1
17.3 The free electron gas model for electrons in a simple metal (Hiroshi Matsuoka)
In the free electron gas model, the conduction electrons in a simple metal are regarded as
non-interacting fermionic particles. “Simple” metals include alkali metals (e.g., Li, Na, K,..) and
noble metals (e.g., Cu, Ag, Au). Our main goal is to drive the molar heat capacity at constant
volume for the conduction electrons:
( !2 " T %
* R$ '
cv = ) 2 $# T F '&
*
+ (3 2) R
(T << TF )
(T >> TF )
where TF is the characteristic temperature or the Fermi temperature for the electrons and
TF ~ O(10 4 ! 105 K ) .
The internal energy and the average number of phonons
r
As the energy eigenvalue ! nr of the electron one-particle state depends on its wave vector k rn :
r
h2 knr 2
,
! nr = ! knr =
2m
( )
we can express the internal energy and the average number of the electrons using summations
over the wave vector:
U = 2$
r
knr
! (k
e
and
N = 2$
r
knr
)
r
n
"{ ! ( k rn ) #µ }
+1
1
! { " ( knr ) #µ }
e
+1
,
where ! " 1 (k BT ) and the factor of 2 accounts for the spin degeneracy as we assume that the
energy eigenvalue ! nr is independent of the electron spin.
In principle, the above equation for N allows us to find the chemical potential per electron as
a function of the temperature T, the volume V, and the number of electrons N for the free electron
gas:
µ = µ (T, V, N ) ,
2
while the equation for U together with this equation for µ lets us find the internal energy as a
function of T, V, and N:
U = U(T ,V, µ (T ,V, N )) = U (T, V, N ) .
In this text, for low temperatures much lower than the Fermi temperature, we will not pursue this
strategy for calculating the internal energy. Instead, we will calculate the heat capacity at
constant volume first and then the internal energy.
The heat capacity at constant volume at low temperatures
Our main goal in this section is to derive the heat capacity at constant volume for the
electrons at low temperatures (i.e., T << TF ) so that T TF <<1) in the free electron gas model.
As T TF is a “small” parameter controlling CV , we should be able to express CV as a power
series in T TF :
! T$
! T $2
! T $2
#
&
#
&
CV = c1 # & + c2 # & + c3 ## && + ''',
" TF %
" TF %
" TF %
0
where we have no constant term (i.e., a term proportional to (T TF ) ) on the right-hand side as
CV = 0 at T = 0 , which is one of the consequences of the third law of thermodynamics.
Using the equations for U and N, we will find CV for low temperatures to be
(*" % 3 ,*
! 2 "$ T %'
T
CV =
nR $ ' + O)$$ '' - ,
2
*+# T F & *.
# TF &
where the Fermi temperature TF is defined by
h2 " 2 N % 2 3
TF =
$ 3!
' ,
2mkB #
V&
so that the corresponding molar heat capacity at constant volume is given by
C
!2
cv = V =
n
2
(*" % 3 ,*
" T%
T
$
'
R$ ' + O)$$ '' - .
*+# TF & *.
# TF&
3
Using this expression for CV , we can calculate the internal energy U for the electrons:
T
U(T ,V,N ) = U (T = 0 K,V ,N ) + " CV (T ,V, N )dT !
0
*,
3 # 2 $& T ')
= nRTF + +
& )
,- 5 4 % T F (
2
.,
*,$ ' 4 .,
T
/ + O+&& )) /
,0
,-% TF ( ,0
(HW#17.3.1: show this)
where U(T = 0 K, V, N ) = (3 5)nRTF , which we will derive shortly. We can also obtain the
entropy S for the electrons from CV :
T
S (T, V, N ) = S(T = 0 K, V, N ) + " dT !
0
CV (T !,V, N )
T!
*,$ ' 3 .,
# 2 $& T ')
T
=
nR& ) + O+&& )) /
2
% TF (
,-% TF ( ,0
(HW#17.3.2: show this)
where we have used the third law of thermodynamics, S (T = 0 K,V ,N ) = 0.
Using the
expressions for U and S, we can find the Helmholtz free energy F:
2)+
)+# & 4 -+
3 " 2 #% T &( +
T
. + O*%% (( . .
