Download 1 III Equilibrium statistical mechanics (Hiroshi Matsuoka) The goal

1
III Equilibrium statistical mechanics
(Hiroshi Matsuoka)
The goal and strategy of statistical mechanics
Statistical mechanics is a theoretical framework, within which we can calculate the
thermodynamic state variables, (e.g., internal energy and entropy) of a microscopic model for a
macroscopic system. Our goal is to reveal a microscopic mechanism behind the universal
properties of these variables shared by the macroscopic systems belonging to a particular
universality class. Our main strategy for reaching this goal is then to construct a “minimal”
microscopic model that also exhibits these universal properties and thus belongs to this universality
class.
12. The canonical ensemble theory based on the Helmholtz free energy
There exist mainly three different approaches for calculating the state variables of a microscopic
model. We will first discuss an approach traditionally called “canonical ensemble theory,” where
we calculate the Helmholtz free energy F for the microscopic model and then calculate other state
variables such as the pressure (i.e., the equation of state), the internal energy, and the entropy of the
model directly from this Helmholtz free energy. This is the most widely used approach, which we
will apply to the ideal gas models for monatomic (Ch.13) and diatomic (Ch.14) low-density gases.
The canonical ensemble theory is built upon another approach called “micro-canonical
ensemble theory,” where we calculate the entropy S for the microscopic model and then calculate
other state variables. We will discuss the micro-canonical ensemble theory in Ch.15.
A problem with the canonical ensemble theory and the grand canonical ensemble theory
A major drawback of this approach is that it is not well-suited for the ideal gas model of a
low-density gas at extremely low temperatures, where the state variables of the gas critically
depend on whether the constituent molecules or atoms are “fermions” or “bosons” (for example,
3
He atoms are fermions whereas 4 He atoms are bosons). It is also difficult to apply this
approach to the free electron model for conduction electrons, which are fermions, in a simple
metal. To treat both fermions and bosons in the same framework, we will adopt a different
approach traditionally called the “grand canonical ensemble theory” based on a quantity called
the “thermodynamic potential” ! of a microscopic model. We will discuss the grand canonical
ensemble theory in Ch.16. In Ch.17, we will then apply this approach to the Debye model for
phonons in a solid and the free electron model for electrons in a simple metal.
12.1 State variables can be calculated directly from the Helmholtz free energy
Entropy, pressure, and the internal energy as functions of T, V, and n
2
As mentioned above, the equation of state, the internal energy, and the entropy of a microscopic
model can be calculated directly from the Helmholtz free energy of a microscopic model. Let’s see
how this is done.
In Sec.8.2, where we have derived the first dS equation, we have defined a new state variable
called the Helmholtz free energy by
F ! U " TS .
Using the fundamental equation of thermodynamics,
dU = TdS ! PdV ,
we have obtained
dF = !SdT ! PdV .
Derivation: dF = d(U ! TS) = dU ! d(TS ) = TdS ! PdV ! (TdS + SdT ) = !SdT ! PdV .
Comparing dF = !SdT ! PdV with the following calculus equation,
" !F %
" !F %
' dT + $
' dV ,
dF = $
# ! T & V ,n
# ! V & T ,n
we have then found
# " F&
S = !% (
$ " T ' V ,n
and
# "F &
( .
P = !%
$ "V ' T, n
The first equation allows us to calculate the entropy of the microscopic model as a function of T,
V, and n, S = S (T, V,n ) , directly from its Helmholtz free energy. Similarly, the second equation
allows us to calculate the equation of state for the model, P = P (T, V,n ) , directly from its
Helmholtz free energy.
We can also calculate the internal energy for the model by
) " # F& ,
U = !T 2 * % ( - .
+ "T $ T ' .V ,n
because
# " F&
) " # F&,
U = F + TS = F ! T % ( = !T 2 *
% (- .
$ " T ' V ,n
+ " T $ T ' .V ,n
3
It is remarkable that a single function, the Helmholtz free energy, carries all the information
about the other state variables of the model so that once we calculate F as a function of T, V, and
n or
F = F (T ,V,n) ,
then we can calculate
P = P (T, V,n ) ,
U = U(T ,V,n) ,
and
S = S (T, V,n ) .
All this follows from dF = !SdT ! PdV , which is a direct consequence of the fundamental
equation, dU = TdS ! PdV . The main task for statistical mechanics is then to calculate the
Helmholtz free energy for a given microscopic model.
Heat capacity at constant volume
We can calculate the heat capacity at constant volume CV from the entropy or the internal
energy by
" ! 2 F%
" !S %
CV = T $ ' = (T $ 2 '
# !T & V ,n
# !T & V ,n
or
" !U %
CV = $
' .
