1 chapter 19 fermi-dirac gases. 2 okay, why should we want to discuss a fermi-dirac gas? gosh, its...
TRANSCRIPT
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CHAPTER 19FERMI-DIRAC GASES
2
Okay, why should we want to discuss a Fermi-Dirac gas?Gosh, its probably because it is necessary for the
understanding of some important problems.OK, what problems?Some major problems that can be tackled with this
formalism are free electrons in conductors, free electrons in white dwarf stars and neutrons in a neutron star.
Fine, lets press on.
Before we do, I would like to mention the Third Law of Thermodynamics.One expression of this law is that as(There are exceptions to this statement, such as with glasses.)Another expression is: It is impossible to reduce the temperature of a system to absolute zero with a finite number of processes.
0S0T
3
We now consider particles with half-integer spin (s=1/2, 3/2, ….) called fermions. These particles obey the Pauli Exclusion Principle.They are indistinguishable in a gas and so obey Fermi-Dirac Statistics.
We have
1
1)()(
)(/)(
kTeg
Nf
)(f is called the fermi function. 1)(0 f)(f gives the probability that a single state will be occupied.
At any T, if 2/1)( f
We define the fermi energy by )0( TF
At T=0
)0(
)0(/)]0([
if
ifkT
Therefore
)0(0
)0(1)(
if
iff
)T(
(not fermi level!)
[The ground state is taken to be at 0eV and so μ(0)>0]
4
The PEP permits only one particle per state so the N particles are crowded into the N lowest energy states. At T=0 only one microstate is possible, so w=1 and S=kln(w)=0. Hence S=0 at T=0.This is in accordance with the 3rd law of thermodynamics.
T=01
0
)(f
F
The density of states is the same as before except that there are two possible values for the quantum number of Hence we have 2 particles per spatial state.Taking this into consideration gives, from our previous calculation of the density of states:
)2/1,2/1(: zs
5
dh
mVdg
2/3
2
24)(
0
)()( dgfN
At T=0 the fermi function is unity up to the fermi energy
2/3
2/3
20
2/3
2
2382
4 Fh
mVd
h
mVN
F
3/22
83
2
V
N
m
hF
The energy of an ordinary gas molecule is of order kT. We define a fermi temperature by
FF kT3/223/22
504.1283
2
V
N
mk
hTor
V
N
mk
hT FF
6
This is analogous to the bose temperature defined earlier. The fermi temperature is a reference temperature, it is not the temperature of the gas. We will calculate the fermi temperature a little later: it is in the range K54 1010
The more difficult situation in when the temperature is greater than zero. Now states above the fermi energy are occupied.
0/)(
2/3
2 1
24 d
eh
mVN
kT
The 1 in the denominator makes the integration difficult. Numerical calculations can be made to determine the chemical potential as a function of the temperature. However approximations can be made, valid for ,givingFTT
FF
TTT
T
)1(12
1)0(22
Notice that in contrast to bosons of whichFTTfor 0 0
7F
)(f
T=0K
T=0.2 TF
)0(97.0
A knowledge of the chemical potential permits the calculation of the fermi distribution function.
8
Next we are going to treat the electrons in a solid as a gas. This may seem unreasonable given the strong coulomb interaction between electrons and the presence of the positively charged lattice sites. The many-body theory of solids shows that these effects can be reasonably ignored, at least to a good 1st
approximation.
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Free electrons in a metal. (This is relevant to conduction in materials, white dwarfs, He-3, and nuclear matter.)To a good first approximation, the electrons confined in the interior of a metal are similar to molecules in a gas, so we speak of an electron gas. Of course this is a simplification because it ignores, among other things, the periodic ionic potential, which leads to band structure so important in semiconductor physics. (An effective mass is often introduced to partially compensate for the simplifications.)
The metal will be at a fixed temperature and there will be a corresponding chemical potential, called the fermi level. The potential energy for the electron gas is as follows:
fermi level
)(T
)(T
=work function (3-4eV)
!
10
An electron near the surface of the metal feels a strong attractive force due to the positive metal ions in its vicinity. To free an electron from a metal one must give an electron at the fermi level a certain energy , called the work function. In the photoelectric effect, this energy is supplied by absorbing a photon.
Note that the fermi level is not the same as the fermi energy but, for FF TTT )0()(
As shown in your textbook, for Ag, the fermi energy is 5.54eV. This is high and the most striking property of a fermi gas. Contrast this with a gas molecule at room temperature, which has an energy of about 0.025eV. The energy of the electrons at the fermi energy, at absolute zero, is about that of a gas molecule at a temperature of about 64,000K.
The energy of an electron gas at T=0K is called the zero point energy.
At room T (Ag) 0047.0104.6
3004
K
K
T
T
F
}kT{ FF
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When the gas is said to be in the degenerate region. This is not to be confused with the degeneracy of an energy level.