F = U ! TS = nRTF * !
%
(
+, 5 4 $ T F ' +/
+,$ T F ' +/
(HW#17.3.3: show this)
Using a relation between the chemical potential per electron µ and F, which we will show later,
we can then calculate µ from F:
2.
*,
*," % 4 .,
" !F %
) 2 "$ T %' ,
T
/ + O+$$ '' /
µ = $ ' = k B TF +1 (
$
'
# ! N & T ,V
,- 12 # T F & ,0
,-# T F & ,0
(HW#17.3.4: show this)
where we have used nR = NkB and TF ! N 2 3 .
4
The Fermi wavenumber, the Fermi energy, and the Fermi temperature
At T = 0 , the Fermi-Dirac distribution function becomes a step function that drops its value
from 1 to 0 at ! = µ (T = 0, V, N ) so that the electron one-particle states with energy less than the
“Fermi energy” defined by ! F " µ (T = 0 K, V,N ) are all occupied by N electrons. The energy of
a one-particle state is proportional to k 2 ,
!=
h 2k 2
" k2 ,
2m
r
so that at T = 0 , all the one-particle states found inside a three-dimensional sphere in the k space with a radius k F are all occupied by two electrons with spin up and down so that
4
"k F 3
Volume
of
a
sphere
with
radius
k
(
V
F)
3
3
N =2
=2
3
3 =
2 kF ,
#
&
(!k )
2"
3"
% (
$ L'
r
3
where the factor of 2 accounts for the “two” spin quantum numbers and (!k ) is the k -space
r
volume per wave vector k . Solving this equation for k F , we find the following expression for
the Fermi wavenumber:
"
N %13
k F = $ 3! 2 ' .
#
V&
Note that (3! 2 N V )
13
!1
has the dimension of (length ) , which is the dimension of wavenumber.
The Fermi energy ! F is then a function of the number density of the electrons:
23
h 2 kF 2 h 2 $ 2 N '
"F =
=
& 3#
) .
2m
2m %
V(
The Fermi energy increases as the number density of the electrons increases. This is because as
! we must fill more one-particle states with higher energies. We define the
we add more electrons,
Fermi temperature TF by
"F
h2 $ 2 N ' 2 3
TF !
=
& 3#
) ,
k B 2mkB %
V(
5
Since
'1 3
! N $ 1 3 ! V $ '1 3 ! Na 3 $
1
&
# & =# & ~ #
= ,
"V%
" N%
" N %
a
where a is the distance between a pair of nearest neighbor atoms in a simple metal, we find
"
N %13 1
k F = $ 3! 2 ' ~ ~ O(101 0 m (1) ,
#
V&
a
so that
$ 10"3 4 J #s 2 101 0 m 2 '
h2 kF 2
(
)(
) ) ~ O(10 "1 8 J ) ~ O*, 10 "1 8 J -/ ~ O(10 eV )
!F =
~ O&&
, "1 9
/
)
2m
+ 10
J eV .
(10 "3 0 kg )
%
(
and
TF =
# 10"1 8 J &
!F
(( ~ O(105 K ) .
~ O%% "2 3
kB
$ 10
J K'
Changing the summation over the wave vectors into an integral
For two neighboring wave vectors that satisfy
dk = k rn ! " knr =
$ 10 '
2#
~ O& "1 ) ~ O(10 2 m "1) ,
% 10 m (
L
we find
% h2 k 2 ( 2dk
% d" (
h2 k
1 %" (
*
!" (knr # ) $ !" ( knr ) = !d" = ! ' * dk = !
dk = ! '
~
' dk * .
& dk )
m
& 2m ) k
k BT & k )
We find the largest value for !" (knr # ) $ !" ( knr ) when the two neighboring wave vectors are near
r
the sphere in the k -space with a radius k F so that
Max !" ( knr# ) $ ! " (knr )
+ (10 $1 8 J )(102 m $1 ) .
% $3 (
1 %' " F (*
0 ~ O' 10 K * .
=
dk
~
O
-, (10$2 3 J K )T(101 0 m $1) 0/
kB T '& kF *)
& T )
For temperatures above 1 K, we therefore find !" (knr # ) $ !" ( knr ) << 1. We can then approximate
r
the summation over the wave vector k rn in the equations for U and N by an integral as follows.