# !T & V ,n
Isothermal compressibility and coefficient of thermal expansion
We can calculate the isothermal compressibility ! T from the Helmholtz free energy by
!T = "
1 $ #V '
1
1
&
) =" $
=
2
$ # F' .
#P '
V % #P ( T ,n
V&
)
V& 2)
% # V ( T ,n
% #V ( T, n
Note that ! T obtained from the Helmholtz free energy is a function of T, V, and n instead of a
function of T, P, and n.
We can also calculate the coefficient of thermal expansion ! from the Helmholtz free energy
by
+- # $ # F ' /, &
) 0
-. #T % #V ( T ,n -1
$ #P '
V ,n
! = "T & ) = *
$ # 2F '
% # T ( V ,n
V& 2 )
% # V ( T ,n
since
4
+- # $ #F ' /! $ #P '
= & ) = *, &
) 0 .
-. # T % # V ( T ,n -1
" T % #T ( V ,n
V, n
Again note that ! obtained from the Helmholtz free energy is a function of T, V, and n instead of
a function of T, P, and n.
F decreases with T while V is kept constant
As the entropy S of any system is positive for a finite temperature T > 0 , we find
" !F %
$ ' = (S < 0
# ! T & V, n
so that the Helmholtz free energy F of any system is a monotonically decreasing function of
temperature. In addition, according to the third law of thermodynamics, the entropy vanishes at
the absolute zero temperature so that the curve for F as a function of T must have a vanishing
slope at T = 0 K . As the heat capacity at constant volume CV of any system is positive, we find
" !2 F %
C
$ 2' = ( V <0.
# ! T & V ,n
T
Therefore, F is concave or bowed-up as a function of temperature.
As the entropy vanishes at the absolute zero temperature, the Helmholtz free energy is
identical to the internal energy at the absolute zero temperature: F = U at T = 0 K .
F decreases with V while T is kept constant
The Helmholtz free energy F also decreases as the volume V is increased since
" !F %
$
' = ( P < 0.
# !V & T ,n
As the isothermal compressibility ! T of any system is positive, we find
" ! 2F %
1
$ 2' =
> 0.
# !V & T ,n V( T
Therefore, F is convex or bowed-down as a function of V.
The Helmholtz free energy of the monatomic ideal gas model
5
In the monatomic ideal gas model, we model all the atoms in the gas to be point particles that
do not exert forces on each other. We will later calculate the Helmholtz free energy of this
model to be
( " v % +
F = !nRT * ln #
& + 1- = nf (T, v ) ,
*) $ vQ (T ) ' -,
where the “quantum molar volume” vQ ( T ) is defined by
$ 2#! 2 ' 3 / 2
vQ (T ) " N Avogadro &
) ,
% mk B T (
where m is the mass of each atom in the gas.
!
As discussed above, f is a decreasing convex or bowed-down function of v.
As we will see later in the next chapter, this expression for the Helmholtz free energy is
based on an approximation that the temperature of the gas is not too low so that we do not need
to take into account whether the atoms in the gas are fermions or bosons. This approximation is
valid as long as the molar volume v of the gas is much larger than the quantum molar volume at
the temperature of the gas:
v >> vQ (T ) ,
which also implies that if the molar volume of the gas is comparable to or lower than the
quantum molar volume, we need to go beyond the approximation and must take into account
whether the atoms in the gas are fermions or bosons. For example, the quantum molar volume at
100 K is on the order of
vQ (100 K ) ~ 10!6 m 3 mol ,
(HW#12.1.1: show this)
6
which implies that at 100 K, as long as the molar volume is much larger than 10 !6 m 3 mol , the
approximation leading to the above Helmholtz free energy is well justified. This is consistent
with our finding in Sec.5.4 and Sec.5.5 that for v > 10 !2 m3 mol , the ideal gas law is a very
good approximation for the equation of state for a steam and other gases. Note that as the
temperature is decreased, the quantum molar volume increases so that at lower temperatures the
molar volume must be very large or the number density of the gas must be very low in order for
the approximation to remain valid. This is exactly why at very low temperatures, we need to
adopt a different approach other the one based on the Helmholtz free energy.
The molar Helmholtz free energy depends on the temperature as
f
T
=!
RTQ
TQ
) "
&3/ 2 ,
+ ln $# T $' + 1.
+ $ T (v ) $
.,
* % Q (
-
where TQ is defined by
! T $ 3/ 2
#
& ' v .
#" TQ ( v) &%
vQ (T )
so that
! vQ (T ) $ 2 /3
4
& T = N 2/ 3 Ttrans( V ) ,
T Q (v ) = #
"# v &%
'
(HW#12.1.2: show this)
where N = nN Avogadro and Ttrans(V ) is the characteristic temperature for the translational motion
introduced in Sec.7.4.1 as
Ttrans
2
!2 # " &
! 2" 2
=
.
