Statistical Mechanics by Konstantin K. Likharev - HTML preview
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ln exp
,
T
N
N
T
k
k
where the summation should be carried out over all possible values of Nk. For the final calculation of this sum, the elementary particle type is essential.
On one hand, for fermions, obeying the Pauli principle, the numbers Nk in Eq. (113) may take only two values, either 0 (the state k is unoccupied) or 1 (the state is occupied), and the summation gives
Nk
k
k
T ln
exp
ln 1 exp
.
(2.114)
k
T
N ,01
T
T
k
66 For a more detailed discussion of this issue, see, e.g., QM Sec. 8.1.
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Now the state occupancy may be calculated from the last of Eqs. (1.62) – in this case, with the (average) N replaced with Nk:
Fermi-
k
1
Dirac
N
(2.115)
k
.
/
distribution
T
k
T , V
e
1
This is the famous Fermi-Dirac distribution, derived in 1926 independently by Enrico Fermi and Paul Dirac.
On the other hand, bosons do not obey the Pauli principle, and for them the numbers Nk can take any non-negative integer values. In this case, Eq. (113) turns into the following equality:
Nk
k
N
k
k
T ln
exp
ln
,
with
exp
.
(2.116)
k
T
N 0
T
N 0
T
k
k
This sum is just the usual geometric series, which converges if < 1, giving 1
T
ln
T ln 1 exp
k
,
for .
(2.117)
k
1
k
T
In this case, the average occupancy, again calculated using Eq. (1.62) with N replaced with Nk , obeys the Bose-Einstein distribution,
Bose-
1
Einstein
k
N
,
for ,
(2.118)
k
k
/ T
distribution
k
T V,
e
1
which was derived in 1924 by Satyendra Nath Bose (for the particular case = 0) and generalized in 1925 by Albert Einstein to the case of arbitrary chemical potential. In particular, comparing Eq. (118) with Eq. (72), we see that harmonic oscillator’s excitations,67 each with energy , may be considered as bosons, with the chemical potential equal to zero. As a reminder, we have already obtained this equality ( = 0) in a different way – see Eq. (93). Its physical interpretation is that the oscillator excitations may be created inside the system, so there is no energy cost of moving them into the system under consideration from its environment.
The simple form of Eqs. (115) and (118), and their similarity (besides “only” the difference of the signs before the unity in their denominators), is one of the most beautiful results of physics. This similarity, however, should not disguise the fact that the energy dependences of the occupancies Nk
given by these two formulas are very much different – see their linear and semi-log plots in Fig. 15.
In the Fermi-Dirac statistics, the level occupancy is not only finite, but is below 1 at any energy, while in the Bose-Einstein it may be above 1, and diverges at k . However, as the temperature is increased, it eventually becomes much larger than the difference ( k – ). In this limit, Nk << 1, so both quantum distributions coincide with each other, as well as with the classical Boltzmann distribution (111) with c = exp{/ T}:
67 As the reader certainly knows, for electromagnetic field oscillators, such excitations are called photons; for mechanical oscillation modes, phonons, etc. It is important, however, not to confuse such mode excitations with the oscillators as such, and be very careful in prescribing to them certain spatial locations – see, e.g., QM Sec. 9.1.
Chapter 2
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Boltzmann
k
N
exp
N
.
(2.119) distribution:
k
,
for
0
T
k
identical
particles
This distribution (also shown in Fig. 15) may be, therefore, understood also as the high-temperature limit for indistinguishable particles of both sorts.
10
0.8
0.6
N
N k
k
1
0.4
0.2
0
0.1
3
2
1
0
1
2
3
3
2
1
0
1
2
3
/ T
/
k
T
k
Fig. 2.15. The Fermi-Dirac (blue line), Bose-Einstein (red line), and Boltzmann (dashed line) distributions for indistinguishable quantum particles. (The last distribution is valid only asymptotically, at N
k << 1.)
A natural question now is how to find the chemical potential participating in Eqs. (115), (118), and (119). In the grand canonical ensemble as such (Fig. 13), with the number of particles variable, the value of is imposed by the system’s environment. However, both the Fermi-Dirac and Bose-Einstein distributions are also approximately applicable (in thermal equilibrium) to systems with a fixed but very large number N of particles. In these conditions, the role of the environment for some subset of N’ << N
particles is essentially played by the remaining N – N’ particles. In this case, may be found by the calculation of N from the corresponding probability distribution and then requiring the result to be equal to the genuine number of particles in the system. In the next section, we will perform such calculations for several particular systems.
For that and other applications, it will be convenient for us to have ready formulas for the entropy S of a general (i.e. not necessarily equilibrium) state of systems of independent Fermi and Bose particles, expressed not as a function of Wm of the whole system, as in Eq. (29), but via the occupancy numbers Nk . For that, let us consider an ensemble of composite systems, each consisting of M >> 1
similar but distinct component systems, numbered by index m = 1, 2, … M, with independent (i.e. not directly interacting) particles – see Fig. 16. Let us assume that though in each of the M component systems, the number N ( m)
()
k
of particles in their k th quantum state may be different, their total number Nk in the composite system is fixed. As a result, the total energy of the composite system is fixed as well, M
M
( m)
( m)
N N const,
E
N
N
,
(2.120)
k
k
k
const
k
k
k
k
m 1
m 1
Chapter 2
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SM: Statistical Mechanics
so an ensemble of many such composite systems (with the same k), in equilibrium, is microcanonical.
According to Eq. (24a), the average entropy Sk per component system in this microcanonical ensemble may be calculated as
ln M
S
k
lim
,
(2.121)
k
M
M
where M
()
k is the number of possible different ways such a composite system (with fixed Nk
) may be
implemented.
component system’s number: 1
2 … m
…
M
k
number of particles in the k th
)
1
(
N
(2)
N
( m)
N
( M )
N
single-particle quantum state:
k
k
k
k
Fig. 2.16. An example of a composite system of N ()
k
particles in
the k th quantum state, distributed between M component systems.
Let us start with the calculation of Mk for Fermi particles – for which the Pauli principle is valid.
Here the level occupancies N ( m)
k
may be only equal to either 0 or 1, so the distribution problem is solvable only if N ()
()
k
M, and evidently equivalent to the choice of Nk balls (in arbitrary order) from the total number of M distinct balls. Comparing this formulation with the definition of the binomial coefficient,68 we immediately get
M
M!
M C
(2.122)
k
N
.
k
( M N )! N !
k
k
From here, using the Stirling formula (again, in its simplest form (27)), we get Fermions:
entropy
S N
ln N N
N
(2.123)
k
k
k
1
k ln1
k ,
where
N
N
k
lim
(2.124)
k
M
M
is exactly the average occupancy of the k th single-particle state in each system, which was discussed earlier in this section. Since for a Fermi system, Nk is always between 0 and 1, its entropy (123) cannot be negative – see the blue line in Fig. 17.
In the Bose case, where the Pauli principle is not valid, the number N ( m) k
of particles in the k th
state in each of the systems is an arbitrary (non-negative) integer. Let us consider N () k
particles and ( M
– 1) partitions (shown by vertical lines in Fig. 16) between M systems as ( M – 1 + N () k
) mathematical
objects ordered along one axis. Each specific location of the partitions evidently fixes all N ( m) k
. Hence
Mk may be calculated as the number of possible ways to distribute the ( M – 1) indistinguishable partitions among these ( M – 1 + N ()
k
) ordered objects, i.e. as the following binomial coefficient:69
68 See, e.g., MA Eq. (2.2).
69 See also MA Eq. (2.4).
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M N 1
( M 1 N )!
M
k
C
k
(2.125)
k
M 1
.
( M )!
1 N !
k
Applying the Stirling formula (27) again, we get the following result,
S N
ln N N
N
(2.126) Bosons:
k
k
k
1
k ln1
k ,
entropy
which again differs from the Fermi case (123) “only” by the signs in the second term, and is valid for any positive Nk - see the red line in Fig. 17.