6
%
V
"( k)
U = 2 & dkk 2 # { " ( k ) $µ }
! 0
e
+1
and
%
V
1
N = 2 & dkk 2 " {# ( k ) $ µ }
.
! 0
e
+1
since
2!
r
k rn
,
,
r
$ L '3
2
2V
V
3
2
2
(
)
=
= 2& )
dk =
3 ! "k
3 + dk4 #k =
2 + dkk ,
+
r
%
(
r
("k ) k rn
2# All k *space ( 2# ) 0
# 0
where we have used
# 2 " & 3 (2 " )3
( =
.
$ L'
V
(!k )3 = %
Integrals with respect to the electron energy and the electron density of states
As the Fermi-Dirac distribution function is a function of the electron energy ! , it would be
more convenient to change the above integrals with respect to the wavenumber k for U and N
into integrals with respect to the electron energy ! . As
!=
h 2k 2
,
2m
we find
! 2m $ 1 2
k = # 2 & '1 2
"h %
and
" dk %
1 " 2m% 1 2 (1 2
dk = $ ' d! = $ 2 ' ! d! .
# d! &
2# h &
We then obtain
1 ! 2m $ 3 2
dkk = # 2 & d'' 1 2
2" h %
2
so that
U=
and
%
%
V
"( k)
V ' 2m * 3 2
"
2
dkk
=
)
,
d"" 1 2 # ( " $ µ )
2 &
2
2
# { " ( k ) $µ }
&
! 0
e
+1
e
+ 1 2! ( h + 0
7
%
%
V
1
V ' 2m * 3 2
1
2
N = 2 & dkk " {# ( k ) $ µ }
=
d## 1 2 "( # $µ )
.
2 ) 2 ,
&
(
+
! 0
e
+1
e
+ 1 2! h
0
By defining a new function De (! ) by
V $ 2m ' 3 2 1 2
De (! ) "
&
) ! ,
2 #2 % h 2 (
we can simplify these expressions as
"
U = # d! De (! ) f + ( ! , µ ,T )!
0
and
"
N = # d!De ( ! ) f + ( ! , µ ,T) ,
0
where f + is the Fermi-Dirac distribution function:
f + ( ! , µ ,T) =
!
" (! # µ )
e
+1
.
D( ! )d! is the number of one-particle states with energy in an interval [ ! , ! + d! ] so that D( ! ) is
the “number density” of one-particle states” and is therefore called “the density of states.”
d! De (! ) f + (! , µ ,T ) then gives the average number of the electrons found in the one-particle
states with energy in the interval [ ! , ! + d! ] . Summing up or integrating d! De (! ) f + (! , µ ,T ) then
gives the average number of electrons N. Each of the electrons in the one-particle states with
energy in the interval [ ! , ! + d! ] has an energy of ! so that summing up or integrating
d! De (! ) f + (! , µ ,T )! gives the average total energy U.
Traditionally, to evaluate the above integrals for U and N at low temperatures, we use the
following “Sommerfeld expansion”:
#
$ d! e (
0
µ
g( ! )
! "µ ) /( k B T )
+1
= $ d!g( ! ) +
0
% 2 dg
(k BT )2 + O (k B T) 4 .
6 d! ! = µ
{
}
However, as mentioned earlier, in this text, instead of using the Sommerfeld expansion, we will
use the integrals for U and N to calculate the heat capacity at constant volume at low
8
temperatures first and then the internal energy. We can thus justify a quick way of finding CV at
low temperatures, which lets us find CV much faster than using the Sommerfeld expansion.
The density of states in terms of the number of the electrons and the Fermi energy
Using the expression for the Fermi energy,
h2 # 2 N & 2 3
!F =
% 3"
( ,
2m $
V'
we can express the constant in front of ! 1 2 in the density of states in terms of N and ! F :
V " 2m % 3 2
3N
=
2 $ 2 '
2! # h &
2( F 3 2
so that
De (! ) =
3N 1 2
! ,
2! F 3 2
which we will find it useful when we calculate the internal energy at T = 0 K . This expression
for the density of states also makes sense dimensionally as the dimension of De (! ) is 1 [ energy ] .
