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9.a. Boltzmann factor and Maxwell-Boltzmann
distribution
9.b. Quantum Statistics
9.c. Bose-Einstein statistics
9.d. Fermi-Dirac statistics
Boltzmann realized that the energy distribution of molecules in gases (or in condensed matter) could take a large number of forms. Some of the possible distributions are much more probable than others. In fact, when the number of molecules in the system increases the range of very probable distributions narrows. When this number reaches values representative to macroscopic systems (Avogadro's number) than the range of distributions has an extremely narrow peak around the most probable distribution. The error of using just the most probable distribution is negligible.
Though Boltzmann also understood to a large extent how systems approach equilibrium, we will be mostly concerned with systems in equilibrium. Equilibrium means that the basic characteristics of the system (temperature, enegy distributions, volume, velocity distributions, etc.) are independent of time. Most of the time we will assume that the system we study is isolated: it does not exchange energy or particles with its surroundings. There is no such thing as complete isolation, though it can be approximated well for practical purposes.
The basic assumption of statistical mechanics is that every state of the same energy has an equal probability. Then the probability of a given configuration is proportional to the number of ways that configuration can be created. Take the example similar to the one in the book but with a lower number of particles and energy. Suppose that 5 noninteracting particles with total energy 8.5 hf are in a harmonic oscillator potential (energy levels of E=(1/2+n)hf, where n=0,1,2,...). In other words the excitation energy is 6hf (the energy in the ground state would be 5x1/2hf = 2.5hf). We wish to count the number of states for each possible configuration. The excitation energy 6hf can be distributed in 10 different ways among the 5 particles (we call these Ci configurations): There are no more possibilities. Particles having different energies are assumed to be distinguishable for the moment. Particles having the same energy will be still assumed to be indistinguishable. This way we obtain what is called classical statistics. Then each of these configurations can be realized in a number of ways. C1 can be realized 5 different ways, because each of the five particles could have the excitation energy 6. This number N1=5 can be obtained as 5! , the number of ways the particles can be ordered (permuted), divided by 4!, the number of ways the particles at identical energy levels can be permuted. Here we use the assumption that particles at identical energy levels cannot be distinguished, but at different levels they can be distinguished. The total number of states is N =N1+N2+...+N10=210. Then the requirement that each state (called microstate in the book) equally probable gives the probability of Ci as pi=Ni/N. Then we obtain for the probablility of configurations:
|
|
|
Boson Configurations |
Fermion configurations | |||
|
Configuration |
Number of microstates |
Probability of Configuration |
# of states |
Probability |
# of states |
Probability |
|
C1=(6,0,0,0,0) |
N1=5! / 4! = 5 |
p1= 5/210 = 0.024 |
1 |
1/10 |
0 |
0 |
|
C2=(5,1,0,0,0) |
N2= 5! / 3! = 20 |
p2 = 20/210 = 0.095 |
1 |
1/10 |
0 |
0 |
|
C3=(4,2,0,0,0) |
N3= 5! / 3! = 20 |
p3 = 20/210 = 0.095 |
1 |
1/10 |
0 |
0 |
|
C4= (4,1,1,0,0) |
N4= 5! / (2!2!) = 30 |
p4 = 30/210 = 0.143 |
1 |
1/10 |
1 |
1/3 |
|
C5= (3,3,0,0,0) |
N5= 5! / (3!2!) = 10 |
p5 = 10/210 = 0.048 |
1 |
1/10 |
0 |
0 |
|
C6= (3,2,1,0,0) |
N6= 5! / 2! = 60 |
p6 = 60/210 = 0.286 |
1 |
1/10 |
1 |
1/3 |
|
C7= (3,1,1,1,0) |
N7= 5! / 3! = 20 |
p7 = 20/210 = 0.095 |
1 |
1/10 |
0 |
0 |
|
C8= (2,2,2,0,0) |
N8= 5! / (3!2!) = 10 |
p8 = 10/210 = 0.048 |
1 |
1/10 |
0 |
0 |
|
C9= (2,2,1,1,0) |
N9= 5! / (2!2!) = 30 |
p9 = 30/210 = 0.143 |
1 |
1/10 |
1 |
1/3 |
|
C10= (2,1,1,1,1) |
N10= 5! / 4! = 5 |
p10 = 5/210 = 0.024 |
1 |
1/10 |
0 |
0 |
|
Total |
N=210 |
1.0 |
10 |
1.0 |
3 |
1.0 |
Having these probabilities we can calculate all kinds of statistical quantities. E.g the average number of particles at the ground state energy level is obtained as we find in each configuration how many particles are at the lowest level and multiply this number with the pi, and add these numbers together. So the average number of particles at the ground state level is
< n0 >= 4p1+3p2+3p3+2p4+3p5+2p6+p7+2p8+p9= 2.002
Similarly we can calculate the number of particles on the first excited state level:
<n1 > = p2+2p4+p6+3p7+2p9+4p10= 1.330. Furthermore
<n2 > = p3+p6+3p8+2p9+p10 = 0.835
< n3 > = 2p5 + p6+ p7 = 0.477
< n4 > = p3 + p4 = 0.238
< n5 > = p2 =0.095
< n6 > = p1 = 0.024.