FTT
Using equation (1) F 99998.00047.012
1)0(2
2
FK 99998.0)300( This is why is often confused with )(T F
We now plot )()( gandf
T=01
0
)(f
0 F
T
)(g
Fermi level Fermi energy
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)()()( gfN
F
)(N
It is the electrons in the tail of this distribution that can be most easily extracted from a metal by various processes.
Internal energy of the gas: dNU
0
)(
0/)(
2/32/3
20 1
24)()( d
eh
mVdgfU
kT
Again we can make approximations to obtain a series solution to this equation.
T=0
13
)2(1612
51
53
4422
FFF T
T
T
TNU
At T=0 FNU 53 This is a large energy. It comes about by
the PEP.
For example, for Ag at T=0,
eVeVN
UF 33.3)54.5(
53
53)0(
For an ordinary gas at T=300K eV025.0
The electrons in an electron gas have much more energy at 0K than the molecules of a gas at room temperature.
14
Specific heat of metals (one of the great accomplishments of the theory).
The law of Dulong and Petit: for all elementary solids at room temperature. (This is an experimental result.)
This law has a simple explanation based on the principle of equipartition of energy. Each atom of the solid is considered to be a linear oscillator with 6 degrees of freedom (vibrating in 3 dimensions and the oscillator has both kinetic and potential energy).
RcV 3
R3cR3T
ucRT3u
nRT3NkT3kT2
16NU
V
V
V
At lower temperatures the specific heat decreases and , at these temperatures, a more sophisticated quantum-mechanical approach is required.
15
Continuing with the electron gas, we associate a specific heat with the free electrons by defining:
V
e dT
dUC
Using equation (2)
)(46
553
342
approxT
T
TT
T
TNC
FFFFFe
)(103
2
322
approxT
T
T
T
T
NC
FFF
Fe
Since FF kT
322
103
2 FFe T
T
T
TNkC
Since we are considering low temperatures it is certainly true that T is very much smaller than the fermi temperature and so the second term is negligible.
)2(1612
51
53
4422
FFF T
T
T
TNU
16
FFe
kTNk
T
TNkC
22
22
(Notice the linear dependence on T)
At room temperature and for Ag
R022.0ceV54.5
eV025.0nR
2C e
2
e
This small value explains a puzzle regarding the specific heat of metals. It was expected that the free electrons, having 3 translational degrees of freedom, would contribute an additional (3/2)R to the specific heat (from equipartition theorem). This is obviouslynot in agreement with experiment. Our calculation shows that, indeed, the electronic contribution is small.
The reason that it is small is as follows. Even though the kinetic energies of the free electrons are much greater than the thermal energy of the particles in a gas, the change in energy (dU/dT) of the electrons is small. Only the electrons near the fermi level can increase their energies because of the availability of unoccupied states, and only a small fraction of the free electrons are near F
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The entropy of the electron gas:
dTCTdS)constantV(dTCQdT
QC VV
V
V
For the free electrons dTT
CS
Te
0
dTTT
T
T
TNkS
T
FF
0
322 1103
2
)3(T
T
10T
TNk
2S
3
F
2
F
2
S=0 at T=0 in accordance with the 3rd law of thermodynamics. This is also true for a Bose gas.
đđ
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Helmholtz thermodynamic potential. F=U-TS
322
4422
102
16125
153
FF
FFF
T
T
T
TNkT
T
T
T
TNkTF
4422
4422
202
803
453
FFF
FFF
T
T
T
TNkT
T
T
T
TNkTF
4422
80453
FFF T
T
T
TNkTF
Eqn(2) –T[eqn(3)]
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The pressure of an electron gas: Now that we have the Helmholtz function, we can calculate the pressure. (potentials yield properties)
NTV
FP
,
Our first chore is then to write F explicitly in terms of the volume!!!
3/2
3/22
83
2
aV
V
N
m
hkTF
Rewriting F
4
44
2
22
80453
F
F
F
FF
T
kTT
T
kTTkTNF
3
444222
)(80)(453
FFF
kT
kT
kT
kTkTNF
3
24443/22223/2
80453
a
VkT
a
VkTaVNF
3
444
3/1
222
3/5 802
4)3/2()3/2(
53
a
VkT
aV
kT
V
aNP
20
3
24443/2222
3/2 40652
a
VkT
a
VkT
V
a
V
NP
3
444222
)(40652
FFF
kT
kT
kT
kTkT
V
NP
4
444
2
222
)(805
)(125
152
FF
F
kT
kT
kT
kT
V
NkTP
4422
16125
152
FF
F
T
T
T
T
V
NkTP
Comparing with equation (2) (slide 15)V
UP
32
Example: For Ag KTmV
NF
3328 10641091.5
Since FTT FkTV
NP
52
3/2
FVkTa and
107Am
kg105.10
3
3
21
PaKK
J
mP 10323
3
28
101.210641038.11091.5
52
atmP 5101.2
In spite of this tremendous pressure, the potential barrier at the surface of the metal keeps the free electrons from evaporating from the surface.