We can derive this expression more directly as follows. We first note that the number
N e (! ) of the one-particle states with energy less than ! is proportional to the volume inside the
r
sphere in the k -space with a radius k so that
N e ! k3 ! " 3 2 ,
where we have used k ! " 1 2 . As D( ! )d! is the number of one-particle states with energy in an
interval [ ! , ! + d! ] , it satisfies
D( ! )d! = N (! + d! ) " N (! ) ,
so that
D( ! ) =
N (! + d! ) " N (! ) d
=
N (! ) # ! 1 2 .
d!
d!
Note also that De (! ) satisfies
!F
N = " d! De (! ) .
0
9
As
!F
" d!!
12
=
0
2 32
! ,
3 F
the constant in front of ! 1 2 in De (! ) must cancel (2 3)! F 3 2 and produce N in the integral of
De (! ) so that this constant must be (3 2)( N ! F 3 2 ) .
The density of states at the Fermi energy
When we calculate the heat capacity at constant volume for low temperatures, we will also
use the following expression for the density of states at the Fermi energy:
De (! F ) =
3N
.
2! F
We can find this more directly by noting N e ! k 3 ! " 3 2 so that
De (! ) =
d
3N e
N e (! ) =
.
d!
2!
Using N e ( ! F ) = N , we then find
De (! F ) =
3N e ( ! F ) 3N
=
.
2!
2 !F
The internal energy at T = 0 K
We will now show that the internal energy has a finite value even at the absolute zero
temperature. This is due to the “Pauli exclusion principle,” because of which at T = 0 K , the
electrons must fill up all the one-particle states up to those with the Fermi energy ! F . We can
directly calculate the internal energy at T = 0 K by
!F
"
U(T = 0 K, V, N ) = # d! De (! ) f + (! , ! F ,T = 0 K )! = # d! De (! )!
0
=
3N
2 !F3 2
0
!F
# d!!
0
32
=
3N $ 2 5 2 ' 3
& ! ) = N!
2! F 3 2 % 5 F ( 5 F
10
where the Fermi-Dirac distribution function at T = 0 K is a step function that takes a value of 1
from ! = 0 to ! = ! F . We then find
U ( T = 0 K) =
3
3
3
N! F = Nk BT F = nRT F .
5
5
5
If you only remember De (! ) " ! 1 2 , then you can still calculate U(T = 0 K, V, N ) by
!F
"
U(T = 0 K, V, N )
=
N
# d! D (! ) f
e
+
(! ,T = 0 K )!
0
"
=
# d!D ( ! ) f
e
+
( ! ,T = 0 K)
0
# d!!
32
# d!!
12
0
!F
2 52
!F
3
= 52
= !F
!F3 2 5
3
0
The internal energy at low temperatures
Earlier, we have calculated the internal energy of the electrons at low temperatures from the
heat capacity at constant volume, which is yet to be calculated:
T
U(T ,V,N ) = U (T = 0 K,V ,N ) + " CV (T ,V, N )dT !
0
*,
3 # 2 $& T ')
= nRTF + +
& )
,- 5 4 % T F (
2
.,
*,$ ' 4 .,
T
/ + O+&& )) /
,0
,-% TF ( ,0
Physically, this result makes sense because at low temperatures, only the electrons just below the
Fermi energy (i.e., electrons with energy roughly between ! F " k BT and ! F ) are excited to the
one-particle states just above the Fermi energy (i.e., states with energy roughly between ! F and
! F + k BT ) so that each of these electrons gains energy on the order of k BT . Roughly,
De (! F )( kB T ) of such electrons exist between ! F " k BT and ! F so that the overall increase in the
internal energy must be on the order of
!U = U( T ) " U (T = 0 K)
% T (2
3N
2
3
# {De ($ F )( kB T )}(k BT ) =
(k T ) = nRT F '' **
2 $F B
2
& TF )
which is not that far off from what we have calculated:
11
# T &2
"2
!U =
nRT F %% (( .
4
$ TF '
The third law puts a constraint on how µ behaves near T = 0 K
Thermodynamics also sets a constraint on how the chemical potential per electron µ should
approach ! F as the temperature is decreased toward zero. More specifically, the third law of
thermodynamics demands that
" !µ %
$ '
=0,
# ! T & V ,N T =0 K
which implies
µ = !F +
1
kB
# "µ &
( kB T ) + O (k BT )2 = ! F + O (k BT )2 .