% ( =
2mk B $ L ' 2mk BV 2 3
As we have found in Sec.7.4.1, Ttrans ~ O(10!1 6 K) in the temperature range where we normally
!
observe a low-density
gas so that
TQ ~ O(1 K) .
(HW#12.1.3: show this)
7
For T > TQ , f decreases as T is increased although near T = 0 f is increasing, which indicates that
the above expression for the Helmholtz free energy is valid only when T >> TQ or equivalently
v >> vQ (T ) . For T > TQ , f is also concave or bowed-up as a function of T as expected from the
discussion in the preceding section..
The state variables of the monatomic ideal gas model, a minimal model for the universality
class of low-density monatomic gases
From this Helmholtz free energy, we can find
# "F &
RT
( =
P = !%
,
$ "V ' T, n
v
(HW#12.1.4: show this)
) " # F& ,
3
U = !T 2 * % ( - = nRT ,
+ "T $ T ' .V ,n 2
(HW#12.1.5: show this)
and
/ ) T ,3 2 5 2
/ ) v , 52
# " F&
S = ! % ( = nR 1 ln *
- + 4 = nR 1 ln *
- + 4 = ns(T, v ) ,
$ " T ' V ,n
2 43
10 + vQ (T ) . 2 43
10 + TQ (v ) .
(HW#12.1.6: show this)
where s(T ,v ) is the molar entropy, which is shown in the figure below. The entropy can be also
found directly from the Helmholtz free energy F and the internal energy U:
) "$ v &$ 5 ,
U! F
' + .,
S=
= nR ++ ln #
.
$
(
)
T
v
T
* % Q $( 2 -
8
where 5 2 = 3 2 + 1 and “ 3 2 “ comes from U T while “1” comes from ! F T .
The entropy does not vanish at the absolute zero temperature because the above expression for
the entropy is valid only for T >> TQ . Using the above expression for the entropy, we can also
express the entropy as
' ! $ 3/ 2 ! $ *
T
V
S = S (T0 ,V 0 ,n) + nR ln )) ## && ## && ,, .
(" T0 % " V 0 % +
(HW#12.1.7: show this)
These results are common or universal among low-density monatomic gases so that the
monatomic ideal gas model can serve as the minimal model for the universality class of lowdensity monatomic gases.
The entropy of the monatomic ideal gas model matches the experimental values of the
entropy for low-density monatomic gases
Experimentally, we can find the value of the entropy of a gas at its boiling point by
sexp( Tboiling,P = 1 atm )
= !ssolid(0 K " Tm ) + !smelting + !sliquid(Tm " Tb ) + !svaporization
where we have used the third law of thermodynamics: ssolid(T = 0 K, P = 1 atm ) = 0 .
evaluate !ssolid and !sliquid , we use
To
9
Tf
!s = s(T f ,1 atm ) " s(Ti ,1 atm ) =
cP (T, 1 atm )
dT .
T
Ti
#
The entropy change due to melting can be evaluated by
!smelting =
l melting
Tm
,
where l melting is the molar latent heat of melting, while the entropy change due to vaporization can
be evaluated by
!svaporization =
l vaporization
Tb
,
where l vaporization is the molar latent heat of vaporization. As l melting > 0 and l vaporization > 0 , the entropy
increases by these phase changes. The table below for three monatomic gases clearly shows an
impressive match between the values for entropy calculated from the equation given above and
the corresponding experimental values.
Gas
Tb (K) sexp ( J / (mol ! K) )
scal ( J / (mol ! K) )
---------------------------------------------------------------------------------Ne
27.2
96.40
96.45
Ar
87.3
129.75
129.24
Kr
119.9
144.56
145.06
Answers for the homework questions in Sec.12.1
HW#12.1.1
# 2"! 2 & 3 / 2
vQ (100 K) = N Avogadro %
(
$ mk B T '
+
2
#
&3 / 2 .
)34
10
J*
s
(
)
( 0
~ O-(10 24 mol)1 )% )27
% (10 kg)(10 )23 J/K)(100 K) ( 0
$
' 0/
-,
~ O(10 )6 m3 /mol)
HW#12.1.2
!
!
" v Q (T ) % 2 / 3
( nN Av + 2 / 3 2.! 2 4 2 / 3
TQ (v ) = $
= N Ttrans (V )
' T =*
) V , mk B .
# v &
10
HW#12.1.3
TQ =
4 2/ 3
2/ 3
N Ttrans ~ O (102 4 ) (10"1 6 K) ~ O(1 K)
!
