the text for the condensation temperature of bosons confined inside a box with rigid walls.
Problem 7.74. Consider a Bose gas confined in an isotropic harmonic trap, as in the previous problem. For this system, because the energy level structure is much simpler than that of a three-dimensional box, it is feasible to carry out the sum in equation 7.121 numerically, without approximating it as an integral. * (a) Write equation 7.121 for this system as a sum over energy taking degeneracy into account. Replace T and /-t with the dimensionless variables t = kTlhl and c /-tlhl.
(b) Program a computer to calculate this sum for any given values of t and c. 15 provided Show that, for N = 2000, equation 7.121 is satisfied at t that c = -10.534. (Hint: You'll need to include approximately the first 200 energy levels in the sum.) (c ) For the same parameters as in part (b), plot the number of particles in each energy level as a function of energy.
(d) Now reduce t to 14, and the value of c until the sum 2000. Plot the number of particles as a function of energy.
equals
(e) Repeat part (d) for t 13, 12, 11, and 10. You should find that the required value of c increases toward zero but never quite reaches it. Discuss the results in some detail. *This problem is based on an article by Martin Ligare, American Journal of Physics 66, 185-190 (1998).
325
326
Chapter 7
Quantum Statistics
Problem 7.75. Consider a gas of noninteracting spin-O bosons at high tempera­ tures, when T »Te. (Note that 'high' in this sense can still mean below 1 K.) (a) Show that, in this limit, the Bose-Einstein distribution function can be written approximately as - nBE -_ e -(E-j.L)/kT [1
+ e -(E-j.L)/kT + ..]
(b) Keeping only the terms shown above, plug this result into equation 7.122 to derive the first quantum correction to the chemical potential for a gas of bosons. (c) Use the properties of the grand free energy (Problems 5.23 and 7.7) to show that the pressure of any system is given by P = (kT jV) In Z, where Z is the grand partition function. Argue that, for a gas of noninteracting particles, In Z can be computed as the sum over all modes (or single-particle states) of In Zi, where Zi is the grand partition function for the ith mode. (d) Continuing with the result of part (c), write the sum over modes as an integral over energy, using the density of states. Evaluate this integral explicitly for a gas of noninteracting bosons in the high-temperature limit, using the result of part (b) for the chemical potential and expanding the logarithm as appropriate. When the smoke clears, you should find
again neglecting higher-order terms. Thus, quantum statistics results in a lowering of the pressure of a boson gas, as one might expect. (e) Write the result of part (d) in the form of the virial expansion introduced in Problem 1.17, and read off the second virial coefficient, B(T). Plot the predicted B(T) for a hypothetical gas of noninteracting helium-4 atoms.
(f) Repeat this entire problem for a gas of spin-1j2 fermions. (Very few mod­ ifications are necessary.) Discuss the results, and plot the predicted virial coefficient for a hypothetical gas of noninteracting helium-3 atoms.
Ten percent or more of a complete stellar inventory consists of white dwarfs, just sitting there, radiating away the thermal (kinetic) energy of their carbon and oxygen nuclei from underneath very thin skins of hydrogen and helium. They will continue this uneventful course until the universe recontracts, their baryons decay, or they collapse to black holes by barrier penetration. (Likely time scales for these three outcomes are 10 14 , 1033 , and 10 1076 -years for the first two and for the third one it doesn't matter.) -Virginia Trimble, SLAC Beam Line 21, 3 (fall, 1991).
8
Systems of Interacting Particles
An ideal system, in statistical mechanics, is one in which the particles (be they molecules, electrons, photons, phonons, or magnetic dipoles) do not exert significant forces. on each other. All of the systems considered in the previous two chapters were 'ideal' in this sense. But the world would be a boring place if everything in it ·were ideal. Gases would never condense into liquids, and no material would magnetize spontaneously, for example. So it's about time we considered some nonideal systems. Predicting the behavior of a nonideal system, consisting of many mutually in­ teracting particles, is not easy. You can't just break the system down into lots of independent subsystems (particles or modes), treat these subsystems one at a time, and then sum over subsystems as we did in the previous two chapters. Instead you have to treat the whole system all at once. Usually this means that you can't calculate thermodynamic quantities exactly- you have to resort to approximation. Applying suitable approximation schemes to various systems of interacting particles has become a major component of modern statistical mechanics. Moreover, analo­ gous approximation schemes are widely used in other research fields, especially in the application of quantum mechanics to multiparticle systems. In this chapter I will introduce just two examples of interacting systems: a gas of weakly interacting molecules, and an array of magnetic dipoles that tend to align parallel to their neighbors. For each of these systems there is an approx­ imation method (diagrammatic perturbation theory and Monte Carlo simulation, respectively) that not only solves the problem at hand, but has also proved useful in tackling a much wider variety of problems in theoretical physics. *
*The two sections of this chapter are independent of each other; feel free to read them in either order. Also, aside from a few problems, nothing in this chapter depends on Chapter 7. 327
328
Chapter 8
Systems of Interacting Particles
8.1 Weakly Interacting Gases In Section 5.3 we made a first attempt at understanding nonideal gases, using the van der Waals equation. That equation is very successful qualitatively, even predicting the condensation of a dense gas into a liquid. But it is not very accurate quantitatively, and its connection to fundamental molecular interactions is tenuous at best. So, can we do better? Specifically, can we predict the behavior of a nonideal gas from first principles, using the powerful tools of statistical mechanics? The answer is yes, but it's not easy. At least at the level of this book, a funda­ mental calculation of the properties of a nonideal gas is feasible only in the limit of low density, when the interactions between molecules are still relatively weak. In this section I'll carry out such a calculation, ultimately deriving a correction to the ideal gas law that is valid in the low-density limit. This approach won't help us understand the liquid-gas phase transformation, but at least the results will be quantitatively accurate within their limited range of validity. In short, we're trading generality for accuracy and rigor. The Partition Function As always, we begin by writing down the partition function. Taking the viewpoint of Section 2.5 and Problem 6.51, let us characterize the 'state' of a molecule by its position and momentum vectors. Then the partition function for a single molecule is Zl
:3 /
3
3
(8.1)
d rd p
where the single integral sign actually represents six integrals, three over the posi­ tion components (denoted d3 r) and three over the momentum components (denoted d3 p). The region of integration includes all momentum vectors, but only those posi­ tion vectors that lie within a box of volume V. The factor of 1/h 3 is needed to give us a unit less number that counts the independent wavefunctions. For simplicity I've omitted any sum over internal states (such as rotational states) of the molecule. For a single molecule with no internal degrees of freedom, equation 8.1 is equiv­ alent to what I wrote in Section 6.7 for an ideal gas (as shown in Problem 6.51). For a gas of N identical molecules, the corresponding expression is easy to write down but rather frightening to look at: /d3 rl .. d3 rN d3PI Z = 1h1 3N
..
d3PN
(8.2)
Now there are 6N integrals, over the position and momentum components of all N molecules. There are also N factors of 1/ h 3 , and a prefactor of 1/N! to account for the indistinguishability of identical molecules. The Boltzmann factor contains the total energy U of the entire system. If this were an ideal gas, then U would just be a sum of kinetic energy terms,
2m
+
2m
+ .. +
2m
(8.3)
8.1
Weakly Interacting Gases
For a nonideal gas, though, there is also potential energy, due to the interactions between molecules. Denoting the entire potential energy as Upot , the partition function can be written as
Z = -1 N!
1
J
d3 rl .. d3 rN d3PI
..
d3 PN e -f31-11 P
2
/2m ••• e -f3I-N P
(8.4) Now the good news is, the 3N momentum integrals are easy to evaluate. Because the potential energy depends only on the positions of the molecules, not on their momenta, each momentum Pi appears only in the kinetic energy Boltzmann factor , and the integral over this momentum can be evaluated exactly as for an ideal gas, yielding the same result: (8.5) Assembling N of these factors gives us
(8.6)
where Zideal is the partition function of an ideal gas, equation 6.85. Thus, our task is reduced to evaluating the rest of this expression, (8.7) called the configuration integral (because it involves an integral over all config­ urations, or positions, of the molecules). The Cluster Expansion
In order to write the configuration integral more explicitly, let me assume that the potential energy of the gas can be written as a sum of potential energies due to interactions between pairs of molecules: Upot
= =
UI2
+ UI3 + .. + UIN + U23 + .. + UN-I,N
L
Uij'
(8.8)
pairs
Each term Uij represents the potential energy due to the interaction of molecule i with molecule j, and I'll assume that it depends only on the distance between these two molecules, Iii - fj I. This is a significant simplification. For one thing, I'm neglecting any possible dependence of the potential energy on the orientation of a molecule. For another, I'm neglecting the fact that when two molecules are close together they distort each other, thus altering the interaction of either of them with
329
330
Chapter 8
SV<:l:t:An1Q
of Interacting Particles
a third molecule. Still, this 'simplification' doesn't make the configuration integral look any prettier; we now have l
Zc
vN /
3 d3rl' ·d rN
n
(8.9)
paIrs
where the IT symbol denotes a product over all distinct pairs i, j. Eventually we'll need to assume an explicit formula for the function Uij. For now, though, all we need to know is that it goes to zero as the distance between molecules i and j becomes large. Especially in a low-density gas, practically all pairs of molecules will be far enough apart that Uij « kT, and therefore the Boltzmann factor e-{3uij is extremely close to 1. With this in mind, the next step is to isolate the deviation of each Boltzmann factor from 1, by writing
(8.10) which defines a new quantity iij, called the Mayer J-function. The product of all these Boltzmann factors is then
II
pairs
II (1 + iij) (8.11)
pairs
If we imagine multiplying out all these factors, the first term will just be 1. Then there will be a bunch of terms with just one i-function, then a bunch with two i-functions, and so on:
(8.12) pairs
pairs
distinct pairs
Plugging this expansion back into the configuration integral yields
1/
d3r, .. d3rN
(1 + I: jij +
I:
lij!.'
+ . -).
(8.13)
paIrs
Our hope is that the terms in this series will become less important as they contain more i-functions, so we can get away with evaluating only the first term or two. The very first term in equation 8.13, with no i-functions, is easy to evaluate: Each d3 r integral yields a factor of the volume of the box, so
V1N / d3 rl
..
d3 r N (1)
1.
(8.14)
In each of the terms with one i-function, all but two of the integrals yield trivial factors of V; for instance, 1 1 / d3rl .. d3rN !I2 = V N V N - 2 j'd3rl d3r2!I2 =
:2 /
(8.15)
d3rl d3r2 !I2,
8.1
Weakly Interacting Gases
because !I2 depends only on f't and is. Actually, since it doesn't matter which molecules we call 1 and 2, everyone of the terms with one i-function is exactly equal to this one, and the sum of all of them is equal to this expression times the number of distinct pairs, N(N - 1)/2: 1N(N 2 V2
(8.16)
Before going on, I'd like to introduce a pictorial abbreviation for this expression, which will also give us a physical interpretation of it. The picture is simply a pair of dots, representing molecules 1 and 2, connected by a line, representing the interaction between these molecules: (8.17)
The rules for translating the picture into the formula are as follows: 1. Number the dots starting with 1, and for each dot i, write down the expression
(l/V) J d3Ti. Multiply by N for the first dot, N - 1 for the second dot, N - 2 for the third dot, and so on.
2. For a line connecting dots i and j, write down a factor iij. 3. Divide by the symmetry factor of the diagram, which is the number of ways of numbering the dots without changing the corresponding product of i-functions. (Equivalently, the symmetry factor is the number of permutations of dots that leave the diagram unchanged.) For the simple diagram in equation 8.17, these rules give precisely the expression written; the symmetry factor is 2, because !I2 = 121. Physically, this diagram represents a configuration in which only two molecules are interacting with each other. Now consider the terms in the configuration integral (8.13) with two i-functions. Each of these terms involves two pairs of molecules, and these pairs could have one molecule in common, or none. In a term in which the pairs share a molecule, all but three of the integrals give trivial factors of V; the number of such terms is N(N - 1)(N 2)/2, so the sum of these terms is equal to the diagram
!
' . =
1 N (N 2
1) (N - 2) Jd3 V3
d3 Tl
d3 T2
T3
i
j 12 23·
(8.18)
This diagram represents a configuration in which one molecule simultaneously in­ teracts with two others. In a term in which the pairs do not share a molecule there are four left-over integrals; the number of such terms is N(N -1)(N - 2)(N - 3)/8, so the sum of these terms is
(I I)
1 N (N - 1) (N
= '8
V4
2) (N - 3) Jd 3
Tl
d3
T2
d3
T3
d3
T4
i 12J34· f
(8.19)
This diagram represents two simultaneous interactions between pairs of molecules. For either diagram, the rules given above yield precisely the correct expression.
331
332
Chapter 8
Systems of Interacting Particles
By now you can probably guess that the entire configuration integral can be written as a sum of diagrams:
Zc = 1
+I +1 +(I I) +A +~ +n +(11)+(1 I 1)+····
(8.20)
Every possible diagram occurs exactly once in this sum, with the constraints that every dot must be connected to at least one other dot, and that no pair of dots can be connected more than once. I won't try to prove that the combinatoric factors work out exactly right in all cases, but they do. This representation of the configuration integral is an example of a diagrammatic perturbation series: The first term, 1, represents the 'ideal' case of a gas of noninteracting molecules; the remaining terms, represented by diagrams, depict the interactions that 'perturb' the system away from the ideal limit. We expect that the simpler diagrams will be more important than the more complicated ones, at least for a low-density gas in which simultaneous interactions of large numbers of molecules should be rare. Although you would never want to calculate the more complicated diagrams, they still give a way to visualize interactions involving arbitrary numbers of molecules. But even for a low-density gas, we can't get away with keeping only the first couple of terms in the diagrammatic expansion of Zc. We'll soon see that even the simplest two-dot diagram evaluates to a number much larger than 1, so the subseries (8.21) does not converge until after many terms, when the symmetry factors grow to be quite large. Physically, this is because simultaneous interactions of many isolated pairs are very common when N is large. Fortunately, though, this sum can be simplified. At least as a first approximation, we can set N = N 1 = N 2 Then, because a disconnected diagram containing n identical subdiagrams gets a symmetry factor of n! in the denominator, this series is simply (8.22) In other words, the disconnected diagrams gather themselves into a simple expo­ nential function of the basic connected diagram. But that's not all. At the next level of approximation, when we keep terms that are smaller by a factor of N, it turns out that the series 8.22 even includes the connected diagram 8.18 (see Prob­ lem 8.7). Similar cancellations occur among the more complicated diagrams, with the end result being that Zc can be written as the exponential of the sum of only those diagrams that are connected, and that would remain connected if any single dot were to be removed: (8.23)
8.1
Weakly Interacting Gases
This formula isn't quite exact, but the terms that are omitted go to zero in the thermodynamic limit, N --+ 00 with N IV fixed. Similarly, at this stage it is valid to set N N - 1 N - 2 = ' in all subsequent calculations. Unfortunately, the general proof of this formula is beyond the scope of this book. Each diagram in equation 8.23 is called a cluster, because it represents a cluster of simultaneously interacting molecules. The formula itself is called the cluster expansion for the configuration integral. The cluster expansion is a well-behaved series: For a low-density gas, a cluster diagram with more dots is always smaller than one with fewer dots. Now let's put the pieces back together. Recall from equation 8.6 that the whole partition function for the gas is equal to the configuration integral times the parti­ tion function for an ideal gas: (8.24) In order to compute the pressure, we really want to know the Helmholtz free energy,
F
= -kTlnZ
-kTlnZideal - kTln
(8.25)
We computed Zideal in Section 6.7; plugging in that result and the cluster expansion for Zc, we obtain
F=-NkTln(:VJ -kT(! + A
+
tl +.-}
(8.26)
The pressure is therefore p
(8.27)
Thus, if we can evaluate some of the cluster diagrams explicitly, we can improve upon the ideal gas law. Problem 8.1. For each of the diagrams shown in equation 8.20, write down the corresponding formula in terms of i-functions, and explain why the symmetry factor gives the correct overall coefficient. Problem 8.2. Draw all the diagrams, connected or disconnected, representing terms in the configuration integral with four factors of Iii' You should find 11 diagrams in total, of which five are connected. Problem 8.3. Keeping only the first two diagrams in equation 8.23, and approx­ imating N : : : : N 1:::::::: N - 2 : : : : ', expand the exponential in a power series through the third power. Multiply each term out, and show that all the numer­ ical coefficients precisely the correct symmetry factors for the disconnected diagrams. Problem 8.4. Draw all the connected diagrams containing four dots. There are six diagrams in total; be careful to avoid drawing two diagrams that look superficially different but are actually the same. Which of the diagrams would remain connected if any single dot were removed?
333
334
Chapter 8
Systems of Interacting Particles
The Second Virial Coefficient Let's now consider just the simplest, two-dot diagram: (8.28) Because the f-function depends only on the distance between the two molecules, let me define '1 '12 - rl and change variables in the second integral from f2 to
r:
I
(8.29)
where f(r) =
1
e-{3u(r)
(8.30)
and u(r) is the potential energy due to the interaction of any pair of molecules, as a function of the distance between their centers. To evaluate the integral over rwe'll have to do a bit of work, but we can say one thing about it already: The result will be some intensive quantity that is independent of rl and of V. This is because f (r) goes to zero when r is only a few times larger than the size of a molecule, and the chance of rl being within this distance of the wall of the box is negligible. So whatever the value of this integral, the remaining integral over rl will simply give a factor of V: (8.31)
Having all the V's written explicitly, we can plug into equation 8.27 for the pressure:
a
(1 J 2
N kT +kT- -NV av 2 V
p
NkT V-
3 ) driI2(r)
J3 21 N V2 d r f( r) + ..
+ ..
2
kT .
(8.32)
NkT( INJ V 1- 2 V d r f(r) + .. 3
)
It is conventional to write this series in the form of the virial expansion, intro­ duced in Problem 1.17:
p
(8.33)
We are now in a position to compute the second virial coefficient, B(T):
B(T) =
-~
J
3
d r f(r).
(8.34)
To evaluate this last triple integral I'll use spherical coordinates, where the measure of the integral is d3 r
= (dr)(rd())(r sin()d¢)
(8.35)
8.1
Weakly Interacting Gases
(see Figure 7.11). The integrand J(r) is independent of the angles () and cp, so the angular integrals give simply 47r, the area of a unit sphere. Even more explicitly, then, (8.36) This is as far as we can go without an explicit formula for the intermolecular potential energy, U (r ) . To model the intermolecular potential energy realistically, we want a function that is weakly attractive at large distances and strongly repulsive at short distances (as discussed in Section 5.3). For molecules with no permanent electric dipole mo­ ment, the long-distance force arises from a spontaneously fluctuating dipole moment in one molecule, which induces a dipole moment in the other and then attracts it; one can show that this force varies as 1/ r 7 , so the corresponding potential energy varies as 1/ The exact formula used to model the repulsive part of the potential turns out not to be critical; for mathematical convenience a term proportional to 1/r 12 is most often used. The sum of these attractive and repulsive terms gives what is called the Lennard-Jones 6-12 potential; with appropriately named constants it can be written
r6.
U(r) = Uo
ro)12 - 2 (ro)6] --;: . [( --;:
(8.37)
Figure 8.1 shows a plot of this function, and of the corresponding Mayer J-function for three different temperatures. The parameter represents the distance between the molecular centers when the energy is a minimum-very roughly, the diameter of a molecule. The parameter Uo is the maximum depth of the potential well. If you plug the Lennard-Jones potential function into equation 8.36 for the second virial coefficient and integrate numerically for various temperatures, you
ro
f(r)
u(r)
kT
= O.5uo
kT= Uo
ro
kT
2ro
= 2uo
r r -uo
-1+----­
Figure 8.1. Left: The Lennard-Jones intermolecular potential function, with a strong repulsive region at small distances and a weak attractive region at somewhat larger distances. Right: The corresponding Mayer f-function, for three different temperatures.
335
336
Chapter 8
Systems of Interacting Particles
o -2 • He (ro
B(T)
---:;r o
o
2.97
A, uo = 0.00057 eV)
A, uo = 0.00315 eV) 3.86 A, uo = 0.0105 eV)
Ne (ro = 3.10
Ar (ro ' H2 (ro = 3.29 A, Uo 0.0027 eV) ~ C02 (ro = 4.35 A, uo = 0.0185 eV) • CH4 (ro 4.36 A, Uo 0.0127 eV)
-8 -10 -12 1
10
100
kT/uo Figure 8.2. Measurements of the second virial coefficients of selected gases, com­ pared to the prediction of equation 8.36 with u(r) given by the Lennard-Jones function. Note that the horizontal axis is logarithmic. The constants ro and Uo have been chosen separately for each gas to give the best fit. For carbon dioxide, the poor fit is due to the asymmetric shape of the molecules. For hydrogen and helium, the discrepancies at low temperatures are due to quantum effects. Data from J. H. Dymond and E. B. Smith, The Viria1 Coefficients of Pure Gases and Mixtures: A Critical Compilation (Oxford University Press, Oxford, 1980).
obtain the solid curve shown in Figure 8.2. At low temperatures the integral of the j-function is dominated by its large upward spike at ro, that is, by the attractive potential well. A positive average j leads to a negative virial coefficient, indicating that the pressure is lower than that of an ideal gas. At high temperatures, however, the negative potential well shows up less dramatically in j, so the integral is dom­ inated by the negative portion of j that comes from the repulsive short-distance interaction; then the virial coefficient is positive and the pressure is greater than that of an ideal gas. At very high temperatures, though, this effect is lessened somewhat by the ability of high-energy molecules to partially penetrate into the region of repulsion. Figure 8.2 also shows experimental values of B(T) for several gases, plotted after choosing ro and Uo for each gas to obtain the best fit to the theoretical curve. For most simple gases, the shape of B(T) predicted by the Lennard-Jones poten­ tial agrees very well with experiments. (For molecules with strongly asymmetric shapes and/or permanent dipole moments, other potential functions would be more appropriate, while for the light gases hydrogen and helium, quantum-mechanical effects become important at low temperatures. *) This agreement tells us that the * As shown in Problem 7.75, the contribution of quantum statistics to B(T) should be negative for a gas of bosons like hydrogen or helium. However, there is another quantum effect not considered in that problem. The de Broglie wave of one molecule cannot pen­
8.1
Weakly Interacting Gases
Lennard-Jones potential is a reasonably accurate model of intermolecular interac­ tions, while the values of TO and Uo that are used to fit the data give us quantitative information about the sizes and polarizabilities of the molecules. Here, as always, statistical mechanics works in both directions: From our theoretical understanding of microscopic physics, it gives us predictions for the bulk behavior of large numbers of bulk matter, it lets us of molecules; and from measurements of the infer a deal about the molecules themselves. In principle, we could nOw go on to compute the third and higher virial coef­ ficients of a low-density gas using the cluster expansion. In practice, though, we would encounter two major problems. The first is that the remaining diagrams in equation 8.27 are very difficult to evaluate explicitly. But worse, the second problem is that when clusters of three or more molecules interact, it is often not valid to write the potential energy as a sum of pair-wise interactions as I did in equation 8.8. Both of these problems can be overcome, * but a proper calculation of the third virial coefficient is far beyond the scope of this book. Problem 8.5. By changing variables as in the text, express the diagram in equa­ tion 8.18 in terms of the same integral as in equation 8.31. Do the same for the last two diagrams in the first line of equation 8.20. Which diagrams cannot be written in terms of this basic integral? Problem 8.6. You can estimate the size of any diagram by realizing that J(r) is of order lout to a distance of about the diameter of a molecule, and J : : : : beyond that. a three-dimensional integral of a product of 1's will generally give a result that is of the order of the volume of a molecule. Estimate the sizes of all the diagrams shown explicitly in equation 8.20, and explain why it was necessary to rewrite the series in exponential form.
°
Problem 8.7. Show that, if you don't make too many approximations, the ex­ ponential series in equation 8.22 includes the three-dot diagram in equation 8.18. There will be some leftover terms; show that these vanish in the thermodynamic limit. Problem 8.8. Show that the nth virial coefficient depends on the in equation 8.23 that have n dots. Write the third virial coefficient, G(T), in terms of an integral of J-functions. Why it would be difficult to carry out this ULV'F,'<>
etrate the physical volume of another, so when the average de Broglie (£Q) becomes larger than the physical diameter (ro), there is more repulsion than there would be The effect of quantum statistics would become dominant only at still lower temperatures, when £Q is comparable to the average distance between molecules. Both hy­ drogen and helium have the inconvenient habit of liquefying before such low temperatures are reached. For a thorough discussion of virial coefficients, including quantum effects, see O. Hirschfelder, Charles F. and R. Byron Bird, Molecular Theory of Gases and Liquids (Wiley, New York, 1954). *For a discussion of the computation of the third virial coefficient and comparisons between theory and experiment, -see Reichl (1998).
337
338
Chapter 8
Systems of Interacting Particles
Problem 8.9. Show that the Lennard-Jones potential reaches its minimum value at r ro, and that its value at this minimum is -Uo. At what value of r does the potential equal zero? Problem 8.10. Use a computer to calculate and plot the for a gas of molecules interacting via the Lennard-Jones kT/ Uo ranging from 1 to 7. On the same graph, plot the in Problem 1.17, choosing the parameters ro and Uo so as
second virial coefficient potential, for values of data for nitrogen given to obtain a good fit.
Problem 8.11. Consider a gas of 'hard spheres,' which do not interact at all unless their separation distance is less than ro, in which case their interaction energy is infinite. Sketch the Mayer I-function for this gas, and compute the second virial coefficient. Discuss the result briefly. Problem 8.12. Consider a gas of molecules whose interaction energy u(r) is infinite for r < ro and negative for r > ro, with a minimum value of -Uo. Suppose further that kT » uo, so you can approximate the Boltzmann factor for r > ro using eX ~ 1 + x. Show that under these conditions the second virial coefficient has the form B(T) b (a/kT), the same as what you found for a van der Waals gas in Problem 1.17. Write the van der Waals constants a and b in terms of ro and u(r), and discuss the results briefly. Problem 8.13. Use the cluster expansion to write the total energy of a monatomic nonideal gas in terms of a sum of diagrams. Keeping only the first diagram, show that the energy is approximately
N 2.3 N kT + V . 271' 10roo r 2 u (r) e -(3u(r) dr. 2
U~
Use a computer to evaluate this integral numerically, as a function of T, for the Lennard-Jones potential. Plot the temperature-dependent part of the correction term, and explain the shape of the graph physically. Discuss the correction to the heat capacity at constant volume, and compute this correction numerically for argon at room temperature and atmospheric pressure. Problem 8.14. In this section I've formulated the cluster expansion for a gas with a fixed number of particles, using the 'canonical' formalism of Chapter 6. A somewhat cleaner approach, however, is to use the 'grand canonical' formalism introduced in Section 7.1, in which we allow the system to exchange particles with a much larger reservoir.
(a) Write down a formula for the grand partition function (Z) of a weakly interacting gas in thermal and diffusive equilibrium with a reservoir at fixed T and /1. Express Z as a sum over all possible particle numbers N, with each term involving the ordinary partition function Z (N).