% (
$ "T ' V , N T = 0 K
[
]
[
]
Derivation
Using the extended form of the fundamental equation, dU = TdS ! PdV + µ dN , we can
show
dF = d(U ! TS) = dU ! (TdS + SdT ) = ! SdT ! PdV + µdN ,
which implies
" !F %
µ=$ '
# ! N & T ,V
and
# " F&
S = !% (
$ " T ' V ,N
so that
" !µ %
" !S %
!2 F
!2 F
$ '
=
=
= ($
' ,
# ! T & V ,N ! T! N ! N!T
# !N & T ,V
which is one of Maxwell’s relations. The third law states that the entropy at T = 0 K
vanishes for any value of V or N:
S (T = 0 K,V ,N ) = 0
so that
12
" !µ %
" !S %
$ '
= ($
'
=0.
# ! T & V ,N T =0 K
# ! N & T, V T =0 K
In this derivation, we have also derived
" !F %
µ=$ ' ,
# ! N & T ,V
which we have used to find µ from F in the above.
The heat capacity at constant volume at low temperatures
Our goal now is to find the heat capacity at constant volume CV at low temperatures
(T << TF ),
(*" % 3 ,*
! 2 "$ T %'
T
CV =
nR $ ' + O)$$ '' - ,
2
*+# T F & *.
# TF &
using the following equations for the internal energy and the number of the electrons:
"
U = # d! De (! ) f + ( ! , µ ,T )!
0
and
"
N = # d!De ( ! ) f + ( ! , µ ,T) .
0
Using the equation for U, we can express CV as
)
" !U %
" !f %
CV = $
'
= * d( De (( ) ( $ + ' .
# !T & V , N 0
# ! T & V ,N
The key observation for this expression of CV is that at low temperatures, (! f + ! T )V ,N as a
function of ! is peaked at ! = µ and quickly decreases towards zero for ! >> µ + k B T and
! << µ " k B T so that the integral for CV gets most of its value from a narrow range centered at
this peak. More specifically, (! f + ! T )V ,N is given by
13
" !f + %
e) ( * (µ )
$
'
= ( ) *(µ
2
# ! T & V ,N
e ( ) +1
(
=
)
+- * ( µ
" !µ % /,(
(
)
$ ' 0
2
# !T & V , N 1.- k BT
+" !µ % /)
,
k
)
*
(
µ
+
(
)
$ ' 0
B
# !T & V , N -1
4cosh 2 () (* ( µ ) 2 ) -.
so that (! f + ! T )V ,N is a function of ! ( " # µ ) . Note also that (!µ ! T)V ,N ~ O[(k BT )] since
[
2
]
µ = ! F + O ( kB T ) .
{
}
h[! ( " # µ )] = 1 4 cosh2 (! (" # µ ) 2) , not only controls the overall size of (! f + ! T )V ,N but
also limits the integral for CV to an interval centered at ! = µ because it is peaked at ! = µ so
that it makes (! f + ! T )V ,N quickly decrease towards zero for ! satisfying ! " µ >> k B T . As can
be seen from the figure below, for x > 10 , h( x ) is practically zero, which implies that
(! f + ! T )V ,N is appreciable only within an interval µ ! 10k BT < " < µ + 10kB T and that the width
of this interval decreases as the temperature is decreased toward zero. Note also that h( x ) is an
even function of x so that for an odd number n, we find
"
# dxh( x )x
n
=0
!"
because h( x ) x n is then an odd function of x
h
0.4
0.3
0.2
0.1
- 10
-5
0
5
Noting that
)
" !N %
" !f %
0=$
'
= * d( De (( )$ + ' ,
# ! T & V, N 0
# !T & V , N
10
x
14
we can rewrite the integral for CV as
*
" !U %
" !N %
" !f %
CV = $
' (µ$
'
= + d) De () )( ) ( µ )$ + ' ,
# !T & V , N
# ! T & V , N (*
# !T & V , N
where the lower limit of the integral is changed from 0 to ! " as (! f + ! T )V ,N is appreciable
only within the interval µ ! 10k BT < " < µ + 10kB T so that changing the lower limit hardly
changes the value of the integral.