In the classical limit when the average occupancies Nk of the state is much smaller than 1, the Fermi and Bose expressions for Sk tend to the same Boltzmann limit: S N
N
N
N
N
(2.127)
k
k ln
k
1
ln
,
for
0.
k
k
k
(The last expression may be also obtained from the functionally similar Eq. (29), by considering an ensemble of systems consisting of just one classical particle each, so Em k and Wm Nk .) 10
S
) k 1
Fig. 2.17. Entropy of particles in
a quantum state as a function of
its average occupancy N k, for
fermions (blue line) and bosons
(red line). The dashed line shows
their common asymptote at N k
0, given by the first of Eqs.
0.1
(127).
0.01
0.1
1
10
100
Nk
Expressions (123) and (126) are valid for an arbitrary (possibly, non-equilibrium) case; they may be also used for an alternative derivation of the Fermi-Dirac (115) and Bose-Einstein (118) distributions, which are valid only in equilibrium. For that, we may use the method of Lagrange multipliers, requiring (just like it was done in Sec. 2) the total entropy of a system of N independent, similar particles, S S ,
(2.128)
k
k
considered as a function of state occupancies Nk, to attain its maximum, under the conditions of the fixed total number of particles N and total energy E:
N N const,
.
(2.129)
k
N
E const
k
k
k
k
The completion of this calculation is left for the reader’s exercise.
Chapter 2
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2.9. Exercise problems
2.1. A famous example of macroscopic irreversibility was suggested in 1907 by P. Ehrenfest.
Two dogs share 2 N >> 1 fleas. Each flea may jump onto another dog, and the rate of such events (i.e.
the probability of jumping per unit time) does not depend either on time or on the location of other fleas.
Find the time evolution of the average number of fleas on a dog, and of the flea-related part of the total dogs’ entropy (at arbitrary initial conditions), and prove that the entropy can only grow.70
2.2. Use the microcanonical distribution to calculate thermodynamic properties (including the entropy, all relevant thermodynamic potentials, and the heat capacity) of a two-level system in thermodynamic equilibrium with its environment, at a temperature T that is comparable with the energy gap . For each variable, sketch its temperature dependence, and find its asymptotic values (or trends) in the low-temperature and high-temperature limits.
Hint: The two-level system is any quantum system with just two different stationary states, whose energies (say, E 0 and E 1) are separated by a gap E 1 – E 0. Its popular (but by no means the only!) example is the spin-½ of a particle, e.g., an electron, in an external magnetic field.71
2.3. Solve the previous problem using the Gibbs distribution. Also, calculate the probabilities of the energy level occupation, and give physical interpretations of your results, in both temperature limits.
2.4. A quantum spin-½ particle with a gyromagnetic ratio is placed into an external magnetic field H = H n z. Neglecting the possible orbital motion of the particle, calculate its average magnetization m z as a function H, and in particular its low-field magnetic susceptibility , in thermal equilibrium at temperature T. Calculate the same characteristics for a classical magnetic dipole m of a fixed magnitude m0, free to change its direction in space, and compare the results.
Hint: The low-field magnetic susceptibility of a single particle is defined72 as
m z
.
H 0
H
2.5.* Calculate the weak-field magnetic susceptibility of a hydrogen atom, at room temperature.
Is this response to the field paramagnetic or diamagnetic? Compare the result with the estimated susceptibility of a hydrogen molecule H2.
2.6. N similar stiff rods of length l are connected
with the joints that allow for free 3D rotation, to form a
T
chain – see the figure on the right. The chain, in thermal
T
l
70 This is essentially a simpler (and funnier :-) version of the particle scattering model used by L. Boltzmann to prove his famous H- theorem (1872). Besides the historical significance of that theorem, the model used in it (see Sec. 6.2 below) is as cartoonish, and not more general.
71 See, e.g., QM Secs. 4.6 and 5.1, in particular, Eq. (4.167).
72 This “atomic” (or “molecular”) susceptibility should not be confused with the “volumic” susceptibility m
Mz/H, where M is the magnetization, i.e. the magnetic moment of a unit volume of a system – see, e.g., EM
Eq. (5.111). For a uniform medium with n N/ V non-interacting dipoles per unit volume, m = n.
Chapter 2
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equilibrium at temperature T, is stretched by a fixed force T. Calculate the spring constant of the chain in the elastic limit T 0.
2.7. Calculate the low-field magnetic susceptibility of a particle with an arbitrary (either integer or semi-integer) spin s, neglecting its orbital motion. Compare the result with the solution of Problem 4.
Hint: Quantum mechanics73 tells us that the Cartesian component m z of the magnetic moment of such a particle, in the direction of the applied field, has (2 s + 1) stationary values: m m ,
m
with
s, s ,...,
1
s ,
1 s ,
z
s
s
where is the gyromagnetic ratio of the particle, and is Planck’s constant.
2.8. Analyze the possibility of using a system of non-interacting spin-½ particles placed into a controllable external magnetic field, for refrigeration.
2.9. A rudimentary “zipper” model of DNA replication is a chain
of N links that may be either open or closed – see the figure on the right.
Opening a link increases the system’s energy by > 0; a link may 1 2 ... n
...
n 1
N
change its state (either open or closed) only if all links to the left of it are open, while all those on the right of it, are closed. Calculate the average number of open links in thermal equilibrium, and analyze its temperature dependence, especially for the case N >> 1.
2.10. Use the microcanonical distribution to calculate the average entropy, energy, and pressure of a classical 3D particle of mass m, with no internal degrees of freedom, free to move in volume V, at temperature T.
Hint: Try to make a more accurate calculation than has been done in Sec. 2.2 for the system of N
harmonic oscillators. For that, you would need to know the volume Vd of a d-dimensional hypersphere of a unit radius. To avoid being too cruel, I am providing it:
d / 2
d
V
d
1,
2
where () is the gamma function.74
2.11. Solve the previous problem using the Gibbs distribution.
2.12. Calculate the average energy, entropy, free energy, and the equation of state of a classical 2D particle (without internal degrees of freedom), free to move within area A, at temperature T, starting from:
(i) the microcanonical distribution, and
(ii) the Gibbs distribution.
Hint: For the equation of state, make the appropriate modification of the notion of pressure.
73 See, e.g., QM Sec. 5.7, in particular Eq. (5.169).
74 For its definition and main properties, see, e.g., MA Eqs. (6.6)-(6.9).
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2.13. A quantum particle of mass m is confined to free motion along a 1D segment of length a.
Using any approach you like, calculate the average force the particle exerts on the “walls” (ends) of such a “1D potential well” in thermal equilibrium, and analyze its temperature dependence, focusing on the low-temperature and high-temperature limits.
Hint: You may consider the series
2
exp
n a known function of . 75
n1
2.14. Rotational properties of diatomic molecules (such as N2, CO, etc.) may be reasonably well described by the so-called dumbbell model: two point particles, of masses m 1 and m 2, with a fixed distance d between them. Ignoring the translational motion of the molecule as a whole, use this model to calculate its heat capacity, and spell out the result in the limits of low and high temperatures. Is your solution valid for the so-called homonuclear molecules consisting of two similar atoms, such as H2, O2, N2, etc.?
2.15.* Modify the solution of the previous problem for homonuclear molecules. Specifically, consider the cases of molecules H2 and N2. For the first of them, compute the equilibrium ratio of the number of the ortho- and parahydrogen molecules as a function of temperature.
Hint: Use the value of d that gives the experimentally observed difference of 1.455 kJ/mol between the ground state energies of these two hydrogen species (“spin isomers”).
2.16. Calculate the heat capacity of a heteronuclear diatomic molecule by using the simple model described in Problem 14, but now assuming that the rotation is confined to one plane.76
2.17. A classical, rigid, strongly elongated body (such as a thin needle) is free to rotate about its center of mass and is in thermal equilibrium with its environment. Are the angular velocity vector and the angular momentum vector L, on average, directed along the elongation axis of the body, or normal to it?