(b) Use equations 8.6 and 8.20 to express Z(N) as a sum of diagrams, then carry out the sum over N, diagram by diagram. Express the result as a sum of similar diagrams, but with a new rule 1 that associates the expression ()./vQ) J d3ri with each dot, where). = e(3/-L. Now, with the awkward factors of N (N 1) .. taken care of, you should find that the sum of all diagrams organizes itself into exponential form, resulting in the formula
8.2
The
Model of a Ferromagnet
Note that the exponent contains all connected diagrams, including those that can be disconnected by removal of a line. (c) Using the properties of the grand partition function (see Problem 7.7), find diagrammatic expressions for the average number of particles and the pressure of this gas. (d) Keeping only the first diagram in each sum, express N(p,) and pep,) in terms of an of the Mayer J-function. Eliminate p, to obtain the same result for the pressure (and the second virial coefficient) as derived in the text. (e) Repeat part (d) keeping the three-dot diagrams as well, to obtain an ex­ pression for the third virial coefficient in terms of an integral of J-functions. You should find that the A-shaped diagram cancels, leaving only the trian­ gle diagram to contribute to G(T).
8.2 The Ising Model of a Ferromagnet In an ideal paramagnet, each microscopic magnetic dipole responds only to the external magnetic field (if any); the dipoles have no inherent tendency to point parallel (or antiparallel) to their immediate neighbors. In the real world, however, atomic dipoles are influenced by their neighbors: There is always some preference for neighboring dipoles to align either parallel or antiparallel. In some materials this preference is due to ordinary magnetic forces between the dipoles. In the more dramatic examples (such as iron), however, the alignment of neighboring dipoles is due to complicated quantum-mechanical effects involving the Pauli exclusion principle. Either way, there is a contribution to the energy that is greater or less, depending on the relative of neighboring dipoles. When neighboring dipoles parallel to each other, even in the absence of an external field, we call the material a ferromagnet (in honor of iron, the most familiar example). When neighboring dipoles align antiparallel , we call the ma­ terial an antiferromagnet (examples include Cr, NiO, and FeO). In this section I'll discuss ferromagnets, although most of the same ideas can also be applied to antiferromagnets. The long-range order of a ferromagnet manifests itself as a net nonzero magneti­ zation. Raising the temperature, however, causes random fluctuations that decrease the overall magnetization. For every ferromagnet there is a certain critical temper­ ature, called the Curie temperature, at which the net magnetization becomes zero (when there is no external field). Above the Curie temperature a ferromagnet becomes a paramagnet. The Curie temperature of iron is 1043 K, considerably higher than that of most other ferromagnets. Even below the Curie temperature, you may not notice that a piece of iron is magnetized. This is because a chunk of iron ordinarily divides itself into domains that are microscopic in size but still contain billions of atomic dipoles. Within each domain the material is magnetized, but the magnetic field created by all the dipoles in one domain neighboring domains a tendency to magnetize in the opposite direction. (Put two ordinary bar magnets side by side and you'll see why.) Because there are so many domains, with about as many pointing one
339
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Chapter 8
Systems of Interacting Particles
way as another, the material as a whole has no net magnetization. However, if you heat a chunk of iron in the presence of an external magnetic field, this field can overcome the interaction between domains and cause essentially all the dipoles to line up parallel. Remove the external field after the material has cooled to room temperature and the ferromagnetic interaction prevents any significant realigning. You then have a 'permanent' magnet. In this section I'd like to model the behavior of a ferromagnet, or rather, of a single domain within a ferromagnet. I'll account for the tendency of neighboring dipoles to align parallel to each other, but I'll neglect any long-range magnetic interactions between dipoles. To simplify the problem further, I'll assume that the material has a preferred axis of magnetization, and that each atomic dipole can only point parallel or antiparallel to this axis. * This simplified model of a magnet is called the Ising model, after Ernst Ising, who studied it in the 1920s. t Figure 8.3 shows one possible state of a two-dimensional Ising model on a 10 x 10 square lattice. Notation: Let N be the total number of atomic dipoles, and let Si be the current state of the ith dipole, with the convention that Si = 1 when this dipole is pointing up, and Si = -1 when this dipole is pointing down. The energy due to the interaction of a pair of neighboring dipoles will be -E when they are parallel and +E when they are antiparallel. Either way, we can write this energy as -ESiSj,
t ~ ~ ~ t ~ ~ ~ ~ t t t t ~ ~ t ~ t t t t t t t t t ~ t t t t t t t ~ ~ t t
t t t t t t
t t t t
t t t t t
~ ~ ~ ~ ~ t ~ t
t t t ~
t ~ t t t ~ t t t ~ ~ t t t t t ~ ~ t t ~ ~ ~
t t t t t ~
t t ~
Figure 8.3. One of the many possible states of a two-dimensional Ising model on a 10 x 10 square lattice.
t t t
*1 should point out that in many respects this model is not an accurate representation of a real ferromagnet. Even if there really is a preferred axis of magnetization, and even if the elementary dipoles each have only two possible orientations along this direction, quantum mechanics is more subtle than this naive model. Because we do not measure the orientation of each individual dipole, it is only the sum of their magnetic moments that is quantized-not the moment of each individual particle. At low temperatures, for instance, the relevant states of a real ferromagnet are long-wavelength 'magnons' (described in Problem 7.64), in which all the dipoles are nearly parallel and a unit of opposite alignment is spread over many dipoles. The Ising model therefore does not yield accurate predictions for the low-temperature behavior of a ferromagnet. Fortunately, it turns out to be much more accurate near the Curie temperature. tFor a good historical overview of the Ising model see Stephen G. Brush, 'History of the Lenz-Ising Model,' Reviews of Modern Physics 39, 883-893 (1967).
8.2
The Ising Model of a J:<elTOjmagw3t
assuming that dipoles i and j are neighbors. Then the total energy of the from all the nearest-neighbor interactions is U
-E
L
SiSj'
(8.38)
neighboring pairs i,j
To predict the thermal behavior of this system, we should try to calculate the partition function, Z=Le-(:JU, (8.39) {sd
where the sum is over all possible sets of dipole alignments. For N dipoles, each with two possible alignments, the number of terms in this sum is 2N , usually a very large number. Adding up all the terms by brute force is not going to be practical. Problem 8.15. For a two-dimensional Ising model on a square lattice, each dipole on the edges) has four 'neighbors'-above, below, and right. (Diagonal are normally not included.) What is the total energy (in terms of E) for the particular state of the 4 x 4 square lattice shown in 8.4?
~ ~it ~ Figure 8.4. One particular state of an Ising model on a 4 x 4 square lattice (Problem 8.15).
t t t t ~ t t t ~ ~ ~ t
Problem 8.16. Consider an Ising model of 100 elementary dipoles. Suppose you wish to calculate the partition function for this system, using a computer that can compute one billion terms of the partition function per second. How long must you wait for the answer? Problem 8.17. Consider an Ising model of just two elementary dipoles, whose mutual interaction energy is ±E. Enumerate the states of this system and write down their Boltzmann factors. Calculate the partition function. Find the proba­ bilities of the dipoles parallel and antiparallel , and plot these probabilities as a function of kT/ E. Also calculate and plot the average energy of the system. At what temperatures are you more likely to find both dipoles up than to find one up and one down?
Exact Solution in One Dimension So far I haven't specified how our atomic dipoles are to be arranged in space, or how I should many nearest neighbors each of them has. To simulate a real arrange them in three dimensions on a crystal lattice. But I'll start with a much simpler with the dipoles strung out along a one-dimensional line (see Figure 8.5). Then each has only two nearest neighbors, and we can actually carry out the partition sum exactly. For a one-dimensional Ising model (with no external magnetic field), the energy is (8.40)
341
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Chapter 8
Systems of Interacting Particles
t!!tt
t
i=1
2
3
4
5
N
=
-1
-1
1
1
1
Si
1
Figure 8.5. A one-dimensional Ising model with N elementary dipoles.
and the partition function can be written (8.41)
where each sum runs over the values -1 and 1. Notice that the final sum, over 8N, is = 2coshpE, (8.42) + regardless of whether 8N-l is +1 or -1. With this sum done, the sum over 8N-l can now be evaluated in the same way, then the sum over 8N-2, and so on down to 82, yielding N 1 factors of 2 cosh PE. The remaining sum over 81 gives another factor of 2, so the partition function is (8.43)
where the last approximation is valid when N is large. So we've got the partition function. Now what? Well, let's find the average energy as a function of temperature. By a straightforward calculation you can show that U
8
8p InZ = -NEtanhpE,
(8.44)
which goes to -NE as T 0 and to 0 as T --t 00. Therefore the dipoles must be randomly aligned at high temperature (so that half the neighboring pairs are parallel and half are antiparallel), but lined up parallel to each other at T = 0 (achieving the minimum possible energy). If you're getting a sense of deja vu, don't be surprised. Yes indeed, both Z and U for this system are exactly the same as for a two-state paramagnet, if you replace the magnetic interaction energy J-LB with the neighbor-neighbor interaction energy E. Here, however, the dipoles like to line up with each other, instead of with an external field. Notice that, while this system does become more ordered (less random) as its temperature decreases, the order sets in gradually. The behavior of U as a function of T is perfectly smooth, with no abrupt transition at a nonzero critical temper­ ature. Apparently, the one-dimensional Ising model does not behave like a real three-dimensional ferromagnet in this crucial respect. Its tendency to magnetize is not great enough, becaus~ each dipole has only two nearest neighbors.
8.2
The Ising Model of a Ferromagnet
So our next step should be to consider Ising models in higher dimensions. Un­ fortunately, though, such models are much harder to solve. The two-dimensional Ising model on a square lattice was first solved in the 1940s by Lars Onsager. On­ sager evaluated the exact partition function as N ~ 00 in closed form, and found that this model does have a critical temperature, just like a real ferromagnet. Be­ cause Onsager's solution is extremely difficult mathematically, I will not attempt to present it in this book. In any case, nobody has ever found an exact solution to the three-dimensional Ising model. The most fruitful approach from here, therefore, is to give up on exact solutions and rely instead on approximations. Problem 8.18. Starting from the partition function, calculate the average energy of the one-dimensional Ising model, to verify equation 8.44. Sketch the average energy as a function of temperature.
The Mean Field Approximation Next I'd like to present a very crude approximation, which can be used to 'solve' the Ising model in any dimensionality. This approximation won't be very accu­ rate, but it does give some qualitative insight into what's happening and why the dimensionality matters. Let's concentrate on just a single dipole, somewhere in the middle of the lattice. I'll label this dipole i, so its alignment is Si which can be -lor 1. Let n be the number of nearest neighbors that this dipole has:
n=
II
12
in in in in in
one dimension;
two dimensions (square lattice);
three dimensions (simple cubic lattice); three dimensions (body-centered cubic lattice); three dimensions (face-centered cubic lattice).
(8.45)
Imagine that the alignments of these neighboring dipoles are temporarily frozen, but that our dipole i is free to point up or down. If it points up, then the interaction energy between this dipole and its neighbors is Ej
=
-E
L
Sneighbor
= -Ens,
(8.46)
neighbors
where
s is
the average alignment of the neighbors (see Figure 8.6). Similarly, if
Figure 8.6. The four neighbors of this particular dipole have an average s value of (+1-3)/4 = -1/2. If the central dipole points up, the energy due to its interactions with its neighbors is +2t, while if it points down, the energy is -2t.
t
t t + t
343
344
Chapter 8
Systems of Interacting Particles
dipole i points down, then the interaction energy is (8.47) The partition function for just this dipole is therefore =
2 cosh(pEns) ,
(8.48)
and the average expected value of its spin alignment is 2 sinh (pEns) 2 cosh(pEns)
tanh(pEns).
(8.49)
Now look at both sides of this equation. On the left is Si, the thermal average value of the alignment of any typical dipole (except those on the edge of the lattice, which we'll neglect). On the right is S, the average of the actual instantaneous alignments of this dipole's n neighbors. The idea of the mean field approxima­ tion is to assume (or pretend) that these two quantities are the same: Si S. In other words, we assume that at every moment, the alignments of all the dipoles are such that every neighborhood is 'typical'-there are no fluctuations that cause the magnetization in any neighborhood to be more or less than the expected ther­ mal average. (This approximation is similar to the one I used to derive the van der Waals equation in Section 5.3. There it was the density, rather than the spin alignment, whose average value was not allowed to vary from place to place within the system.) In the mean field approximation, then, we have the relation
s = tanh(pEns),
(8.50)
where s is now the average dipole alignment over the entire system. This is a transcendental equation, so we can't just solve for s in terms of pEn. The best approach is to plot both sides of the equation and look for a graphical solution (see Figure 8.7). Notice that the larger the value of pEn, the steeper the slope of the hyperbolic tangent function near s = O. This means that our equation can have either one solution or three, depending on the value of pEn. {3€n
<1
{3€n
>1
'­ Stable I
.
-+----+~--_E_---+----+-111--
S
Stable solution
Unstable
Figure 8.7. Graphical solution of equation 8.50. The slope of the tanh function at the origin is {3€n. When this quantity is less than 1, there is only one solution, at S 0; when this quantity is greater than 1, the s = 0 solution is unstable but there are also two nontrivial stable solutions.
8.2
The Ising Model of a Ferromagnet
When f3Ert < 1, that is, when kT > rtE, the only solution is at s 0; there is no net magnetization. If a thermal fluctuation were to momentarily increase the value of s, then the hyperbolic tangent function, which dictates what s should be, would be less than the current value of s, so s would tend to decrease back to zero. The solution s = 0 is stable. When f3Ert > 1, that is, when kT < rtE, we still have a solution at s 0 and we also have two more solutions, at positive and negative values of s. But the solution at s 0 is unstable: A small positive fluctuation of s would cause the hyperbolic ta]lg~mt function to exceed the current value of s, driving s to even higher values. The stable solutions are the other two, which are symmetrically located because the has no inherent tendency toward positive or magnetization. Thus, the system will acquire a net nonzero magnetization, which is equally likely to be positive or negative. When a system has a built-in such as this, yet must choose one state or another at low temperatures, we say that the symmetry is spontaneously broken. The critical temperature below which the system becomes magnetized is (8.51 ) proportional to both the neighbor-neighbor interaction energy and to the number of neighbors. This result is no surprise: The more neighbors each dipole has, the greater the tendency of the whole system to magnetize. Notice, though, that even a one-dimensional Ising model should magnetize below a temperature of 2E/k, according to this analysis. Yet we already saw from the exact solution that there is no abrupt transition in the behavior of a one-dimensional model; it magnetizes only as the temperature goes to zero. Apparently, the mean field approximation is no good at all in one dimension. * Fortunately, the accuracy improves as the dimensionality increases. Problem 8.19. The critical temperature of iron is 1043 K. Use this value to make a rough estimate of the dipole-dipole interaction energy E, in electron-volts. Problem 8.20. Use a computer to plot s as a function of kTIE, as predicted by mean field theory, for a two-dimensional Ising model (with a square lattice). Problem 8.21. At T 0, equation 8.50 says that s 1. Work out the first temperature-dependent correction to this value, in the limit pEn » 1. Compare to the low-temperature behavior of a real ferromagnet, treated in Problem 7.64. Problem 8.22. Consider an Ising model in the presence of an external magnetic field B, which gives each dipole an additional energy of if it points up and +MBB if it points down (where MB is the dipole's magnetic moment). Analyze this system using the mean field approximation to find the analogue of equation 8.50. Study the solutions of the equation graphically, and discuss the magnetization of this system as a function of both the external field and the temperature. Sketch the region in the T-B plane for which the equation has three solutions. *There do exist more complicated versions of the mean field approximation that lack this fatal flaw, predicting correctly that the one-dimensional model magnetizes only at T = O. See, for example, Pathriar (1996).
345
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Systems of Interacting Particles
Problem 8.23. The Ising model can be used to simulate other systems besides ferromagnets; examples include antiferromagnets, binary alloys, and even fluids. The Ising model of a fluid is called a lattice gas. We imagine that space is divided into a lattice of sites, each of which can be either occupied by a gas molecule or unoccupied. The system has no kinetic energy, and the only potential energy comes from interactions of molecules on adjacent sites. Specifically, there is a contribution of ~Uo to the energy for each pair of neighboring sites that are both occupied. (a) Write down a formula for the grand partition function for this system, as a function of uo, and J-t.
(b) Rearrange your formula to show that it is identical, up to a multiplicative factor that does not depend on the state of the system, to the ordinary partition function for an Ising ferromagnet in the presence of an external magnetic field B, provided that you make the replacements Uo .... 4E and J-t .... 2J-tBB - 8E. (Note that J-t is the chemical potential of the gas while J-tB is the magnetic moment of a dipole in the magnet.) (c ) Discuss the implications. Which states of the magnet correspond to low­ density states of the lattice gas? Which states of the magnet correspond to high-density states in which the gas has condensed into a liquid? What shape does this model predict for the liquid-gas phase boundary in the P-T plane? Problem 8.24. In this problem you will use the mean field approximation to analyze the behavior of the Ising model near the critical point. (a) Prove that, when x
«
1, tanh x ::::::: x
(b) Use the result of part (a) to find an expression for the magnetization of the Ising model, in the mean field approximation, when T is very close to the critical temperature. You should find M ex (Te T)f3, where f3 (not to be confused with l/kT) is a critical exponent, analogous to the f3 defined for a fluid in Problem 5.55. Onsager's exact solution shows that f3 = 1/8 in two dimensions, while experiments and more sophisticated approximations show that f3 ::::::: 1/3 in three dimensions. The mean field approximation, however, predicts a larger value. (c) The magnetic susceptibility X is defined as X (8M/8B)T' The behavior of this quantity near the critical point is conventionally written as X ex (T where '( is another critical exponent. Find the value of '( in the mean field approximation, and show that it does not depend on whether T is slightly above or slightly below Te. (The exact value of '( in two dimensions turns out to be 7/4, while in three dimensions '( ::::::: 1.24.)
Monte Carlo Simulation Consider a medium-sized, two-dimensional Ising model on a square lattice, with 100 or so elementary dipoles (as shown in Figure 8.3). Although even the fastest computer could never compute the probabilities of all the possible states of this sys­ tem, maybe it isn't necessary to consider all of them-perhaps a random sampling of only a million or so states would be enough. This is the idea of Monte Carlo summation (or integration), a technique named after the famous European gam­ bling center. The procedure is to generate a random sampling of as many states as
8.2
The Ising Model of a Ferromagnet
possible, compute the Boltzmann factors for these states, and then use this random sample to compute the average energy, magnetization, and other thermodynamic quantities. Unfortunately, the procedure just outlined does not work well for the Ising model. Even if we consider as many as one billion states, this is only a tiny fraction-about one in 1021 -of all the states for a modest 10 x 10 lattice. And at low temperatures, when the system wants to magnetize, the important states (with nearly all of the dipoles pointing in the same direction) constitute such a small fraction of the total that we are likely to miss them entirely. Sampling the states purely at random just isn't efficient enough; for this reason it's sometimes called the naive Monte Carlo method. A better idea is to use the Boltzmann factors themselves as a guide during the random generation of a subset of states to sample. A specific algorithm that does this is as follows: Start with any state whatsoever. Then choose a dipole at random and consider the possibility of flipping it. Compute the energy difference, D-.U, that would result from the flip. If D-.U :s; 0, so the system's energy would decrease or remain unchanged, go ahead and flip this dipole to generate the next system state. If b.U > 0, so the system's energy would increase, decide at random whether to flip the dipole, with the probability of the flip being e- llU /kT. If the dipole does not get flipped, then the new system state will be the same as the previous one. Either way, continue by choosing another dipole at random and repeat the process, over and over again, until every dipole has had many chances to be flipped. This algorithm is called the Metropolis algorithm, after Nicholas Metropolis, the first author of a 1953 article* that presented a calculation of this type. This technique is also called Monte Carlo summation with importance sampling. The Metropolis algorithm generates a subset of system states in which low­ energy states occur more frequently than high-energy states. To see in more detail why the algorithm works, consider just two states, 1 and 2, which differ only by the flipping of a single dipole. Let U1 and U2 be the energies of these states, and let us number the states so that U1 :s; U2 . If the system is initially in state 2, then the probability of making a transition to state 1 is liN, simply the probability that the correct dipole will be chosen at random among all the others. If the system is initially in state 1, then the probability of making a transition to state 2 is (1/N)e-(U2-ud/kT, according to the Metropolis algorithm. The ratio of these two transition probabilities is therefore P(l
(1/N)e-(U2- Ud/ kT
P(2
(liN)
(8.52)
simply the ratio of the Boltzmann factors of the two states. If these were the only *N. Metropolis, A. W. Rosenbluth, M. N. Rosenbluth, A. H. Teller, and E. Teller, 'Equation of State Calculations for Fast Computing Machines,' Journal of Chemical Physics 21, 1087-1092 (1953). In this article the authors use their algorithm to calculate the pressure of a two-dimensional gas of 224 hard disks. This rather modest calculation required several days of computing time on what was then a state-of-the-art computer.
347
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Chapter 8
Systems of Int,en:tct:mg Particles
two states available to the system, then the frequencies with which they occur would be in exactly this ratio, as Boltzmann statistics demands. * Next consider two other states, 3 and 4, that differ from 1 and 2 by the flipping of some other dipole. The system can now go between 1 and 2 through the indirect process 1 3 4 2, whose forward and backward rates have the ratio (-t
P(l P(2
(-t
(-t
-----+
3
-----+
4
e-U3/kT e- U4 / kT
-----+
4
-----+
3
e- UdkT e- U3 / kT
-=--:-c-=
(8.53)
again as demanded by Boltzmann statistics. The same conclusion applies to transi­ tions involving any number of steps, and to transitions between states that differ by the flipping of more than one dipole. Thus, the ~1etropolis algorithm does indeed O'prlpr:~~p states with the correct Boltzmann probabilities. Strictly speaking, this conclusion applies only after the algorithm has been running infinitely so that every state has been generated many times. We want to run the algorithm for a relatively short time, so that most states are never generated at all! Under these circumstances we have no guarantee that the subset of states actually will accurately represent the full collection of all system states. In fact, it's hard to even define what is meant by an 'accurate' representation. In the case of the Ising model, our main concerns are that the randomly generated states an accurate picture of the expected energy and magnetization of the system. The most noticeable exception in practice will be that at low temperatures, the Metropolis algorithm will rapidly push the system into a 'metastable' state in which nearly all of the dipoles are parallel to their neighbors. Although such a state is quite probable according to Boltzmann statistics, it may take a very long time for the algorithm to generate other probable states that differ significantly, such as a state in which every dipole is flipped. (In this way the Metropolis algorithm is analogous to what happens in the real world, where a large all possible microstates, and the relaxation time never has time to for achieving true thermodynamic equilibrium can sometimes be very long.) With this limitation in let's now go on and implement the ~1etropolis algorithm. The algorithm can be programmed in almost any traditional computer language, and in many nontraditional languages as well. Rather than singling out one particular language, let me instead present the algorithm in 'pseudocode,' which you can translate into the language of your choice. A pseudocode program for a basic two-dimensional simulation is shown in 8.8. This program produces only graphical output, showing the lattice as an array of colored squares~ one color for dipoles pointing up, another color for dipoles pointing down. Each time a dipole is flipped the color of a square changes, so you can see exactly what sequence of states is being generated. The program uses a two-dimensional array called s (i , j) to store the values of the spin orientations; the indices i and j each go from 1 to the value of size, which can be changed to simulate lattices of different sizes. The temperature T, *When the transition rates between two states have the correct ratio, we say that the transitions are in detailed balance.
8.2
The Ising Model of a Ferromagnet
program ising
Monte Carlo simulation of a 2D Ising model using the Metropolis algorithm
size = 10 T = 2.5 initialize for iteration = 1 to 100*size-2 do i = int(rand*size+1) j = int(rand*size+1) deltaU(i,j,Ediff) if Ediff <= 0 then s(i,j) = -s(i,j) colorsquare Ci , j )
else
if rand < exp(-Ediff/T) then s(i,j) = -s(i,j) colorsquare(i,j)
end if end if next iteration end program
Width of square lattice Temperature in units of Elk
subroutine deltaU(i,j,Ediff)
Main iteration loop Choose a random row number
and a random column number
Compute flU of hypothetical flip
If flipping reduces the energy ..
then flip it!
otherwise the Boltzmann factor
gives the probability of flipping
Now go back and start over ..
Compute flU of flipping a dipole (note periodic boundary conditions)
if i 1 then top = s(size,j) else top = s(i-1,j)
if i size then bottom = s(1,j) else bottom = s(i+1,j)
if j 1 then left = s(i,size) else left = s(i,j-1)
if j size then right = s(i,1) else right = s(i,j+1)
Ediff = 2*s(i,j)*(top+bottom+left+right)
end subroutine subroutine initialize for i = 1 to size for j = 1 to size
if rand < .5 then s(i,j) colorsquare(i,j)
next j next i end subroutine subroutine colorsquare(i,j)
Initialize to a random array
1 else s(i,j)
-1
Color a square according to 8 value (implementation depends on system)
Figure 8.8. A pseudocode program to simulate a two-dimensional Ising model,
using the Metropolis algorithm.
349
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Chapter 8
Systems of Interacting Particles
measured in units of Elk, can also be changed for different runs. After setting these two constants, the program calls the subroutine initialize to assign the initial value of each s randomly. * The heart of the program is the 'main iteration loop,' which executes the Metropolis algorithm 100 times per dipole so that each dipole will have many chances to be flipped. The value 100 can be changed as appropriate. (Note that * represents multiplication, while represents exponentiation.) Within the loop, we first choose a dipole at random; the function rand is assumed to return a random real number between 0 and 1, while int () returns the largest integer less than or equal to its argument. The subroutine deltaU, defined later in the program, computes the energy change upon hypothetically flipping the chosen dipole; this energy change (in units of E) is returned as Ediff. If Ediff is negative or zero, we flip the dipole, while if Ediff is positive, we use it to compute a Boltzmann factor and compare this to a random number to decide whether to flip the dipole. If the dipole gets flipped, we call the subroutine colorsquare to change the color of the corresponding square on the screen. The subroutine deltaU requires further explanation. There is always a problem, when a simulation uses a relatively small lattice, in dealing with 'edge effects.' In the Ising model, dipoles on the edge of the lattice are less constrained to align with their neighbors than are dipoles elsewhere. If we're modeling a very small system whose size is the same as that of our simulated lattice, then we should treat the edges as edges, with fewer neighbors per dipole. But if we're really interested in the behavior of much larger systems, then we should try to minimize edge effects. One way to do this is to make the lattice 'wrap around,' treating the right edge as if it were immediately left of the left edge and the bottom edge as if it were immediately above the top edge. Physically this would be like putting the array of dipoles on the surface of a torus. Another interpretation of this wrapping is to imagine that the lattice is flat and infinite in all directions, but that its state is always perfectly periodic, so that moving up, down, left, or right by a certain amount (the value of size) always takes you to an equivalent place where the dipoles have exactly the same alignments at all times. Based on this latter interpretation, we say that we are using periodic boundary conditions. Back to the subroutine deltaU, notice that it correctly identifies all four nearest neighbors, whether or not the chosen dipole is on an edge. The change in energy upon flipping is then twice the product of s (i, j) with s of the neighbor, summed over the four neighbors. To convert my pseudocode into a real program that runs on a real computer, you first need to pick a computer system and a programming language. The syntax for arithmetic operations, variable assignments, if-then constructions, and for-next loops will vary from language to language, but almost any common programming language should provide easy ways to do these things. Some languages require that A
*In principle, the initial state can by anything. In practice, the choice of initial state can be important if you don't want to wait forever for the system to equilibrate to a 'typical' state. A random initial state works well at high temperatures; a completely magnetized initial state would work better at low temperatures.