(
)
HW#12.1.4
# "F &
) "
,
# "
&
nRT
P = !%
( = !*
!nRT lnV + (the rest )]- = nRT %
ln V ( =
[
$ "V ' T, n
+"V
.T, n
$ "V
' T ,n
V
=
RT
v
HW#12.1.5
) " # F& ,
)"
,
U = !T 2 * % ( - = !T 2 * [ !nRln T 3/ 2 + ( the rest )]+ "T $ T ' .V ,n
+" T
.V ,n
= T2
&
3 # "
3 1 3
nR%
lnT ( = T 2 nR = nRT
' V, n
2 $ "T
2 T 2
HW#12.1.6
)+ 0
)+ # v & -+3 -+
# " F&
" 2
( + 1.5 .
S = !% ( = ! *
!nRT *ln %%
$ " T ' V ,n
+, "T 21
+, $ v Q (T )(' +/54 +/
V ,n
)+ # v & -+
)
( + 1. + nRT * " [ lnT 3/ 2 + ( the rest )].
= nR*ln %%
, "T
/V ,n
+, $ vQ (T ) (' +/
)+ # v & -+
#
&
( + 1. + nRT 3 % " ln T(
= nR*ln %%
' V ,n
+, $ vQ (T ) (' +/
2 $ "T
)+ #
)+ #
v &( + 3
v &( 5 +
%
%
= nR*ln %
+ 1. + nR = nR *ln %
+ .
+, $ vQ (T ) (' +/ 2
+, $ vQ (T ) (' 2 +/
32
0 )
3
+ T -+
5
2
. + 5 = ns(T ,v)
= nR2 ln *
2 54
1 +, TQ (v ) +/
HW#12.1.7
(* " v % 5 ,*
(* " v %
0
$
'
'+
S (T, V,n ) ! S( T0 ,V 0 , n) = nR )ln $
+ - ! nR)ln $$
*+ # vQ (T ) '& 2 *.
*+ # vQ (T 0 ) '&
/" 1 V % " v ( T ) n % 2
4
' $$ Q 0
= nRln 11 $$
' # 1 V ''& 4
(
)
v
T
n
&
0# Q
0 3
/ " % 3/ 2 " % 2
T
V
= nRln 11 $$ '' $$ '' 44
0# T0 & # V 0 & 3
5 ,*
2 *.
11
(" % 3/ 2 " % +
T
V
! S(T ,V,n) = S(T0 ,V0 ,n) + nRln **$$ '' $$ '' -)# T0 & # V 0 & ,
12.2 A microscopic model for a system in thermal equilibrium with a heat reservoir
As presented above, when the Helmholtz free energy F of a microscopic model is regarded as
a function of its temperature T, volume V, and mole number n, we can calculate its equation of
state, internal energy, and entropy directly from its Helmholtz free energy. We therefore need to
specify each equilibrium state of the model by its temperature T, volume V, and mole number n.
To control the temperature of the model, we attach it to a heat reservoir at a given
temperature T so that the model and the heat reservoir remain in thermal equilibrium with each
other and share the same value for their temperatures while we keep its volume and mole number
to remain constant:
T = const ,
V = const ,
and
n = const .
On the macroscopic level, the state variables of the model system remain constant
On the macroscopic level, the state variables such as the internal energy U and the pressure P
of the model system remain constant while the model and the heat reservoir remain in thermal
equilibrium with each other so that we can regard U and P as functions of T, V, and n.
On the microscopic level, energy is being exchanged between the model system and the heat
reservoir
On the microscopic level, the model is exchanging energy with the heat reservoir through the
thermal contact so that the total energy and the pressure of the model are fluctuating around their
average values, U and P, which we measure as the internal energy and the pressure of the model
system on the macroscopic level. Strictly speaking, the atoms or molecules in the model,
especially those near the boundary between the model and the reservoir, are exchanging energy
with the atoms or molecules in the reservoir through inter-atomic or inter-molecular forces. We
assume that the energy exchanged between the system and the reservoir is much smaller than the
total energy of the model or the total energy of the reservoir.
12
12.3 Macro from micro: the Helmholtz free energy of a microscopic model from its
partition function
The link between a microscopic model and its Helmholtz free energy F is provided by what
we call the “partition function” Z, which is a function of the temperature T, the volume V, and
the number of particles N in the model and is directly related to the Helmholtz free energy by
F( T, V,n ) = !k BT ln Z(T ,V,nN Avogadro) .
We will later derive this relation based on the microscopic definition of entropy. This
extraordinary relation or a “bridge” between the macroscopic quantity, the Helmholtz free
energy, and the microscopic quantity, the partition function, is no doubt one of the most
important equations in statistical mechanics.