As De (! ) " ! 1 2 , De (! ) is a slowly varying function near ! = µ so that we can approximate it
by
De (! ) " De (µ ) +
dDe
d!
(! # µ )
!=µ
so that
*
$ #f '
dD
CV = De (µ ) + d! (! " µ )& + ) + e
% # T ( V , N d!
"*
*
2 $ #f + '
d
!
(
!
"
µ
)
&
) ,
+
% #T ( V , N
! = µ "*
To find the linear term in k BT for CV , we can neglect the second integral as it is at least on the
3
order of (k BT ) because
dDe
d!
*
2$ # f + '
d
!
!
"
µ
(
)
&
)
+
% #T ( V , N
! =µ "*
dDe
=
d!
2
{, (! " µ )} 354k ,( ! " µ ) + $& #µ ')
(k T ) + d{, (! " µ )}
- , (! " µ ) 0 56
% #T (
2
4cosh /
2
*
B
! =µ
B
2
"*
/.
dDe
=
d!
*
$ #µ '
(k BT ) &% )( + d{, (! " µ )}
# T V, N " *
! =µ
2
[
= O (k BT )
3
75
8
5
V, N 9
21
2
{, (! " µ )}
2
- , (! " µ ) 0
2
4 cosh2 /
/.
21
2
]
3
where an integral of h[! ( " # µ )]{! (" # µ )} vanishes as it is an odd function of ! ( " # µ ) , and
we have also used (!µ ! T)V ,N ~ O[(k BT )] and
dDe
d!
=
! =µ
3N "1 2
3N
2
=
32 µ
3 2 ! F + O ( kB T )
4! F
4 !F
{
[
]}
"1 2
=
dDe
d!
{1+ O[(k T ) ]} .
2
B
!=!F
15
The first integral for CV then becomes
*
$ #f '
3
CV = De (µ ) + d! (! " µ )& + ) + O ( kB T )
% # T ( V, N
"*
[
De (µ ) *
=
+ d {,( ! " µ )}
, "*
35
$ #µ ' 75
, (! " µ )
3
4
k
,
!
"
µ
+
(
)
& ) 8 + O (k B T )
B
0
% # T ( V ,N 59
, ( ! " µ ) 56
2
4cosh 2 /
/.
21
2
[
{, (! " µ )}
*
D (µ )
= e
+ d {,( ! " µ )}
, "*
*
= De (µ )kB 2 T + dx
"*
=
]
3
2
[
( )
0 + O k BT
2 ,( ! " µ )
4cosh /
2
21
./
2
x2
[
$ ' + O (k B T )
2 x
4cosh & )
% 2(
:2
3
De ( µ )kB2 T + O ( kB T )
3
[
]
3
3
]
]
]
where an integral of h[! ( " # µ )]{! (" # µ )} vanishes as it is an odd function of ! ( " # µ ) , and we
have also used
"
x2
$2
# dy 4cosh 2( x 2) = 3 .
!"
As
De (µ ) = De ( ! F ) +
[
2
dDe ( µ )
(µ " ! F ) = De (! F ) + O (k BT )2 ,
dµ µ =! F
[
]
]
where µ ! " F = O ( kB T ) , we finally obtain
CV =
=
"2
" 2 $ 3N ' 2
3
3
2
De (#F ) kB T + O ( k B T ) = &
) kB T + O ( k B T )
3
3 % 2#F (
[
$T'
"2
3
NkB & ) + O ( k B T )
2
% TF (
[
"2 $ T '
3
=
nR& ) + O ( kB T )
2
% TF (
[
!
]
]
]
[
]
16
where we have used ! F = k BTF , Nk B = nR, and
3N
.