8.2
The Ising Model of a Ferromagnet
variables be declared and given a type (such as integer or real) at the beginning of the program. Variables that are accessed both in the main program and in subroutines may require special treatment. The least standardized element of all is the handling of graphics; the contents of the subroutine colorsquare will vary wildly from system to system. Nevertheless, I hope that you will have little trouble implementing this program on your favorite computer and getting it to run. fun: You get to watch the squares con­ Running the ising program is stantly changing colors as the system tries to find states with relatively Boltz­ mann factors. It is tempting, in fact, to imagine that you are watching a simulation of what really happens in a magnet, as the dipoles change their alignments back and forth with the passage of time. Because of this similarity, a Monte Carlo program importance sampling is usually called a Monte Carlo simulation. But please remember that we have made no attempt to simulate the real time-dependent be­ havior of a magnet. Instead we have implemented a 'pseudodynamics,' which flips only one dipole at a time and otherwise ignores the true time-dependent dynamics of the The only realistic property of our pseudodynamics is that it gener­ ates states with probabilities proportional to their Boltzmann factors, just as the real dynamics of a magnet presumably does. Figure 8.9 shows some graphical output from the ising program for a 20 x 20 lattice. The first image shows a random initial state generated by the program, while the remaining images each show the final state at the end of a run of 40,000 iterations (100 per dipole), for various temperatures. Although these snapshots are no substitute for watching the program in action, they do show what a typical state at each temperature looks like. At T 10 the final state is still almost random, with only a slight tendency for dipoles to align with their neighbors. At successively lower temperatures the dipoles tend to form larger and clusters* of positive and negative magnetization until, at T = 2.5, the clusters are about as as the lattice itself. At T = 2 a single cluster has taken over the whole lattice, and we would say that the is 'magnetized.' Small clusters of dipoles will still occasionally flip, but they don't last long; we would have to wait a very long time for the whole lattice to flip to a (just as probable) state of opposite magnetization. The T = 1.5 run happens to have settled into the opposite magnetization, and at this temperature fluctuations of individual dipoles are becoming uncommon. At T 1 we might expect the to magnetize completely and stay that way, and indeed, sometimes it does. About half the time, however, it instead becomes stuck in a metastable state with two domains, one positive and the other negative, as shown in the figure. Based on these results, we can conclude that this system has a critical tem­ perature somewhere between 2.0 and 2.5, in units of Elk. Recall that the mean field approximation predicts a critical temperature of 4Elk~not bad qualitatively, though off by nearly a factor of 2. But a 20 x 20 lattice is really quite small; what *I'm making no attempt here to define a 'cluster' -just look at the pictures and use your intuition. A careful definition of the 'size' of a cluster is given in Prob­ lem 8.29.
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Random initial state
-
T
T
10
3
T
2.5
I I I
~~
I
I­
I
•
I
Ii
i
. I
I
• I
tt# •
•
. I I
-
I I I
. I
I
I
I
T
=::
2
T 1.5
T
=::
1
Figure 8.9. Graphical output from eight runs of the ising program, at succes­ sively lower temperatures. Each black square represents an 'up' dipole and each white square represents a 'down' dipole. The variable T is the temperature in units of Elk.
happens in larger, more realistic simulations? The answer isn't hard to guess. As long as the temperature is sufficiently high, so that the size of a typical cluster is much smaller than the size of the lattice, the behavior of the system is pretty much independent of the lattice size. But a larger lattice allows for larger clusters, so near the critical temperature we should use as large a lattice as possible. With sufficiently long runs with large lattices one can show that the size of the largest clusters approaches infinity at a temperature of 2.27E/k (see Figure 8.10). This, then, is the true critical temperature in the thermodynamic limit. And indeed, this result agrees with Onsager's exact solution
8.2
The Ising Model of a Ferromagnet
Figure 8.10. A typical state generated by the ising program after a few billion iterations on a 400 x 400 lattice at T = 2.27 (the critical temperature). Notice that there are clusters of all possible sizes, from individual dipoles up to the size of the lattice itself.
of the two-dimensional Ising model. Similar simulations have been performed for the three-dimensional Ising model, although this requires much more computer time and the results are harder to dis­ play. For a simple cubic lattice one finds a critical temperature of approximately 4.5E/k, again somewhat less than the prediction of the mean field approximation. The Monte Carlo method can also be applied to more complicated models of ferro­ magnets and to a huge variety of other systems including fluids, alloys, interfaces, nuclei, and subnuclear particles. Problem 8.25. In Problem 8.15 you manually computed the energy of a particular state of a 4 x 4 square lattice. Repeat that computation, but this time apply periodic boundary conditions. Problem 8.26. Implement the ising program on your favorite computer, using your favorite programming language. Run it for various lattice sizes and temper­ atures and observe the results. In particular:
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.
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= 10, 5, 4, 3, and 2.5, for at least 100 iterations per dipole per run. At each temperature make a rough estimate of the size of the largest clusters.
(a) Run the program with a 20 x 20 lattice at T
(b) Repeat part (a) for a 40 x 40 lattice. Are the cluster sizes any different? Explain. (c) Run the program with a 20 x 20 lattice at T = 2, 1.5, and 1. Estimate the average magnetization (as a percentage of total saturation) at each of these temperatures. Disregard runs in which the system gets stuck in a metastable state with two domains.
(d) Run the program with a 10 x 10 lattice at T
2.5. Watch it run for 100,000 iterations or so. Describe and explain the behavior.
(e) Use successively larger lattices to estimate the typical cluster size at tem­ peratures from 2.5 down to 2.27 (the critical temperature). The closer you are to the critical temperature, the larger a lattice you'll need and the longer the program will have to run. Quit when you realize that there are better ways to spend your time. Is it plausible that the cluster size goes to infinity as the temperature approaches the critical temperature? Problem 8.27. Modify the ising program to compute the average energy of the system over all iterations. To do this, first add code to the initialize subroutine to compute the initial energy of the lattice; then, whenever a dipole is flipped, change the energy variable by the appropriate amount. When computing the average energy, be sure to average over all iterations, not just those iterations in which a dipole is actually flipped (why?). Run the program for a 5 x 5 lattice for T values from 4 down to 1 in reasonably small intervals, then plot the average energy as a function of T. Also plot the heat capacity. Use at least 1000 iterations per dipole for each run, preferably more. If your computer is fast enough, repeat for a 10 x 10 lattice and for a 20 x 20 lattice. Discuss the results. (Hint: Rather than starting over at each temperature with a random initial state, you can save time by starting with the final state generated at the previous, nearby temperature. For the larger lattices you may wish to save time by considering only a smaller temperature interval, perhaps from 3 down to 1.5.) Problem 8.28. Modify the ising program to compute the total magnetization (that is, the sum of all the S values) for each iteration, and to tally how often each possible magnetization value occurs during a run, plotting the results as a histogram. Run the program for a 5 x 5 lattice at a variety of temperatures, and discuss the results. Sketch a graph of the most likely magnetization value as a function of temperature. If your computer is fast enough, repeat for a 10 x 10 lattice. Problem 8.29. To quantify the clustering of alignments within an Ising magnet, we define a quantity called the correlation function, c(r). Take any two dipoles i and j, separated by a distance r, and compute the product of their states: SiSj' This product is 1 if the dipoles are parallel and -1 if the dipoles are antiparallel. N ow average this quantity over all pairs that are separated by a fixed distance r, to obtain a measure of the tendency of dipoles to be 'correlated' over this distance. Finally, to remove the effect of any overall magnetization of the system, subtract off the square of the average s. Written as an equation, then, the correlation function is
c(r)
=
-
-2
Si ,
8.2
The Ising Model of a Ferromagnet
where it is understood that the first term averages over all pairs at the fixed distance r. Technically, the averages should also be taken over all possible states of the system, but don't do this yet. (a) Add a routine to the ising program to compute the correlation function for the current state of the lattice, averaging over all pairs separated either vertically or horizontally (but not diagonally) by r units of distance, where r varies from 1 to half the lattice size. Have the program execute this routine periodically and plot the results as a bar graph. (b) Run this program at a variety of temperatures, above, below, and near the critical point. Use a lattice size of at least 20, preferably larger (especially near the critical point). Describe the behavior of the correlation function at each temperature. (c) Now add code to compute the average correlation function over the dura­ tion of a run. (However, it's best to let the system 'equilibrate' to a typical state before you begin accumulating averages.) The correlation length is defined as the distance over which the correlation function decreases by a factor of e. Estimate the correlation length at each temperature, and plot a graph of the correlation length vs. T. Problem 8.30. Modifiy the ising program to simulate a one-dimensional Ising model. (a) For a lattice size of 100, observe the sequence of states generated at various temperatures and discuss the results. According to the exact solution (for an infinite lattice), we expect this system to magnetize only as the tem­ perature goes to zero; is the behavior of your program consistent with this prediction? How does the typical cluster size depend on temperature? (b) Modify your program to compute the average energy as in Problem 8.27. Plot the energy and heat capacity vs. temperature and compare to the exact result for an infinite lattice. (c) Modify your program to compute the magnetization as in Problem 8.28. Determine the most likely magnetization for various temperatures and sketch a graph of this quantity. Discuss. Problem 8.31. Modify the ising program to simulate a three-dimensional Ising model with a simple cubic lattice. In whatever way you can, try to show that this system has a critical point at around T = 4.5. Problem 8.32. Imagine taking a two-dimensional Ising lattice and dividing the sites into 3 x 3 'blocks,' as shown in Figure 8.11. In a block spin transforma­ tion, we replace the nine dipoles in each block with a single dipole, whose state is determined by 'majority rule': If more than half of the original dipoles point up, then the new dipole points up, while if more than half of the original dipoles point down, then the new dipole points down. By applying this transformation to the entire lattice, we reduce it to a new lattice whose width is 1/3 the original width. This transformation is one version of a renormalization group trans­ formation, a powerful technique for studying the behavior of systems near their critical points. *
* For more about the renormalization group and its applications, see Kenneth G. Wilson, 'Problems in Physics with Many Scales of Length,' Scientific American 241, 158-179 (August, 1979).
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Figure 8.11. In a block spin transformation, we replace each block
of nine dipoles with a single dipole whose orientation is determined by
'majority rule.'
(a) Add a routine to the ising program to apply a block spin transformation to the current state of the lattice, drawing the transformed lattice alongside the original. (Leave the original lattice unchanged.) Have the program execute this routine periodically, so you can observe the evolution of both lattices.
(b) Run your modified program with a 90 x 90 original lattice, at a variety of temperatures. After the system has equilibrated to a 'typical' state at each temperature, compare the transformed lattice to a typical 30 x 30 piece of the original lattice. In general you should find that the transformed lattice resembles an original lattice at a different temperature. Let us call this temperature the 'transformed temperature.' When is the transformed temperature greater than the original temperature, and when is it less? (c) Imagine starting with a very large lattice and applying many block spin transformations in succession, each time taking the system to a new effec­ tive temperature. Argue that, no matter what the original temperature, this procedure will eventually take you to one of three fixed points: zero, infinity, or the critical temperature. For what initial temperatures will you end up at each fixed point? [Comment: Think about the implications of the fact that the critical temperature is a fixed point of the block spin transformation. If averaging over the small-scale state of the system leaves the dynamics unchanged, then many aspects of the behavior of this sys­ tem must be independent of any specific microscopic details. This implies that many different physical systems (magnets, fluids, and so on) should have essentially the same critical behavior. More specifically, the different systems will have the same 'critical exponents,' such as those defined in Problems 5.55 and 8.24. There are, however, two parameters that can still affect the critical behavior. One is the dimensionality of the space that the system is in (3 for most real-world systems); the other is the dimensional­ ity of the 'vector' that defines the magnetization (or the analogous 'order parameter') of the system. For the Ising model, the magnetization is one­ dimensional, always along a given axis; for a fluid, the order parameter is also a one-dimensional quantity, the difference in density between liquid and gas. Therefore the behavior of a fluid near its critical point should be the same as that of a three-dimensional Ising model.]
A
Elements of Quantum Mechanics
You don't need to know any quantum mechanics to understand the basic principles of thermal physics. But to predict the detailed thermal properties of specific phys­ ical systems (like a gas of nitrogen molecules or the electrons in a chunk of metal), you do need to know what are the possible 'states' and corresponding energies of these systems. The states and their energies are determined by the principles of quantum mechanics. Even so, you don't need to know much quantum mechanics in order to read this book. At each point in the text where a quantum mechanics result is needed I have summarized that result, in enough detail for the calculation at hand. If you don't care where the result comes from, then you need not read this appendix. At some point, however, you may want to see a more systematic overview of the quantum mechanics that is used in this book. This appendix is intended to provide that overview, whether you choose to read it before or after reading the main text. *
A.I Evidence for Wave-Particle Duality The historical roots of quantum mechanics are intimately entwined with the de­ velopment of statistical mechanics around the turn of the 20th century. Especially important in this history was the breakdown of the equipartition theorem, both for electromagnetic radiation (the 'ultraviolet catastrophe' described in Section 7.4) and for the vibrational energy of solid crystals (evidenced by anomalously low heat capacities at low temperature, as investigated in Problems 3.24 and 3.25 and in *There are many good quantum mechanics textbooks that you may wish to consult for a less superficial treatment of the subject. I especially recommend An Introduction to Quantum Physics by A. P. French and Edwin F. Taylor (Norton, New York, 1978), and Introduction to Quantum Mechanics by David J. Griffiths (Prentice-Hall, Englewood Cliffs, NJ, 1995).
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Section 7.5). But there is also plenty of more direct evidence for quantum mechan­ ics, that is, for the idea that neither a wave model nor a particle model is adequate to understand matter and energy at the atomic scale. In this section I will briefly describe some of this evidence.
The Photoelectric Effect If you shine light on a metal surface, it can knock some electrons out of the metal and send them flying off the surface. This phenomenon is called the photoelectric effect; it is the basic mechanism of video cameras and a variety of other electronic light detectors. To study the photoelectric effect quantitatively, you can put the piece of metal (called a photocathode) into a vacuum tube with another piece of metal (the anode) to catch the ejected electrons. Then you can measure either the voltage that builds up as electrons collect on the anode, or the current produced as these electrons run around the circuit back to the cathode (see Figure A.I). The current is a measure of how many electrons (per unit time) are ejected from the cathode and collected on the anode. Not surprisingly, the current increases if the light source is made more intense: Brighter light ejects more electrons. The voltage, on the other hand, is a measure of the energy that an electron needs to cross the gap between the cathode and the anode. Initially the voltage is zero, but as electrons collect on the anode they create an electric field that pushes other electrons back toward the cathode. Before long the voltage stabilizes at some final value, indicating that no electrons are ejected with sufficient energy to cross. Voltage is just energy per unit charge, so if the final voltage is V, then the maximum kinetic energy of the ejected electrons (as they leave the cathode) must be Kmax = eV, where e is the magnitude of the electron's charge.
Voltmeter
Ammeter
Figure A.I. Two experiments to study the photoelectric effect. When an ideal voltmeter (with essentially infinite resistance) is connected to the circuit, electrons accumulate on the anode and repel other electrons; the voltmeter measures the energy (per unit charge) that an electron needs in order to cross. When an ammeter is connected, it measures the number of electrons (per unit time) that collect on the anode and then circulate back to the cathode.
A.1
Evidence for Wave-Particle Duality
Here's the surprise: The final voltage, and hence the maximum electron kinetic energy, is independent of the brightness of the light source. Brighter light ejects more electrons, but does not give an individual electron any more energy than faint light. On the other hand, the final voltage does depend on the color of the light, that is, on the wavelength (,) or frequency (f = c/ '). In fact, there is a linear relation between the maximum kinetic energy of the ejected electrons and the frequency of the light: (A.l) Kmax = hf - cp, where h is a universal constant called Planck's constant and cp is a constant that depends on the metal. This relation was first predicted in 1905 by Einstein, extrapolating from Planck's earlier explanation of blackbody radiation. Einstein's interpretation of the photoelectric effect is simple: Light comes in tiny bundles or particles, now called photons, each with energy equal to Planck's constant times the frequency of the light: Ephoton
= h f.
(A.2)
A brighter beam of light contains more photons, but the energy of each photon still depends only on the frequency, not on the brightness. When light hits the photocathode, each electron absorbs the energy of just one photon. The constant cp (called the work function) is the minimum energy required to get an electron out of the metal; once the electron is free, the maximum energy it can have is the photon's energy (hf) minus cp. We don't normally notice that light comes in discrete bundles, because the bundles are so small: The value of Planck's constant is only 6.63 x 10- 34 J·s, so visible-light photons have energies of only about two or three electron-volts. A typical light bulb gives off 10 20 photons per second. But the technology needed to detect individual photons (from photomultiplier tubes in physics laboratories to CCD cameras for astronomy) is rather commonplace today. Problem A.I. Photon fundamentals. (a) Show that he
= 1240 eY·nm.
(b) Calculate the energy of a photon with each of the following wavelengths: 650 nm (red light); 450 nm (blue light); 0.1 nm (x-ray); 1 mm (typical for the cosmic background radiation). (c) Calculate the number of photons emitted in one second by a 1-milliwatt red He-Ne laser (, = 633 nm). Problem A.2. Suppose that, in a photoelectric effect experiment of the type described above, light with a wavelength of 400 nm results in a voltage reading of 0.8 Y. (a) What is the work function for this photocathode? (b) What voltage reading would you expect to obtain if the wavelength were changed to 300 nm? What if the wavelength were changed to 500 nm? 600 nm?
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Electron Diffraction If light, which everyone thought was a wave, can behave like a stream of particles, then perhaps it's not so surprising that electrons, which everyone thought were particles, can behave like waves. But let me back up a bit. What do we mean when we say that light behaves like a wave? We don't actually see anything waving (as with water waves or waves on a guitar string). What we can observe is diffraction and interference effects, when light passes through a narrow opening or around a small obstacle. Perhaps the simplest example is two-slit interference, in which monochromatic light from a single source passes through a pair of closely spaced slits and forms a pattern of alternating light and dark spots on a viewing screen some distance away (see Figure A.2). Well, electrons do the same thing: Take a beam of electrons (as in a TV picture tube or an electron microscope) and aim it at a pair of very closely spaced slits. On the viewing screen (a TV screen or some other detector) you get an interference pattern, exactly as with light (see Figure A.3). The wavelength of the electron beam can be determined from the slit spacing and the size of the interference pattern, just as for light. It turns out that the wavelength is inversely proportional to the momentum of the electrons , and the constant of proportionality is Planck's constant:
A= '3:. p
(A.3)
This famous relation was predicted in 1923 by Louis de Broglie. It holds for photons too, and there it is a direct consequence of the Einstein relation E = hf and the relation p = E / c between the energy and momentum of anything that travels at the speed of light. De Broglie guessed correctly that electrons (and all other 'particles' ) have wavelengths that are related to their momenta in the same way. (The Einstein
Figure A.2. In a two-slit interference experiment, monochromatic light (often from a laser) is aimed at a pair of slits in a screen. An interference pattern of dark and light bands app~ars on the viewing screen some distance away.
A.l
Evidence for Wave-Particle Duality
Figure A.3. These images were produced using the beam of an electron micro­ the scope. A positively charged wire was placed in the path of the beam, electrons to bend around either side and interfere as if they had passed through a double slit. The current in the electron beam increases from one image to the next, Sh()Wlllle: that the interference pattern is built up from the statistically distributed flashes of individual electrons. From P. G. Merli, G. F. Missiroli, and G. American .Tournal of Physics 44, 306 (1976).
relation E hf also turns out to apply to electrons and other particles, but this relation is not as useful because the 'frequency' of an electron is not directly measurable. ) The fact that both electrons and photons can act like waves and produce inter­ ference raises some tricky questions. Each individual particle (electron or photon) can land in only one spot on the viewing screen, so if you send the particles the apparatus slowly enough, the pattern builds up gradually, dot by dot, as shown in A.3. Apparently, the place where each particle lands is random, with the probability varying across the screen as determined by the brightness of the final This means that each photon or electron must somehow pass through both slits and then interfere with itself to determine the probability dis­ tribution for where it will finally land. In other words, the particle behaves like a through the slits, and the amplitude of this wave at the location wave when of the screen determines the probabilities that govern the final position. (More prethe probability of landing in a particular place is proportional to the square of the final wave amplitude, just as the brightness of an electromagnetic wave is proportional to the square of the electric field amplitude.) UUi.'-JU.'-U
Problem A.3. Use the Einstein relation E = hf and the relation E that the de Broglie relation A.3 holds for photons.
pc to show
Problem A.4. Use the relativistic definitions of energy and momentum to show that E pc for any particle traveling at the speed of light. (For electromagnetic waves this relation can also be derived from Maxwell's equations, but this is much Problem A.5. The electrons in a television picture tube are typically accelerated to an energy of 10,000 e V. Calculate the momentum of such an electron, and then use the de relation to calculate its wavelength. Problem A.6. In the experiment shown in Figure A.3, the effective slit spacing was 6 /-Lm and the distance from the 'slits' to the detection screen was 16 cm. The sp:aClng between the center of one bright line and the next (before magnification) was 100 nm. From these parameters, determine the wavelength of the electron beam. What voltage was used to accelerate the electrons?
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Problem A.7. The de Broglie relation applies to all 'particles,' not just electrons and photons. (a) Calculate the wavelength of a neutron whose kinetic energy is 1 eV. (b) Estimate the wavelength of a pitched baseball. (Use any reasonable values for the mass and speed.) Explain why you don't see baseballs diffracting around bats.
A.2 Wavefunctions Given that individual particles can behave like waves, we need a way of describing particles that allows for both particlelike and wavelike properties. For this purpose physicists have invented the quantum-mechanical wavefunction. In describing the 'state' of a particle, the wavefunction serves the same purpose in quantum mechanics that the position and momentum vectors serve in classical mechanics: It tells us everything there is to know about what the particle is doing at some particular instant. The usual symbol for a particle's wavefunction is W, and it is a function of position, or of the three coordinates x, y, and z. It's simpler, though, to begin with wavefunctions for particles constrained to move in just the x direction. In this case W at any given time is a function only of x. A particle can have all sorts of wavefunctions. There are narrow, spiky wave­ functions, corresponding to states in which the particle's position is well defined (see Figure A.4). There are also broad, oscillating wavefunctions, corresponding to states in which the particle's momentum is well defined (see Figure A.5). In this w(x)
w(x)
~----~~--------------~x
x=a
-+----------J+I----4.. X x=b
Figure A.4. Wavefunctions for states in which a particle's position is well de­ fined (at x = a and x = b, respectively). When a particle is in such a state, its momentum is completely undefined.
W(X)~
W(X)h
A
I .
IVV
x
AAAAAAAAA . x VVVVVVVV
Figure A.5. Wavefunctions for states in which a particle's momentum is well defined (with small and large values, respectively). When a particle is in such a state, its position is completely undefined.
A.2
Wavefunctions
latter case, the momentum p of the particle is related to the wavelength A by the de Broglie relation, p = hi A. Actually the wavelength of the wavefunction tells us only the magnitude of the particle's momentum. Even in one dimension, Px could be positive or negative, and you can't tell which it is from looking at Figure A.5. In order for II to completely determine the state of the particle, we need to make it a two-component object, that is, a pair of functions. For a particle with well-defined momentum, the second component has the same wavelength as the first component but is out of phase by 90° (see Figure A.6). For the case shown, the momentum is in the +x direction. To give the particle the opposite momentum, we just flip the second component upside down, so it's 90° out of phase in the other direction. The two components of II are normally represented by a single complex-valued function, whose 'real part' is the first component and whose 'imaginary part' (which is no less, or more, real) is the second component. If you want, you can imagine plotting the imaginary part of II along an axis that points up out of the page. Then the three-dimensional graph of a definite-momentum wavefunction is a helix or corkscrew, with a right-handed twist for positive Px and a left-handed twist for negative Px. Besides definite-position wavefunctions and definite-momentum wavefunctions, there are all sorts of others (see Figure A.7). For any wavefunction, though, there is a precise interpretation that's very important. First, take your wavefunction and compute its square modulus: (A.4)
w(x)
Real part
Figure A.6. A more complete illustration of the wavefunction of a particle with well-defined momentum, showing both the 'real' and 'imaginary' parts of the function.
w(x)
w(x)
+---~--~----+---~--~x
+---~----~~~-,----~x
Figure A. 7. Other possible wavefunctions for which neither the position nor the momentum of the particle is well defined.
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This function, when between any two points x 1 and X2, gives you the probability of finding the particle somewhere between those two points if you were to measure its position at that time. (So 1]i 12 is a function whose purpose in life is to be integrated.) Qualitatively, you are more likely to find the particle where its wavefunction is large in magnitude, and less likely to find the particle where its wavefunction is small in magnitude. For a narrow spiky wavefunction you're certain to find the particle at the location of the spike, while for a definite-momentum wavefunction you could find the particle absolutely anywhere. There's also a way to compute the probabilities of getting various outcomes if you were to measure a particle's momentum. Unfortunately, the procedure is mathematically intricate: First you have to take the 'Fourier transform' of the wavefunction, which is a function of the 'wavenumber' k = 27r/ A. Then change variables to Px = hk/27r, and square this flmction to get a function which, when integrated between two values of Px, gives the probability of getting a momentum within that range. Qualitatively, you can usually tell just by looking at a wavefunction whether its momentum is reasonably well defined. A perfectly sinusoidal wavefunction (with the proper relation between real and imaginary parts) has a perfectly precise wavelength and therefore a perfectly well-defined momentum, while a definite-position wavefunction has no wavelength at all: If you were to measure the momentum of such a particle, you could get any result whatsoever. Problem A.S. A definite-momentum wavefunction can be expressed by the for­ mula W(x) = A (cos kx + i sin kx), where A and k are constants.
(a) How is the constant k related to the particle's momentum? (Justify your answer.)
(b) Show that, if a particle has such a wavefunction, you are equally likely to find it at any position x. (c) Explain why the constant A must be valid for all x.
if this formula is to be
(d) Show that this wavefunction satisfies the differential equation dw/ dx -ikw. (e) Often the function cos f) + i sin f) is written instead as e ifJ . Treating the i as an ordinary constant, show that the function Aeikx obeys the same differential equation as in part (d).