Example: The monatomic ideal gas model as the minimal model for the universality class of
low-density monatomic gases. For the monatomic ideal gas model, we will later
calculate its partition function to be
N
1 !# V %#
"
& ,
Z(T ,V, N ) =
N! #$ V Q (T ) #'
where the “quantum volume” is defined by
# 2"! 2 & 3 / 2
VQ (T ) = %
( .
$ mk B T '
The Helmholtz free energy of the model is then given by
!
) "$ v &$ ,
' + 1. ,
F = !nRT ++ ln #
$
(
)
v
T
* % Q $( .(HW#12.3.1: show this using ln N! ! N ln N " N )
where the quantum molar volume is related to the quantum volume by
# 2"! 2 & 3 / 2
vQ (T ) = N AvogadroVQ (T ) = N Avogadro %
( .
$ mk B T '
!
13
Answers for the homework questions in Sec.12.3
HW#12.3.1
) "
)
,
&N ,
"$ V &$
+ 1 $# V $' .
+
.
#
'
F = !k B T ln Z = !k BT ln +
. = !k B T + N ln $ V (T ) $ ! ln N!.
$
$
(
)
N!
V
T
%
(
%
(
Q
*
Q
*
)
,
) "$ V
"$ V &$
&$ ,
' ! N ln N + N . = ! Nk BT + ln #
' + 1.
/ !k BT + N ln #
+*
.+* $% V Q (T )N $( .$% VQ ( T ) $(
) "$
1
V &$ ,.
= !nRT + ln #
' +1
+* $% N AvogadroV Q (T ) n $( .) "$ v &$ ,
' + 1.
= !nRT ++ ln #
$
* % vQ ( T ) $( .where we have used N = nN Avogadro.
12.4 Energy eigenvalues of a microscopic model
Before we calculate the partition function for a microscopic model, we need to obtain the
energy eigenvalues for the model by solving the following Schrödinger equation for the model:
! !
!
! !
!
H"s ( r1, r2 ,..., rN ) = E s (V,N )"s ( r1, r2 ,..., rN )
where the index s is the label or “the quantum number” for the model’s energy eigenstates, each
! !
!
of which is !
represented by the quantum wave function "s ( r1, r2 ,..., rN ) , which depends on the
positions of the N microscopic particles in the model and is accompanied by a corresponding
energy eigenvalue Es (V, N ) , which depends on the volume V of the model system and the
number N of the microscopic particles. The !Hamiltonian operator H is an operator that
operates on the wave function and corresponds to the total energy of the model:
N
! !
!
H = " Ti + #( r1, r2 ,..., rN ) ,
i=1
where Ti is the kinetic energy operator for the i-th particle and is given by
!
!2 # " 2
"2
"2 &
Hi = !
% 2 + 2 + 2(,
2m $ "x i
"yi
" zi '
14
! !
!
while "( r1, r2 ,..., rN ) is the potential energy function for the particles. The potential energy
consists of two parts, one of which is due to the inter-molecular forces among the constituent
molecules while the other part is used to keep the molecules inside the box of volume V.
!
It is extremely difficult to solve the above Schrödinger equation exactly except for a limited
number of models including the ideal gas model we will examine below. The best we can hope
for is to find the energy eigenvalues approximately and we are still looking for a better way of
calculating the energy eigenvalues with a better accuracy.
To calculate the partition function, the only information we need from the microscopic model
is its energy eigenvalues. This is not surprising because the internal energy U, together with the
volume V and the mole number n, can specify each equilibrium state on the macroscopic level
and the internal energy we measure in the lab is a statistical average of the energy eigenvalues,
which are measured values for the total energy of the model system on the microscopic level.
The key question is how we can estimate the average value of the energy eigenvalues whose
number is typically “astronomically” large.
12.5 The partition function from the Boltzmann factor
The partition function is based on a quantity called “Boltzmann factor” and is defined as
Z(T ,V, N ) =
" E (V, N ) %
exp$ ! s
',
kBT &
#
s: all the energy
(
eigenstates
where the index s is the label or “the quantum number” for the energy eigenstates of a
microscopic model and runs over all the energy eigenstates each with its corresponding energy
eigenvalue Es , which normally depends on the volume V and the number N of microscopic
particles (e.g., atoms) in the model. The exponential factor, exp (! Es / k BT ) , is called “the
Boltzmann factor” so that
Z = the sum of the Boltzmann factors for all the energy eigenstates
Note that the partition function is dimensionless because the Boltzmann factor is
dimensionless.