2.18. Two similar classical electric dipoles, of a fixed magnitude d, are separated by a fixed distance r. Assuming that each dipole moment vector d may point in any direction and that the system is in thermal equilibrium, write general expressions for its statistical sum Z, average interaction energy E, heat capacity C, and entropy S, and calculate them explicitly in the high-temperature limit.
2.19. A classical 1D particle of mass m, residing in the potential well U x
x ,
with 0 ,
is in thermal equilibrium with its environment, at temperature T. Calculate the average values of its potential energy U and the full energy E:
(i) directly from the Gibbs distribution, and
(ii) by using the virial theorem of classical mechanics77 and the equipartition theorem.
75 It may be reduced to the so-called elliptic theta-function 3( z, ) for a particular case z = 0 – see, e.g., Sec. 16.27
in the Abramowitz-Stegun handbook cited in MA Sec. 16(ii). However, you do not need that (or any other) handbook to solve this problem.
76 This is a reasonable model of the constraints imposed on small atomic groups (e.g., ligands) by their atomic environment inside some large molecules.
Chapter 2
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2. 20. For a slightly anharmonic classical 1D oscillator with mass m and potential energy
U x
2
3
x x
2
with a small coefficient , in thermal equilibrium with its environment, calculate: (i) the statistical average of the coordinate x, and
(ii) the deviation of the heat capacity from its basic value C =1,
in the first (linear) approximation in low temperature T.
2.21. A small conductor (in this context, usually called the
single-electron island) is placed between two conducting
C
electrodes, with voltage V applied between them. The gap between
one of the electrodes and the island is so narrow that electrons may V
"island"
tunnel quantum-mechanically through this gap (the “weak tunnel
Q -ne
junction”) – see the figure on the right. Calculate the average
n
tunnel
C
junction
charge of the island as a function of V at temperature T.
0
Hint: The quantum-mechanical tunneling of an electron
through a weak junction78 between two macroscopic conductors and its subsequent energy relaxation, may be treated as a single inelastic (energy-dissipating) event, so the only energy essential for the thermal equilibrium of the system is its electrostatic potential energy.
2.22. A lumped LC circuit (see the figure on the right) is in thermodynamic equilibrium with its environment. Calculate the r.m.s. V
C
L
fluctuation V V 21/2 of the voltage across it, for an arbitrary ratio T/, where = ( LC)-1/2 is the resonance frequency of this “tank circuit”.
2.23. Derive Eq. (92) from simplistic arguments, by representing the blackbody radiation as an ideal gas of photons treated as classical ultra-relativistic particles. What do similar arguments give for an ideal gas of classical but non-relativistic particles?
2.24. Calculate the enthalpy, the entropy, and the Gibbs energy of blackbody electromagnetic radiation in thermal equilibrium with temperature T inside volume V, and then use these results to find the law of temperature and pressure drop at an adiabatic expansion.
2.25. As was mentioned in Sec. 6(i), the relation between the temperature T of the visible Sun’s surface and that ( T o) of the Earth’s surface follows from the balance of the thermal radiation they emit.
Prove that the experimentally observed relation indeed follows, with good precision, from a simple model in which the surfaces radiate as perfect black bodies with constant temperatures.
Hint: You may pick up the experimental values you need from any reliable source.
77 See, e.g., CM Problem 1.12.
78 In this particular context, the adjective “weak” denotes a junction with a tunneling transparency so low that the tunneling electron’s wavefunction loses its quantum-mechanical coherence before the electron has a chance to tunnel back. In a typical junction of a macroscopic area, this condition is fulfilled if its effective resistance is much higher than the quantum unit of resistance (see, e.g., QM Sec. 3.2), R Q /2 e 2 6.5 k.
Chapter 2
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2.26. If the surface is not perfectly radiation-absorbing (“black”), the electromagnetic power of its thermal radiation differs from the Planck radiation law by a frequency-dependent factor < 1, called emissivity. Prove that such a surface reflects the (1 – ) fraction of the incident radiation.
2.27. If two black surfaces, facing each other, have different
temperatures (see the figure on the right), then according to the Stefan radiation law (89), there is a net flow of thermal radiation, from the hotter T
net
P
T T
surface to the colder one:
1
2
1
net
P
4
4
T T .
1
2
A
For many applications, notably including most low-temperature
experiments, this flow is detrimental. One way to suppress it is to reduce the emissivity (for its definition, see the previous problem) of both surfaces – say by covering them with shiny metallic films.
An alternative way toward the same goal is to place, between the surfaces, a thin layer (usually called the thermal shield), with a low emissivity of both surfaces – see the dashed line in Fig. above. Assuming that the emissivity is the same in both cases and neglecting its possible dependence on the angle and frequency, find out which way is more efficient.
2.28. Two perfectly reflecting parallel plates of area A are separated by a free-space gap of a constant thickness t << A 1/2. Calculate the energy of the thermally-induced electromagnetic field inside the gap in thermal equilibrium, with temperature T in the range
c
c
T
.
A 1/ 2
t
Does the field push the plates apart?
2.29. Use the Debye theory to estimate the specific heat of aluminum at room temperature (say, 300 K) and express the result in the following popular units:
(i) eV/K per atom,
(ii) J/K per mole, and
(iii) J/K per gram.
Compare the last number with the experimental value (from a reliable source).
2.30. Low-temperature specific heat of some solids has a considerable contribution from the thermal excitation of spin waves, whose dispersion law at 0 scales as k 2.79 Neglecting anisotropy, calculate the temperature dependence of this contribution to CV at low temperatures, and discuss conditions of its experimental observation.
Hint: Just as the photons and phonons discussed in section 2.6, the quantum excitations of spin waves (called magnons) may be considered non-interacting bosonic quasiparticles with zero chemical potential, whose statistics obeys Eq. (2.72).
79 Note that the same dispersion law is typical for bending waves in thin elastic rods – see, e.g., CM Sec. 7.8.
Chapter 2
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2.31. Derive a general expression for the specific heat of a very long
d
d
straight chain of similar particles of mass m, confined to move only in the m
m
m
direction of the chain and elastically interacting with effective spring
constants – see the figure on the right. Spell out the result in the limits of very low and very high temperatures. Would using the Debye approximation change these results?
2
2
d
Hint: You may like to use the following integral:80
.
sinh 2
6
0
2.32. Use the Debye approximation to obtain a general expression for the longitudinal phonon contribution to the specific heat of a stand-alone monatomic layer of an elastic material (such as graphene). Find its explicit temperature dependence at T 0.
2.33. Calculate the r.m.s. thermal fluctuation of an arbitrary point of a uniform guitar string of length l, stretched by force T, at temperature T. Evaluate your result for l = 0.7 m, T = 103 N, and room temperature.
sin2
n 1
Hint: You may like to use the following series:
, for 0 1.
2
n 1
n
2
2.34. Use the general Eq. (123) to re-derive the Fermi-Dirac distribution (115) for a system in equilibrium.
2.35. Each of two identical particles, not interacting directly, may be in any of two quantum states, with the single-particle energies equal to 0 and . Write down the statistical sum Z of the system, and use it to calculate its average total energy E at temperature T, for the cases when the particles are:
(i) distinguishable (say, by their spatial positions);
(ii) indistinguishable fermions;
(iii) indistinguishable bosons.
Analyze and interpret the temperature dependence of E for each case, assuming that > 0.
2.36. Each of N >> 1 indistinguishable fermions has two non-degenerate energy levels separated by gap . Calculate the chemical potential of their system in thermal equilibrium at temperature T, if the direct interaction of the particles is negligible.
80 It may be reduced, via integration by parts, to the table integral MA Eq. (6.8d) with n = 1.
Chapter 2
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Chapter 3. Ideal and Not-So-Ideal Gases
In this chapter, the general principles of thermodynamics and statistics, discussed in the previous two chapters, are applied to examine the basic physical properties of gases, i.e. collections of identical particles (for example, atoms or molecules) that are free to move inside a certain volume, either not interacting or weakly interacting with each other. We will see that due to the quantum statistics, properties of even the simplest, so-called ideal gases, with negligible direct interactions between particles, may be highly nontrivial.