The Uncertainty Principle Another important type of wavefunction is what I sometimes call a 'compromise' wavefunction, more often called a wavepacket. A wavepacket is approximately sinusoidal within a certain region but then dies out beyond so it's still reasonably localized in space Figure A.8). For such a wavefunction both x and Px are defined approximately, but neither is defined precisely. If we were to measure the position of a particle in such a state, we could a range of values. If we had a million particles, all in this same state, and we measured their positions, the values would center arOlHld some average with a spread that we could quantify by
A.2 Imaginary part
Real part
w(x)
Wavefunctions
I~
x
Figure A.S. A wavepacket, for which both x and Px are defined approximately but not precisely. The 'width' of the wavepacket is quantified by .6.x, technically the standard deviation of the square of the wavefunction. (As you can see, .6.x is actually a few times smaller than the 'full' width.)
taking the standard deviation of all the values obtained. I'll refer to this standard deviation as ~x; it is a rough measure of the width of the wavepacket. Similarly, if we had a million particles all in the same state and we measured their momenta, the values would center around some average with a spread that we could quantify by taking the standard deviation. I'll refer to this standard deviation as ~Px; it is a rough measure of the width of the wavepacket in 'momentum space.' We can easily construct a wavepacket with a smaller ~x, just by making the oscillations die out more rapidly on each side. But there is a price: We then get fewer complete oscillations, so the wavelength and momentum of the particle become more poorly defined. By the same token, to construct a wavepacket with a precisely defined momentum we have to include many oscillations, resulting in a large ~x. There is an inverse relation between the width of a wavepacket in position space and its width in momentum space. To be more precise about this relation, suppose we make a wavepacket so narrow that it includes only one full oscillation before dying out. Then the spread in position is roughly one wavelength, while the spread in momentum is quite large, comparable to the momentum itself: ~Px
Because a smaller
~Px
rv
Px
implies a larger
h
A
rv
~x,
h ~x'
(A.5)
the relation (A.6)
actually applies to any wavepacket, not just a very narrow one. More generally, one can use Fourier analysis to prove that for any wavefunction whatsoever, (A.7)
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This is the famous Heisenberg uncertainty principle. It says that if you pre­ pare a million particles with identical wavefunctions, then measure the positions of half of the particles and the momenta of the other half and compute the standard deviations, the product of the standard deviations can't be less than h/47r. There­ fore, no matter how you prepare a particle, you can't put it into a state in which both ~x and ~px are arbitrarily small. A properly constructed wavepacket can attain the best-case limit where the product equals h/47r; for most wavefunctions, however, the product is greater. Problem A.9. The formula for a 'properly constructed' wave packet is
w(x) = AeikOxe-o.x2, where A, a, and ko are constants. (The exponential of an imaginary number is defined in Problem A.7. In this problem, just assume that you can manipulate the i like any other constant.) (a) Compute and sketch [W(x)12 for this wavefunction. (b) Show that the constant A must equal (2a/7r)1/4. (Hint: The probability of finding the particle somewhere between x = -00 and x = 00 must equal 1. See Section 1 of Appendix B for help with the integral.) (c ) The standard deviation ~x can be computed as and the average value of x 2 is just the sum of all values of x 2 , weighted by their probabilities:
- 1
00
x2 =
-00
x 2 [w(x)1 2 dx.
Use these formulas to show that for this wavepacket,
~x =
1/(2Va).
(d) The Fourier transform of a function w(x) is defined as
~(k)
w(x) dx.
Show that ~(k) (A/$a) exp[-(k-ko)2 /4a] for a properly constructed wavepacket. Sketch this function.
(e) Using formulas analogous to those in part (c), show that, for this wave­ function, ~k Va. (Hint: The standard deviation does not depend on ko, so you can simplify the calculation by setting ko = 0 from the start.) ~Px for this wavefunction, and check whether the uncertainty principle is satisfied.
(f) Compute
Problem A.lO. Sketch a wavefunction for which the product (~x)(~Px) is much greater than h / 47r. Explain how you would estimate ~x and ~Px for your wave­ function.
Linearly Independent Wavefunctions As you can tell from the preceding illustrations, the number of possible wavefunc­ tions that a particle can have is enormous. This poses a problem in statistical mechanics, where we need to count how many states are available to a particle. There is no sensible way to count all the wavefunctions; what we need is a way to count independent wavefunctions, in a sense that I will now make precise.
A.3
Definite-Energy Wavefunctions
If a wavefunction '11 can be written in terms of two others '11 1 and '11 2 ,
(A.8) for some (complex) constants a and b, then we say that '11 is a linear combination of WI and '11 2 . On the other hand, if there are no constants a and b for which equation A.8 is true, then we say that '11 is linearly independent of WI and '11 2 , More generally, if we have some set of functions '11 n (x), and '11 (x) cannot be written as a linear combination of the '11 n 's, then we say that '11 is linearly independent of the '11 n 'so And if none of the wavefunctions in a collection can be written as a linear combination of the others, then we say they are all linearly independent. What we want to do in statistical mechanics is count the number of linearly independent wavefunctions available to a particle. If the particle is confined within a finite region and its energy is limited, this number is always finite. Even so, there are many different sets of linearly independent wavefunctions that we can work with. In Section 2.5 I used wavepackets, approximately localized in both position space and momentum space. Usually, though, it is more convenient to use wavefunctions with definite energy, to be discussed in the next section. Problem A.1I. Consider the functions Wl(X) = sin(x) and W2(X) = sin(2x), where x can range from 0 to 1r. Write down formulas for three different nontrivial linear combinations of WI and W2, and sketch each of your three functions. For simplicity, keep your functions real-valued.
A.3 Definite-Energy Wavefunctions Among all the possible wavefunctions a particle can have, the most important are the wavefunctions with definite total energy. Total energy is kinetic plus potential; for a nonrelativistic particle in one dimension, 2
E
=
;~ + V(x),
(A.9)
where the potential energy function V (x) can be practically anything. In the special 0, the total energy is the same as the kinetic energy, which case where V (x) depends only on momentum, and so any definite-momentum wavefunction is also a definite-energy wavefunction. When V (x) is nonzero, however, the potential energy is not well defined for a definite-momentum wavefunction, so the definite-energy wavefunctions will be different. To find the definite-energy wavefunctions for a given potential energy V (x) you have to solve a differential equation, called the time-independent Schrodinger equation. * This equation and methods of solving it are discussed at length in quan­ *There's also a time-dependent Schrodinger equation, whose purpose is completely dif­ ferent: It tells you how any wavefunction changes with the passage of time. Definite-energy wavefunctions oscillate from real to imaginary and back again with frequency f = E / h, while other wavefunctions evolve in more complicated ways. Because definite-energy wave­ functions have the simplest possible time dependence, there is a close mathematical rela­ tion between the two Schrodinger .equations.
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tum mechanics textbooks; here I'll just describe the solutions in a few important special cases.
The Particle in a Box The simplest nontrivial potential energy function is the 'infinite square well,'
Vex)
{~
for 0 < x < L, elsewhere.
(A.I0)
This idealized potential confines the particle to the region between 0 and L, a one­ dimensional 'box' (see Figure A.9). Within the box there is no potential energy, while outside the box the particle cannot exist since it would have to have infinite energy. This potential energy function is so simple that we can find the definite-energy wavefunctions without bothering to solve the time-independent Schrodinger equa­ tion. Every allowed wavefunction must be zero outside the box, while inside the box, where there is no potential energy, a definite-energy wavefunction will have def­ inite kinetic energy and therefore definite momentum. Well, almost. The definite­ energy wavefunctions need to go continuously to zero at x 0 and at x = L, since a discontinuity would introduce an infinite uncertainty in the momentum (that is, zero-wavelength Fourier components). But the definite-momentum wavefunctions don't go to zero anywhere. To make a wavefunction that does have zeros (nodes), we need to add together two wavefunctions with equal and opposite momenta to make a 'standing wave.' Such a wavefunction will still have definite kinetic energy, since kinetic energy depends only on the square of the momentum. A few of the definite-energy wavefunctions are shown in Figure A.9. In order for the wavefunction to go to zero at both ends of the box, only certain wavelengths are permitted: 2L, 2L/2, 2L/3, and so on. For each of these wavelengths we can Vex) E4~------------------r
E3~------------------r
1jJ3(X)~ • x
• x
E2~------------------~
El~------------------~
- t - - - - - - - - - - + - - - - 1.. x
a
L
Figure A.9. A few of the lowest energy levels and corresponding definite-energy wavefunctions for a particle in a one-dimensional box.
A.3
Definite-Energy Wavefunctions
use the de Broglie relation to find the magnitude of the momentum, then compute the energy as p2/2m. Thus the allowed energies are
2 ( n )2
h 2m
1 (h)2 2m An
2m
(A.II)
2L
where n is any positive integer. Notice that the energies are quantized: Only certain discretely spaced energies are possible, because the number of half wavelengths that fit within the box must be an integer. More generally, any time a particle is confined within a limited region, its wavefunction must go to zero outside this region and have some whole number of 'bumps' inside, so its energy will be quantized. Definite-energy wavefunctions are important not just because they have defi­ nite energy, but also because any other wavefunction can be written as a linear combination of definite-energy wavefunctions. (In the case of the particle-in-a-box wavefunctions, this statement is the same as the theorem of Fourier analysis that says that any function within a finite region can be written as a linear combination of sinusoidal functions.) Furthermore, the definite-energy wavefunctions are all lin­ early independent of each other (at least for a particle in one dimension that is confined to a limited region). So counting the definite-energy wavefunctions gives us a convenient way to count 'all' the states available to a particle. For a particle confined inside a three-dimensional box, we can construct a definite-energy wavefunction simply by multiplying together three one-dimensional definite-energy wavefunctions:
(A.I2)
where 'l/Jx, 'l/JYl and 'l/Jz can each be any of the sinusoidal wavefunctions for a one­ dimensional box. (For definite-energy wavefunctions it's conventional to use a lower­ case 'I/J.) These products aren't all the definite-energy wavefunctions, but the others can be written as linear combinations of these, so counting wavefunctions that decompose in this way suffices for our purposes. The total energy also decomposes nicely into a sum of three terms:
E
2
2)
+Py +Pz
1 [(hnx)2
= 2m
2Lx
hny ( 2Ly
)2 + (hnz )2] 2Lz '
(A.I3)
where L y , and are the dimensions of the box and n x , ny, and n z are any three positive integers. If the box is a cube, this formula reduces to
E
h2
2
8mL2 (nx
2
2
+ ny + n z )·
(A.I4)
Each triplet of n's yields a distinct linearly independent wavefunction, but not every triplet yields a distinct energy: 1Vlost of the energy levels are degenerate, corre­ sponding to multiple linearly independent states that must be counted separately in statistical mechanics. (The number of linearly independent states that have a given energy is called the degeneracy of the level.)
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Problem A.12. Make a rough estimate of the minimum energy of a proton confined inside a box of width 10- 15 m (the size of an atomic nucleus). Problem A.13. For ultrarelativistic particles such as photons or high-energy electrons, the relation between energy and momentum is not E = p2/2m but rather E = pc. (This formula is valid for massless particles, and also for massive particles in the limit E » mc 2 .) (a) Find a formula for the allowed energies of an ultrarelativistic particle con­ fined to a one-dimensional box of length L. (b) Estimate the minimum energy of an electron confined inside a box of width 10- 15 m. It was once thought that atomic nuclei might contain electrons; explain why this would be very unlikely. (c) A nucleon (proton or neutron) can be thought of as a bound state of three quarks that are approximately massless, held together by a very strong force that effectively confines them inside a box of width 10- 15 m. Estimate the minimum energy of three such particles (assuming all three of them to be in the lowest-energy state), and divide by c2 to obtain an estimate of the nucleon mass. Problem A.14. Draw an energy level diagram for a nonrelativistic particle con­ fined inside a three-dimensional cube-shaped box, showing all states with energies below 15· (h 2/8mL2). Be sure to show each linearly independent state separately, to indicate the degeneracy of each energy level. Does the average number of states per unit energy increase or decrease as E increases?
The Harmonic Oscillator
Another important potential energy function is the harmonic oscillator potential,
(A.15)
where ks is some 'spring constant.' The definite-energy wavefunctions for a particle subject to this potential are not easy to guess, but can be found by solving the time­ independent Schrodinger equation. A few of them are shown in Figure A.IO. These are not sinusoidal functions, but they still have an approximate local 'wavelength,' which is smaller near the middle (where there is less potential energy and hence more kinetic) and larger off to either side (where there is very little kinetic energy). As with the particle in a box, the definite-energy wavefunctions for a quantum harmonic oscillator must go to zero at each side, with an integer number of 'bumps' in between, so the energies are quantized. This time the allowed energies turn out to be
E
~hj, ~hj, ~hj, .. ,
(A.16)
where j = 2~ Jks/m is the natural frequency of the oscillator. The energies are equally spaced, instead of getting farther apart as you go up as they do for a particle in a box. (This is because the harmonically oscillating particle can 'travel' farther to either side if its energy is larger, allowing more space for the wavefunction and hence longer wavelengths.) Often it is convenient to measure all energies relative to the ground-state energy, so that the allowed energies become
E = 0, hj, 2hj, ..
(A.17)
A.3
Definite-Energy Wavefunctions
¢2(XfL
V(x)
V
E5 E4
¢l(X)j
E3 E2
¢O(X)!
El
Eo x
~ • x
~
V
~
• x
• x
Figure A.10. A few of the lowest energy levels and corresponding wavefunctions for a one-dimensional quantum harmonic oscillator.
Shifting the zero-point in this way has no effect on thermal interactions. See Prob­ lem A.24, however, for a situation in which the zero-point energy does matter. Many real-world systems oscillate harmonically, at least to a first approximation. A good example of a quantum oscillator is the vibrational motion of a diatomic molecule such as N2 or CO. The vibrational energies can be measured by looking at the light emitted as the molecule makes a transition from one state to another; an example is shown in Figure A.II. Problem A.15. A CO molecule can vibrate with a natural frequency of 6.4 x 10 13 s- 1 . (a) What are the energies (in eV) of the five lowest vibrational states of a CO molecule?
(b) If a CO molecule is initially in its ground state and you wish to excite it into its first vibrational level, what wavelength of light should you aim at it?
Problem A.16. In this problem you will analyze the spectrum of molecular nitrogen shown in Figure A.l1. You may assume that all of the transitions are correctly identified in the energy level diagram. (a) What is the approximate difference in energy between the upper and lower electronic states, neglecting any vibrational energy (aside from the zero­ point energies ~hf)?
(b) Determine the approximate spacing in energy between the vibrational lev­ els, for both the lower and upper electronic states. (c) Repeat part (b) using a different set of spectral lines, to verify that the diagram is consistent. (d) How can you tell from the spectrum that the vibrational levels (for either electronic state) are not quite evenly spaced? (This is an indication that the potential energy function is not exactly quadratic.)
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300 I
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310
320
I
I
Wavelength (nm)
330 340 350 I
I
360
I
I
370 380 390 400 410
I
I
I
I
I
I
4 3 2
1
o ,
7 6
5
,
4
3 2 1
o
,
,
,
.
Figure A.l1. A portion of the emission spectrum of molecular nitrogen, N2. The energy level diagram shows the transitions corresponding to the various spectral lines. All of the lines shown are from transitions between the same pair of electronic states. In either electronic state, however, the molecule can also have one or more 'units' of vibrational energy; these numbers are labeled at left. The spectral lines are grouped according to the number of units of vibrational energy gained or lost. The splitting within each group of lines occurs because the vibrational levels are spaced farther apart in one electronic state than in the other. From Gordon M. Barrow, Introduction to Molecular Spectroscopy (McGraw-Hill, New York, 1962). Photo originally provided by J. A. Marquisee. (e) For the lower electronic state, what is the effective 'spring constant' of the bond that holds the two nitrogen atoms together? (Hint: First determine the spring constant for each half of the spring, by considering each atom to be oscillating relative to the fixed center of mass. Then think carefully about how the spring constant (force per amount of stretch) of a whole spring is related to the spring constant of each half.)
Problem A.l 7. A two-dimensional harmonic oscillator can be considered as a system of two independent one-dimensional oscillators. Consider an isotropic two-dimensional oscillator, for which the natural frequency is the same in both directions. Write a formula for the allowed energies of this system, and draw an energy level diagram showing the degeneracy of each level. Problem A.lS. Repeat the previous problem for a three-dimensional isotropic oscillator. Find a formula for the number of degenerate states with any given energy.
I
A.3
Definite-Energy Wavefunctions
The Hydrogen Atom A third important potential energy function is V(r) = _ k e e r
2 ,
(A.18)
the Coulomb potential experienced by the electron in a hydrogen atom. (Here e is the charge of the proton and ke is the Coulomb constant, 8.99 x 10 9 N·m 2 jC2.) This is a three-dimensional problem, and solving the time-independent Schrodinger equation is a bit of a chore, but the resulting formula for the energy levels is quite simple: 13.6 eV (A.19) n2 for n = 1, 2, 3, .. The number of linearly independent wavefunctions correspond­ ing to level n is n 2 : 1 for the ground state, 4 for the first excited state, and so on. In addition to these negative-energy states, there can also be states with any positive energy; for these states the electron is ionized, no longer bound to the proton. An energy level diagram for the hydrogen atom is shown in Figure A.12. The definite-energy wavefunctions are interesting and important, but hard to draw in a small space because they depend on three variables (and there are so many of them). You can find pictures of the wavefunctions (or, more commonly, the squares of the wavefunctions) in most textbooks of modern physics or introductory chemistry. V(r)
Continuum ~r-----------------------------------~r
E3
= -1.5 eV =t:;;::::::::::::;;~:::;;;;:;:;=---9 states
E2 = -3.4 eV ~---.',c-
El = -13 .6 eV
Figure A.12. Energy level diagram for a hydrogen atom. The heavy curve is the potential energy function, proportional to -l/r. In addition to the discretely spaced negative-energy states, there is a continuum of positive-energy (ionized) states.
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Problem A.19. Suppose that a hydrogen atom makes a transition from a high-n state to a low-n state, emitting a photon in the process. Calculate the energy and wavelength of the photon for each of the following transitions: 2 ---+ 1, 3 ---+ 2, 4 ---+ 2, 5 ---+ 2.
A.4 Angular Momentum Besides position, momentum, and energy, we might also want to know a particle's angular momentum (about some origin). More specifically, we might want to know the magnitude of its angular momentum vector, ILl, or equivalently, the square of this magnitude, ILI2. In addition, we might want to know the three components of the angular momentum, Lx, L y , and L z . Here again, there are special wavefunctions for which any particular variable is well defined. However, there are no wavefunctions for which more than one of the components Lx, L y , and L z is well defined (except in the trivial case where all three components are zero). The best we can do, with any particular wavefunction, is to specify the value of ILI2 and also the value of anyone component of L; usually we call it the z component. The angular momentum of a particle determines only the angular dependence of its wavefunction-in spherical coordinates, the dependence on the angular variables () and cp. Wavefunctions with well-defined angular momentum turn out to be various sinusoidal functions of these variables. For the wavefunction to be single-valued, these sinusoidal functions must go through a whole number of oscillations when you go around any complete circle. Thus only certain 'wavelengths' of oscillation are allowed, corresponding to certain quantized values of the angular momentum. The allowed values of ILI2 turn out to be £(£ + 1)n2 , where £ is any nonnegative integer and n is an abbreviation for h/27r. For a given value of £, the allowed values of Lz (or Lx or Ly) are mn, where m is any integer from -£ to £. One way to visualize these states is shown in Figure A.13.
tz I-Lz I
=0
:-Lz = 0
Figure A.13. A particle with well-defined ILl and Lz has completely undefined Lx and L y , so we can visualize its angular momentum 'vector' as a cone, smeared over all possible Lx and Ly values. Shown here are the allowed states for f = 1 and f = 2.
A.4
Angular Momentum
Angular momentum is most important in problems with rotational symmetry. rota­ Then, classically, angular momentum is conserved. In quantum tional symmetry implies that we can find definite-energy wavefunctions that also have definite angular momentum. An important example is the hydrogen atom, for which the ground state must have Ill2 = 0, the first excited state (n 2) can have Ill2 = 0 or Ill2 = 21i? (that is, f 0 or f = 1), and so on. (The rule for the hydrogen atom is that f must be less than n, the integer that determines the energy. By the way, like n, f, and m are called quantum numbers.) Problem A.20. A very naive, but partially correct, way to understand quanti­ zation of angular momentum is as follows. Imagine that a particle is confined to travel around a circle of radius r. Its angular momentum about the center is then ±rp, where p is the magnitude of its linear momentum at any moment. Let s be a coordinate that labels the position of the particle around the so that s ranges from 0 to 21rr. The wavefunction of this particle is a function of s. Now suppose that the wavefunction is sinusoidal, so that p is well defined. Using the fact that the wavefunction must undergo an integer number of complete oscilla­ tions over the entire circle, find the allowed values of p and the allowed values of the angular momentum. Problem A.21. Enumerate the quantum numbers (n, l, and m) for all the independent states of a hydrogen atom with definite E, Il12, and up to n = 3. Check that the number of independent states for level n is equal to n 2 .
Rotating Molecules An important application of angular momentum in thermal physics is to the rota­ tion of molecules in a gas. The analysis divides conveniently into the three cases of monatomic, diatomic, and polyatomic molecules. A monatomic molecule (that is, a single atom) doesn't really have any rotational states. It's true that the electrons in the atom can carry angular momentum, and this angular momentum can have various orientations (all with the same energy if the atom is isolated). But to change the magnitude of the electrons' angular mo­ mentum would require putting them into excited states, which typically requires a few electron-volts of energy, more than is available at ordinary temperatures. In any case, these excited states are classified as electronic states, not molecular rota­ tional states. The nucleus, in addition, can possess an intrinsic angular momentum ('spin'), which can have various orientations, but changing the magnitude of its angular momentum requires huge amounts of energy, typically 100,000 eV. More generally, when we talk about the rotational states of a molecule, we are not interested in the rotation of individual nuclei, nor are we interested in excited electronic states. It's appropriate, therefore, to model the nuclei as point masses, and to neglect the electrons entirely since they're merely for the ride.' I'll make both of these simplifications throughout this discussion. In a diatomic molecule, the bond holding the two atoms together is normally quite stiff, so we can picture the two nuclei being held together by a rigid, massless rod. Let's suppose that the center of mass of this 'dumbbell' is at rest (that is, we'll neglect any translational motion). Then classically, the configuration of the system
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in space depends only on two angles, () and cp, specifying the direction toward one of the nuclei in spherical coordinates. (The position of the second nucleus is then completely determined, on the side opposite the first.) The energy of the molecule is determined by its angular momentum vector, conventionally called J (instead of l) in this case. More precisely, the energy is just the usual rotational kinetic energy, (A.20) where 1 is the moment of inertia about the center of mass (that is, 1 = mlr?+m2r~, where mi is the mass and ri is the distance from the rotation axis for the ith nucleus). Quantum mechanically, the wavefunction of this system is a function only of the angles () and cp. Therefore, specifying the angular momentum state (lJ12 and Jz) is sufficient to specify the entire wavefunction, and the number of independent wavefunctions available to the molecule is the same as the number of such angular momentum states. Furthermore, the value of IJI 2 determines the molecule's rota­ tional energy, according to equation A.20. Quantization of 1112 therefore implies quantization of energy, with the allowed energies being E rot = '--'-'----:..-
(A.21)
Here j is basically the same as the quantum number f used above, with the allowed values 0, 1, 2, .. The degeneracy of each energy level is simply the number of distinct Jz values for that value of j, namely 2j + 1. An energy level diagram for a rotating diatomic molecule is shown in Figure 6.6. What I've just said, however, applies only to diatomic molecules made of dis­ tinguishable atoms: CO, or CN, or even H2 where the two hydrogen atoms are of different isotopes. If the two atoms are indistinguishable, then there are only half as many distinct states, because interchanging the two atoms with each other results in exactly the same configuration. Basically this means that half of the j values in equation A.21 are allowed and half are not; Problem 6.30 explains how to figure out which half is which. For any diatomic molecule, the spacing between rotational energy levels is pro­ portional to 1i 2 /21. This quantity is largest when the moment of inertia is small, but even for the smallest molecules it turns out to be less than 1/100 eV. Generally, therefore, the rotational energy levels of a molecule are much more closely spaced than the vibrational levels (see Figure A.14). Because kT » 1i 2 /21 for nearly all molecules at room temperature, rotational 'degrees of freedom' normally hold a significant amount of thermal energy. A linear polyatomic molecule, like CO 2, is similar to a diatomic molecule in that its rotational configuration can be specified in terms of only two angles. The rotational energies of such a molecule are therefore again given by equation A.21. Most polyatomic molecules, however, are more complicated. For example, the orientation of an H 20 molecule is not completely specified by the position of one of the hydrogen atoms relathle to the center of mass; even holding the hydrogen atoms
A.4
Angular Momentum
Figure A.14. Enlargement of a portion of the N2 spectrum shown in Figure A.11, covering approximately the range 370-390 nm. Each of the broad lines is actually split into a 'band' of many narrow lines, due to the multiplicity of rotational levels for each vibrational level. From Gordon M. Barrow, Introduction to Molecular Spectroscopy (McGraw-Hill, New York, 1962). Photo originally provided by J . A . Marquisee.
fixed, the oxygen atom can still travel around in a little circle. So to specify the orientation of a nonlinear polyatomic molecule we need a third angle. This means that the rotational wavefunctions are now functions of three variables instead of two, and the total number of states available is greater than for a diatomic molecule. The energy level structure is usually quite complex, because the moments of inertia about the three possible rotation axes are usually all different. At reasonably high temperatures, where many rotational states are available, the number of such states is enough to count as three 'degrees of freedom.' Beyond this important fact , the detailed behavior of polyatomic molecules is beyond the scope of this book. * Problem A.22. In Section 6.2 I used the symbol E as an abbreviation for the constant 1i 2 /21. This constant is ordinarily measured by microwave spectroscopy: bombarding the molecule with microwaves and looking at what frequencies are absorbed. (a) For a CO molecule , the constant E is approximately 0.00024 eV. What microwave frequency would induce a transition from the j = 0 level to the j = 1 level? What frequency would induce a transition from the j = 1 level to the j = 2 level?
(b) Use the measured value of
E
to calculate the moment of inertia of a CO
molecule. (c) From the moment of inertia and the known atomic masses, calculate the 'bond length,' or distance between the nuclei, for a CO molecule.
Spin In addition to angular momentum due to its motion through space, a quantum­ mechanical particle can have an internal or 'intrinsic' angular momentum, called spin. Sometimes, if you look closely enough, the particle turns out to have internal structure and its spin is merely the result of the motion of its constituents. But in the case of 'elementary' particles like electrons and photons, it's best to just * More details on polyatomic molecules can be found in physical chemistry textbooks such as Atkins (1998).
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think of spin as an intrinsic angular momentum that can't be visualized in terms of internal structure. As with other forms of angular momentum, the magnitude of the spin angular momentum can have only certain values: s(s + l)n, where s is a quantum number analogous to.e. However, it turns out that s does not have to be an integer; it can also be a half-integer, that is, 1/2 or 3/2 or 5/2, etc. Each species of elementary particle has its own value of s, which is fixed once and for all. For electrons, protons, neutrons, and neutrinos, s = 1/2; for photons, s = 1. For composite particles there are various rules for combining the spins and orbital angular momenta of the constituents to obtain the total spin; a helium-4 atom in its ground state turns out to have s 0, for instance, because the spins of its constituents cancel each other out. (By the way, when the angular momentum of a system comes from a combination of orbital motion and spin, or when we don't want to commit ourselves as to which it we normally call it J, and use the quantum number j instead of.e or s.) Given the value of s for a particle, the component of its spin angular momentum along the z axis (or any arbitrary axis) can still take on several possible values. Just as with orbital angular momentum, these values range from down to in integer steps. So if s 1/2, for instance, the z component of the angular momentum can be +n/2 or -n/2. If s = 3/2, there are four possible values for the z component: 3n/2, n/2, -n/2, and -3n/2. For a massless particle, there's one more twist to the rules: Only the two most extreme values of the z component are allowed, for instance, ±n for the photon (s = 1), and ±2n for the graviton (s = 2). A spinning charged particle acts like a little bar magnet, whose strength and orientation are characterized by a magnetic moment vector, [1. For a macro­ scopic loop of electric current, 1;11 is the product of the amount of current and the area enclosed by the loop, while the direction of [1 is determined by a right-hand rule. This definition isn't very useful for microscopic magnets, though, so it's best to just define [1 in terms of the energy needed to twist the particle when it's in an external magnetic field 13. The energy is lowest when ;1 is parallel to 13 and highest when it is antiparallel; if we take the energy to be zero when [1 and 13 are perpendicular, the general formula is
V
sn
rJIDlagnetic
= - [1 .