12.6 Macroscopic state variables as statistical averages of microscopic quantities
As discussed in Sec.4.2, the macroscopic state variables such as the internal energy and the
pressure of a macroscopic system must be statistical averages of their corresponding microscopic
quantities.
15
Internal energy as a statistical average of energy eigenvalues
We can show this connection between the internal energy and the microscopic energy
eigenvalues of a microscopic model by substituting the relation between its Helmholtz free
energy and partition function,
F( T, V,n ) = !k BT ln Z(T ,V,nN Avogadro)
into the equation for the internal energy in terms of the Helmholtz free energy,
) " # F& ,
2
U = !T * % ( - .
+ "T $ T ' .V ,n
We then obtain
U=
#1
&
1
E s exp[! Es (k BT )] = " Es $ exp[ ! E s (k B T )]'
"
%Z
(
Z s
s
(HW#12.6.1: show this)
which can be expressed as
U = ! ps Es " Es ,
s
where ps is defined by
ps !
1
exp[" Es (k BT )] ,
Z
which implies that the internal energy is a “statistical average” or mean Es
of the energy
eigenvalue Es over all the energy eigenstates, each of which contributes to the average according
to statistical weight ps , which is proportional to the Boltzmann factor. Note also that
1
1
! p = ! Z exp[ " E ( k T )] = Z ! exp[" E (k T )] = 1 ,
s
s
s
s
B
s
B
s
which further supports our interpretation of ps as the statistical weight since the sum of all the
statistical weights for all the eigenstates must be one.
We can also obtain U directly from the partition function Z by
U=
% $
(
1
E se !"Es = ! '
ln Z* ,
#
Z s
& $"
) V, N
16
where we have used the following useful shorthand notation:
!"
1
.
k BT
Heat capacity at constant volume and the variance for the system’s energy
We can also show that the heat capacity at constant volume CV of the system is proportional to
the “variance” for the system’s energy Es or the mean square deviation from the mean Es = U :
CV =
1
k BT 2
(E
s
! Es
)
2
=
1
k BT 2
(E s ! U )2
.
We can then define the spread !E of the energy values around the average value Es = U by
!E "
(E
s
# Es
)
2
and find it to depend on CV as
!E = k B T 2 CV .
Derivation
% -/
" !U %
-+ ! " 1
1
()E
CV = $
'
= , $$ * Ese s '' 0 =
# !T & V , N . !T # Z s
& 1V ,N k BT 2
(E
s
( Es
)
2
(! " 1 (k T ))
B
(HW#12.6.2: show this)
As both CV and U = Es are extensive, they are proportional to N,
CV ! N
and
U !N,
so that
!E
=
U
k B T 2 CV
1
"
,
U
N
which implies that the energy spread normalized by U decreases as the number of atoms or
molecules in the system increases or
17
!E
" 0.
U
(as N ! " )
For 1 mole of molecules, we then find
CV ~ O( nR) ~ O( N Avogadrok B )
and
U ~ O(CV T ) ~ O( NAvogadrok B T )
so that
!E
=
U
"
%
k B T 2 CV
1
' ~ O(10(1 2 )
~ O$$
'
U
# NAvogadro &
(HW#12.6.3: show this)
As we have claimed in Sec.4.2, the standard deviation for the energy of the system is therefore
very minute and unobservable on the macroscopic level and this is why the internal energy U,
which is the statistical average of Es , is practically constant and therefore well-defined on the
macroscopic level.
Macroscopic pressure as a statistical average of microscopic pressure
We can also show that the macroscopic pressure is a statistical average of its microscopic
counterpart:
P = ! ps Ps " Ps ,
(HW#12.6.4: show this)
s
where the “microscopic” pressure Ps of the model system that is in energy eigenstate s is defined
by
$ #E '
Ps ! " & s ) .
% #V ( N
We can also show that the standard deviation for the pressure of the system is related to its
isothermal compressibility ! T and is also very minute and unobservable on the macroscopic
level, which is why the pressure P, which is the statistical average of Ps , is practically constant
and therefore well-defined on the macroscopic level.. For more detail on this, see Appendix#2 at
the end of this chapter.