3.1. Ideal classical gas
Direct interactions of typical atoms and molecules are well localized, i.e. rapidly decreasing when the distance r exceeds a certain scale r 0. In a gas of N particles inside volume V, the average distance r ave between the particles is ( V/ N)1/3. As a result, if the gas density n N/ V = ( r ave)-3 is much lower than r -3
3
0 , i.e. if nr 0 << 1, the chance for its particles to approach each other and interact significantly is rather small. The model in which such direct interactions are completely ignored is called the ideal gas.
Let us start with a classical ideal gas, which may be defined as the ideal gas in whose behavior the quantum effects are also negligible. As was discussed in Sec. 2.8, the condition of that is to have the average occupancy of each quantum state low:
N
1.
(3.1)
k
It may seem that we have already found all properties of such a system, in particular, the equilibrium occupancy of its states – see Eq. (2.111):
k
N
const exp
.
(3.2)
k
T
In some sense this is true, but we still need, first, to see what exactly Eq. (2) means for the gas, i.e. a system with an essentially continuous energy spectrum, and, second, to show that, rather surprisingly, the particles’ indistinguishability affects some properties of even classical gases.
The first of these tasks is evidently easiest for gas out of any external fields and with no internal degrees of freedom.1 In this case, k is just the kinetic energy of the particle, which is an isotropic and parabolic function of p:
p
p 2
2
p 2 p 2
x
y
z
.
(3.3)
k
2 m
2 m
Now we have to use two facts from other fields of physics, hopefully well known to the reader. First, in quantum mechanics, the linear momentum p is associated with the wavevector k of the de Broglie wave, 1 In more realistic cases when particles do have internal degrees of freedom, but they are all in a certain (say, ground) quantum state, Eq. (3) is valid as well, with k counted from the fixed (e.g., ground-state) internal energy.
The effect of thermal excitation of the internal degrees of freedom will be discussed at the end of this section.
© K. Likharev
SM: Statistical Mechanics
p = k. Second, the eigenvalues of k for any waves (including the de Broglie waves) in free space are uniformly distributed in the momentum space, with a constant density of states, given by Eq. (2.82): dN
gV
dN
gV
states
states
,
i.e.
,
(3.4)
3
3
3
3
d k
(2 )
d p
(2
)
where g is the degeneracy of the particle’s internal states (for example, for all spin-½ particles, the spin contribution to the internal degeneracy g = 2 s + 1 = 2). Even regardless of the exact proportionality coefficient between dN states and d 3 p, the very fact that this coefficient does not depend on p means that the probability dW to find the particle in a small region d 3 p = dp 1 dp 2 dp 3 of the momentum space is proportional to the right-hand side of Eq. (2), with k given by Eq. (3): 2
2
2
2
p
p p p
Maxwell
3
1
2
3
dW C exp
d p C exp
dp dp dp .
(3.5)
distribution
1
2
3
2 mT
2 mT
This is the famous Maxwell distribution.2 The normalization constant C may be readily found from the last form of Eq. (5), by requiring the integral of dW over all the momentum space to equal 1.
Namely, such integral is evidently a product of three similar 1D integrals over each Cartesian component pj of the momentum ( j = 1, 2, 3), which may be readily reduced to the well-known dimensionless Gaussian integral,3 so we get
3
3
2
p
j
2
C exp
dp
(3.6)
j
2 mT 1/2 e
d 2 mT 3/2.
2 mT
As a sanity check, let us use the Maxwell distribution to calculate the average energy corresponding to each half-degree of freedom:
2
2
2
2
2
2
p
p
p
p
p
j
j
j
j
j
1/ 3
dW C
exp
1/ 3
dp j C
exp
'
dp j
2 m
2 m
2 m
2 mT
2
'
mT
(3.7)
T
2
2
e
d.
1/ 2
The last, dimensionless integral equals 1/2/2,4 so, finally,
2
2
p
mv
j
j
T
(3.8)
2 m
2
2
2 This formula had been suggested by J. C. Maxwell as early as 1860, i.e. well before the Boltzmann and Gibbs distributions were developed. Note also that the term “Maxwell distribution” is often associated with the distribution of the particle momentum (or velocity) magnitude,
2
2
p
mv
dW 4
2
Cp exp
dp 4
3 2
Cm v exp
dv,
0
with p, v ,
2 mT
2
T
which immediately follows from the first form of Eq. (5), combined with the expression d 3 p = 4 p 2 dp due to the spherical symmetry of the distribution in the momentum/velocity space.
3 See, e.g., MA Eq. (6.9b).
4 See, e.g., MA Eq. (6.9c).
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This result is (fortunately :-) in agreement with the equipartition theorem (2.48). It also means that the r.m.s. velocity of each particle is
1/ 2
1/ 2
3
1/ 2
1/ 2
T
2
2
v
v
v
3 2
v
(3.9)
j
j
3 .
j 1
m
For a typical gas (say, for N2, our air’s main component), with m 28 m p 4.710-26 kg, this velocity, at room temperature ( T = k B T K k B300 K 4.110-21 J) is about 500 m/s, comparable with the sound velocity in the same gas – and well above with the muzzle velocity of a typical handgun bullet. Still, it is measurable using even simple table-top equipment (say, a set of two concentric, rapidly rotating cylinders with a thin slit collimating an atomic beam emitted at the axis) that was available in the end of the 19th century. Experiments using such equipment gave convincing early confirmations of the Maxwell distribution.
This is all very simple (isn’t it?), but actually the thermodynamic properties of a classical gas, especially its entropy, are more intricate. To show that, let us apply the Gibbs distribution to a gas portion consisting of N particles rather than just one of them. If the particles are exactly similar, the eigenenergy spectrum { k} of each of them is also exactly the same, and each value Em of the total energy is just the sum of particular energies k( l) of the particles, where k( l), with l = 1, 2, … N, is the number of the energy level on which the l th particle resides. Moreover, since the gas is classical, Nk
<< 1, the probability of having two or more particles in any state may be ignored. As a result, we can use Eq. (2.59) to write
E
m
1
k ( l )
Z exp
exp
...
exp
,
(3.10)
k ( l )
m
T k( l)
T l
k )1
( k (2)
k ( N )
T
l
where the summation has to be carried over all possible states of each particle. Since the summation over each set { k( l)} concerns only one of the operands of the product of exponents under the sum, it is very tempting to complete the calculation as follows:
N
Z Z
exp
k )
1
(
exp
k (2)
...
exp
k ( N )
exp
k
, (3.11)
dist
k )
1
(
T k(2)
T k( N)
T k
T
where the final summation is over all states of one particle. This formula is indeed valid for distinguishable particles.5 However, if the particles are indistinguishable (again, meaning that they are internally identical and free to move within the same spatial region), Eq. (11) has to be modified by what is called the correct Boltzmann counting:
N
Correct
1
Boltzmann
Z
exp
k
,
(3.12)
counting
N!
k
T
that considers all quantum states different only by particle permutations, as the same state.
5 Since, by our initial assumption, each particle belongs to the same portion of gas, i.e. cannot be distinguished from others by its spatial position, this requires some internal “pencil mark” for each particle – for example, a specific structure or a specific quantum state of its internal degrees of freedom.
Chapter 3
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This expression is valid for any set { k} of eigenenergies. Now let us use it for the translational 3D motion of free particles, taking into account that the fundamental relation (4) implies the following rule for the replacement of a sum over quantum states of such motion with an integral:6
gV
3
gV
...
... dN
...
...
.
(3.13)
states
d k
3
3
d 3 p
k
(2 )
(2 )
In application to Eq. (12), this rule yields
N
3
2
1 gV
p
j
Z
exp
.
(3.14)
3
dp
j
N! (2 )
2
mT
The integral in the square brackets is the same one as in Eq. (6), i.e. is equal to (2 mT)1/2, so finally N
N
1
3 / 2
gV
1
3 / 2
mT
Z
(2 mT )
gV
.
(3.15)
N
3
! (2 )
N
!