-sn
(A.22)
It's usually most convenient to call the direction of 13 the z direction, in which case (A.23) Now since ;1 is proportional to a particle's angular momentum, a quantum­ mechanical particle has quantized values of flz: two possible values for a spin-1/2 particle, three for a spin-1 particle, and so on. For the important case of spin-1/2 particles introduced in Section 2.1, I've written (A.24)
A.5
Systems of Many Particles
so J1 lJ1z I. But while this notation is convenient, please note that this J1 is not the same as Ittl, any more than IJzl is the same as 111 for a quantum-mechanical angular momentum. Problem A.23. Draw a cone diagram, as in Figure A.13, showing the spin states of a particle with s = 1/2. Repeat for a particle with s = 3/2. Draw both diagrams on the same scale, and be as accurate as you can with magnitudes and directions.
A.5 Systems of Many Particles A system of two quantum-mechanical particles has only one wavefunction. In one is a function of two spatial dimension, the wavefunction of a two-particle variables, Xl and X2, corresponding to the positions of the two particles. More precisely, if you integrate the square of the wavefunction over some range of Xl values and over some range of X2 values, you get the probability of finding the first particle within the first range and the second particle within the second range. Some two-particle wavefunctions can be factored into a product of single-particle wavefunctions:
(A.25) This is an enormous simplification, which is valid only for a tiny fraction of all two­ particle wavefunctions. Fortunately, though, all other two-particle wavefunctions can be written as linear combinations of wavefunctions that factor in this way. So if we're only interested in counting linearly independent wavefunctions, we're free to consider only those wavefunctions that factor. (The preceding statement is true whether or not the two particles interact with each other. But if they do not interact, there is a further simplification: The total energy of the system is then the sum of the energies of the two particles, and if we take II a and II b to be the appropriate single-particle definite-energy wavefunctions, then their product will be a definite-energy wavefunction for the combined system.) If the two particles in question are distinguishable from each other, there's not much more to say. But quantum mechanics also allows for particles to be absolutely indistinguishable, so that no possible measurement can reveal which is which. In this case, the probability of finding particle 1 at position a and particle 2 at position b must be the same as the probability of finding particle 1 at position b and particle 2 at position a. In other words, the square of the wavefunction must be unchanged under the operation of interchanging its two arguments:
(A.26) This almost implies that II itself is unchanged under this operation, but not quite; another possibility is for II to change sign:
(A.27)
(Since interchanging the arguments a second time must restore II to its original form, these are the only two possibilities; multiplying by i or some other complex number won't do.)
379
380
Appendix A
Elements of Quantum Mechanics
It turns out that nature has taken advantage of both possible signs in equation A.27. For some types of particles, called hosons, [1 is unchanged under interchange of its arguments. For other particles, called fermions, [1 changes sign under this operation: for bosons, (A.28) for fermions.
(In the first case we say that [1 is 'symmetric,' while in the second case we say that [1 is 'antisymmetric.') Examples of bosons include photons, pions, and many types of atoms and atomic nuclei. Examples of fermions include electrons, protons, neutrons, neutrinos, and many other types of atoms and nuclei. In fact, it turns out that all particles with integer spin (or more precisely, integer values of the quantum number s) are bosons, while all particles with half-integer spin (that is, s = 1/2, 3/2, etc.) are fermions. The most straightforward application of rule A.28 is to the case where both particles are in the same single-particle state: (A.29) For bosons, this equation guarantees that [1 will be symmetric under interchange of Xl and X2. For fermions, however, such a state isn't possible at all, because such a function cannot be equal to minus itself (unless it is zero, which isn't allowed). (When we take the spin orientation of a particle into account, the situation is actually a bit more complex. The 'state' of a particle includes not just its spatial wavefunction but also its spin state, so for a fermion, the spatial part of the wavefunction can be symmetric as long as the spin part is antisymmetric. For the important case of spin-1/2 particles, the bottom line is that any given spatial wavefunction can be occupied by at most two such particles of the same species, provided that they are in what's called an antisymmetric spin configuration.) All of the statements and formulas of this section generalize in a natural way to systems of three or more particles. The wavefunction of a system of several identical bosons must be unchanged under the interchange of any pair of the corresponding arguments, while the wavefunction of a system of several identical fermions must change sign under the interchange of any pair. Any given single-particle state (where 'state' means both the spatial wavefunction and the spin configuration) can hold arbitrarily many identical bosons, but at most one fermion.
A.6 Quantum Field Theory Classical mechanics deals not just with systems of pointlike particles, but also with continuous systems: strings, vibrating solids, and even 'fields' such as the electromagnetic field. The usual approach is to first pretend that the continuous object is really a bunch of point particles connected together by little springs, and eventually take the limit where the number of particles goes to infinity and the space between them goes to zero. The result is generally some kind of partial
A.6
Quantum Field Theory
differential equation (for instance, the linear wave equation or Maxwell's equations) that governs the motion as a function of place and time. When this partial differential equation is linear, it is most easily solved by Fourier analysis. Think of the initial shape of the system (say a string) as a su­ perposition of sinusoidal functions of different wavelengths. Each of these 'modes' oscillates sinusoidally in time, with its own characteristic frequency. To find the shape of the string at some future time, you first figure out what each mode will look like at that time, then add them back up in the same proportions as initially. So much for classical continuum mechanics; what if we now want to apply quantum mechanics to a continuous system? Here again, the most fruitful approach is usually to work with the Fourier modes of the system. Each mode behaves as a quantum harmonic oscillator, with quantized energy levels determined by the natural frequency of oscillation:
E = !hj, ~hj, ~hj, ..
(A.30)
So for any given mode, there is a 'zero-point' energy of !hj (which we usually neglect), plus any integer number of energy units with size hf. Since different modes have different frequencies, the system as a whole can have energy units with lots of different sizes. In the case of the electromagnetic field, these units of energy are called photons. Being discrete entities, they behave very much like particles that are in definite­ energy wavefunctions. And since definite-energy states are not the only states of the field, we can even mix different modes together to make a 'photon' that is localized in space. Thus, we are again confronted with wave-particle duality, this time in an even richer context. We started with a classical field, spread out in space. By applying the principles of quantum mechanics to this system, we see that the field acts in many ways like a collection of discrete particles. But now we have a system in which the number of particles is dependent on the current state, rather than being built in and fixed from the outset. In fact, the field can be in states for which the number of particles is not even well defined. These are just the features required to build an accurate model of the quantum fluctuations of the electromagnetic field, and to describe reactions in which photons are created and destroyed. * Analogously, the units of vibrational energy in a solid crystal are called pho­ nons. Like photons, they can be localized or spread-out, and can be created and destroyed in various reactions. Fundamentally, of course, phonons are not 'real' particles: Their wavelengths and energies are limited by the nonzero atomic spacing in the crystal lattice, and they behave simply only when their wavelengths are much larger than this distance. For this reason, phonons are called quasiparticles. Still, *For a good, brief treatment of the quantized electromagnetic field, see Ramamurti Shankar, Principles of Quantum Mechanics, second edition (Plenum, New York, 1994), Section 18.5. For a more complete introduction to quantum field theory, a good place to start is F. Mandl and G. Shaw, Quantum Field Theory, second edition (Wiley, Chichester, 1993).
381
382
Appendix A
Elements of Quantum Mechanics
the phonon picture provides a beautifully accurate description of the low-energy excitations of a crystal. Furthermore, the low-energy excitations of many other materials, from magnetized iron to liquid helium, can be similarly described in terms of various types of quasiparticles. At a more fundamental level, we can use quantum fields to describe all the other species of 'elementary' particles found in nature. Thus there is a 'chromo dynamic' field, for the force that holds the proton together, manifested as particles called 'gluons.' There is also an electron field, a muon field, various neutrino fields, quark fields, and so on. Fields corresponding to particles that are fermions need to be set up rather differently, so that each mode can hold only zero units of energy or one, not an unlimited number as with bosons. Fields corresponding to charged particles (electrical or otherwise) turn out to have two types of excitations, one for the particle and another for its 'antiparticle,' a particle with the same mass but opposite charge. Quite generally, the quantum theory of fields seems to include all the features needed to build an accurate model of elementary particle physics as we now understand it. However, it seems very likely that at sufficiently short wavelengths and high energies, this model will break down, just as the phonon model breaks down when the wavelength becomes comparable to the atomic spacing. Perhaps someday we will discover a new level of structure at some very small length scale, and conclude that all the 'particles' of nature are actually quasiparticles. Problem A.24. According to equation A.30, each mode of a quantum field has a 'zero-point' energy of ~hf even when no further units of energy are present. If the field is really a vibrating string or some other material object, this isn't a problem because the total number of modes is finite: You can't have a mode whose wavelength is shorter than half the atomic spacing (see Section 7.5). But for the electromagnetic field and other fundamental fields corresponding to elementary particles, there is no obvious limit on the number of modes, and the zero-point energies can add up to something embarassing. (a) Consider just the electromagnetic field inside a box of volume L 3 . Use the methods of Chapter 7 to write down a formula for the total zero-point energy of all the modes of the field inside this box, in terms of a triple integral over the mode numbers in the x, y, and z directions. (b) There are good reasons to believe that most of our current laws of physics, including quantum field theory, break down at the very small length scale where quantum gravity becomes important. By dimensional analysis, you can guess that this length scale is of order Gn I c3 , a quantity called the Planck length. Show that the Planck length indeed has units of length, and calculate it numerically.
J
(c) Going back to your expression from part (a), cut off the integrals at a mode number corresponding to a wavelength of the Planck length. Then evaluate your expression to obtain an estimate of the energy per unit volume in empty space, due to the zero-point energy of the electromagnetic field. Express your answer in J 1m3 , then divide by c2 to obtain the equivalent mass density of empty space (in kg/m3 ). Compare to the average mass density of ordinary matter in the universe, which is roughly equivalent to one proton per cubic meter. [Comment: Since most physical effects depend only on differences in energy, and since the zero-point energy never
A.6
Quantum Field Theory
changes, a large energy density in 'empty' space would be harmless as far as most of the laws of physics are concerned. The only exception, and it's a big one, is gravity: Energy gravitates, so a large energy density in empty space would affect the expansion rate of the universe. The energy density of empty space is therefore known as the cosmological constant. From the observed expansion rate of the universe, cosmologists estimate that the actual cosmological constant cannot be any greater than 10- 7 J /m 3 . The discrepancy between this observational bound and your calculated value is one of the greatest paradoxes in theoretical physics. (The obvious solution would be to find some negative contribution to the energy density coming from some other source. In fact, fermionic fields give a negative contribution to the cosmological constant, but nobody knows how to make this negative contribution cancel the positive contribution from bosonic fields to the required precision. *) 1
*For more about the cosmological constant paradox and various proposed solutions, see Larry Abbott, Scientific American 258, 106-113 (May, 1988); Ronald J. Adler, Brendan Casey, and Ovid C. Jacob, American Journal of Physics 63,620-626 (1995); and/or Steven Weinberg, Reviews of Modern Physics 61, 1-82 (1989).
383
B
Mathematical Results
Although no mathematics beyond multivariable calculus is needed to understand the material in this book, in a few places I have quoted mathematical results that are not normally derived in a first course in calculus. The purpose of this appendix is to derive those results. If you're willing to take the results on faith (or better, check them approximately or in special cases), then there's no need to read this appendix. But some of the tools used in the derivations are more broadly applicable in theoretical physics, while all of the derivations themselves are quite lovely. So I hope you'll read on and enjoy this excursion along some of the less-traveled (but very scenic) byways of calculus.
B.l Gaussian Integrals The function (called a Gaussian) has an antiderivative, but there's no way to express that antiderivative in terms of familiar functions (like roots and powers and exponentials and logs). So if you're confronted with an integral of this function you'll probably just want to evaluate it numerically. you're in luck. It turns out However, if the limits of the integral are 0 or that the integral of from -00 to 00 is exactly equal to Vii, (B.l) and since the integrand is an even function (see Figure B.l), the integral from 0 to 00 is just half of this, Vii /2. The proof of this simple result makes use of a two-dimensional integral in polar coordinates. Let me define 00
1=1
-00
384
(B.2)
B.1
Gaussian Integrals
1
~------~~----------~·-----+-----=.-------~x
-2
-1
o
Figure B.1. The Gaussian function e- x
1 2
,
2
whose integral from
-00
to
00
is
.Jii.
The trick is to square this quantity:
(B.3) where I've carefully renamed the integration variable to y in the second factor so as not to confuse it with the integration variable in the first factor. Now the second factor is just a constant, so I can move it inside the x integral. And the function x2 eis independent of y, so I can move it inside the y integral:
What we now have is the integral over all of two-dimensional space of the function e-(x2+y2). I'll carry out this integral in polar coordinates, rand ¢ (see Figure B.2). r2 The integrand is simply e- , while the region of integration is r from 0 to 00 and ¢ from 0 to 27r. Most importantly, the infinitesimal area element dx dy becomes (dr) (r d¢) in polar coordinates, as shown in the figure. Therefore our double integral becomes
(B.5) verifying formula B.l.
Figure B.2. In polar coordinates, the infinitesimal area element is (dr) (r d<jJ).
385
386
Appendix B
Mathematical Results
From equation B.l you can perform a simple substitution to get the more general result
f'
e-
ax
'
dx =
~
If,
(B.6)
where a is any positive constant. And from this equation we can get another useful result by differentiating with respect to a: (B.7)
On the left-hand side we can move the derivative inside the integral, where it hits e- ax2 and brings down a factor of _x 2 • Evaluating the right-hand side and canceling the minus signs gives (B.8)
This trick of 'differentiating under the integral' is an incredibly handy way to evaluate all sorts of definite integrals of transcendental functions multiplied by powers of x. (The alternative is to integrate by parts, but that's much slower.) Integrals of Gaussian functions come up all the time in physics and mathematics, so you may want to make yourself a small reference table of the results of this section (including the problems below). In statistical mechanics, Gaussian integrals arise most commonly as integrals of a Boltzmann factor, where the energy is a quadratic function of the integration variable (as in Sections 6.3 and 6.4). Problem B.1. Sketch an antiderivative of the function Problem B.2. Take another derivative of equation B.8 to evaluate
iotx)x 4 e-ax
2
dx.
2
Problem B.3. The integral of xne- ax is easier to evaluate when n is odd. (a) Evaluatei: xe-
ax2
dx. (No computation allowed!)
(b) Evaluate the indefinite integral (Le., the antiderivative) of simple substitution. (c) Evaluate
ior= xe- ax
, using a
2
dx.
(d) Differentiate the previous result to evaluate
1=
2
x 3 e -ax dx.
Problem B.4. Sometimes you need to integrate only the 'tail' of a Gaussian function, from some large x up to infinity:
1=
t2
e- dt
?
when
x:» 1.
Evaluate this integral approximately as follows. First, change variables to 8 t 2 , to obtain a simple exponential times something proportional to 8 -1 /2. The integral is dominated by the region near its lower limit, so it makes sense to expand 8- 1 / 2
B.2
The Gamma Function
in a Taylor series about that point, keeping only the first few terms in the series. Do this to obtain a series expansion for the integral. Evaluate the first three terms of the series explicitly to obtain
=e _t
1
2
dt
_x2 (
e
1 2x
1 4x3
+
3
_ ..).
Note: When x is fairly large, the first few terms of this series will converge very rapidly toward the exact answer. However, if you calculate too many terms, the coefficients in the numerators will eventually start to grow more rapidly than the denominators and the series will diverge. This happens sooner or later no matter how large x is! Series expansions of this type are called asymptotic expansions. They're incredibly useful, though they make me rather queasy. Problem B.5. Use the methods of the previous problem to find an asymptotic t2 expansion for the integral of t 2 e- , from x to 00, when x » 1. 2
Problem B.6. The antiderivative of e- x ,set equal to zero at x = 0 and multiplied by 2/ Vii, is called the error function, abbreviated erf x: erf x (a) Show that erf(±oo)
(b) Evaluate
fox t 2 e -
2
t
(X
2
10
2
e - t dt.
±1.
dt in terms of erf x.
(c) Use the result of Problem B.4 to find an approximate expression for erf x when x» 1.
B.2 The Gamma Function If you start with the integral
1=
e- ax dx
a-I
(B.9)
and repeatedly differentiate with respect to a, you'll eventually be convinced that
1=
Setting a
= (n!)a-(n+!).
(B.IO)
xne- X dx.
(B.II)
xne- ax dx
I then gives a formula for n!:
n!
1=
I'll use this formula in the following section to derive Stirling's approximation for
nL The integral B.II can be evaluated (not necessarily analytically) even when n isn't an integer, so it gives us a way to generalize the factorial function to nonin­ tegers. The generalization is called the gamma function, denoted r(n), and for some reason it's defined with an offset of I in its argument: (B.12)
387
388
Appendix B
Mathematical Results
So for integer arguments,
f(n+l)
n!.
(B.13)
Perhaps the handiest property of the gamma function is the recursion relation
f(n
+ 1) =
nf(n).
(B.14)
When n is an integer, this formula is essentially the definition of a factorial. But it works for noninteger n too, as you can show from the definition B.12. From either the definition B.12 or the recursion formula B.14 you can see that f(n) blows up at n O. When the argument of the gamma function is negative, the definition B.12 continues to diverge, but we can still define the gamma function (for noninteger arguments) by the recursion formula B.14. A plot of the gamma function for both positive and negative arguments is shown in Figure B.3. The gamma function gives meaning to some ambiguous expressions for factorials that occur in the text of this book, for instance, O!
f(l)
(~ - I)! = f(~).
= 1;
(B.15)
The gamma function also arises in the evaluation of many definite integrals that occur in theoretical physics. We'll see it again in Section B.4. Problem B.7. Prove the recursion formula B.14. Do not assume that n is an integer. Problem B.8. Evaluate f(~). (Hint: Change variables to convert the integrand to a Gaussian.) Then use the recursion formula to evaluate f( ~) and f( - ~ ).
u (
-4
I ~-6 Figure B.3. The gamma function, f(n). For positive integer arguments, f(n) = (n-l)!' For positive nonintegers, f(n) can be computed from equation B.12, while for negative nonintegers, f(n) can be computed from equation B.14.
B.3
Stirling's Approximation
Problem B.9. Carry out the integral B.12 numerically to evaluate r( §-) and r( ~). A useful identity whose proof is beyond the scope of this book is 7r
r(n)r(l-n) = -'-(-)' sm
n7r
Check this formula numerically for n = 1/3.
B.3 Stirling's Approximation In Section 2.4 I introduced Stirling's approximation, (B.I6) which is accurate when n » 1. Since this formula is so important, I'll derive it not once but twice. The first derivation is easier, but not as accurate. Let's work with the natural log of n!: In n! = In[n· (n-I) . (n-2)·· ·IJ (B.I7) = In n + In(n-I) + In(n-2) + .. + In 1. This sum of logarithms can be represented as the area under a bar graph (see Figure B.4). Now if n is fairly large, the area under the bar graph is approximately equal to the area under the smooth curve of the logarithm function. Therefore,
In n! ' 1nlnxdx = (x lnx - x)
I:
=
nln n - n.
(B.I8)
In other words, n! ~ (n/e)n. This result agrees with equation B.I6, aside from the final factor of J27rn. When n is sufficiently large, as is nearly always the case in statistical mechanics, that factor can be omitted so this result is all we need.
Figure B.4. The area under the bar graph, up to any integer n, equals In n!. When n is large, this area can be approximated by the area under the smooth curve of the logarithm function.
389
390
Appendix B
Mathematical Results
To derive a more accurate formula for n!, you can repeat the previous calculation but choose the limits on the integral more carefully (see Problem B.lO). But to really get it right, I'll use a completely different method, starting from the exact formula B.ll: (B.19)
Let's think about the integrand, xne- x , when n is large. The first factor, x n , rises very rapidly as a function of x, while the second factor, e- x , falls very rapidly to zero. The product is a function that rises and then falls, as shown in Figure B.5. You can easily show that the maximum value is reached precisely at the point x = n (see Problem B.ll), and that the height of the peak is nne-no What we want is the area under the graph, and to estimate this area we can approximate the function as a Gaussian. To find the Gaussian function that best fits the exact function near x = n, let me first write the function as a single exponential: (B.20) Next, define y the logarithm:
x ­ n, rewrite the exponent in terms of y, and get ready to expand n In x
x
n In( n
+ y)
n- y
= n In [n (1 + ~)] ­ n -
(B.2l)
y
nlnn-n+nln(l+~)
y.
Near the peak of the graph, y is much less than n so we can expand the logarithm in a Taylor series:
(
Y) Y 1(y)2 .
In 1+; ~;-2 ;
(B.22)
The linear term is canceled by the final -y in equation B.2!. Putting everything
- - - - - Gaussian approximation
~----------~~----~------~~----------. .
n
x
Figure B.5. The function xne- x (solid curve), plotted for n = 50. The area under this curve is nL The dashed curve shows the best Gaussian fit, whose area gives Stirling's aPJ.?roximation to n!.
BA
Area of ad-Dimensional Hypersphere
else together, we obtain the approximation with y
x
n.
(B.23)
This is the best Gaussian approximation to the exact integrand in equation B.19; it is shown as the dashed curve in Figure B.5. To get n!, we want to integrate this -00, function from x = 0 to x = 00. But we might as well start the integral at x since the function is negligible at negative x values anyway. Using the integration formula B.6, we obtain (B.24) in agreement with equation B.16. Problem B.lO. Choose the limits on the integral in equation B.I8 more carefully, to derive a more accurate approximation to nL (Hint: It's the upper limit that is more critical. There's no obvious best choice for the lower limit, but do the best you can.) Problem B.ll. Prove that the function xne- x reaches its maximum value at x n. Problem B.12. Use a computer to plot the function xne- x , and the Gaussian approximation to this function, for n = 10, 20, and 50. Notice how the relative width of the peak (compared to n) decreases as n increases, and how the Gaussian approximation becomes more accurate as n increases. If your computer software permits it, try looking at even higher values of n. Problem B.13. It is possible to improve Stirling's approximation by keeping more terms in the expansion of the logarithm (B.22). The exponential of the new terms can then be expanded in a Taylor series to yield a polynomial in y multiplied by the same Gaussian as before. Carry out this procedure, consistently keeping all terms that will end up being smaller than the leading term by one power of n. (Since the Gaussian cuts off when y is of order yfii, you can estimate the sizes of various terms by setting y yfii.) When the smoke clears, you should find nl
~ nne- n V21Tn( 1 +
1
Check the accuracy of this formula for n 1 and for n 10. (In practice, the correction term is rarely needed. But it does provide a handy way to estimate the error in Stirling's approximation.)
B.4 Area of ad-Dimensional Hypersphere In Section 2.5 I claimed that the surface 'area' of ad-dimensional 'hypersphere' of radius r is 27r d / 2 d-l --r (B.25) r(~) For d
= 2 this formula
giv~s
the circumference of a circle, A2 (r)
= 27rr) while for
391
392
Appendix B
Mathematical Results
rsinO
•
I
-
•
-_-__
0
-~
-rdO Figure B.6. To calculate the area of a sphere, divide it into loops and integrate. To calcu­ late the area of a hypersphere, do the same thing.
= 3 it gives the surface area of a sphere, A3(r) = 41Tr2. (For d = 1 it gives Al(r) = 2, the number of points bounding a line segment.) Before proving equation B.25 in general, let's warm up by considering just the case d = 3, a true three-dimensional sphere. The surface of the sphere can be built out of loops, as shown in Figure B.6. Each loop has width r de and circumference A2 (r sin e) = 21Tr sin e, so the total area of the sphere is d
A3(r) =
l'
A 2(rsinO) rdO = 21fr21' sinOdO = 41fr2,
(B.26)
By a completely analogous calculation, we can prove equation B.25 for any d, assuming by induction that it holds for d-l. Imagine building the surface of a d-dimensional sphere out of (d-l )-dimensional 'loops,' each with width r de and with 'circumference' A d - 1 (r sin e). The total 'area' is again the integral from 0 to 1T:
(B.27) In Problem B.14 you can show that
rr(. e)n de = Vir(~ + ~) sm r (~ + 1) ,
}0
(B.28)
so that (B.29) as claimed.
B.5
Integrals of Quantum Statistics
393
Problem B.l4. The proof of formula B.28 is by induction. (a) Check formula B.28 for n
0 and for n
1.
(b) Show that
(Hint: First write (sinO)n as o)n-2(1 - cos 2 0). Integrate the second term by parts, differentiating one factor of cos 0 and integrating everything else. ) (c) Use the results of parts (a) and (b) to prove formula B.28 by induction.
Problem B.l5. A cleaner, but much trickier, derivation of formula B.25 is similar to the method used in Section B.l to evaluate the basic Gaussian integral. The 2 trick is to consider the integral ofthe function e- r ,over all space, in d dimensions. (a) First evaluate this integral in rectangular coordinates. You should obtain Jrd/2.
(b) Because the integral has spherical symmetry, you can also evaluate it in d­ dimensional spherical coordinates. Explain why the angular integrals must give a factor of Ad(l), the area of a d-dimensional unit hypersphere. Thus, r2 oo d 1 show that the integral is equal to Ad(l) . r - e- dr.
Jo
(c) Evaluate the integral over r in terms of the gamma function, and thus derive equation B.25.
Problem B.l6. Derive a formula for the volume of ad-dimensional hypersphere.
B.5 Integrals of Quantum Statistics In quantum statistics (Chapter 7) we frequently encounter integrals of the form xn --±-dx, o eX 1
1
00
(B.30)
when summing over states for a system of bosons ( - in the denominator) or fermions in the denominator). These integrals can, of course, be done numerically. When n is an odd however, the answer can be expressed exactly in terms of 1T. The first is to rewrite the integral as an infinite series. J1omentarily putting aside the factor of x n , note that the rest of the integrand can be written as a geometric series: e- X
1 eX
1
e- x =F (e- x )2
1 ± e- X = e- x =F
+ e-
3x
=F
+
=F' .
(B.31)
+ ..
Now it's easy to multiply by xn and integrate term by term. For the case n = 1 we obtain 00 2x 00 x dx =F xe­ + o eX ± 1 0 (B.32)
1
1
1
394
Appendix B
Mathematical Results
This type of infinite series comes up a lot in mathematics, so mathematicians have given it a name. The Riemann zeta function, ((n), is defined as
((n)
1
1
1 + -2n + -3n + ..
=
1
= ''-.
L..J kn
(B.33)
k=l
Therefore we can write simply
=- -x d x = ((2). o eX - 1
1
(B.34)
When the integrand has a plus in the denominator the series alternates, so we need to do a bit of manipulation:
1 = +
1 1 ) -2 ( -1 + -1+ -1+ .. )
- -xd x = ( 1+-+-+··· o eX 1 22 32 22 4 2 6 2 2 ( 1 + 22 1 + 31 + .. )
2
= ((2) - 22
= ((2) =
~((2)
~((2).