If we add up all these numbers than we get 5, the total number of particles. Then the probability that a particle is on the ground state level is P(0) = n0/5 = 2.002/5 = 0.4. These probabilities tell us that what is the average number of particles on a subsystem of the total system. If we plot these numbers as a function of energy, E=(n+1/2) hf, then we obtain a close to exponential drop of the probability.
One can prove in general, if the total number of particles, N and the total energy, ET, are very large then the number of particles in a subsystem of energy E (in thermal equilibrium with the large system) can be shown to be equal to
n(E) = g(E) A e-E/kT,
where g(E) is the degeneracy of the level, T is the absolute temperature, and A i s a normalization constant. the factor e-E/kT is proportional to the probability that the subsystem is in a particular state of energy E (The degeneracy level is the number of different physical configurations that have the same energy). This factor is called the Boltzmann factor and k=kB=8.617¥10-5 eV/K. is Boltzmann's constant.
Example: Determination of the approximate temperature of the model system we considered.
Solution: Calculate the probability of finding a particle on the 7 available levels of the harmonic oscillator. These are obtained as Pi=<ni>/N = <ni>/5 (Do not mix Pi with the previously defined pi!!!). and are equal to 0.4044, 0.266, 0.167, 0.0954, 0.0476, 0.0190, 0.0048. These numbers should approximately be equal to probabilities given by the Maxwell Boltzmann distribution with some temperature, T:
Pi = e-E(i)/kT / Sj e-E(j)/kT,
where E(i) = (j+1/2) hf.
Obviously, the distribution will only depend on the ratio hf/kT. This is the quantity we wish to determine. In other words, if the harmonic oscillator energy is provided one can determine the temperature.
One could fit the numbers we obtained for Pi in our model system to the above formula. That would be laborious. Instead of this we could fix the constant hf/kT by matching average energies.
The average energy from our model distribution is<E> = kT [ 1/2 x 0.4044 + 3/2 x 0. 0.266 + 5/2x 0. 167, + 7/2 x 0. 0954 + 9/2 x 0.0476 + 11/2 x 0.0190 + 13/2 x 0.0048 ] = hf 1.6865
We can also calculate the average energy from the Maxwell Boltzmann distribution as well. By equating these two we can determine the ratio hf/kT, or knowing f, the temperature.
<E> = Sj E(j) e-E(j)/kT / Sj e-E(j)/kT
It is easy to calculate the denominator. It is a geometric series. Summing it we obtain
Sj e-E(j)/kT = e-hf/2kT / (1 - e-hf/kT).
We can get the numerator if we differentiate the denominator with respect to -1/kT. Then we obtain
Sj E(j) e-E(j)/kT = hf [ 1/2 e-hf/2kT / (1 - e-hf/kT) + e-hf/kT / (1 - e-hf/kT)2 ]
This gives for <E>
<E> = hf [ 1/2 + e-hf/2kT / (1 - e-hf/kT) ]
Comparing with our value obtained for the model system we obtain the equation
1/2 + e-hf/2kT / (1 - e-hf/kT) = 1.6865 1.4054.