2
2
Now, assuming that N >> 1,7 and applying the Stirling formula, we can calculate the gas’ free energy: 1
V
F T ln
NT ln
Nf ( T ),
(3.16a)
Z
N
with8
3 / 2
mT
Sackur –
f ( T ) T ln g
.
(3.16b) Tetrode
2
1
2
formula
The first of these relations exactly coincides with Eq. (1.45), which was derived in Sec. 1.4 from the equation of state PV = NT, using thermodynamic identities. At that stage, this equation of state was just postulated, but now we can derive it by calculating the pressure from Eq. (16a), using the second of Eqs. (1.35):
F
NT
P
.
(3.17)
V
V
T
So, the equation of state of the ideal classical gas, with density n N/ V, is indeed given by Eq. (1.44): NT
P
nT .
(3.18)
V
Hence we may use Eqs.(1.46)-(1.51), derived from this equation of state, to calculate all other thermodynamic variables of the gas. For example, using Eq. (1.47) with f( T) given by Eq. (16b), for the internal energy and the specific heat of the gas we immediately get
6 As a reminder, we have already used this rule twice in Sec. 2.6, with particular values of g.
7 For the opposite limit when N = g = 1, Eq. (15) yields the results obtained, by two alternative methods, in the solutions of Problems 2.8 and 2.9. Indeed, for N = 1, the “correct Boltzmann counting” factor N! equals 1, so that the particle distinguishability effects vanish – naturally.
8 This formula was derived (independently) by O. Sackur and H. Tetrode as early as in 1911, i.e. well before the final formulation of quantum mechanics in the late 1920s.
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df T 3
C
E
V
1
3
E N f T
T
NT ,
c
,
(3.19)
V
dT 2
N
N T
2
V
in full agreement with Eq. (8) and hence with the equipartition theorem.
Much less trivial is the result for entropy (essentially, conjectured in Sec. 1.4):
F
V
df T
( )
S
N ln
.
(3.20)
T V
N
dT
This formula provides the means to resolve the following gas mixing paradox (sometimes called the
“Gibbs paradox”). Consider two volumes, V 1 and V 2, separated by a partition, each filled with the same gas, with the same density n, at the same temperature T, and hence with the same pressure P. Now let us remove the partition and let the gas portions mix; would the total entropy change? According to Eq.
(20), it would not, because the ratio V/ N, and hence the expression in the square brackets is the same in the initial and the final state, so the entropy is additive, as any extensive variable should be. This makes full sense if the gas particles in both parts of the volume are truly identical, i.e. the partition’s removal does not change our information about the system. However, let us assume that all particles are distinguishable; then the entropy should clearly increase because the mixing would decrease our information about the system, i.e. increase its disorder. A quantitative description of this effect may be obtained using Eq. (11). Repeating for Z dist the calculations made above for Z, we readily get a different formula for entropy:
df ( T )
3 / 2
mT
dist
S
N ln V
,
f ( T ) T ln g
.
(3.21)
dist
dist
dT
2
2
Please notice that in contrast to the S given by Eq. (20), this entropy includes the term ln V
instead of ln( V/ N), so S dir is not proportional to N (at fixed temperature T and density N/ V). While for distinguishable particles this fact does not present any conceptual problem, for indistinguishable particles it would mean that entropy was not an extensive variable, i.e. would contradict the basic assumptions of thermodynamics. This fact emphasizes again the necessity of the correct Boltzmann counting in the latter case.
Using Eq. (21), we can calculate the change of entropy due to mixing two gas portions, with N 1
and N 2 distinguishable particles, at a fixed temperature T (and hence at unchanged function f dist): V V
V V
S
N N ln( V V ) N ln V N ln V N ln
N ln
,
(3.22)
dist
1
2
1
2
1 1
2
2
1
2
1
2
1
2
V
V
1
2
so the change is positive even for V 1/ N 1 = V 2/ N 2. Note that for a particular case, V 1 = V 2 = V/2, Eq. (22) reduces to the simple result, S dist = ( N 1 + N 2) ln2, which may be readily understood in terms of the information theory. Indeed, allowing each particle of the total number N = N 1 + N 2 to spread to a twice larger volume, we lose one bit of information per particle, i.e. I = ( N 1 + N 2) bits for the whole system.
Let me leave it for the reader to show that Eq. (22) remains valid if particles in each sub-volume are indistinguishable from each other but different from those in another sub-volume, i.e. for mixing of two different gases.9 However, it is certainly not applicable to the system where all particles are identical, 9 By the way, if an ideal classical gas consists of particles of several different sorts, its full pressure is a sum of independent partial pressures exerted by each component – the so-called Dalton law. While this fact was an Chapter 3
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SM: Statistical Mechanics
stressing again that the correct Boltzmann counting (12) does indeed affect the gas entropy, even though it may be not as consequential as the Maxwell distribution (5), the equation of state (18), and the average energy (19).
In this context, one may wonder whether the change (22) (called the mixing entropy) is experimentally observable. The answer is yes. For example, after free mixing of two different gases, and hence increasing their total entropy by S dist, one can use a thin movable membrane that is semipermeable, i.e. whose pores are penetrable for particles of one type only, to separate them again, thus reducing the entropy back to the initial value, and measure either the necessary mechanical work
W = T S dist or the corresponding heat discharge into the heat bath. Practically, measurements of this type are easier in weak solutions 10 – systems with a small relative concentration c << 1 of particles of one sort ( solute) within much more abundant particles of another sort ( solvent). The mixing entropy also affects the thermodynamics of chemical reactions in gases and liquids.11 Note that besides purely thermal-mechanical measurements, the mixing entropy in some conducting solutions ( electrolytes) is also measurable by a purely electrical method, called cyclic voltammetry, in which a low-frequency ac voltage, applied between two solid-state electrodes embedded in the solution, is used to periodically separate different ions, and then mix them again.12
Now let us briefly discuss two generalizations of our results for ideal classical gases. First, let us consider such gas in an external field of potential forces. It may be described by replacing Eq. (3) with 2
p
k
U (r ) ,
(3.23)
k
2
k
m
where r k is the position of the k th particular particle, and U(r) is the potential energy of the particle. If the potential U(r) is changing in space sufficiently slowly,13 Eq. (4) is still applicable, but only to small volumes, V → dV = d 3 r whose linear size is much smaller than the spatial scale of substantial variations of the function U(r). Hence, instead of Eq. (5), we may only write the probability dW of finding the particle in a small volume d 3 rd 3 p of the 6-dimensional phase space:
3
p 2
3
U (r)
dW w(r, p) d rd p,
w(r, p) const exp
.
(3.24)
2 mT
T
important experimental discovery in the early 1800s, for statistical physics this is just a straightforward corollary of Eq. (18) because in an ideal gas, the component particles do not interact.
10 The statistical mechanics of weak solutions is very similar to that of ideal gases, with Eq. (18) recast into the following formula (derived in 1885 by J. van ’t Hoff), PV = cNT, for the partial pressure of the solute. One of its corollaries is that the net force (called the osmotic pressure) exerted on a semipermeable membrane is proportional to the difference in the solute concentrations it is supporting.
11 Unfortunately, I do not have time for even a brief introduction to this important field and have to refer the interested reader to specialized textbooks – for example, P. A. Rock, Chemical Thermodynamics, University Science Books, 1983; or P. Atkins, Physical Chemistry, 5th ed., Freeman, 1994; or G. M. Barrow, Physical Chemistry, 6th ed., McGraw-Hill, 1996.
12 See, e.g., either Chapter 6 in A. Bard and L. Falkner, Electrochemical Methods, 2nd ed., Wiley, 2000 (which is a good introduction to electrochemistry as a whole); or Sec. II.8.3.1 in F. Scholz (ed.), Electroanalytical Methods, 2nd ed., Springer, 2010.
13 Quantitatively, the spatial scale of substantial variations of the potential, U(r)/ T-1, has to be much larger than the mean free path l of the gas particles, i.e. the average distance a particle passes between successive collisions with its counterparts. (For more on this notion, see Chapter 6 below.)