(B.35)
It's only a little harder (see Problem B.17) to derive the more general results
=
1 1 o
xn -x-
e - 1
dx = f(n+l)((n+1);
= -xn- dx
o eX
+1
=
(1 1- - ) f(n+1)((n+1).
(B.36)
2n
(When n is an integer, f(n+1) = nL) Now the problem is 'simply' to sum the infinite series B.33 that defines the Riemann zeta function. Unfortunately, getting a simple answer is not a simple task at all. I'll do it in a very tricky, roundabout way that uses a Fourier series. * The trick is to consider a square-wave function, with period 21r and amplitude 1r/4 (see Figure B.7). Fourier's theorem states that any periodic function can be written as a linear superposition of sines and cosines. For an odd function such as ours only sines are necessary, so we can write
f(x) =
L= ak sin(kx),
(B.37)
k=l
*How anyone ever thought of this method in the first place is beyond me. I learned it from Mandl (1988).
B.5
Integrals of Quantum Statistics
f(x) 7r 4
I I I I
-
7r
I I I I
... x
27r
7r 4
-
Figure B.7. A square-wave function with period 27r and amplitude 7r/4. The Fourier series for this function yields values of ((n) when n is an even integer.
for some set of coefficients ak. Notice that the first sine wave in the sum has the same period as f(x), while the rest have periods of 1/2, 1/3, 1/4 as much, and so on. To solve for the coefficients we can use 'Fourier's trick': Multiply by sin(jx) (where j is any positive integer) and integrate over one period of the function:
f27r
10
L 00
f(x) sin(jx) dx
ak
127r
k=l
sin(kx) sin(jx) dx.
(B.38)
0
The integral on the right-hand side is zero except when k = j, when it equals Keeping only this nonzero term and renaming j k, we obtain, for any k,
1127r f(x) sm(kx) . dx 1r
-217r f(x) sin(kx) dx.
0
1r
(B.39)
0
This formula the Fourier coefficients of any odd function f(x) with period For our square-wave function, the coefficients are ak
2 f7r 1r = 1r 10 '4 sin(kx) dx =
Therefore, for 0 < x <
{
l/k 0
1r.
for k
= 1, 3, 5, .. ,
for k
2, 4, 6, .. .
21r.
(BAO)
1r,
'' sin( kx) 6 k .
4
(BAl)
odd k
The final trick is to integrate this expression successively with respect to x, then evaluate the result at 1r /2. Integrating carefully from x 0 to x = x' gives 1rX'
14
=
1 k
L odd k
and plugging in x'
1r /2
{Xl
10
sin(kx)dx
0
L
=
1
(1- COSkX') ,
(BA2)
odd k
yields simply 1
8
L k k2'
odd
(BA3)
395
396
Appendix B
Mathematical Results
But ((2) is the sum over all positive integers:
((2) =
L
1
odd k
+
L :2 even k
(B.44)
In other words,
~3 7r8 = 2
1(2) = ':,
6 .
(B.45)
This result suffices to evaluate our original integral (B.30) for the case n 1, with either sign in the denominator. For higher odd values of n the procedure is to take more derivatives of equation B.42, then again evaluate the result at 7r /2 and manipulate the series slightly (see Problem B.19). Unfortunately, this method does not yield any values of ((n) for odd n; in fact, these cannot be written in terms of 7r so they must be evaluated numerically. Problem B.17. Derive the general integration formulas B.36. Problem B.1S. Use a computer to plot the sum of sine waves on the right-hand side of equation B.41, terminating the sum first at k = 1, then at k = 3, 5, 15, and 25. Notice how the series does converge to the square-wave function that we started with, but the convergence is not particularly fast. Problem B.19. Integrate equation B.42 twice more, then plug in x 7r/2 to obtain a formula for L:odd(l/k4 ). Use this formula to show that ((4) = 7r4 /90, and thus evaluate the integrals B.36 for the case n = 3. Explain why this procedure does not yield a value for ((3). Problem B.20. Evaluate equation B.41 at x 7r/2, to obtain a famous series for 7r. How many terms in this series must you evaluate to obtain 7r to three significant figures? Problem B.21. In calculating the heat capacity of a degenerate Fermi gas in Section 7.3, we needed the integral 1r2
3 To derive this result, first show that the integrand is an even function, so it suffices to integrate from 0 to 00 and then multiply by 2. Then integrate by parts to relate this integral to the one in equation B.35.
Problem B.22. Evaluate ((3) by numerically summing the series. How many terms do you need to to get an answer that is accurate to three significant figures?
Suggested Reading
Undergraduate Thermal Physics Texts Callen, Herbert B., Thermodynamics and an Introduction to Thermostatistics, second edition (Wiley, New York, 1985). Develops thermodynamics from an abstract, logically rigorous approach. The application chapters are somewhat easier, and clearly written. Carrington, Gerald, Basic Thermodynamics (Oxford University Press, Oxford, 1994). A nice introduction that sticks to pure classical thermodynamics. Kittel, Charles, and Herbert Kroemer, Thermal Physics, second edition (W. H. I-<'rt:.arrHH1 San Francisco, 1980). An insightful text with a great of modern applications. Statistical Physics, second edition (Wiley, 1988). A clearly written Mandl, text that emphasizes the statistical approach. Fundamentals of Statistical and Thermal Physics (McGraw-Hill, New York, Reif, 1965). More advanced than most undergraduate texts. Emphasizes the statistical approach and includes extensive chapters on transport theory. Stowe, Keith, Introduction to Statistical Mechanics and Thermodynamics (Wiley, New 1984). Perhaps the easiest book that takes the statistical approach. Very well written, but unfortunately marred by an incorrect treatment of chemical potential. Zemansky, Mark W., and Richard H. Dittman, Heat and Thermodynamics, seventh edition (McGraw-Hill, New York, 1997). A classic text that includes good descriptions of experimental results and Earlier editions contain a wealth of material that didn't make it into the most recent edition; I especially like the fifth edition (1968, written by Zemansky alone).
Graduate-Level Texts Chandler, David, Introduction to Ivfodern Statistical Mechanics (Oxford University New York, 1987). My favorite advanced text: short and well written, with lots of inviting problems. A partial solution manual is also available. Landau, L. D., and E. M. Statistical Physics, third edition, Part I, trans. J. B. and M. J. Kearsley Press, Oxford, 1980). An authoritative classic. Pathria, R. K., Statistical second edition (Butterworth-Heinemann, Oxford, 1996). Good systematic coverage of statistical mechanics. Pippard, A. B., The Elements of Classical Thermodynamics (Cambridge University Cambridge, 1957). A concise summary of the theory as well as several applications. Reichl, A Modern Course in Statistical Physics, second edition (Wiley, New York, 1998). Encyclopedic in coverage and very advanced.
397
398
Suggested Reading Introductory Texts Ambegaokar, Vi nay, Reasoning About Luck: Probability and its Uses in Physics (Cam­ bridge University Press, Cambridge, 1996). An elementary text that teaches proba­ bility theory and touches on many physical applications. Fenn, John B., Engines, Energy, and Entropy: A Thermodynamics Primer (W. H. Free­ man, San Francisco, 1982). A introduction to classical thermodynamics, em­ phasizing everyday applications and featuring cartoons of Charlie the Caveman. Feynman, Richard P., Robert B. Leighton, and Matthew Sands, The Feynman Lectures on Physics (Addison-vVesley, Reading, MA, 1963). Chapters 1, 3, 4, 6, and 39-46 treat topics in thermal physics, with Feynman's incredibly high density of deep insights per page. Moore, Thomas A., Six Ideas that Shaped Physics, Unit T (McGraw-Hill, New York, 1998). This very clearly written text inspired my approach to the second law in Sections 2.2 and 2.3. Reif, F., Statistical Physics: Berkeley Physics Course-Volume 5 (McGraw-Hill, New York, 1967). A rather advanced introduction, but much more leisurely than Reif (1965). Popularizations Atkins, P. VV., The Second Law (Scientific American Books, New York, 1984). A nice coffee-table book with lots of pictures. Goldstein, Martin, and lnge F. Goldstein, The Refrigerator and the Universe (Harvard University Press, Cambridge, MA, 1993). An extensive tour of thermal physics and its diverse applications. Zemansky, Mark W., Temperatures Very Low and Very High (Van Nostrand, Princeton, 1964; reprinted by Dover, New York, 1981). A short paperback that focuses on physics at extreme temperatures. Very enjoyable reading, except when the author slips into textbook mode. Engines and Refrigerators Moran, Michael J., and Howard N. Shapiro, Fundamentals of Engineering Thermodynam­ third edition (Viley, New York, 1995). One of several good encyclopedic texts. Whalley, P. Basic Engineering Thermodynamics (Oxford University Oxford, 1992). Refreshingly concise. Chemical Thermodynamics Atkins, P. V., Physical Chemistry, sixth edition (W. H. Freeman, New York, 1998). One of several good physical chemistry texts, packed with information. Findlay, Alexander, Phase Rule, ninth edition, revised by A. N. Campbell and N. O. Smith (Dover, New York, 1951). Everything you ever wanted to know about phase diagrams. Haasen, Physical Metallurgy, third edition, trans. Janet Mordike (Cambridge Uni­ versity Press, Cambridge, 1996). An authoritative monograph that doesn't shy away from the physics. Rock, Peter A., Chemical Thermodynamics (University Science Books, Mill Valley, CA, 1983). A well-written introduction to chemical thermodynamics with plenty of inter­ esting applications. Smith, E. Brian, Basic Chemical Thermodynamics, fourth edition (Oxford University Press, Oxford, 1990). A nice short book that covers the basics.
Suggested Reading
Biology Asimov, Isaac, Life and Energy (Doubleday, Garden City, NY, 1962). A popular account
of thermodynamics and its applications in biochemistry. Old but still very good.
Stryer, Lubert, Biochemistry, fourth edition (W. H. Freeman, New York, 1995). Mar­
velously detailed, though not as quantitative as one might like. Tinoco, Ignacio, Jr., Kenneth Sauer, and James C. Wang, Physical Chemistry: Principles and Applications in Biological Sciences, third edition (Prentice-Hall, Englewood Cliffs, NJ, 1995). Less comprehensive than a standard physical chemistry text, but with many more biochemical applications.
Earth and Environmental Science Anderson, G. M., Thermodynamics of Natural Systems (Wiley, New York, 1996). A prac­ tical introduction to chemical thermodynamics, with a special emphasis on geological applications. Bohren, Craig F., Clouds in a Glass of Beer: Simple Experiments in Atmospheric Physics (Wiley, New York, 1987). Short, elementary, and delightful. Begins by observing that 'a glass of beer is a cloud inside out.' Bohren has also written a sequel, What Light Through Yonder Window Breaks? (Wiley, New York, 1991). Bohren, Craig and Bruce A. Albrecht, Atmospheric Tlwrmodynamics (Oxford Univer­ sity Press, New York, 1998). Though intended for meteorology students, this textbook will appeal to anyone who knows basic physics and is curious about the everyday world. Great fun to read and full of food for thought. Harte, John, Consider a Spherical Cow; A Course in Environmental Problem Solving (University Science Books, Sausalito, CA, 1988). A wonder:fy.Ybook that applies undergraduate-level physics and mathematics to dozens of interesting environmental problems. Kern, Raymond, and Alain Weisbrod, Thermodynamics for Geologists, trans. Duncan McKie (Freeman, Cooper and Company, San Francisco, 1967). Features a nice selec­ tion of worked examples. Nordstrom, Darrell Kirk, and James L. Munoz, Geodlemical Thermodynamics, second edition (Blackwell Scientific Publications, Palo Alto, CA, 1994). A well-written ad­ vanced textbook for serious geochemists.
Astrophysics and Cosmology Carroll, Bradley W., and Dale A. Ostlie, An Introduction to !vfodern Astrophysics (Addi­ son-Wesley, Reading, MA, 1996). A clear, comprehensive introduction to astrophysics at the intermediate undergraduate level. Peebles, P. J. E., Principles ofPhysical Cosmology (Princeton University Press, Princeton, NJ, 1993). An advanced treatise on cosmology with a detailed discussion of the thermal history of the early universe. Shu, Frank H., The Physical Universe: An Introduction to Astronomy (University Science Books, Mill Valley, CA, 1982). An astrophysics book for physics students, disguised as an introductory astronomy text. Full of physical insight, this book portrays all of astrophysics as a competition between gravity and the second law of thermodynamics. Weinberg, Steven, The First Three Minutes (Basic Books, New York, 1977). A classic account of the history of the early universe. Written for lay readers, yet gives a physicist plenty to think about.
399
400
Suggested Reading
Condensed Matter Physics Ashcroft, Neil W., and N. David Mermin, Solid State Physics (Saunders College, Philadel­ phia, 1976). An excellent text that is somewhat more advanced than Kittel (below). Collings, Peter J., Liquid Crystals: Nature's Delicate Phase of Matter (Princeton Uni­ versity Press, Princeton, NJ, 1990). A short, elementary overview of both the basic physics and applications. Goodstein, David L., States of Jl.latter (Prentice-Hall, Englewood Cliffs, NJ, 1975; re­ printed by Dover, New York, 1985). A well written graduate-level text that surveys the properties of gases, liquids, and solids. Gopal, E. S. R., Specific Heats at Low Temperatures (Plenum, New York, 1966). A nice short monograph that emphasizes comparisons between theory and experiment. Kittel, Charles, Introduction to Solid State Physics, seventh edition (Wiley, New York, 1996). The classic undergraduate text. Wilks, J., and D. S. Betts, An Introduction to Liquid Helium, second edition (Oxford University Press, Oxford, 1987). A concise and reasonably accessible overview. Yeomans, J. M' Statistical Mechanics of Phase 'Transitions (Oxford University Press, Oxford, 1992). A brief, readable introduction to the theory of critical phenomena.
Computer Simulations Gould, Harvey, and Jan Tobochnik, An Introduction to Computer Simulation Methods, second edition (Addison-Wesley, Reading, MA, 1996). Covers far-ranging applications at a variety of levels, including plenty of statistical mechanics. Vhitney, Charles A., Random Processes in Physical Systems: An Introduction to Prob­ ability-Ba.ged Computer Simulations (Wiley, New York, 1990). A good elementary textbook that takes you from coin flipping to stellar pulsations.
History and Philosophy Bailyn, Martin, A Survey of Thermodynamics (American Institute of Physics, New York, 1994). A textbook that gives a good deal of history on each topic covered. Brush, Stephen G., The Kind of !vfotion We Call Heat: A History of the Kinetic Theory of Gases in the 19th Century (North-Holland, Amsterdam, 1976). A very scholarly treatment. Kestin, Joseph (ed.), The Second Law of Thermodynamics (Dowden, Hutchinson & Ross, Stroudsburg, PA, 1976). Reprints (in English) of original papers by Carnot, Clausius, Thomson, and others, with helpful editorial comments. Leff, Harvey S., and Andrew F. Rex (eds.), Maxwell's Demon: Entropy, Information, Com­ puting (Princeton University Press, Princeton, NJ, 1990). An anthology of important papers on the meaning of entropy. Mendelssohn, K., The Quest for Absolute Zero, second edition (Taylor & Francis, London, 1977). A popular history of low-temperature physics, from the liquefaction of oxygen to the properties of superfluid helium. Von Baeyer, Hans Christian, !vfaxwell's Demon: Why Warmth Disperses as Time Passes (Random House, New York, 1998). A brief popular history of thermal physics with an emphasis on the deeper issues. Highly recommended.
T
Suggested Reading
Tables of Thermodynamic Data Keenan, Joseph Frederick G. Keyes, Philip G. Hill, and Joan G. Moore, Steam Tables (S.!' Units) (Wiley, New York, 1978). Fascinating. Lide, David R. (ed.), CRC Handbook of Chemistry and Physics, 75th edition (Chemical Rubber Company, Boca Raton, FL, 1994). Cumbersome but widely available. Editions published since 1990 are better organized and use more modern units. National Research Council, International Critical Tables of Numerical Data (McGraw-Hill, New York, 1926-33). A seven-volume compendium of a great variety of data. Reynolds, William C., Thermodynamic Properties in S1 (Stanford University Dept. of Mechanical Engineering, Stanford, CA, 1979). A handy compilation of properties of 40 important fluids. Vargaftik, N. B., Handbook of Physical Properties of Liquids and Gases (Hemisphere, Washington, DC, 1997). Detailed property tables for a variety of fluids. Woolley, Harold W., Russell B. Scott, and F. G. Brickwedde, 'Compilation of Thermal Properties of Hydrogen in its Various Isotopic and Ortho-Para Modifications,' Journal of Research of the National B~ of Standards 41, 379-475 (1948). Definitive but not very accessible.
*
*
*
An awkward aspect of reading any new textbook is getting used to the notation. For­ tunately, many of the notations of thermal physics have become widely accepted and standardized through decades of use. There are several important exceptions, however, including the following: This book Total energy Multiplicity Helmholtz free energy Gibbs free energy Grand free energy Partition function Maxwell speed distribution Quantum length Quantum volume Fermi-Dirac distribution Density of states
U
n
Other E W, 9
F
A
G
F
1> Z V(v)
Q, q
n P(v)
A, AT vQ
A}, l/nQ
nFD(E) g(E)
f(E) D(E)
401
Reference Data
Physical Constants 10- 23 J /K = 8.617 x 10- 5 eV/K NA = 6.022 X 10 23
k = 1.381
R= h= = c=
X
8.315 J/mol·K 6.626 X 10- 34 J·s 4.136
X
10- 15 eV·s
2.998
X
10 8 m/s
10- 11 N ·m2/kg2 e = 1.602 x 10- 19 C me = 9.109 x 10- 31 kg 27 kg mp = 1.673 x 10-
G = 6.673
X
U nit Conversions
= 1.013 bar = 1.013 x 105 N /m 2 = 14.7lb/in2 = 760 mmHg (T in °C) = (T in K) - 273.15 1 atm
(T in OF) = ~(T in °C) 1 oR =
Q 9
+ 32
K
1 cal = 4.186 J 1 Btu = 1054 J 1 eV = 1.602 x 10- 19 J 1 u = 1.661 X 10- 27 kg
402
·.at
The atomic number (top left) is the number of protons in the nucleus. The atomic mass (bottom) is weighted by isotopic abundances in tll€ earth's surface. Atomic masses are relative to the mass of the carbon-12 isotope, defined to be exactly 12 unified atomic mass units (u). Uncertainties ran~e from 1 to 9 in the last digit quoted. Relative isotopic abundances often vary considerably, both in natural and commercial samples. A nLunber III parentheses is the mass of the longest-lived isotope of that element-no stable isotope exists. However, although Th, Pa, and U have no stable isotopes, they do have characteristic terrestrial compositions, and meaningful weighted masses can be given. For elements 110-112, the mass numbers of known isotopes are given. From the Review of Particle Physics by the Particle Data Group, The European Ph'ysical Journal C3, 73 (1998). 18 VillA
IA
Helium
Hydrogen
2
13
14
15
16
17
1.00794
IIA
IliA
IVA
VA
VIA
VilA
.
Be
Li 4
3
Lithium
Beryllium
6.941
9.012182
11
Na
12
Magnesium
22,989770
24.3050
K 20
Ca
IIIB 21
5
IVB
Sc
22
VB
Ti
7
6 VIB
VIIB
Scandium Titanium Vanadium Chromium Manganese
39,0983
40,078
44.955910
. - 47.867
50,9415
Rb
38
39
V 40
41
Nb
Iron
42
Mo
10
43
Tc
VIII ----,
IB
Cobalt
Ni
28
Nkkel
44

An Introduction To Thermal Physics Schroeder Pdf free. download full

Ru
45
Rh
Pd
46
Yttrium
Zirconium
Niobium
Molybd.
Technet.
Ruthen.
Rhodium Palladium
85.4678
87.62
88,90585
91.224
92.90638
95,94
(97,907215)
101.07
102,90550
Cs
56
Ba
Barium
132,90545
137.327
87
.
F, 88
Francium
Radium
(223.019731) (226,025402)
10.811
2,0107
14,00674
15,9994
18.998403:2
57-71 Lantha-
72
Hf
Hafnium
73
Ta
74
W 75
Tantalum Tungsten
Re
Os
76
Rhenium
Osmium
77
Ir
Iridium
106.42
Pt
78
Platinum
Seaborg.
Bohrium
Hassium .
.
.----'-~- ~---.
Nd I 61 Pm 162 Sm 163 Prometh.
Samarium
28.0855
15
Phospho
30.973761
Sulfur
Eu 164
Europium
Ga
32
Ge
33
As
32.066
-
34
Se
35
B(
Arsenic
Selenium
Bromine
63,546
65.39
69.723
72,61
74.92160
78,96
79,904
Ag
47
Cd
48
In
49
Sn
50
Silver
Cadmium
Indium
Tin
107.8682
112.411
114.818
118.710
Au
79
Hg
80
81
.J:,
Gold
(272)
Gd 165
TI
Pb
82
51
Sb
52
Te
Antimony Tellurium
121.760 83
Bi
127.60 84
Po
Polonium
Ar
Argon
36
Kr
83,80
Iodine
126,90441 85
20,1797 18
Krypton
I
53
Ne
39.948
35.4521
i­
German.
At
Astatine
54
Xe
Xenon
131.29 86
Rn
Radon
Mercury
Thallium
Lead
Bismuth
200,59
204.3833
207,2
208.98038 (208,982415) (209,98713l) (222,017570)
(277)
---'-'---'­
Gadolin,
31
Chlorine
Gallium
Meitncr.
(261.1089) (262.1144) (263.1186) (262.1231) (265,1306) (266,1378) (269. 273)
.
Silicon
Zinc
--~.
Actinides Rutherford Dubnium
Si
14
p- c16- - S­17-C'
Copper
178,49 180,9479 183.84 186,207 190,23 192,217 195,078 196.96655 _ - ­nides Ra 89-103 104 Rf 105 Db 106 Sg 107 Bh 108 Hs 109 Mt 110 11!
' ­.
Fluorine
AI
10
Neon
Oxygen
26.981538
Zn
30
C--.~
Strontium
Cesium
Cu
29
f
9
Nitrogen
Aluminum
liB
0
8
'. Carbon
.'~r--.-
Co
Rubidium
55
12
11
55,845 58.933200 58,6934 . -!---oo_ i­
51.9961 54.938049
~-
Zr
r-­
. - - - - r . - - . - - - - - . 23 V 24 Cr 25 Mn 26 Fe 27
Calcium
Sr
9
8
N
C 7
6
4.002602
Boron
13
4
3
Potassium
37
B
5
Periodic Table of the Elements
Mg
Sodium
19
He
2
H
1
.