Using the notation z = e-hf/2kT we obtain
1/2 + z /(1-z2) = 1.6865,
or z2 + 0.8428 z - 1 = 0, z = 0.6638.
From here we get hf/kT = -2 log 0.6638 = 0.8195. In other words, whatever f is we obtain the temperature as T = 1.22 h f / k, where k is Boltzmann's constant.
Example: Velocity distribution of molecules of ideal gas (Maxwell veocity distribution): The velocity space element is dvx dvy dvz = 4 p v2 dv = g(E)dE. For free particles E= mv 2 / 2, so we have
n(E) dE = A v2 dv exp{ - mv2/2kT }.
Notice that here the distribution is assumed to be continuous unlike in the example we discussed earlier. Just apply our knowledge concerning continuous distributions to build up the machinery needed to calculate physical quantities.
First determine A from the condition N/V = Ú dE n(E) = Ú A v2 dv exp{ - mv2/2kT }. Using Ú x2 dx exp{ - x2 } = p1/2/4, we obtain
N/V = A[2kT/m]3/2 p1/2/4,
A = (N/V) 4 p [m/2 p kT]3/2,
so the final form of the distribution is
n(v) dv = (N/V) 4 p [m/2 p kT]3/2 v2 exp{ - mv2/2kT } dv.
This is the distribution of velocities of molecules in an ideal gas.
Example: Find the average velocity of molecules in an ideal gas.
<v> = Ú v n(v) dv / Ú n(v) dv = Ú 4 p [m/2 p kT]3/2 v3 exp{ - mv2/2kT } dv. Before performing the calculation lets make a guess what the average velocity could be. The probability distribution contains only one significant constant of the dimension of velocity, v0 = [2kT/m]1/2. The overall "normalization constant" is dimensional, but irrelevant for averages. One can also say that roughly up to v0 the distribution is increasing due to the multiplier of v2 but beyond that value, v > v0 the distribution is sharply cutoff. Thus the average must be close to v0 . In fact the exact calculation gives [4/ p]1/2 v0 . In a similar way the average v2 must be proportional to v0 2 = 2kT/m. In fact we obtain
<v2> = Ú v2 n(v) dv / Ú n(v) dv = 3/2 v0 2
The root of this quantity, vrms = [3kT/m]1/2 is called the root mean squared velocity.
Now the average energy (equals to the average kinetic energy) can be written as
<E> = <K> = m <v2>/2 = m ( vrms)2 / 2 = 3kT/2.
This means that for each degree of freedom one has an energy of kT/2. This is called the equipartition theorem.
When the particles are not free but sitting in an oscillator potential then there are additional terms of the energy, such as m w2 / 2 (x2 + y2 + z2). The equipartition theorem is valid for these degrees of freedom as well and each term of this potential energy contribute to the expectation value of the energy with similar amounts, so the average total energy of particles is 6 x (1/2) kT = 3kT.
Example: Intensity of emission lines. The occupation level of excited states of hydrogen gas can be obtained from the Maxwell-Boltzmann statistics. What is the probability that at room temperature the atoms in the Hydrogen gas are in excited states?
Solution: The ratio of probabilities of being in the first excited state and in the ground state is
P2 / P2 = g2 exp{-E2/kT} / g1 exp{-E1/kT},
where g1=2, and g2 = 6, are the degeneracies of the first two levels and E1= - 13.6eV, and E2 = -3.4 eV are the energies of these levels. Then
P2 / P1 = 3 exp{ -10.2 eV / (8.617 x 10-5 eV/K 300K)} = 4 e-400
This is an extremely small number, much smaller then Avogadro's number. In other words if the gas is isolated and is in equilibrium then the probability that even one of the atoms is in the excited state is remote. In practice this will never happen. The gas is bombarded by natural radioactivity, by photons and by cosmic rays that will keep exciting atoms into excited states.
The probabilities we derived for the five particle system were
applicable for classical distinguishable particles. If the particles
satisfy the laws of quantum physics then the situation changes.