Chapter 3
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Hence, the Maxwell distribution of particle velocities is still valid at each point r, so the equation of state (18) is also valid locally. A new issue here is the spatial distribution of the total density, n(r) N w(r, p d 3
)
p ,
(3.25)
of all gas particles, regardless of their momentum/velocity. For this variable, Eq. (24) yields14
U (r)
n(r) n( )
0 exp
,
(3.26)
T
where the potential energy at the origin (r = 0) is used as the reference for U. The local gas pressure may be still calculated from the local form of Eq. (18):
U (r)
P(r) n(r T
) P( )
0 exp
.
(3.27)
T
A simple example of numerous applications of Eq. (27) is an approximate description of the Earth’s atmosphere. At all heights h << R E ~ 6106 m above the Earth’s surface (say, above the sea level), we may describe the Earth’s gravity effect by the potential U = mgh, and Eq. (27) yields the so-called barometric formula
Barometric
h
T
k T
P( h) P( )
0 exp
formula
,
h
B K
with
.
(3.28)
h
0
mg
mg
0
For the same N2, the main component of the atmosphere, at T K = 300 K, h 0 ≈ 7 km. This result gives the correct order of magnitude of the atmosphere’s thickness, though the exact law of the pressure change differs somewhat from Eq. (28) because electromagnetic radiation flows result in a relatively small deviation of the atmospheric air from the thermal equilibrium, namely a drop of its temperature T with height, with the so-called lapse rate of about 2% (~6.5 K) per kilometer.
The second generalization I need to discuss is to particles with internal degrees of freedom. Now ignoring the potential energy U(r), we may describe them by replacing Eq. (3) with p 2
' ,
(3.29)
k
2 m
k
where k’ describes the internal energy spectrum of the k th particle. If the particles are similar, we may repeat all the above calculations, and see that all their results (including the Maxwell distribution, and the equation of state) are still valid, with the only exception of Eq. (16), which now becomes
mT 3/ 2
'k
f T
( ) T ln g
1 ln
.
(3.30)
2
exp
2
'
T
k
As we already know from Eqs. (1.50)-(1.51), this change may affect both specific heats of the ideal gas – though not their difference, cV – cP = 1. They may be readily calculated for usual atoms and molecules, at not very high temperatures (say the room temperature of ~25 meV), because in these conditions, k’ >> T for most of their internal degrees of freedom, including the electronic and 14 In some textbooks, Eq. (26) is also called the Boltzmann distribution, though it certainly should be distinguished from Eq. (2.111).
Chapter 3
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SM: Statistical Mechanics
vibrational ones. (The typical energy of the lowest electronic excitations is of the order of a few eV, and that of the lowest vibrational excitations is only an order of magnitude lower.) As a result, these degrees of freedom are “frozen out”: they are in their ground states, so their contributions exp{- k’/ T} to the sum in Eq. (30), and hence to the heat capacity, are negligible. In monoatomic gases, this is true for all degrees of freedom besides those of the translational motion, already taken into account by the first term in Eq. (30), i.e. by Eq. (16b), so their specific heat is typically well described by Eq. (19).
The most important exception is the rotational degrees of freedom of diatomic and polyatomic molecules. As quantum mechanics shows,15 the excitation energy of these degrees of freedom scales as
2/2 I, where I is the molecule’s relevant moment of inertia. In the most important molecules, this energy is rather low (for example, for N2, it is close to 0.25 meV, i.e. ~1% of the room temperature), so at usual conditions they are well excited and, moreover, behave virtually as classical degrees of freedom, each giving a quadratic contribution to the molecule’s kinetic energy. As a result, they obey the equipartition theorem, each giving an extra contribution of T/2 to the energy, i.e. ½ to the specific heat.16 In polyatomic molecules, there are three such classical degrees of freedom (corresponding to their rotations about the three principal axes17), but in diatomic molecules, only two.18 Hence, these contributions may be described by the following generalization of Eq. (19):
,
3/2
gases,
monoatomic
for
c
(3.31)
V
5/2,
of
gases
for
molecules,
diatomic
3,
of
gases
for
molecules.
polyatomic
Please keep in mind, however, that as the above discussion shows, this simple result is invalid at very low and very high temperatures. In the latter case, the most frequent violations of Eq. (31) are due to the thermal activation of the vibrational degrees of freedom; for many important molecules, it starts at temperatures of a few thousand K.
3.2. Calculating
Now let us discuss the properties of ideal gases of free, indistinguishable particles in more detail, paying special attention to the chemical potential – which, for some readers, may still be a somewhat mysterious aspect of the Fermi and Bose distributions. Note again that particle indistinguishability is conditioned by the absence of thermal excitations of their internal degrees of freedom, so in the balance of this chapter such excitations will be ignored, and the particle’s energy k will be associated with its
“external” energy alone: for a free particle in an ideal gas, with its kinetic energy (3).
Let us start from the classical gas, and recall the conclusion of thermodynamics that is just the Gibbs potential per unit particle – see Eq. (1.56). Hence we can calculate = G/ N from Eqs. (1.49) and (16b). The result,
15 See, e.g., either the model solution of Problem 2.14 (and references therein) or QM Secs. 3.6 and 5.6.
16 This result may be readily obtained again from the last term of Eq. (30) by treating it exactly like the first one was and then applying the general Eq. (1.50).
17 See, e.g., CM Sec. 4.1.
18 This conclusion of the quantum theory may be interpreted as the indistinguishability of the rotations about the molecule’s symmetry axis.
Chapter 3
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SM: Statistical Mechanics
3 / 2
2
V
N 2
T ln
f T T T ln
,
(3.32a)
N
gV mT
which may be rewritten as
3 / 2
2
N 2
exp
,
(3.32b)
T gV mT
gives us some idea about not only for a classical gas but for quantum (Fermi and Bose) gases as well.
Indeed, we already know that for indistinguishable particles, the Boltzmann distribution (2.111) is valid only if Nk << 1. Comparing this condition with the quantum statistics (2.115) and (2.118), we see again that the condition of the gas behaving classically may be expressed as
k
exp
1
(3.33)
T
for all k. Since the lowest value of k given by Eq. (3) is zero, Eq. (33) may be satisfied only if exp{/ T} << 1. This means that the chemical potential of a classical gas has to be not just negative, but also “strongly negative” in the sense
T.
(3.34a)
According to Eq. (32), this condition may be represented as
T T ,
(3.34b)
0
with T 0 defined as
2 / 3
2 / 3
2
2
2
Quantum
N
n
scale of
T
,
(3.35)
0
2 / 3
2
temperature
m gV
m g
g
mr ave
where r ave is the average distance between the gas particles:
1/ 3
1
V
r
.
(3.36)
ave
1/ 3
n
N
With the last form of T 0, the condition (34) is very transparent physically: disregarding the factor g 2/3 (which is typically of the order of 1), it means that the average thermal energy of a particle, which is always of the order of T, has to be much larger than the energy of quantization of particle’s motion at the length r ave. An alternative form of the same condition is
1
/ 3
r
g
r ,
where r
.
(3.37)
ave
c
c
1/ 2
( mT )
In quantum mechanics, the parameter r c so defined is frequently called the correlation length.19 For a typical gas (say, N2, with m 14 m p 2.310-26 kg) at the standard room temperature ( T = k B300K
4.110-21 J), the correlation length r c is close to 10-11 m, i.e. is significantly smaller than the physical size r 0 ~ 310-10 m of the molecule. This estimate shows that at room temperatures, as soon as any practical gas is rare enough to be ideal ( r ave >> r 0), it is classical. Thus, the only way to observe quantum effects in the translational motion of molecules is by using very deep refrigeration. According to Eq. (37), for 19See, e.g., QM Sec. 7.2 and in particular Eq. (7.37).
Chapter 3
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SM: Statistical Mechanics
the same nitrogen molecule, taking r ave ~ 102 r 0 ~ 10-8 m (to ensure that direct interaction effects are negligible), the temperature should be well below 1 mK.