-'.­
Tb 166
Terbium
Dy 167
Dyspros,
Ho 168
Holmium
Er 169 Tm 170
Erbium
Thulium
g:;
Vb I 71
Ytterbium
til' '1
(1)
103
Lr
LaW-rene,
::::! n
(1)
V
~
Q:)
~
o
(J:I
404
Reference Data
Thermodynamic Properties of Selected Substances All of the values in this table are for one mole of material at 298 K and 1 bar. Following the chemical formula is the form of the substance, either solid (s), liquid (1), gas (g), or aqueous solution (aq). When there is more than one common solid form, the mineral name or crystal structure is indicated. Data for aqueous solutions are at a standard concentration of 1 mole per kilogram water. The enthalpy and Gibbs free energy of formation, AfH and AfG, represent the changes in Hand G upon forming one mole of the material starting with elements in their most stable pure states (e.g., C (graphite), 02 (g), etc.). To obtain the value of AH or AG for another reaction, subtract Af of the reactants from A f of the products. For ions in solution there is an ambiguity in dividing thermodynamic quantities between the positive and negative ions; by convention, is assigned the value zero and all others are chosen to be consistent with this value. Data from Atkins (1998), Lide (1994), and Anderson (1996). Please note that, while these data are sufficiently accurate and consistent for the examples and problems in this textbook, not all of the digits shown are necessarily significant; for research purposes you should always consult original literature to determine experimental uncertainties. (kJ)
S (J/K)
Cp (J/K)
V (cm 3 )
0 -2594.29 -2590.27 -2587.76
0 -2443.88 -2442.66 -2440.99
28.33 83.81 93.22 96.11
24.35 121.71 122.72 124.52
9.99 44.09 51.53 49.90
0
0
154.84
20.79
C (graphite) C (diamond) CH 4 (g) C2 H 6 (g) C3H S (g) C2 H 5 0H (1) C6H1206 (glucose) CO (g) C02 (g) H2C03 (aq) HC0 3 (aq)
0 1.895 -74.81 -84.68 -103.85 -277.69 -1268 -110.53 -393.51 -699.65 -691.99
0 2.900 -50.72 -32.82 -23.49 -174.78 -910 -137.17 -394.36 -623.08 -586.77
5.74 2.38 186.26 229.60 269.91 160.7 212 197.67 213.74 187.4 91.2
8.53 6.11 35.31 52.63 73.5 111.46 115 29.14 37.11
(aq) CaC03 (calcite) CaC03 (aragonite) CaCl2 (s)
-542.83 -1206.9 -1207.1 -795.8
-553.58 -1128.8 -1127.8 -748.1
-53.1 92.9 88.7 104.6
81.88 81.25 72.59
36.93 34.15 51.6
Cl2 CI- (aq)
0 -167.16
0 -131.23
223.07 56.5
33.91 -136.4
17.3
Cu (s)
0
0
33.150
24.44
7.12
Fe (s)
0
0
27.28
25.10
7.11
Substance (form) Al Al2Si05 (kyanite) AbSi05 (andalusite) AbSi05 (sillimanite) Ar (g)
5.30 3.42
58.4
Reference Data
(kJ)
S (JjK)
Op (JjK)
0 217.97 0 -285.83 -241.82
0 203.25 0 -237.13 -228.57
130.68 114.71 0 69.91 188.83
28.82 20.78 0 75.29 33.58
He (g)
0
0
126.15
20.79
Hg (I)
0
0
76.02
27.98
0 -46.11
0 -16.45
191.61 192.45
29.12 35.06
-240.12 -411.15 -3935.1 -3030.9
-261.91 -384.14 -3711.5 -2852.1
59.0 72.13 207.40 133.5
46.4 50.50 205.10 160.0
0
0
146.33
20.79
02 (g) 02 (aq) OH- (aq)
0 -11.7 -229.99
0 16.4 -157.24
205.14 110.9 -10.75
29.38 -148.5
Pb (s) Pb02 (s) PbS04 (s)
0 -277.4 -920.0
0 -217.33 -813.0
64.81 68.6 148.5
26.44 64.64 103.2
SO~- (aq) HSOi (aq)
-909.27 -887.34
-744.53 -755.91
20.1 131.8
-293 -84
-910.94 -1449.36
-856.64 -1307.67
41.84 215.13
44.43 468.98
Substance (form) H2 (g) H (g) H+ (aq) H20 (I) H2 0 (g)
N2 (g) NH3 (g) Na+ (aq) NaCI (s) NaAISi308 (albite) NaAISi206 (jadeite) Ne (g)
Si02 (a quartz) H4Si04 (aq)
V (cm3)
18.068
14.81
-1.2 27.01 100.07 60.40
18.3
22.69
405
Index
Abbott, Larry, 383
Absolute temperature scale, 4-5, 129
Absolute zero, 4-5, 94-95, 146, 148
Absorption, by a surface, 303
by an atom, 293-294
Absorption refrigerator, 130
Abt, Helmut A., 226
Accessible states, 57-58, 67, 76, 225
Acids, 215, 217
Adiabat, 25
Adiabatic compression/expansion, 24-27,
125-126, 159, 175
Adiabatic cooling, 27, 142, 146, 177-178
Adiabatic exponent, 26, 132
Adiabatic rate, 27, 177-178
Adiabatic, relation to isentropic, 112
Adler, Ronald J., 383
Adsorption, 259-260
Age of the universe, 58
Ahlborn, B., 127
7-8, 17,39,43,45,47
liquefaction of, 186, 193-194
Air conditioners, 127-129, 137-138, 141
Albite, 176, 195
Albrecht, Bruce A., 48, 399
Alloys, 186, 191, 194, 198-200, 346, 353
Q (expansion coefficient), 6
Altitude, effect on boiling water, 175
effect on speed of sound, 27
Aluminum, 30, 97
Aluminum silicate, 172, 176
Ambegaokar, Vi nay, 398
Ammonia, 137, 152, 210-213
Andalusite, 172, 176
Anderson, G. M., 399, 404 _
105, 234,374­ Angular momentum, 379
Anharmonic oscillator, 233, 371
Annihilating a system, 33, 149-150
Anorthite, 195
Antiferromagnet, 339, 346
Antifreeze, 198-200
Antiparticles, 297-300, 382
Approximation schemes, 285, 327
Aqueous solutions, 202, 216
Aragonite, 171, 176
Area of a hypersphere, 70-71, 391~393
Argon, 152, 241, 336
Arnaud, J., 270
Ashcroft, Neil W., 272, 400
Asimov, Isaac, 399
Astronomical numbers, 84
Astrophysics, see also Black holes, Cos­ mology, Stars, Sun
Asymptotic expansion, 387
Atkins, P. W., 377, 398, 404
atm (atmosphere), 7, 402
Atmosphere, density of, 8, 120, 228
molecules escaping from, 246
opacity of, 306
solar, 226-227
temperature gradient of, 27, 177-178
Atmospheric pressure, 7, 402
Atomic mass (unit), 8
Atoms, excitation of, 226-227, 293
ATP (adenosine triphosphate), 156
Automobile engine, 131-133
Available work, 150
Average energy, 12-15, 229-231
number of particles, 261, 266-268
pressure, 11 ~ 12
406
Index Average speed, 13, 245-256
values, 11-13, 229-231
Avogadro, Amedeo, 44
Avogadro's number, 7, 44, 61, 67, 119,
210,402
Azeotrope, 195-196
B (bulk modulus), 27
B (magnetic field), 98
B(T) (second virial coefficient), 9
Bagpipes, 27
Baierlein, Ralph, 67
Bailyn, Martin, 400
Balloon, hot-air, 2-3, 8
Balmer lines, 226-227
bar (unit of pressure), 7
Barometric equation, 8, 178
Barrow, Gordon M., 372, 377
Baseball, wavelength of, 362
Basic solution, 215
Battery, 19, 154-155
Battlefield Band, 27
Beckenstein, Jacob, 84
Bed-spring model of a solid, 16
Beginning of time, 83
Benefit/cost ratio, 123, 128
Bernoulli, Daniel, 10
f3 (l/kT), 229
f3 (critical exponent), 186, 346
f3 (expansion coefficient), 6
Betelgeuse, 307
Betts, D. S., 321, 400
Bicycle tire, 14, 26
Big bang, see Early universe
Billiard-ball collision, 246
Binomial distribution, see Two-state sys­
tems
Binomial expansion, 9
Biological applications, 36, 47, 97, 156,
204-205, 259-260, 304, 399
Bird, R. Byron, 337
Black holes, 83-84, 92, 304, 326
Blackbody radiation, 302-307, 359
Block spin transformation, 355-356
Blood, 259-260
Body temperature, 5
Body, radiation from, 304
Bohr magneton, 105, 148, 234, 313
Bohr radius, 227
Bohren, Craig F., 48, 399
Boiling point, effect of solute, 206-.208
Boiling water, 33-35, 175
Boltzmann distribution, 223, 268-269
Boltzmann factor, 223-256, 321-322, 347
Boltzmann, Ludwig, 129
Boltzmann statistics, 220-256, 265, 270
applicability of, 264-265, 268·-269, 271
Boltzmann's constant, 7, 12-13, 75, 402
Bose gases, 290-326
Bose, Satyendra Nath, 263
Bose-Einstein condensation, 144, 315-325
Bose-Einstein distribution, 268-271, 290,
308, 315-316
Bosons, 238, 263, 265, 267-271, 290, 315,
326, 380-383
Boundary between phases, 178
Bowen, N. L., 195
Box, particles in, 252, 255, 290, 368-370
Brass, 191
Breaking even, 124
Brick, dropping, 162
Brillouin function, see Paramagnet
Brush, Stephen G., 340,400
Btu (British thermal unit), 39
Bubbles expanding, 26
Bulk modulus, 27, 159, 275-276
Bull, H. 205
c (specific heat), 28
C (heat capacity), 28
Cailletet, Louis, 141
Calcium carbonate (calcite), 171, 176
Calculus, ix, 384
Callen, Herbert B., 397
Calorie, 19
Campbell, A. J. R. and A. N., 190
Canonical distribution, 223
Canonical ensemble, 230
Carbon dioxide, 16, 137, 167-168,
217,237,306,336
Carbon monoxide, 95, 235-236, 371, 377
poisoning, 259-260
Carbon, phases of, 170-171, 173-174
Carbonic acid, 217
Cards, playing, 52
shuffling, 75-77
Carnot cycle, 125-126, 128
Carnot, Sadi, 125, 129, 148
Carrington, Gerald, 160, 397
Carroll, Bradley W., 37, 399
Casey, Brendan, 383
Cells, biological, 47, 202, 204
407
408
Index Celsius temperature 3-6
Centered-difference approximation, 102
Centigrade, 3
Chandler, David, 397
Chemical equilibrium, 208-219, 290
Chemical potential, 115-120
and Gibbs free energy, 157, 164-165
at equilibrium, 210-211
in quantum statistics, 267-269
of Bose gases, 315-319, 324
of an Einstein solid, 117-119
of Fermi gases, 272, 281-288
of an ideal gas, 118-120, 165-166, 255
of multiparticle system, 251, 270
of gases in blood, 259
of a photon gas, 290
of solute, solvent, 201-202
Chemical thermodynamics, 149, 398
Chemical work, 117
Chlorofluorocarbons, 137-138
Choosing n from N, 51
Chu, Steven, 147
Classical partition function, 239, 256
Classical physics, 239--240, 288-289
Classical thermodynamics, 120
Clausius, Rudolf, 95, 129
Clausius-Clapeyron relation, 172-179
Clay-firing temperature, 4, 293
Clouds, 47,177-179,305
Cluster expansion, 332-333
Clusters of dipoles, 351-355
Clusters of molecules, 181, 332-333
Coefficient of performance, 128, 130-131,
138, 140-141
Coin flipping, 49-52, 67
Colligative properties, 208
Collings, Peter J., 400
Collision time, 42
Collisions, 11, 41-42, 246
Combinations, 51
Combinatorics, 49--55, 92, 279, 322-323,
331-332
Complex functions, 363-366
Composite systems, 249-251
Compressibility, 32, 159, 171, 186
Compression factor, 185
Compression ratio, 131-133
Compression work, 20-26
Computer problems, ix
Computer simulations, 346-356, 400
Computing, entropy creatiDn during, 98
Concentration, and chemical potential,
118, 201-202
Concentration, standard, 155, 404
Condensate, 318
Condensation temperature, 317-320, 325
Condensed matter physics, 400
see also Critical point, Helium, Low
temperatures, Magnetic systems,
Phase transformations, Quantum
statistics, Solid state physics
Condenser, 134-135, 138
Conduction electrons, 38, 271-288, 311
Conduction, of heat, 19, 37-44
Conductivity, electrical, 287
Conductivity, thermal, 38-44
Cone diagrams, 374, 379
Cones, potter's, 4
Configuration integral, 329-333
Conservation of energy, 17-19
Constants, 402
Convection, 19, 27, 37, 177-178, 306
Conversion factors, 402
Cookie-baking temperature, 294
Cooking pasta, 31, 175, 208
Cooking time, at high altitude, 175
Coolant, 198-200
Cooling, see Adiabatic cooling, Evapora­ tive cooling, Refrigerators
COP (coefficient of performance), 128
Copper, 39, 276, 278, 311, 313
Corak, William 311
Corn flakes, 36
Corrections to this book, ix
Correlation function, 354-355
cosh function, 104
Cosmic background radiation, 228, 295­ 300, 359
Cosmological constant, 383
Cosmology, 83, 228, 295-300, 399
Costly energy, 150
Coulomb potential, 373
Counting arrangements, 49-59, 68-71,
262-263, 321-323
Counting wavefunctions, 367, 369
Cox, Keith G., 174
Creating a system, 33, 149-150
Critical exponents, 186, 346, 356
Critical point, 167-169, 184-186, 339
of Ising model, 343, 345-346, 351-356
Crystal structures, 166, 343
Cumulus clouds, 177-178
Index Curie temperature, 169, 339-340, 345
Curie's law, 105
Curie, Pierre, 105
Curtiss, Charles F., 337
Curzon, F. L., 127
Cutoff ratio, 132- 133
Cyanogen, 228
Cyclic processes, 23, 122-141
Dark matter , see Neutrinos
Data, tables of, 136, 140, 143, 401-405
De Broglie wavelength, 252- 253, 264,
336-337, 360-361, 369
De Broglie, Louis, 360
Debye temperature, 310-314
Debye theory, 307- 313
Debye, Peter, 309
Decay, atomic, 293-294
Deficit, national, 84
Definite-energy wavefunctions, 252, 367­ 376, 379
Definition, operational, 1
Degeneracy pressure, 275-277
Degenerate Fermi gases, 272-288
Degenerate states, 221, 224, 227-228, 243,
369, 372
Degrees of freedom , 14- 17, 29, 72, 92,
238-240, 290- 291 , 310, 376-377
Degrees (temperature), 4
b (critical exponent) , 186
~Go (standard ~G), 211
Demon, Maxwell's, 76-77
Density of energy in the universe, 296
Density of states, 280- 282, 316
in a harmonic trap, 325
in a magnetic field, 288
in a semiconductor, 286-287
in two dimensions, 282
of a relativistic gas, 293
Derivatives, numerical, 102
Desalination, 202, 205
Detailed balance, 57, 348
Deuterium (HD , D2), 237-238
Dew point, 177- 178
Dewar, James, 142
Diagrammatic perturbation series 327
331-333, 338 ' ,
Dial thermometer, 4, 6
Diamond, 170- 171, 173- 174, 176
entropy of, 114, 176
heat capacity of, 30, 114, 312 _
Diamond, formation of, 174
Diatomic gases, 15-16, 29--30, 233-238,
255, 371-372, 375-377
Diesel engine, 26, 132-133
Differentials, 18
Differentiating under the integral, 386
Diffraction, 360- 362
Diffusion, 46-48, 67
Diffusive equilibrium, 2-3, 115-116, 120
Dilute solutions, 200-210, 214-217
Dilution refrigerator, 144-145, 320
Dimensional analysis , 70, 83, 278, 285,
294, 302
Dimensionless variables 108 185 246
, , ,
286, 297, 323- 324' Dipoles, magnetic, 52-53, 98- 107, 232­ 234, 339- 356, 378-379
field created by, 148
interactions among, 146, 148, 339- 356
Dirac, Paul A. M., 263
Disorder, 75
Dissociation of hydrogen, 30, 233, 256
Dissociation of water , 208- 210, 214-215
Distinguishable particles, 70, 80- 81, 235­ 237, 249-250, 262, 321-322, 376, 379
Distribution functions, in quantum statis­ tics, 266- 269
Distribution of molecular speeds, 242-247
Dittman, Richard H., 198, 397
Domains, 339, 351-352
Doppler effect , 147, 295
DPPH, 105-107
Dry adiabatic lapse rate, 27
Dry ice, 141, 167
Du Pont, 137
Dulong and Petit, rule of, 29
Dumbbell model of a molecule, 375
Dymond, J. H., 336
e (efficiency), 123
e (fundamental charge) , 373
E (energy of a small system), 223, 230
Early universe, 228, 295- 300, 304
Earth, heat flow from, 40
surface temperature of, 305-306
Earth science, see Geological applications,
Meteorology
Economical numbers, 84
Edge effects, 350
Efficiency, of engines, 123- 127, 132-137
of fuel cells, 154
409
410
Index Efficiency of human body, 36, 156
of incandescent bulb, 304
see also Coefficient of performance
Effusion, 14, 46
Einstein A and B coefficients, 293-294
Einstein, Albert, 263, 293, 359
Einstein relation, 291, 359-361, 363
Einstein solid, 53-55, 107-108, 117-119,
233, 307, 312
in high-T limit, 63-66, 75, 91, 92, 231
in low-T limit, 64, 91, 93, 307, 312
pair of, exchanging energy, 56-60, 64­ 66, 77, 86-88, 224
Electrical conductivity, 38, 287
Electrical work, 19, 21, 152-156
Electrolysis, 152-153
Electromagnetic field, 288-289, 380-381
Electron-positron pairs, 298-300
Electron-volt, 13
Electronic states, 251-252, 375
see also Hydrogen
Electrons, as magnetic dipoles, 52, 105­ 107, 145, 288
diffraction of, 360-361
in chemical reactions, 154-156
in metals, 38, 271-288
in semiconductors, 261, 286-288
wave nature of, 360-361
Elementary particles, 382
Emission, see Blackbody radiation, Spec­
trum
Emissivity, 303-304
Energy, 1-383
capacity, 228
definition of, 17
conservation of, 17-19
exchange of, 2, 56-60, 64-66, 72, 85-91
fluctuations of, 65-66, 72, 231
in quantum mechanics, 367-383
of interacting particles, 17, 240-241,
329, 335, 341
of mixing, 188-192, 195-196
of the vacuum, 382-383
relation to temperature, 2--3, 10, 12-17,
85-92, 229-231
relativistic, 240, 276, 291, 299, 370
tendency to decrease, 162
see also Free energy, Heat, Work
Energy levels, of a harmonic oscillator,
53-54, 289, 370-371
of a hydrogen atom,-221, 373
Energy levels, of a magnetic dipole, 99,
228, 232, 234,378
of a particle in a box, 252, 368-369
of a rotating molecule, 234-238, 376
probabilities of being in, 220-224
Engines, 122-137, 175, 398
English units, 39
Enlightened behavior, 90
Ensembles, 230, 258
Enthalpy, 33, 149-160
capacity, 34
for magnetic systems, 160
in refrigeration, 138-144
in a throttling process, 139-144
of formation, 35, 404
of hydrogen, 143
of nitrogen, 143
of water and steam, 135-137
Entropy, 75-84
alternative definition, 249
analogy to happiness, 89-90
and human action, 76
and information, 76, 98
and phase changes, 171-176, 179
fluid model of, 96-97
in cyclic processes, 122-130
measurement of, 93-95, 112-114, 255
of system plus environment, 161-162
of mixing, 79-81, 187-188, 194
order-of-magnitude estimates of, 79
origin of term, 129
original definition of, 95, 129
relation to heat, 78-79, 92-98, 112-115,
129
relation to multiplicity, 75
relation to pressure, 108-110
relation to temperature, 85-92
residual (at T = 0), 94-95
tabulated values, 95, 136, 140, 404
see also entries under various systems
Environment, 161-162
Environmental science, 399
Enzymes, 156, 212
€ (small amount of energy), 88, 266, 340
EF (Fermi energy), 272
Equation of state, 9, 180
Equilibrium, 2-3, 66, 72-74, 85, 212
diffusive, 115-116
chemical, 208-219
in contrast to kinetics, 37, 44, 213
internal, 20-21
Index Equilibrium, mechanical, 108-110
thermal, 2-3, 85, 110
Equilibrium constant, 212-217, 256
Equipartition theorem, 14~-17, 25, 29-30,
91, 238-240, 290-291, 307, 311, 357
Error function, 387
Errors in this book, ix
Escape of gas from a hole, 14
Escape of photons from a hole, 300-302
Escape velocity, 300
Ethylene glycol, 198-200
Eutectic, 197-200
Evaporation, 176-177
Evaporative cooling, 36, 124, 144
Evaporator, 138
Everest, Mt., 9
Exchanged quantities, 2, 85, 120
Exclusion principle, 263, 275, 339, 380
Exhaling water vapor, 177
Expansion of a gas, 24, 31, 78
Expansion of the universe, 295
Expansion, thermal, 1-6, 28
Expansion work, 21-26
Explosives, 212
Exponential atmosphere, 8, 120, 228
Exponential function, 61-62, 364
Extensive quantities, 163-164, 202
Extent of reaction, 209
I I
(number of degrees of freedom), 15-16
(frequency), 53, 370
I-function, Mayer, 330-335, 339
F (Helmholtz free energy), 150
Factorials, 51, 53, 62, 387-391
Fahrenheit scale, 5
Feldspar, 194-195
Fenn, John B., 398
Fermi-Dirac distribution, 267-288
Fermi energy, 272-288
Fermi, Enrico, 263
Fermi gases, 271-288, 326
relativistic, 276-277, 298-300
two-dimensional, 282, 285
Fermi temperature, 275
Fermions, 237, 263, 265-288, 297-300,
321, 326, 380-383
Ferromagnets, 52, 169, 179, 339-359
low-energy excitations of, 313-314
Fertilizers, 212
Feynman, Richard P., 55, 84, 398
Fiberglass batting, 40
Fick, Adolph Eugen, 47
Fick's laws, 47-48
Field theory, quantum, 380-383
Field, electromagnetic, 290-291, 380-383
Findlay, Alexander, 398
First law of thermodynamics, 19, 123­ 124, 128
First-order phase transition, 169
Flipping coins, 49-52, 67
Fluctuations, 66, 230--231, 261, 344
Fluid, energy as, 17
entropy as, 96-97
Ising model of, 346
near critical point, 168, 356
van der Waals model of, 180-186
Flux of energy, 48
Flux of particles, 47
Food coloring, 46-48
Force, see Interactions, Pressure, Tension
Formation, enthalpy of, 35, 404
free energy of, 152, 404
Fourier, J. B. J., 38
Fourier analysis, 365-366, 381, 394-396
Fourier heat conduction law, 38, 43
Fowler, Ralph, 89
Frautschi, Steven, 83
Free energy, 149-165
see also Gibbs f. e., Helmholtz f. e.
Free expansion, 78--79, 113
Free particles, 272
density of states for, 280
Freezing out, 16, 29-30, 95, 240, 255, 290,
308,310
see also Third law of thermodynamics
Freezing point, of a solution, 208
French, A. P., 357
Freon, 137-138
Frequency, 53, 307-308, 370
in quantum mechanics, 361, 367
Friction, during compression, 21
Fried, Dale G., 323
Frying pan, 40
Fuel cell, 154-156, 158
Fundamental assumption, 57, 270, 323
Fundamental particles, 382
Furnace, electric, 130
g(E) (density of states), 280
G (Gibbs free energy), 150
'I (adiabatic exponent), 26
'I (critical exponent), 186, 346
411
412
Index Gamma function, 387-389
Gas thermometer, 4
Gases, diffusion through, 47
liquefaction of, 141-144
nonideal, 9, 180-186, 328-339
thermal conductivities of, 39-43
virial coefficients of, 336
viscosity of, 45-46
weakly interacting, 328-339
see also Ideal gas
Gasoline engine, 131-133
Gasoline, combustion of, 36
Gaussian functions, 65, 240, 244, 390
integrals of, 384-387, 393
General Motors, 137
Generosity, 89-90, 101
Geological applications, 40, 170-176, 194­ 195, 200, 217, 399
Geometric series, 233, 234, 267, 289, 393
Geothermal gradient, 40
Gibbs factor, 258-260, 262, 266
Gibbs free energy, 150-217
for magnetic systems, 179, 160
how to measure, 151-152
of a dilute solution, 201
of mixtures, 187-192
of reaction, 153, 211
of van der Waals fluid, 182
pressure dependence of, 170-171
tabulated values of, 404
temperature dependence of, 171-172
tendency to decrease, 162
Gibbs, J. Willard, 81
Gibbs paradox, 81
Gibbs sum, 258
Glacier, 174
Glass, thermal conductivity of, 39
Glucose, metabolism of, 36, 156
Gluons, 382
Goates, J. Rex, 199
God, 323
Goldstein, 1. F. and M' 219, 398
Goodstein, David L., 55, 84, 400
Gopal, E. S. R., 255,400
Gould, Harvey, 400
Grand canonical ensemble, 258, 338
Grand free energy, 166, 262, 326
Grand partition function, 258-260, 262,
266-267, 338-339, 346
Grand potential, 166, 262, 326
Graphite, 114, 170-171, 173-174, 176
Grass, 97
Gravitationally bound systems, 36-37,
83-84, 90-92, 97, 276-277
Greediness, 89, 101
Greenhouse effect, 306-307
Griffin, 321
Griffiths, David J., 160, 323, 357
Grimes, Patrick, 154
Grobet, 106
Ground state energy, 53, 224-225, 315,
370-371, 381-383
Ground state occupancy, 315-325
Gutfreund, H., 205
h (Planck's constant), 53, 359
n (h/27r) , 53, 374
H (enthalpy), 33
1t (magnetic field), 160
H20, 166-168
see also Ice, Steam, Water
Haasen, Peter, 398
Haber, Fritz, 212
Hailstorm, 14
Hakonen, Pertti, 102, 146
Hampson, William, 142
Hampson-Linde cycle, 142144
Happiness, entropy as, 89-90
Hard disks, gas of, 347
Hard spheres, gas of, 338
Hardy, G. H., 279
Harmonic oscillator, 16, 53-55, 107-108,
233, 288, 370-372
free energy of, 249
in field theory, 288-289, 381-382
partition function of, 233
two- and three-dimensional, 372
see also Einstein solid, Vibration
Harmonic trap, 265, 270-271, 325
Harte, John, 399
Hawking radiation, 304
Hawking, Stephen, 84
Haystack, needle in, 188
Heat capacity, 28-34
at absolute zero, 95
empirical formula for, 114
measurement of, 31
negative, 36-37, 90
of a Bose gas, 324
of an Einstein solid, 92-93, 107-108,
233, 307, 312
of a Fermi gas, 277--279, 396
Index Heat capacity, of a ferromagnet, 313, 354
of an ideal gas, 29-30, 92, 254
of the Ising model, 354
of nonideal gases, 338
of a paramagnet, 102-103, 105
of a photon gas, 295
of solids, 29-30, 93, 97, 278, 311-312,
357
predicting, 92-93
relation between Cp and Cv, 159
relation to energy fluctuations, 231
relation to entropy, 93-94, 114
rotational, 30, 236-238
tabulated values, 112, 404
vibrational (of gases), 30, 108, 233
Heat, 2, 17-20, 49
cause of, 56, 59
during compression, 23-24
flowing out of earth, 40
rate of flow, 37-44, 126
relation to entropy, 92-98, 112-115
reversible, 82
waste, 122-124, 154
Heat conduction, 37-44
Heat death of the universe, 83
Heat engines, 122-137, 175, 398
Heat equation, 40, 48
Heat exchanger, 142, 144-145
Heat pump, 130
Heisenberg uncertainty principle, 69-70,
364-366
Helium (general and 4He), 17, 22, 43-44,
78, 118, 181, 246, 326, 336, 382
cooling with, 144-145
isotopes of, 94, 168-169
liquefaction of, 141-142
phases of, 168-169
phonons in liquid, 313
superfluid, 168-169, 320-321, 323
Helium-3, 144-145, 168-169, 175, 278­ 279, 285, 288, 320-321
Helium dilution refrigerator, 144-145, 320
Helmholtz free energy, 150-152, 155-165,
224
in the early universe, 299
of a Bose gas, 324
of a harmonic oscillator, 249
of an ideal gas, 254
of a magnetic system, 160
of a photon gas, 290, 297
of a multiparticle system,·251
Helmholtz free energy, cont.
of a nonideal gas, 333
of a van der Waals fluid, 185
relation to Z, 247-248
tendency to decrease, 161-163
Helmholtz, Hermann von, 19
Helsinki University, 146
Hemoglobin, 205, 259-260
Henry's law, 217
Henry, W. E., 106
HFC-134a, 138, 140-141
High-temperature limit, see Equipartition
theorem
Hiking, 36
Hirschfelder, Joseph 0., 337
Historical comments, ix, 4, 10, 19, 76-77,
95, 129, 141-142, 213, 357
Historical references, 400
Hole, escape of gas from, 14
escape of photons from, 300-303
Hooke's law, 115
Human intervention, 76
Humidity, 177, 179
Humpty Dumpty, 83
Hydrofluorocarbons, 138
Hydrogen, 13, 35, 152-155, 158, 319, 323,
336, 401
atomic excitations of, 163, 221, 226­ 227, 373-375
dissociation of, 30, 36, 256
heat capacity of, 29-30, 254
ionization of, 166, 218-219, 227, 260­ 261
liquefaction of, 141-143
normal, 238
rotation of, 236-238
vibration of, 108, 233
Hyperbolic functions, 104
Hypersphere, 70-71, 391-393
Ice, 33, 94, 174-175
Ice cream making, 199
Ideal gas, 6-17, 68-74, 93,121,139,251­ 256
chemical potential of, 118-120, 165-166,
255
diffusion in, 48
energy of, 12, 15-17, 91, 254
entropy of, 77-78, 255
free energy of, 254
heat capacity of, 29-30, 92, 254
413
414
Index Ideal gas, cont.