Imagine the first configuration. The wavefunction corresponding to
that would be y6(x1) y0(x2) y0(x3) y0(x4) y0(x5). If the particles are
regarded as classical, i.e. no symmetrization or antisymmettrization
is needed then there are alltogether distinct states obtained by
permuting the coordinates. On the other hand if these particles are
bosons then there is only a single state obtained by the
symmetrization (sum) of these five wavefunctions. If the particles
are fermions then this configuration is completely forbidden. We
could place two particles at the ground state level that have
opposite orientation of spin, but no more. We can go through all ten
configuration and conclude that if the particles are bosons the each
of the configuartions are allowed but each one of them corresponds to
a single state only. Then the probability of the system being in any
of these states is 1/10. If the particles are fermions then the only
configurations that are admissible are the ones which do not contain
more than two particles on the same level: (4,1,1,0,0), (3,2,1,0,0),
and (2,2,1,1,0). The probability that the system is in each of these
states is 1/3. Then e.g. for fermions the average number of particles
on level 0 is <n0> = 2 x 1/3 + 2 x 1/3 + 1/3 = 5/3.
Furthermore, <n1> = 2 x 1/3 + 1/3 + 2 x 1/3 = 5/3,
<n2> = 1/3 + 2 x 1/3 = 1, <n3> =
1/3, <n4> = 1/3, <n5> =
<n6> = 0. For bosons we have < n0
>=
4p1+3p2+3p3+2p4+3p5+2p6+p7+2p8+p9=
21/10 =2.1,
<n1 > =
p2+2p4+p6+3p7+2p9+4p10=
13/10 = 1.3,
<n2 > =
p3+p6+3p8+2p9+p10
= 8/10 = 0.8
< n3 > = 2p5 + p6+
p7 = 2/10 = 0.2
< n5 > = p2 = 1/10 = 0.1
< n6 > = p1 = 1/10 = 0.1.
These expectation values are also decreasing functions of energy (Note that < n0 > corresponds to the ground state, E= 1/2 hf, < nk > gives the number of particles in the kth excited state with E k = (k+1/2)hf ). Note that these expecttaion values are rapidly decreasing functions of the energy both for bosons and for fermions. In fact, their exact form can be derived in a way similar to that of the Maxwell-Boltzmann statistics.
The quatum statistics can also be given exact forms when the number of particles is very large. Their form is different than that of the Maxwell-Boltzmann statistics. They all have the form of
n(E) dE = g(E) f(E) dE,
where n(E) dE is the number of particles occupying levels between E and E+dE, g(E) dE is the number of states between E and E+dE, and f(E) is the average occupancy of a level of energy E at temperature T. g(E) depends on the physical system. For free particles the density of states can be shown to be obtained from the expression (which has the dimension of 1/V, as it should be)
g(E) dE = dpx dpy dpz / h3 = 4 p p2 dp / h3 = 8 p 21/2 m3/2 E1/2 dE / h3.
If the free particle has spin s, then the above density of states should be multiplies by 2s+1, the number of possible spinstates for each energy. Translated to velocities this density of states is uniform in velocity space, i.e. each elementary cube has the same number of velocity states. This is a statement analogous to the statement that in homogeneous coordinate space the number of points is the same in every volume element of identical size.
The function f(E) can be derived to have the following form for the three types of distributions:
|
Statistics |
Maxwell-Boltzmann |
Bose-Einstein |
Fermi-Dirac |
|
Applicable for |
Classical, or high E |
Quantum, bosons |
Quantum, fermions |
|
f(E) |
A e-E/kT |
1 / [ e E/kT / A - 1] |
1 / [ e E/kT / A +1] |
|
A |
normalize |
e-m/kT |
exp{-EF / k T } |
Comments:
Finally, before discussing quantum statistics in more details it
is worth while to consider the condition for their application. In
principle every time when one considers systems of elementary
partricles, atoms, or molecules one should use quantum statistics. At
the same time, as we saw above, at high energies all three
distributions are alike and it is perfectly admissible to use
Maxwell-Boltzmann statitics. The condition for using classical
statistics is determined by certain parameters. These are
characteristic length, or characteristic times, or sometimes
characteristic velocities, etc. When there is a transition between
two different regimes then there are two characteristic parameters of
the same dimension the relative magnitude of which determines the
best , most economic description. In statistical systems one obvious
parameter of dimension of length is the average distance between
particles. That can easily be expressed by the number density: V/N is
the average volume alloted to each particle. If we imagine this
volume as a cube then an edge of this cube d = [V/N]1/3 is
the average distance among particles. Now we need to decide at what
distance the quantum nature of particles, i.e. the necessity to use
the symmetrization or antisymmetrization of wave function (and with
that, the use of quantum statistics), is needed to be taken into
account. It is easy to show that the symmetrization (or
antisymmetrization, for fermions) of wavefunctions is irrelevant if
they do not overlap. In other words quantum statistics, which differs
from Maxwell-Boltzmann statistics exactly in the symmetrization
properties, need not be taken into account if on the average
wavefunctions of neighboring particles do not overlap. The extent of
wavefunctions is the uncertainty of the coordinate, Dx. Thus we arrive at the condition of the
necessity to apply quantum statistics as Dx > d. Maxwell-Boltzmann statistics can be
used if d >> Dx. Now , the
uncertainty Dx can be easily estimated.