In order to analyze quantitatively what happens with gases when T is reduced to such low values, we need to calculate for an arbitrary ideal gas of indistinguishable particles. Let us use the lucky fact that the Fermi-Dirac and the Bose-Einstein statistics may be represented with one formula: 1
N
,
(3.38)
/ T
e
1
where (and everywhere in the balance of this section) the top sign stands for fermions and the lower one for bosons, to discuss fermionic and bosonic ideal gases in one shot.
If we deal with a member of the grand canonical ensemble (Fig. 2.13), in which not only T but also is externally fixed, we may use Eq. (38) to calculate the average number N of particles in volume V. If the volume is so large that N >> 1, we may use the general state counting rule (13) to get 3
2
gV
gV
d p
gV
4 p
dp
N
N 3
d k
.
(3.39)
2 3
23
[ ( p)]/ T
e
1 23 [ ( p)]/ T
0 e
1
In most practical cases, however, the number N of gas particles is fixed by particle confinement (i.e. the gas portion under study is a member of a canonical ensemble – see Fig. 2.6), and hence rather than N
should be calculated. Let us use the trick already mentioned in Sec. 2.8: if N is very large, the relative fluctuation of the particle number, at fixed , is negligibly small ( N/ N ~ 1/ N << 1), and the relation between the average values of N and should not depend on which of these variables is exactly fixed.
Hence, Eq. (39), with having the sense of the average chemical potential, should be valid even if N is exactly fixed, so the small fluctuations of N are replaced with (equally small) fluctuations of .
Physically, in this case the role of the -fixing environment for any sub-portion of the gas is played by the rest of it, and Eq. (39) expresses the condition of self-consistency of such chemical equilibrium.
So, at N >> 1, Eq. (39) may be used to calculate the average as a function of two independent parameters: N (i.e. the gas density n = N/ V) and temperature T. For carrying out this calculation, it is convenient to convert the right-hand side of Eq. (39) to an integral over the particle’s energy ( p) =
p 2/2 m, so p = (2 m)1/2, and dp = ( m/2)1/2 d, getting 3 / 2
1/ 2
gVm
d
Basic
N
.
(3.40) equation
2
3 (
2
)/ T
for
0 e
1
This key result may be represented in two other, sometimes more convenient forms. First, Eq.
(40), derived for our current (3D, isotropic and parabolic-dispersion) approximation (3), is just a particular case of the following self-evident state-counting relation
N g() N
d ,
(3.41)
0
where
g dN
d
(3.42)
states
is the temperature-independent density of all quantum states of a particle – regardless of whether they are occupied or not. Indeed, according to the general Eq. (4), for the simple isotopic model (3), Chapter 3
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SM: Statistical Mechanics
3 / 2
dN
d gV 4
gVm
g
states
3
1/ 2
g ( )
p
,
(3.43)
3
d
d 2
3
2
3
3
2
so plugging it into Eq. (41), we return to Eq. (39).
On the other hand, for some calculations, it is convenient to introduce the following dimensionless energy variable: / T, to express Eq. (40) via a dimensionless integral: 3 / 2
1/ 2
gV ( mT )
d
N
.
(3.44)
2
3
2
/ T
0 e
1
As a sanity check, in the classical limit (34), the exponent in the denominator of the fraction under the integral is much larger than 1, and Eq. (44) reduces to
gV ( mT 3/ 2
)
1/2 d
gV ( mT )3/ 2
N
.
(3.45)
2
3
exp 1/2 e d,
at T
/
2
3
2
0 e
T
2
T 0
By the definition of the gamma function (),20 the last integral is just (3/2) = 1/2/2, and we get 3 / 2
2
3
2
2
T
0
exp N
2
,
(3.46)
3 / 2
T
gV ( mT )
T
which is exactly the same result as given by Eq. (32), which was obtained earlier in a rather different way – from the Boltzmann distribution and thermodynamic identities.
Unfortunately, in the general case of arbitrary , the integral in Eq. (44) cannot be worked out analytically.21 The best we can do is to use the T 0 defined by Eq. (35), to rewrite Eq. (44) in the following convenient, fully dimensionless form:
2
/ 3
1/ 2
T
1
d
,
(3.47)
2
/ T
T
0
2
0 e
1
and then use this relation to calculate the ratios T/ T 0 and / T 0 (/ T)( T/ T 0), as functions of / T
numerically. After that, we may plot the results versus each other, now considering the former ratio as the argument. Figure 1 below shows the resulting plots for both particle types. They show that at high temperatures, T >> T 0, the chemical potential is negative and approaches the classical behavior given by Eq. (46) for both fermions and bosons – just as we could expect. However, at temperatures T ~ T 0 the type of statistics becomes crucial. For fermions, the reduction of temperature leads to changing its sign from negative to positive and then approaching a constant positive value called the Fermi energy,
F 7.595 T 0 at T 0. On the contrary, the chemical potential of a bosonic gas stays negative and then turns into zero at a certain critical temperature T c 3.313 T 0. Both these limits, which are very important for applications, may (and will be :-) explored analytically, separately for each statistics.
20 See, e.g., MA Eq. (6.7a).
21 For the reader’s reference only: for the upper sign, the integral in Eq. (40) is a particular form (for s = ½) of a special function called the complete Fermi-Dirac integral Fs, while for the lower sign, it is a particular case (for s
= 3/2) of another special function called the polylogarithm Li s. (In what follows, I will not use these notations.) Chapter 3
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10
F
8
fermions
T 0
6
4
2
0
Fig. 3.1. The chemical potential of an ideal
T 0 2
gas of N >> 1 indistinguishable quantum
particles, as a function of temperature at a
4
bosons
T
fixed gas density n N/ V (i.e. fixed T
6
0 n 2/3),
c
for two different particle types. The dashed
8
line shows the classical approximation (46),
10
0
1
2
3
4
5
6
7
8
9
10
valid only asymptotically at T >> T 0.
T / T 0
Before carrying out such analyses (in the next two sections), let me show that rather surprisingly, for any non-relativistic ideal quantum gas, the relation between the product PV and the energy, 2
PV E ,
(3.48) Ideal gas:
3
PV vs. E
is exactly the same as follows from Eqs. (18) and (19) for the classical gas, and hence does not depend on the particle statistics. To prove this, it is sufficient to use Eqs. (2.114) and (2.117) for the grand thermodynamic potential of each quantum state, which may be conveniently represented by a single formula,
T
ln 1 ( )/ T
k
e
,
(3.49)
k
and sum them over all states k, using the general summation formula (13). The result for the total grand potential of a 3D gas with the dispersion law (3) is
3 / 2
gV
2
p
m T
gVm
T
ln1
(
/ 2 ) /
e
4
2
p dp T
ln
(
)/
e
T d
(3.50)
3
1
1/2
2
.
0
2 2 3
0
Working out this integral by parts, exactly as we did it with the one in Eq. (2.90), we get 2
3 / 2
3 / 2
gVm
d
2
g ( )
N d
(3.51)
T
3
.
3 2 2 3
(
) /
0 e
1
3 0
But the last integral is just the total energy E of the gas:
2
gV
p
4 2
3 / 2
3 / 2
p dp
gVm
d
Ideal
E
g ( )
N d
gas:
(3.52)
p T
T
3
3
2
,
2
[ ( )
]/
(
) /
m e
energy
0
1
2 2 3
0 e
1 0
so for any temperature and any particle type, = –(2/3) E. But since, from thermodynamics, = – PV, we have Eq. (48) proved. This universal relation22 will be repeatedly used below.
22 For gases of diatomic and polyatomic molecules, whose rotational degrees of freedom are usually thermally excited, Eq. (48) is valid only for the translational motion’s energy.
Chapter 3
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SM: Statistical Mechanics
3.3. Degenerate Fermi gas
Analysis of low-temperature properties of a Fermi gas is very simple in the limit T = 0. Indeed, in this limit, the Fermi-Dirac distribution (2.115) is just the step function:
,
1 for ,
N
(3.53)
,
0
for ,
– see the bold line in Fig. 2a. Since the function = p 2/2 m is isotropic in the momentum space, in that space the particles, at T = 0, fully occupy all possible quantum states inside a sphere (called either the Fermi sphere or the Fermi sea) with some radius p F (Fig. 2b), while all states above the sea surface are empty. Such degenerate Fermi gas is a striking manifestation of the Pauli principle: though in thermodynamic equilibrium at T = 0 all particles try to lower their energies as much as possible, only g of them may occupy each translational (“orbital”) quantum state. As a result, the sphere’s volume is proportional to the particle number N, or rather to their density n = N/ V.