mixing of, 79
multiplicity of, 68--72
pair of, interacting, 72
partition function of, 251--254
pressure of, 6--7, 110, 255
thermal conductivity of, 41 ~44
viscosity of, 45~46
Ideal gas law, 6~7, 12
correction to, 9, 333~336
derivation of, 110, 255
Ideal mixture, 81, 187~188, 191, 202
Ideal paramagnet, 98, 146, 232-234, 339
Ideal systems, 327
Identical particles, 70~71, 80~81, 236,
250~251, 262-265, 322-323, 376, 379
Igneous rocks, 186, 194-195
Immiscible mixtures, 144, 189-192
Importance sampling, 347
Incandescent light, 303-304
Independent wavefunctions, 69, 367, 379
Indistinguishable, see Identical
Inert gas, effect on vapor pressure, 176
Inexact differentials, 18
Infinite temperature, 101, 103
Infinitesimal changes, 18, 21
Information, and entropy, 76, 84, 98

An Introduction To Thermal Physics Schroeder Pdf Free Download For Windows 10

Inhomogeneities, 181
Initial conditions, 59
in Monte Carlo, 350
of the universe, 83
Insulator, 286
Integrals, 384-396
see also Sums
Intensive quantities, 163-164
Interactions (between systems), 56-60,
72,85-87, 108-110, 115-116, 120,
161-162, 220-223, 257~258
Interactions (between particles), 57, 146,
148, 180-181, 320, 327-356
Interchanging particles, 70, 81, 201, 250
Interference, two-slit, 360-361
Intermolecular potential energy, 17, 241,
329-330, 334~338
Internal combustion engines, 122, 131-133
Internal energy, of gas molecules, 251, 254
Internal partition function, 251 ~256
Internal temperature variations, 93
Inversion temperature, 142-143
Ionization of hydrogen, 166, 218, 227, 297
Ions in solution, 208~210, 214:-215
Iron ammonium alum, 106, 148
Iron, 39, 40, 169, 313~314, 339, 345
Irreversible processes, 49, 56, 59, 82--83
Isentropic, 112
Isentropic compressibility, 159
Ising, Ernst, 340
Ising model, 340-356
exact solution of, 341-343
in one dimension, 341-343, 345, 355
in two dim., 340--341, 343, 346-356
in three dimensions, 343, 353, 355
with external magnetic field, 345
ising program, 348-356
Isotherm, 24--25, 181-186
Isothermal compressibility, 32, 159
Isothermal processes, 24~25, 78, 125-126
Isotopes, 13, 94~95, 237-238, 376
Jacob, Ovid C., 383
Jadeite, 176
Joule (unit), 19
Joule, James P., 19
Joule-Thomson process, 139
k (Boltzmann's constant), 7
K (equilibrium constant), 212
Kartha, Sivan, 154
Keenan, Joseph H., 401
Kelvin temperature scale, 4~6
Kelvin, Baron, 4
Kern, Raymond, 399
Kestin, Joseph, 400
Kiln, 293, 294, 297, 300
Kilowatt-hour, 40
Kinetic energy, 12
Kinetic theory, 43-44
Kinetics, 37
Kittel, Charles, 272, 397, 400
Kroemer, Herbert, 397
Kyanite, 172, 176
£Q (quantum length), 253
L (latent heat), 32, 173
Lamb, John D., 199
Laminar flow, 44-45
Landau, L. D., 397
Langmuir adsorption isotherm, 260
Large numbers, 61, 84
Large systems, 60-67
Laser, 293, 359, 360
Index Laser cooling, 147-148, 319
Latent heat, 32-33, 36, 167, 173-178, 186
Lattice gas, 346
Lattice vibrations, 38, 102
see also Einstein solid, Debye theory
Law of mass action, 212
Le Chatelier's principle, 212
Lead, 5, 17,30, 169, 198-199,312
Lead-acid cell, 154-155
Leadville, Colorado, 9
Harvey S., 20, 76, 400
Legendre transformation, 157
Leighton, Robert B., 398
Lennard-Jones potential, 241, 335-338
Lenz-Ising model, 340
Lever rule, 195
License plates, 55
Lide, David R., 143, 167, 194, 401, 404
Life, and entropy, 76, 97
Lifshitz, E. M., 397
Ligare, Martin, 325
Light bulb, 303-304, 359
Linde, Carl von, 142
Linear independence, 366-367, 369
Linear thermal expansion, 6, 241
Liquefaction of gases, 141-144, 186, 194
Liquid crystals, 168, 198
Liquid helium, see Helium
Liquids, 16-17, 46, 166-208
Liter, 7
Lithium, 102, 107, 228, 319
Logarithm function, 61-63, 389
Lounasmaa, OUi V., 102, 144, 146
Low temperatures, 106-107, 144-148,
169, 319-320, 323
see also Bose-Einstein condensation,
Helium, Paramagnet
Low-energy excitations, 382
Luminosity, of the sun, 305
m (mass), 11
. m (molality), 202
Iv! (magnetization), 99
Macrostate, 50, 56, 59, 74
Magician, 33, 150
Magnetic cooling, 144-146, 148
Magnetic systems, 160, 169, 179
see also Ferromagnets, Paramagnets
Magnetic moment, 99, 234, 321, 378 379
Magnetization, 99, 160
of a ferromagnet, 313-31.4
Magnetization, of the Ising model, 354
of a paramagnet, 99, 102-106, 233-234
Magnons, 313--314, 340, 382
Mallinckrodt, A. John, 20
Mandl, F., 160, 381, 394, 397
Marquisee, J. A., 372, 377
Massalski, Thaddeus, 198
Mather, J. C., 296
Maxwell, James Clerk, 76, 244
Maxwell construction, 183-185
Maxwell relations, 158-159
Maxwell speed distribution, 242-247
Maxwell's Demon, 76-77
Mayer i-function, 330-335, 339
Mayer, Robert, 19
Mean field approximation, 343-346
Mean free path, 41-44, 67
Mechanical interaction, 2-3, 108-110, 120
Melting ice, 33,167-168,174-175
Membrane, semipermeable, 202-205
Mendelssohn, K., 400
Mercury, 1, 3, 6, 32, 159
Merli, P. G., 361
Mermin, N. David, 272, 400
Metabolism of food, 36, 76, 156
Metallurgy, 199
Metals, conduction electrons in, 280
heat capacity of, 311-312
heat conduction in, 38
Metastable states, 166, 170, 348, 351
Meteorology, 27, 177-179, 399
Methane, 35-36, 155-156, 336
Metropolis algorithm, 347-350
Metropolis, Nicholas, 347
Microcanonical ensemble, 230
Microstate, 50
Microwave oven, 18, 20
Millikan, Robert, 44
Miscible mixtures, 187-188, 192-196
Miserly behavior, 9(}---91
Missiroli, G. F., 361
Mixing, 46-48, 186-192
energy of, 188-192, 195-196
entropy of, 79-81, 94-95, 187-188, 209
Mixing clouds, 177
Mixtures, 120, 158, 186--219
ideal, 81, 187-188, 191, 202
miscible, 187-188, 192-196
phase changes of, 186--200, 206-208
nonideal, 144, 188-192
Modes of vibration, 16, 308--309
415
416
Index Molality, 202
Mole, 7,119
Mole fraction, 120
Molecular clouds, 228
Molecular weights, 205
Moment of inertia, 376-377
Momentum, quantum, 363-364
Momentum space, 68-71, 320, 365-366
Momentum transport, 44-46
Monatomic gas, see Gases, Ideal gas
Money, 89-90
Monte Carlo simulation, 327, 346-356
Moon, atmosphere of, 246
Moore, Thomas A., 56, 398
Moran, Michael J., 130, 140, 398
Morrison, Philip and Phylis, 227
Motor oil, 45
Mt. Everest, 9

An Introduction To Thermal Physics Schroeder Pdf Free Download For Mac

Mt. Ogden, 36
Mt. Whitney, 9
J1, (magnetic moment), 99, 378-379
J1,B (Bohr magneton), 105
J1, (chemical potential), 116
J1,0 (standard chem. potential), 165, 202
Multiparticle systems, 249-251, 262-265,
379-382
Multiplicity, 50-75, 92-94, 247
of a classical system, 72
of an Einstein solid, 55
of an ideal gas, 68-72
of interacting systems, 56-60, 64-66, 72
of a paramagnet, 53, 99
of a two-state system, 51
Munoz, James L., 399
Muscle, energy conversion in, 156
Myoglobin, 259-260
Myosin, 156
Neutron, 8, 228, 362, 370
Neutron star, 277
Never, 58, 66
Newbury, N. R., 147
Newton's laws, 11
Nitrogen (N2), 5, 8, 9, 13, 168, 181, 245­ 246, 255, 338
enthalpy table, 143
liquefaction of, 141-142
mixed with oxygen, 193-194
spectrum of, 372, 377
vibrational states of, 371-372
Nitrogen fixation, 211-213
Noble gas solids, 241
Nonideal gases, 180-186, 328-339
Nonideal mixtures, 188-192
Nonideal paramagnets, 146, 148, 339
Nonideal systems, 327-356
Non-quasistatic processes, 112-113
Noodles, cooking, 31, 175, 208
Nordstrom, Darrell Kirk, 399
Normal behavior, 89-90
Normal gas, 264-265
Normal hydrogen, 238
Notation, 18, 401
Nuclear paramagnets, 101-102, 106-107,
146,228
Nuclear power plants, 137
Nuclear spin entropy, 95
Nucleation, 178
Nucleon, 228, 370
Nucleus, as a box of particles, 276, 370
Nucleus, rotation of, 375
Numerical integration, 246, 285, 297, 299,
304, 311, 323, 335
Numerical summation, 325
n (number of moles), 6
Occam's Razor, 323
Occupanc~ 266-269
Ocean thermal gradient, 124-125
Ogden, Mt., 36
Ogden, Utah, 9
Olivene, 194
n (multiplicity), 50
Onnes, Heike Kamerlingh, 142
Onsager, Lars, 343
Onsager's solution, 343, 346, 352
Operational definition, 1, 3-4
Order parameter, 356
Orthohydrogen, 237-238
n (quantum number), 252, 369-375
n (number of particles in a mode), 266
n (average occupancy), 266
N (number of particles), 7
N (number of atoms), 16, 307
N (number of oscillators), 54
NA (Avogadro's number), 7
n-space, 273
Needle in a haystack, 188
Negative temperature, 101-102, 228
Neon, 43, 336
Neutrinos, 263, 297-300, 387
Index Oscillator, see Anharmonic oscillator,
Einstein solid, Harmonic oscillator
Osmosis and osmotic pressure, 203-205
Ostlie, Dale A., 37, 399
Other work, 34, 151
Ott, J. Bevan, 199
Otto cycle, 131-133
Otto, Nikolaus August, 131
Oven, radiation in, 289
Overbar notation, 11-12
Overcounting, see Interchanging particles
Oxygen, 13, 255, 236, 238
binding to hemoglobin, 259-260
dissolving in water, 216-217
liquefaction of, 141-142
mixed with nitrogen, 192-194
Ozone layer, 138
p (momentum), 68
p (pressure), 6
po (standard pressure), 165
Pv (vapor pressure), 167
P (probability), 222
Pa (pascal), 7
Pairs, electron-positron, 298-300
Parahydrogen, 237-238
Paramagnet(s), 52-53, 98-107, 145-146,
232-234, 342
electronic, 105, 145, 148
entropy of, 100-101, 107, 145-146, 148

An Introduction To Thermal Physics Schroeder Pdf free. download full

Fermi gas as, 288
heat capacity of, 102-103, 105
ideal, 98, 146, 232-234, 339
magnetization of, 102-106, 233-234
multi-state, 106, 145, 228, 234
multiplicity of, 53, 64, 67, 99
nonideal, 146, 148, 339
nuclear, 101-102, 106-107, 146, 228
pair of, interacting, 60
Paramagnetic salts, 106, 148
Partial derivatives, 28, 31, 32, 111, 158­ 159
Partial pressure, 120, 165, 211-213, 260
Particles, exchange of, 2-3, 115-120, 257
Particles, quantum, 358, 381-382
Particles, transport of, 46--48
Partition function, 225, 247-251
internal, 251-252, 254-256
of a composite system, 249-251
of a harmonic oscillator, 233, 289
of an ideal gas, 251-256
Partition function, cont.
of the Ising model, 341-342
of a magnetic dipole, 232-234
of a multiparticle system, 262, 321-322
of a weakly interacting gas, 328-333
rotational, 236-238
thermodynamic properties from, 231,
247-248
translational, 251-253, 256
Partition, insertion of, 77, 81
Partitions (combinatoric) 279
Pascal, 7
Pasta, cooking, 31, 175, 208
Pathria, R. K., 345, 397
Pauli exclusion principle, 263, 339, 380
Pauli, Wolfgang, 263
Payne, Cecilia, 227
Peebles, P. J. E., 399
Perfume, 48
Periodic boundary conditions, 350, 353
Periodic table, 403
Peritectic point, 200
Perturbation series, 331- 333
pH, 215, 217
Phase, 166
Phase diagrams, 166-169
carbon, 174
carbon dioxide, 167
ferromagnet, 169
H 2 0, 167
helium, 168
magnetic systems, 179
mixtures, 193-200
nitrogen + oxygen, 194
plagioclase feldspar, 195
showing solubility gap, 190, 192
superconductor, 169
tin + lead, 198
van der Waals fluid, 184
water phenol, 190
Phase transformations, 15, 32-33, 166­ 200, 206-208, 279, 346
classification of, 169
Phenol, 190
(grand free energy), 166
Phonons, 308-313, 381-382
Photoelectric effect, 358-359
Photon gas, 292-297
Photons, 290-304, 359, 371-374, 381-382
Physical constants, 402
1r, series for, 396
417
418
Index Pippard, Ao Bo, 160, 397
pK,215
Plagioclase feldspar, 194-195
Planck distribution, 289-291, 308
Planck length, 382
Planck, Max, 290, 359
Planck spectrum, 292-294, 296
Planck's constant, 53, 359
Playing cards, 52, 75, 77
Plumber's solder, 199
Poise, 45
Poker hands, 52
Polar coordinates, 385
Polarization, of photons, 291
Polarization, of sound waves, 308
Polyatomic molecules, 16, 376-377
Polymers, 114-115
Pomeranchuk cooling, see Helium-3
Porous plug, 138-139
Position space, 68-69
Positrons, 298-300
Potential energy, 14-16, 117,276
intermolecular, 17, 139-142, 329-330,
334-338
Potentials, thermodynamic, 151
Pound, Ro Vo, 102, 107
Power, maximizing, 127
Power of radiation emitted, 302-303
Power plants, 124, 137
Pozzi, Go, 361
Pressure, 10-12, 108-110, 120
constant, 28-35, 149, 162
degeneracy, 275
due to intermolecular forces, 180
effect on equilibrium, 215-216
of a photon gas, 297
of an ideal gas, 6-12,110,255
of a nonideal gas, 333-336
of quantum gases, 324, 326
of a two-dimenisional gas, 347
partial, 120
under ground, 171
variation with altitude, 8
Probability, 49-52, 58, 220-228, 242-247,
257-258
in quantum mechanics, 361-364
relation to entropy, 249
Proteins, 156, 205
Proton, 8, 44, 228, 370
Purcell, Edward Mo, 102
Purcell-Pound experiment, 102, 107, 228
Purpose in 292, 364
of a function, 243, 280,
q (number of energy units), 55
Q (heat), 18
Quadratic energies, 14-17,238-240
Quantization of energy, 369-370, 376
Quantum field theory, 380-383
Quantum gas, 264-265
Quantum length, 253, 337
Quantum mechanics, 53, 69,240,268,
280, 289, 323, 327, 336, 357-383
Quantum numbers, 375, 378
Quantum statistics, 262-326, 336
Quantum volume, 253-255, 264-265
Quarks, 370
Quartz, 8, 176, 217
Quasiparticles, 381-382
Quasistatic, 21, 82, 112-113
R (gas constant), 6-7
R value of insulation, 39-40
Rabbit, 33, 150, 163
Radiation, 2, 19, 37, 288-307
Rainwater, 215, 217
Ramanujan, Srinivasa, 279
Random fluctuations, 66
Random, meaning of, 11
Random processes, 11,49, 57-58
Random walk, 67
Randomness, in quantum mechanics, 361
Rankine cycle, 134-137
Rankine temperature scale, 4-5
Raoult's law, 207-208
Rates of processes, 37-48, 126-127, 212,
300-304
Reference point (for energies), 152
Reference data, 402-405
Reflection of radiation, 303, 305
Refrigerants, 137-138, 140
Refrigeration cycle, 138, 140-141
Refrigerators, 127-131, 137-148, 398
Regenerator, 133-134
Reichl, 337, 397
Reif, 42, 397, 398
Relative humidity, 177, 179
Relativistic energy and momentum, 240,
291, 299, 361, 370
Relativistic gases, 256, 276-277, 298-300
Relaxation time, 2, 5, 102, 348
Index Renormalization group, 355
Reservoir, 122, 161, 221, 247, 257, 266
Reverse osmosis, 205
Reversible processes, 82
Rex, Andrew F., 76,400
Reynolds, William C., 167, 401
Rhodium, magnetic cooling of, 146
Riemann zeta function, 393-396
Rock, density of, 171
thermal conductivity of, 40
Rock, Peter A., 255, 398
Room temperature, 4, 13, 16
Root-mean-square deviation, 231, 366
Root-mean-square speed, 13, 245-246
Rosenbluth, A. W. and M. N., 347
Rotation, molecular, 14-16,228,234-238,
254, 375-377
Rotational heat capacity, 29-30, 236-238
partition function, 236-238
Rotini Tricolore, 31
Royal flush, 52
Rubber, 114-115
Rubidium, 148, 319-320, 323, 325
Rumford, Count, 19
s (state of a system), 222
S (entropy), 75
Sackur-Tetrode equation, 78-81
Saha equation, 218, 260-261
Salt water, 198, 200
Sands, Matthew, 398
Saturation, of air, 177
Sauer, Kenneth, 399
Schrodinger equation, 367-368, 370
Scuba tanks, 24
Seawater, 202, 205, 207-208
Second law of thermodynamics, 59, 74,
76, 85, 120-121
applied to an engine, 123
applied to refrigerator, 128
early versions of, 129
paraphrase of, 124
violation of, 74, 76-77, 81, 97, 303
Second virial coefficient, 9, 326, 334-339
Second-order phase transition, 169
Seismic waves, 308
Semiconductors, 47, 261, 286-288
Semipermeable membrane, 202-205
Shakespeare, 121
Shankar, Ramamurti, 381
Shapiro, Howard N., 130).140, 398
Shaw, G., 381
Shear stress, 45
Shu, Frank H., 399
Shuffling cards, 75-77
SI units, 4
Sign convention for Q, W, 18, 122
Silica, dissolved, 217
Silicon, 261, 287-288
Sillimanite, 172, 176
Silly analogy, 89-90
Silver, heat capacity of, 311
Simulation, Monte Carlo, 346-356
Single-particle state, 251
sinh function, 104
Sirius A and B, 277, 306
Small numbers, 61
Smith, E. Brian, 336, 398
Snoke, D. W., 321
Snow, C. P., 121
Snow, melting of, 33
Sodium, 319
Solar constant, 305
Solder, 198-199
Solid solution, 194
Solid state physics, 272, 312, 400
Solids, 16, 29-30, 38, 46, 54, 107-108
see also Conduction electrons, Debye
theory, Einstein solid, Magnetic
systems, Phase transformations
Solubility gap, 189-192
Solute, 200
Solutions, 194, 200-210, 214-217
Solvent, 200
Sommerfeld expansion, 282-285
Sommerfeld, Arnold, 282
Sound, 21, 308, 312
Space travelers, 14
Specific heat capacity, 28
Spectrum, of nitrogen, 372, 377
of thermal radiation, 292, 300, 303-307
of sunlight, 226-227, 295
of stars, 226-228
Speed of sound, 21, 27, 308, 312
Speeds, of gas molecules, 13, 242-247
Spherical coordinates, 274-275, 291, 300­ 301, 334, 374
in d dimensions, 393
Spin, 52, 95, 105, 227, 234, 261, 263, 266,
377-379
see also Dipoles
Spin waves, 313-314
419
420
Index Spontaneous emission, 293-294
Spontaneous processes, 2-3, 59, 76, 162
Spontaneously broken symmetry, 345
Spreadsheet program, 58
Spring, energy stored in, 14-16, 53, 370,
372
Stability, of mean field solutions, 344-345
of 182-183, 185, 189­ 192, 197
of thermal equilibrium, 91
Standard deviation, 231, 261, 365-366
Standard states, 214, 404
Standing wave, 368
Star 36, 90
Stars, 90, 97, 226-228, 276­ 277, 306-307
State, in quantum mechanics, 357, 362
micro- vs. macro-, 50
of a gas molecule, 69, 252, 328
TQrTlr>lc. vs. system, 250~251
mE~ch,aIll.cs, vii, 121, 220, 337
122, 134-137
134-137
Steel, expansion of, 6
Stefan's law, 302
Stefan-Boltzmann constant, 302
Stiffness, see Bulk modulus
Stimulated 293-294
Stirling 133-134
Stirling's approximation, 62-63, 389-391
Stoichiometric coefficient, 210
Stowe, 397 Stryer, Sublimation, 167
Subnuc1ear 353
~ublscripts, 28
Sulfuric acid, 215
Summary, 120-121
Sums approximated as integrals, 235,
291,316-317,389-391 239-240, Sun, 79, 305~306
conditions at center, 37, 276, 285
energy output of, 304-305
energy received from, 33, 97, 305
life expectancy of, 36, 83
spectrum of, 226-227, 305
surface 219, 226-227, 295
Superconductors, 169, 179, 321
Supercooling, 166
Superfiuid, 168-169, 320-.321
Superfiuid, cont. see also Helium, Helium-3
Surface tension, 178-179
Susceptibility, magnetic, 346
Susskind, Leonard, 84
Symmetry, in Ising model, 345
266
System, in quantum System, magnetic, 160
System state, 251, 321-323
Sze, S. M., 287
t (time), 38
T (temperature), 6
Tc (critical temperature), 184
To (De bye temperature), 310
TF (Fermi 275
Tables of data, 136, 140, 143, 167,401­ 405
tanh function, 104, 346
357
Taylor, Edwin Taylor series, 63, 283, 315
Tea, sipping temperature, 33
347
Teller, A. H. and Temperature, 8592, 120, 129
analogy to ge:ne1:OSlty 89-90 held constant, 161-162,
220-223, 247
infinite, 101, 103
negative, 101-102, 107,228
relation to energy, 28, 49, 85-90
relation to entropy, 85-90, 102
Tension force, 115
Thaddeus, Patrick, 228
Therm,40
Thermal conductivity, 169
Thermal contact, 1-2
Thermal energy, 15
Thermal equilibrium, 85, 91, 110
Thermal excitation of atoms, 226-227
Thermal expansion, 32, 159, 241
Thermal interaction, 85, 120
Thermal physics, vii
Thermal pollution, 124
Thermocouple, 4
Thermodynamic 136, 140, 143, 167,
401, 404-405
Thermodynamic 111-115, 117,
119-120, 257
for G, and 157-158 for <1>, 166
Thermodynamic limit, 66
Index Thermodynamic potentials, 151, 166
Thermodynamics, vii, 37, 120, 219
Thermometers, 1-6, 48, 88-89
Thermos bottle, 2
Third law of thermodynamics,
102, 148, 159, 278
Third virial coefficient, 337, 339
Thompson, Benjamin, 19
Thomson, William, 4, 19, 77
Three-component systems, 199
Throttling process, 138-144
Time and rates of processes, 37-48
Time, beginning of, 83
Time between collisions, 42
Time scales, 2, 56, 58, 76, 77, 102, 326
Tin + lead, 198-199
Tinoco, Ignacio, 399
Tobochnik, Jan, 400
Toilet seats, 48
Touloukian, Y. S., 30
Transcendental equation, 344
Transitions, atomic, 226, 293, 374
molecular, 371-372
Translational motion and energy, 12, 30,
251-256
Transport theory, 37
Trimble, Virginia, 326
Triple point, 167, 176
Troposphere, 27
Tungsten filament, 303-304
Turbine, 134-137
Two Cultures, The, 121
Two-dimensional systems 72 79 121
247, 347 ' , , ,
Bose gas, 325
Fermi gas, 282, 285
Ising model, 340-341, 343, 346-356
magnet, 314
solid, vibrations of, 313
Tw?-particle system, quantum, 379-380
Two-slit interference, 360-361
Two-state systems, 49-53
u (atomic mass unit), 8
U (energy of a large 15, 18, 230
Ultraviolet catastrophe, 288-290, 357
Uncertainty principle, 69-70, 364-366
Unit conversions, 402
Unit Police, 4
Universality, see Critical exponent~
Universe, 58, 83
see also Early Universe
Unstable states, see Stability
Uranium hexafluoride, 13
Utility rates, 40
11Q (quantum volume), 253
V (volume), 6
Vacuum, energy of, 382-383
Van den Bosch, A., 106
Van der Waals, Johannes, 180
Van der Waals 9, 180-186, 328,
338, 344
Van Gerven, 106
Van't Hoff equation, 213
Van't Hoff's formula, 204
Van't Hoff, Jacobus Hendricus 204
Vapor pressure, 167, 175-178, '184
effect of an inert gas, 176
efFect of a solute, 207208
Vapor pressure equation, 175
Vargaftik, N. 143, 401
Velocity space, 243 244
Venus, 307
Verne, Jules, 14
Very large 61, 326
Vibration, 14-17
of gas molecules, 16-17,29 30,54,108,
228, 233, 254, 371 ~372
of a solid, 16, 29-30, 38, 54, 307-313
Violation of the second law, 74, 7677,
81, 97, 303
Virial 9, 326, 334-336. 339
Virial theorem, 37, 97 '
Viscosity, 169
Voltage, 154 155,358-359
Volume, of, 2-3, 72-73, 108­ 110, 120
fluctuations in, 73
held constant, 161-162
of a molecule, 180, 337
of a hydrogen atom, 223, 227
tabulated values, 404
Von Hans Christian, 400
W (work), 18
Wang, James C., 399
Waste heat, 122-124, 154
421
422
Index Water, 3, 6, 8, 19, 23, 28, 32, 35, 39, 45,
47, 181
electrolysis of, 152-153
individual molecule, 17, 228
phases of, 167-168
vapor pressure of, 167, 175-178
Water vapor, as greenhouse gas, 306
Wave equation, 381
Wave-particle duality, 357-362, 381
Wavefunctions, 69, 264, 362-380
during compression, 82
for multiparticle systems, 379-380
ground-state, 320
in a box, 252, 272, 368
Wavelength, in quantum mechanics, 360-­ 363, 368-370
Wave packet , 364-366
Weakly coupled systems, 56
Weakly interacting gases, 328-339
Web site, ix
Weberium, 231
Weinberg, Steven, 300, 383, 399
Weisbrod, Alain, 399
Wet adiabatic lapse rate, 178
Whalley, P. B., 398
Wheatley, John C., 144
White dwarf stars, 276-277, 306, 326
Whitney, Mt., 9
Wieman, Carl E., 147, 319-320
Wien's law, 293
Wiggling atoms, 307
Wilks, J., 321, 400
Wilson, Kenneth G., 355
Window, heat conduction through, 38-39
Window, pressure on, 14
Woolley, Harold W., 30, 143, 401
Work, 17-26
chemical, 117
different definitions of, 20
during heating, 28
electrical, 19, 21, 152-156
magnetic, 160
other, 34-35
to create a system, 150
to make room, 33, 35
Work function, 359
Working substance, 125
World-Wide Web, ix
Yeomans, J. M., 400
Young, David A., 174
Z (partition function), 225
Z (grand partition function), 258
Zemansky, Mark W., 102,198,397,398
Zero-point energy, 53, 371, 381-383
Zeroth law of thermodynamics, 89
Zeta function, 393-396
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If you are fortunate enough not to be a physicist, you will toss this book away in frustration at the lack of proper derivations and explanations. If, however, you are fortunate enough to _be_ a physicist (or even an aspiring one), you will toss this book away in frustration at the flagrant absurdity to the textbook's approach. To put it another way, if you are the type of person who knows material without ever being taught it, _even then_ this book is not for you.. for why would such a person need a textbook?
Should you be in the unfortunate situation that your school requires you to use this text, be sure to frequent the index such that you may find necessary background information in the chapters ahead of what you are presently reading. Under no circumstance should you attempt the homework problems without doing so. This author provides no clear examples, no solutions, and I have found many problems which, to one who reads the book 'linearly,' cannot be solved without information from future problems and sections in the text.