Dx µ
h / 2 Dp =
h / 2 [
<px2>]1/2 = h
/ 2 [ <2mK/3>]1/2 = h / 2 [
mkT]1/2. Here we neglected interactions which can in
principle modify the estimate. Consequently, the condition for using
classical statistics is
N/V << 8 [ mkT]3/2/
h3
The quantity h3 / 8 [
mkT]3/2 is called the quantum volume. We conclude that the
classical statistics can be used at low density or/and at high
temperature.
Example: Gases at room temperature and 1 atm pressure are very well described by classical statistics. They behave as ideal gases. They satisfy d >> Dx. kT = 1/40 eV at room temperature, while mc2 > 2 x 109 eV (the value for the lightest gas, hydrogen). Then 8 [ mc2kT]3/2/ (
hc)3 = 8 [ (1/40) 2 x 109 ]3/2 / [ 197 nm ]3 = 1.3 x 1026cm-3, which means a thousand moles/ cm3, certainly much larger than the density, N/V, of any gas at ordinary pressure (23 liters/mole). As is shown in the book the electron gas in metals must be described by Fermi-Dirac statistics because Dx > d.Example: Bose-Einstein condensation.
For Bose-Einstein statistics the normalization of the distribution is equivalent to the determination of the chemical potential. In other words, one needs to determine m from the equation
N / V = Ú 8 p 21/2 (m3/2 / h3) E1/2 dE / [ e (E-m)/kT - 1]
introducing integrattion variable x=E/kT we obtain
N / V = 8 p 21/2 (mkT/h2)3/2 Ú x1/2 dx / [ A ex - 1],
where A = e -m/kT>1 so that the denominator would be positive even at x=0. For a given number of particles in a given volume and at a given temperature, A, or in other words the chemical potential m must be determined from the above equation. Suppose now we lower and lower T at fixed N and V. Then the integral on the right hand side must increase. That can only happen if A decreases, because the integral does not depend on anything else. But A cannot be smaller then one, as we remarked earlier, so its minimal value is 1. But the integral is still convergent at A=1 so one gets a chritical temperature Tc below which one does not have a solution for A. This temperature is determined by the relation
N / V = 8 p 21/2 (mkTc/h2)3/2 Ú x1/2 dx / [ ex - 1].
So what is the meaning of being below Tc? Then the equation
N1 / V = 8 p 21/2 (mkT/h2)3/2 Ú x1/2 dx / [ ex - 1],
cannot account for all particles N, but rather only N1<N. Notice however that the summation (integration) over states misses states with energy E=0, due to the fact that the density of states is proportional to E1/2 dE. Many bosons can be in the same E=0 state. Th eonly resolutio of the contradiction is that the number of bosons not in the ground state is N1, wile the rest, N- N1 are all in the ground state. This is a macroscopic fraction of the total number at T< Tc. This phenomenon is called Bose-Einstein condensation. The fraction of particles that is in the ground state has entirely new properties, such as suprfluidity. The condensation temperature in liquid Helium is below 3oK.
A typical example of a system composed of bosons is a gas of photons, the system one needs to deal with when studying black body radiation. Again, for bosons (if the chemical potential vanishes, as it does for photons)
n(E) dE = g(E) dE / [ eE/kT - 1 ].