2! F
De (! F ) =
Knowing that various steps for this derivation of CV can be justified, the next time you derive
CV for the electrons in the free electron gas model, you can quickly derive CV as follows:
*
" !U %
" !U%
" !N %
" !f %
CV = $
'
=$
'
( µ$
'
= + d) De () )( ) ( µ )$ + '
# !T & V , N # ! T & V , N
# ! T & V, N 0
# !T & V , N
*
, De ( ) F )( kB T ) + d {- () ( µ )}
(*
.0
" !µ % 20
- () ( µ )
/
k
)
(
µ
+
(
) $# '& 30
4 cosh2 (- () ( µ ) 2) 01 B
! T V, N 4
{- () ( µ )}
*
, De ( ) F )k T + d{- () ( µ )}
2
B
(*
*
= De ( ) F )k B2 T + dx
(*
2
4 cosh2 (- () ( µ ) 2)
x2
52
,
D () )k 2 T
4cosh 2 (x 2 ) 3 e F B
The heat capacity at constant volume at high temperatures (T >> TF )
At high temperatures, we are in the classical regime so that the Fermi-Dirac distribution
function is reduced to the Boltzman distribution function and we find the internal energy per
electron to be
$
U
=
N
% d!
0
$
% d!
0
!32
e"
( ! #µ )
!
e
+1
12
" ( ! #µ )
+1
&
%
%
$
0
$
0
d!! 3/2e
# ( ! #µ ) /( k B T )
(
d!! 1/ 2 e # ! #µ
) / ( k BT )
( 3'1/ 2 + ( ' 1/ 2 + #1
3
=*
-*
- k B T = k BT
2
) 4 ,) 2 ,
so that
U!
$
( kB T )5/ 2 e µ / ( k T ) %0 dt t 3/ 2 e #t
=
$
(k BT )3/ 2 e µ / ( k T ) %0 dt t 1/ 2 e# t
3
3
NkB T = nRT
2
2
B
B
17
The heat capacity at constant volume is then given by
" !U %
3
CV = $
'
( nR ,
# !T & V , N 2
so that the molar heat capacity at constant volume is given by
cv =
CV 3
! R.
n
2
Answers for the homework questions in Sec.17.3
HW#17.3.1
T
# !U &
U(T ,V,N ) = U (T = 0 K,V ,N ) + ) %
( dT "
$ !T " ' V , N
0
T
T
+-# T & 4 /3
3
* 2 #% T &(
= nRT F + ) CV (T ,V ,N ) dT " = nRT F + )
nR % ( dT " + O,%% (( 0
5
5
2
-.$ TF ' -1
$ TF '
0
0
+-# & 4 /# T &2
3
*2
T
= nRT F +
nRTF %% (( + O,%% (( 0
5
4
$ TF '
.-$ TF ' 12/
++-# & 4 /3 * 2 #% T &( T
= nRTF , +
0 + O,%% (( 0
%
(
-. 5 4 $ T F ' -1
-.$ TF ' -1
HW#17.3.2
T
$ T! ' #2 $ T '
CV (T !,V, N )
#2
S (T, V, N ) = S(T = 0 K, V, N ) + " dT !
= " dT !
nR&& )) =
nR && ))
T!
2T ! % TF ( 2
%TF(
0
0
T
HW#17.3.3
22)+
)+
# T &2
3 " 2 #% T &( + " 2
3 " 2 #% T &( +
%
(
.!
.
F = U ! TS = nRTF * +
nRT F % ( = Nk B TF * !
+, 5 4 %$ T F (' +/ 2
+, 5 4 %$ TF (' +/
$ TF'
18
HW#17.3.4
" !F %
µ=$ '
# ! N & T ,V
2.
*,
*," % 4 .,
*, 3 " 2T % ) 2 " (1 % " 2T % .,
3 ) 2 "$ T %' ,
T
2
F
F
$$ 2 '' $
/ + Nk B + $
= kB TF + (
'(
' T / + O+$$ '' /
$
'
,- 5 # 3N & 4 # TF & # 3N & ,0
,- 5 4 # TF & ,0
,-# T F & ,0
*, ) 2 " T % 2 .,
*," T % 4 .,
$ ' / + O+$$ '' /
= kB TF +1 (
,- 12 $# TF '& ,0
,-# TF & ,0
where we have used nR = NkB and TF ! N 2 3 so that
" !T F %
2T F
$
' =
# ! N & T,V
3N
and
(* ! " 1 % ,*
"
%"
%
$$ '' - = $$ / 1 2 '' $ 2T F ' .
)
*+ ! N # TF & *.
# TF & # 3N &
T, V