N
(a)
pz
(b)
1
T 0
Fig. 3.2. Representations of the
0
p
Fermi sea: (a) on the Fermi
p
y
F
T
p
F
x
distribution plot, and (b) in the
momentum space.
0
F
Indeed, the radius p F may be readily related to the number of particles N using Eq. (39), with the upper sign, whose integral in this limit is just the Fermi sphere’s volume:
p F
gV
gV
4
2
3
N
p dp
p
.
(3.54)
2
4
3
2
3
0
3
F
Now we can use Eq. (3) to express via N the chemical potential (which, in the limit T = 0, bears the special name of the Fermi energy F)23:
2 / 3
1/ 3
2
2
4
Fermi
p
N
9
energy
F
2
6
T 595
.
7
T ,
(3.55a)
F
T 0
0
0
2 m
2 m
gV
2
where T 0 is the quantum temperature scale defined by Eq. (35). This formula quantifies the low-temperature trend of the function ( T), clearly visible in Fig. 1, and in particular, explains the ratio F/ T 0
mentioned in Sec. 2. Note also a simple and very useful relation,
3
N
3
N
,
i.e. g ( )
,
(3.55b)
F
2 g ( )
3
F
2
3
F
F
that may be obtained immediately from the comparison of Eqs. (43) and (54).
The total energy of the degenerate Fermi gas may be (equally easily) calculated from Eq. (52): 23 Note that in the electronic engineering literature, is usually called the Fermi level, for any temperature.
Chapter 3
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SM: Statistical Mechanics
p
gV
F p 2
4
3
2
gV
p 5
E
F
,
(3.56)
2
4 p dp
3
2 m
2
2
5
5
0
N
3
m
F
showing that the average energy, E/ N, of a particle inside the Fermi sea is equal to 3/5 of that (F) of the particles in the most energetic occupied states, on the Fermi surface. Since, according to the basic formulas of Chapter 1, at zero temperature H = G = N, and F = E, the only thermodynamic variable still to be calculated is the gas pressure P. For it, we could use any of the thermodynamic relations P =
( H – E)/ V or P = –( F/ V) T, but it is even easier to use our recent result (48). Together with Eq. (56), it yields
1/ 3
2 E
2 N
36 4
2
5 / 3
n
P
F
P 3.035 P ,
where P nT
.
(3.57)
3 V
5 V
125 0
0
0
0
2 / 3
mg
From here, it is straightforward to calculate the isothermal bulk modulus (reciprocal compressibility),24
P
2
N
K V
,
(3.58)
T
V
F
3
V
T
which is frequently simpler to measure experimentally than P.
Perhaps the most important example25 of the degenerate Fermi gas is the conduction electrons in metals – the electrons that belong to the outer shells of isolated atoms but become shared in solid metals, and as a result, can move through the crystal lattice almost freely. Though the electrons (which are fermions with spin s = ½ and hence with the spin degeneracy g = 2 s + 1 = 2) are negatively charged, the Coulomb interaction of the conduction electrons with each other is substantially compensated by the positively charged ions of the atomic lattice, so they follow the simple model discussed above, in which the interaction is disregarded, reasonably well. This is especially true for alkali metals (forming Group 1
of the periodic table of elements), whose experimentally measured Fermi surfaces are spherical within 1% – even within 0.1% for Na.
Table 1 lists, in particular, the experimental values of the bulk modulus for such metals, together with the values given by Eq. (58) using the F calculated from Eq. (55) with the experimental density of the conduction electrons. The agreement is pretty impressive, taking into account that the simple theory described above completely ignores the Coulomb and exchange interactions of the electrons. This agreement implies that, surprisingly, the experimentally observed firmness of solids (or at least metals) is predominantly due to the kinetic energy (3) of the conduction electrons, rather than any electrostatic interactions – though, to be fair, these interactions are the crucial factor defining the equilibrium value 24 For a general discussion of this notion, see, e.g., CM Eqs. (7.32) and (7.36).
25 Recently, nearly degenerate gases (with F ~ 5 T) have been formed of weakly interacting Fermi atoms as well –
see, e.g., K. Aikawa et al., Phys. Rev. Lett. 112, 010404 (2014), and references therein. Another interesting example of the system that may be approximately treated as a degenerate Fermi gas is the set of Z >> 1 electrons in a heavy atom. However, in this system the account of electron interaction via the electrostatic field they create is important. Since for this Thomas-Fermi model of atoms, the thermal effects are unimportant, it was discussed already in the quantum-mechanical part of this series (see QM Chapter 8). However, its analysis may be streamlined using the notion of the chemical potential, introduced only in this course – the problem left for the reader’s exercise.
Chapter 3
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SM: Statistical Mechanics
of n. Numerical calculations using more accurate approximations (e.g., the Density Functional Theory26), which agree with experiment with a few-percent accuracy, confirm this conclusion.27
Table 3.1. Experimental and theoretical parameters of electrons’ Fermi sea in some alkali metals28
Metal
F (eV)
K (GPa)
K (GPa)
(mcal/moleK2) (mcal/moleK2)
Eq. (55)
Eq. (58)
experiment
Eq. (69)
experiment
Na
3.24
923
642
0.26
0.35
K
2.12
319
281
0.40
0.47
Rb
1.85
230
192
0.46
0.58
Cs
1.59
154
143
0.53
0.77
Looking at the values of F listed in this table, note that room temperatures ( T K ~ 300 K) correspond to T ~ 25 meV. As a result, most experiments with metals, at least in their solid or liquid form, are performed in the limit T << F. According to Eq. (39), at such temperatures, the occupancy step described by the Fermi-Dirac distribution has a non-zero but relatively small width of the order of T
– see the dashed line in Fig. 2a. Calculations for this case are much facilitated by the so-called Sommerfeld expansion formula 29 for the integrals like those in Eqs. (41) and (52):
2
d
2
( )
Sommerfeld
I ( T ) ( ) N d ( ) d
T
, for T ,
expansion
(3.59)
6
d
0
0
where () is an arbitrary function that is sufficiently smooth at = and integrable at = 0. To prove this formula, let us introduce another function,
df
f
( )
( ' ) d ' ,
that
so
,
(3.60)
d
0
and work out the integral I( T) by parts:
df
I ( T )
N
d
N
df
d
0
0
N
N f
f d N f
d .
(3.61)
0
0
0
26 See, e.g., QM Sec. 8.4.
27 Note also a huge difference between the very high bulk modulus of metals ( K ~ 1011 Pa) and its very low values in usual, atomic gases (for them, at ambient conditions, K ~105 Pa). About four orders of magnitude of this difference is due to that in the particle density N/ V, but the balance is due to the electron gas’ degeneracy. Indeed, in an ideal classical gas, K = P = T( N/ V), so that the factor (2/3) F in Eq. (58), of the order of a few eV in metals, should be compared with the factor T 25 meV in the classical gas at room temperature.
28 Data from N. Ashcroft and N. D. Mermin, Solid State Physics, W. B. Saunders, 1976.
29 Named after Arnold Sommerfeld, who was the first (in 1927) to apply quantum mechanics to degenerate Fermi gases, in particular to electrons in metals, and may be credited for most of the results discussed in this section.
Chapter 3
Page 15 of 34
SM: Statistical Mechanics
As evident from Eq. (2.115) and/or Fig. 2a, at T << the function – N()/ is close to zero for all energies, besides a narrow peak of the unit area, at . Hence, if we expand the function f() in the Taylor series near this point, just a few leading terms of the expansion should give us a good approximation:
2
df
d f
N
I ( T )
1
f ()
2
d
2
d
2
d
0
N
N
( ' ) d '
d ()