Now g(E) dE = 2 dpx dpy dpz / h3 = 8 p p2 dp / h3 = 8 p E2 dE / (hc)3,
where the extra multiplier 2 takes care of the two possible polarization (spin orientation) of the photon. This formula exactly coincides with the formula we derived for the number of modes of blackbody radiation. In fact, leaving g(E) undetermined and comparing the above expression and the one derived earlier (Planck distribution) we can arrive to the formula g(p) dp = dpx dpy dpz / h3 , that we postulated earlier for the number of modes. Combining g(E) and f(E) we have
n(E) dE = g(E) f(E) dE = g(E) dE / [ eE/kT - 1 ] = 8 p E2 dE / [(hc)3( eE/kT - 1 )],
Then u(E), the energy density (the contribution of photon modes of energy between E and E+dE to the total energy of unit volume ) is
u(E) dE = E n(E) dE = 8 p E3 dE / [(hc)3( eE/kT - 1 )].
This is exactly the Planck blackbody radiation formula if we use freqency , rather than energy, E=hf,
u(f,T) = (8 p h f3 / c3 ) / [ ehf/kT - 1 ].
Another application of Bose-Einstein statistics is the study of lattice vibrations. In a solid, e.g in a metal, in addition to the electron gas, there are ions, which are more or less unmovable. At nonzero temperature they undergo oscillations. Neglecting interactions of ions, and regarding atoms as tiny three dimensional oscillators, each of them have six degrees of freedom (kinetic + potential energy). hen the eqipartition theorem says that the average energy of atoms is 3kT. In one mole there are NA atoms so the total oscillation energy represented by one mole is U = 3NA kT. One defines the specific heat as C = dU(T)/dT. The meaning of the specific heat is the energy that is needed to be added to the system to raise its temperature by one degree. For our simple model the specific heat would be
C = 3NA k = 3 R = 3 x 1.99 cal / mole K. This prediction says that the specific heat is independent of the temperature.
The specific heat can be easily measured. While at high temperature (hundreds of degrees) the specific heat indeed tends to this constants, in vanishes rapidly if the temperature is lowered toward zero. Einstein proposed that at low temperature interactions among atoms cannot be neglected. Ions in a solid interact strongly with each other, therefore they are not able to oscillate alone, but they transfer the oscillation energy to each other. Just like segment of violin string or the oboe do not oscillate alone, the atoms oscillate in unison. They have collective modes of oscillation, called phonons (phonos=sound), quanta of lattice vibrations. They have energy E=hf, just like any other particle, according to the deBroglie relation. In fact at nonzero temperature there is a large numer of phonon quanta present, we can speak of a phonon gas. Neglecting interactions of the phonon gas one has the following energy carried by a mode of energy hf
u(f) = hf / [ ehf/kT - 1 ]
Einstein argued that in one mole gas the total energy is
U = 3NA hf / [ ehf/kT - 1 ],
because there are NA atoms and each has 3 modes of oscillation. Then the specific heat is
C = dU(T)/dT = 3NA k (hf / kT)2 ehf/kT / [ ehf/kT - 1 ]2.
At high temperature this expression tends to 3NA k = 3R in agreement with experiments, while at low temperatures drops very rapidly. Writing hf= k TE, we can define the Einstein temperature. This formula agrees well with the observed drop of the specific heat for temperatures above 0.2 TE. The Einstein temperature for diamond is about 1300K.
If one goes to temperatures much lower then the Einstein temperature then the Einstein formula does not agree with observation.. The Einstein formula for specific heat predicts that for very low temperatures
C µ e-hf/kT/T2.
The observedbehavior is rather C µ T3. Debye explained that in the following way. The Einstein frequency, which is obtained from fitting to data gives only the maximal frequency of phonons in the solid. There is a continuum of phonon energies available down to zero energy and these form a phonon gas which can be regarded as noninteracting. Then the energy frequency density is described by the Planck formula, the same way as for photons (phonons are also massless, or they are usually called gapless excitations). Then the energy density for phonons in the freequency range f and f+df is
u(f,T) = (8 p h f3 / c3 ) / [ ehf/kT - 1 ],
and the total energy is
U(T) = Ú df (8 p h f3 / c3 ) / [ ehf/kT - 1 ] = (8 p h f3 / c3 )(kT / h)4 Ú dx x3 / [ ex - 1 ].