18
Answers for the homework questions in Sec.12.6
HW#12.6.1
) " # F& ,
# "
&
1 # "Z &
U = !T 2 * % ( - = k BT 2 %
ln Z(
= kBT 2 % (
+ "T $ T ' .V ,N
$ "T
' V ,N
Z $ "T ' V , N
=
kB T 2 # " ! Es
/% e
Z s $ "T
( kB T ) &
=
1
E s exp[! Es (k BT )]
Z/
s
Es
2 1
(
= k BT
exp[ ! Es (k BT )] ,
/
' V ,N
Z s k BT 2
HW#12.6.2
+- ! " 1
% -/
" !U %
CV = $
= , $$ * Ese ()E s '' 0
'
# !T & V , N -. !T # Z s
& 1-V ,N
=
" ! ( Es
1
Es $
e
*
# !T
Z s
=
1
Es 2 ( Es
e
*
Z s k BT 2
=
1 1
Es2 e ( Es
kB T 2 Z *
s
=
1
kB T 2
(E
2
1 " !Z%
( 2 $ ' * Es e( )Es
'
& V , N Z # !T & V , N s
( kB T )
( Es
s
(k B T )%
(
1 " ! ( Es
*$ e
Z s # !T
( kB T )
2
( Es
( k BT ) %
'
E
& V ,N s
1
Es (E s
e
*
Z s kB T 2
(k B T )
)
On the other hand,
1
k BT 2
=
(E
s
! Es
1
Es 2
2
kB T
[
1
2
E s2 ! 2 Es Es + Es
2
k BT
2
1
! 2 Es Es + Es =
E s2 ! Es
2
kBT
)
2
=
(
]
2
so that
CV =
1
Es 2 ! Es
2
k BT
(
2
) = k 1T
2
(E
s
! Es
)
2
B
HW#12.6.3
!E
=
U
"
2
k B T 2 CV
$ k BT N AvogadrokB
~ O$
U
# NAvogadrok B T
" 1 %
' ~ O(10 (1 2)
~ O$$
24 '
# 10 &
%
"
%
1
'
$
'
~
O
'
$ N
'
Avogadro&
#
&
)
19
HW#12.6.4
# "F &
# "
&
1 # "Z &
P = !%
ln Z(
= kBT % (
( = k BT %
$ "V ' T, N
$ "V
' T ,N
Z $"V 'T,N
k T
1 *, # " Es & ., ! Es
+! %
(
= B )
( /e
' T ,N
Z s kB T ,- $ " V ' N ,0
=
kB T # " !E s
)% e
Z s $ "V
(k B T )&
=
# E &
1 ,*+ # " Es & ,./
!%
exp %% ! s ((
(
)
Z s ,- $ "V ' N ,0
$ kB T '
(k B T )
,
12.7 The Boltzmann factor represents the effect of the heat reservoir
The Boltzmann factor is proportional to the statistical weight ps for eigenstate s, which we
can interpret as the “probability” of finding the system to be in eigenstate s when we observe the
system in the lab:
exp (! Es / k BT ) ! (probability of finding the system in eigenstate s),
which implies that at a given temperature T, the Boltzmann factor favors the eigenstates with
lower energy eigenvalues so that the system is more likely to be found among these eigenstates
with lower energy eigenvalues. Note that the Boltzmann factor depends only on the value of the
energy eigenvalue so that the energy eigenstates that share the same value for their energy
eigenvalues share the same value for the Boltzmann factor as well as for the statistical weight ps .
When the energy eigenvalue for energy eigenstate s is much higher than k BT or Es >> k BT ,
ps !
1
exp[" Es (k BT )] << 1
Z
so that one of the main roles of the Boltzmann factor is to suppress these energy eigenstates
whose energy eigenvalues are much higher than k BT . In fact, the Boltzmann factor or the
statistical weight ps for energy eigenstate s represents the likeliness for the heat reservoir to give
up the energy Es to the system. Recall that the temperature of the system is controlled by the
heat reservoir through the thermal equilibrium between them.
At a temperature very close to T = 0, the heat reservoir is very unlikely to provide any
positive amount of energy to the system so that the system gets stuck with near certainty, in its
ground state with the lowest energy eigenvalue. At low temperatures, the heat reservoir is most
likely to provide a small amount of energy to the system so that the system is most likely to be
found in one of the eigenstates with low energy eigenvalues, whereas at high temperatures, the
heat reservoir is likely to provide almost any amount of energy to the system so that the system is
likely to be found among more and more eigenstates with a wide range of energy eigenvalues.
20
Note that as the temperature approaches infinity, the statistical weight ps becomes almost
constant for a wide range of energy eigenvalues.
12.8 A sharp peak of the statistical weight for energy value at E = U
In the equation for the internal energy expressed as the statistical average of the energy
eigenvalues,
U = ! ps Es " Es ,
s
the statistical weight ps suppresses the energy eigenstates whose energy eigenvalues are much
higher than k BT . Experiments have also shown that whenever we measure the internal energy of
a macroscopic system, its value is practically the same, which suggests that the system is most
likely to be found in energy eigenstates whose energy eigenvalues are very close to the value of
the internal energy so that the statistical weight for energy value, instead of that of the energy
eigenstate, must be sharply peaked around the value U of the internal energy.