This pressure tends to increase the volume. This is balanced by the interaction between the electrons and the ions.
If we were to continue with a description of solids we would then choose a more realistic potential. The regular spacing of the ions is described by a periodic potential. This would lead to an energy level diagram which has bands of energy states separated by gaps in which there are no states. This structure permits us to understand electrical conduction in semiconductors.
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> assume(x>=0);> int((x^2)/(exp(x)-1),x=0..infinity);
2*Zeta(3)> evalf(%);2.404113806
Here is how you can coax MAPLE to give you a numerical answer to an integral.
23
White Dwarf Stars. These stars have masses comparable to the mass of the sun and radii comparable to the radius of the Earth. Therefore they are extremely dense. The core temperature is of the order ofUnder these conditions the atoms are completely ionized so we have nuclei and an electron gas. At the white dwarf stage the H has been used up in thermonuclear reactions, fusion is greatly reduced, the star cools and begins to collapse. This collapse is stopped by the pressure of the electron gas.
In an elementary discussion of white dwarfs, several approximations are made. Relativistic effects are not usually dominant so it is not too egregious to make a non-relativistic calculation. All densities are assumed uniform (perhaps the most serious approximation).
The first white dwarf discovered was Sirius B, so we shall use it as an example.
K710
KTmVmRkgM 7320630 101023.71057.51009.2
24
The white dwarf consists of light elements and hence 2Z
A
Let Nnuc be the number of nuclei in the star and A the average mass number. The total mass, M, is then M=NnucAMH
Since the atoms are completely ionized the number of electrons is:N=ZNnuc
56
27
30
HHH
nuc 1030.6N)kg1066.1(2
kg1009.2
M2
M
M
M
A
Z
AM
MZZNN
3/2
320
56
31
2343/22
1023.78
1030.63
)1011.9(2
)1063.6(
8
3
2
mkg
sJ
V
N
m
hF
MeVJF 333.01034.5 14
The fermi temperature can now be calculated.
KT
KJJ
kT F
FF
9
23
14
1087.31038.1
1034.5
25
The temperature of the star is much less than the fermi temperature and so we have a degenerate electron gas.
Now we can calculate the pressure of this degenerate gas.
PaJmV
NP F
2214320
56
1086.11034.51023.71030.6
52
52
From slide 20
Is this pressure, due to the degenerate electron gas, sufficient to prevent future collapse?
Condition for stability: U=minimumU has two terms: electron gas and gravitational Ge UUU
atmP 171086.1
26
We will write U=U( R ) and determine the radius at equilibriumThe gravitational U can be obtained by building up the star
by bringing in mass elements from infinity. We obtain
22
53
53
GMaR
aU
R
GMU GG
Since and using the expression for the energy on slide 15FTT
3/2
3
3/223/22
34
83
253
83
253
53
R
N
m
hN
V
N
m
hNNU Fe
27
2
3/2
2
e
2
2
3/2
e
2
e R
1
32
N9
m2
Nh
5
3
R
1
4
3
8
N3
m2
hN
5
3U
3/5
3/2
2
2
2 329
103
Nm
hb
R
bU
ee
2R
b
R
aU For equilibrium 0
dR
dU
)(2
02
32mequilibriu
a
bR
R
b
R
a
3/2
22
3/52
23/5
3/2
2
2
329
35
329
103
2
GMm
Nh
GMN
m
hR
ee
3/2
2230
2
21131
3/556234
329
)1009.2(1067.6)1011.9(
)1030.6()1063.6(
kg
kg
mNkg
sJR
28
mR 61015.7
This is equal to the observed radius, to within the accuracy of the calculation. Hence it is reasonable to assume that Sirius B is now a stable white dwarf. However it will eventually become invisible.In most known white dwarfs the core contains C and O with an outer layer which consists of H or He or both. The star cools slowly. It is estimated that white dwarfs take about 1010 years to become invisible. Since the age of the universe is about years, most white dwarfs are still visible.10104.1
)m1057.5R measured( 6
29
If the mass of a star is sufficiently large so that the electrons are relativistic, a stable equilibrium is not possible. The largest possible mass is called the Chandrasekhar limit. This limit is about 1.44 solar masses. A star exceeding this mass collapses, the density approaches that of nuclear matter and the electrons combine with the protons (inverse β-decay) to form a neutron star. Now a degenerate neutron gas stabilizes the star. Again there is a limiting mass, which is about three solar masses. More massive stars collapse to form black holes.