This is obviously proportional to T4. Then the specific heat is the derivative of U(T) with respect to T, that is proportional to T3. The integral is really cut off at ta value corresponding to the Einstein temperature, xmax= TE/T. At very low temperature this can be regarded as infinity, but close to the Einstein temperature it yields the Einstein formula.
Let us examine the shape of the Fermi-Dirac distribution. If the fermi energy is much larger then the temperature, which is certainly true for a metal at room temperature, then at very low energy the denominator of the Fermi distribution is
e E/kT e -EF/kT +1 µ e -EF/kT +1 µ 1
Note that EF = 3.23 eV for Na, while room temperature is 1/40 eV. Consequently, the occupancy of levels is 1. Everey level is occupied at very low energy. if one goes to higher energies, comparable to the Fermi energy then the occupancy starts to decrease, perceptibly if (E- EF )/kT µ 1, or in other words, if
E µ EF - kT.
Then at E = EF the occupancy is exactly 1/2. If one goes above the Fermi level then the occupancy drops further and it becomes very small if E >> EF + kT. In other words the approximate width of the transition region around the Fermi energy (a few eV) where the occupancy goes from 1 to 0 is 2kT, which is about 1/20 eV. The occupancy is almost a rectangular function of energy. It becomes exactly rectangular when the length of the transition region, 2kT, goes to zero, i.e. at zero temperature. Then every level is occupied below the Fermi energyand no level is occupiee above it. Increasing the temperature adds energy to some of the electrons below the Fermi energy and they rise above it and the sharp edge f the distribution is smeared.
Earlier we derived the form of the degeneracy for massive nonrelativistic particles. We obtained
g(E) dE = 8 p 21/2 m3/2 E1/2 dE / h3.
Then the number of fermions in a Fermi gas between energies E and E+dE is
n(E) dE = g(E) f(E) dE = 8 p 21/2 m3/2 h-3 E1/2 dE / ( exp{ (E- EF )/kT} + 1 ).
At zero temperature this reduces to
n(E) dE = 8 p 21/2 m3/2 h-3 E1/2 dE, if E < EF
and n(E) dE = 0 if E > EF.
Integrating this distribution we obtain
N/V = Ú n(E) dE = 8 p 21/2 m3/2 h-3 (2/3 ) EF3/2.
Expressing EF we obtain
EF(0) = ( h2 / 2 m) (3 N / 8 p V)2/3.
The argument of the Fermi energy indicates temperature. This is the value of the Fermi energy at zero temperature. A similar calculation at nonzero temperature gives a slightly different Fermi energy. One can also define a Fermi temperature: kTF = EF.
We calculated above the contribution of lattice vibrations to the specific heat of metals. The natural question arises: What is the contribution of the electron gas to the specific heat? Since experiments agree with predictions based on lattice vibrations alone electrons should not much contribute. Physically, electrons should contribute because part of the heat transfered to a metal goes into the electron gas and electrons are raised from levels below the Fermi energy to above the Fermi energy. If all electrons contributed to the specific heat then the contribution would be, first for the energy, using equipartition: U = NA3kT/2 = 3RT/2. This would give an unacceptable large contribution to th especific heat (increasing it by 50%). The truth is that only a small fraction of the electrons having energies near the Fermi energy contribute to the specific heat. These are the electrons that are removed from below the fermil level to above it. Their number is the density of states at the Fermi level, multplied by the thickness of the layer, kT. Then we obtain the fraction of electrons contributing to the specific heat by dividind gy the total number. We obtain:
fraction = [kT g(EF)]/[N/V] = 3kT/ 2EF = 3T / 2TF.
The energy these electrons gain is only kT, so the total energy contributing to the specific heat is
U = 3kNAT2 / 2TF,
so the specific heat is
C = dU/dT = 3R T/TF µ 0.03 R,
negligible compared to the specific heat due to lattic vibrations.