To find the statistical weight for energy value, let us rewrite the above expression for the
internal energy as follows:
U = ! ps Es =
s
#
= $ dE
E0
1
1 #
exp
"
E
k
T
E
=
(
)
[ s B ] s Z $ dED( E,V ,N ) exp[ " E (k BT )]E
Z!
s
E0
D( E,V ,N )
exp[ " E (k BT )]E
Z(T ,V, N )
where E0 is the lowest energy eigenvalue and the summation over all the energy eigenstates is
replaced with an integral over all possible energy values. D( E,V ,N ) dE is the number of energy
eigenstates with their energy eigenvalues in an interval [ E,E + dE ] so that D is the “number
density” of energy eigenstates. This replacement of the summation by the integral is well
justified because for any macroscopic system D( E,V ,N ) turns out to be a smooth function of E
as a difference between adjacent energy eigenvalues is minuscule compared to the magnitude of
E, which is a macroscopic quantity. We can then define the statistical weight g for energy value
E by
g( E,T, V, N ) !
D( E,V, N )
exp[" E (k B T)]
Z(T ,V, N )
so that we can express the internal energy as the statistical average of energy value according to
the statistical weight g as
21
!
U = " dEg (E,T, V, N ) E .
E0
Being the statistical weight, g must satisfy a normalization condition:
!
" dEg( E,T, V, N ) = 1 ,
E0
which follows from the normalization condition for ps :
!
!
1
1
"E dEg( E,T, V, N ) = Z E" dED( E, V, N ) exp[ # E ( kB T )] = Z $s exp[ # E ( k BT )] = $s ps = 1
0
0
We can calculate the function D for various microscopic models and we always find it to be a
rapidly increasing function of energy E. The statistical weight g is therefore a product of the
fast increasing D and the fast decreasing Boltzmann factor exp[ ! E ( kB T )] Z so that it must have
a peak. To show that g is sharply peaked around E = U , where U is the value of the internal
energy, we use the following relationship between W ( E, V, N ) ! D(E,V, N )dE and the entropy
of the system:
S (E,V, N ) = k B lnW ( E,V ,N ) ,
where k B is the Boltzmann constant. As we will discuss in Ch.15, this relation is the “foundation
stone” for the entire statistical mechanics and F = !k B T ln Z , the central equation in the
canonical ensemble theory, follows from this relation.
Note that
! dE $
S = k B lnW = k B ln [D( E )dE ] = kB ln[ D( E ) E ] + kB ln # & ' kB ln[ D(E ) E ] ,,
" E%
where the term ln (dE E ) is negligible compared to the ln[ D( E) E ] since ln[ D( E) E ] ~ O( N )
while ln( dE E ) ~ O(1 ! 10) . We can then express D in terns of S as
D( E ) =
so that we can rewrite g as
! S$
1
exp## &&
E
" kB %
22
g( E,T, V, N ) =
1
exp {S (E,V, N ) kB ! E (k B T )}
Z( T, V, N )E
In Appendix#1 at the end of this chapter, we will then show that g is well approximated by
the following Gaussian:
g( E,T, V, N ) !
&$
&(
( E # U )2
1
%
),
exp
#
2
&' 2kB T CV ( T, V, N ) &*
2"k BT 2 CV (T, V, N )
where CV is the heat capacity at constant volume of the system and we find U as a function of T,
V, and N by solving
" !S %
$ '
# !E & V , N
=
E =U
1
,
T
This Gaussian approximation for g clearly shows that it has a very sharp peak at E = U . As
!1 2
e = 0.607 , we can define the peak “half-width” !E by
1
1
!E 2 =
2
2kB T CV
2
so that
!E = k B T 2 CV =
(E
s
" Es
)
2
.
The half-width !E of the peak in the statistical weight g is then nothing but the spread of the
energy values around the average value U = Es defined in Sec.12.6.
To see that g converges to a delta function as N approaches infinity, we consider g as a
function of a new variable x defined by x ! E U :
g( E,T, V, N )dE !
%' ( x $ 1)2 )'
1
*dx , gˆ( x,T, V, N )dx ,
exp &$
'( 2( #E U )2 '+
2" (#E U)
where we have used dE = Udx . As we have found in Sec.12.6,
!E
=
U
k B T 2 CV
1
"
.
U
N
23
so that gˆ( x,T, V ,N ) has a peak at x = 1 and as N approaches infinity, the peak half-width !E U
2 ! ( "E U ) approaches infinity. In
of gˆ( x,T, V ,N ) approaches zero while its peak height 1
{
}
fact, gˆ( x,T, V ,N ) approaches a delta function:
lim gˆ ( x, T ,V, N ) = # ( x $ 1) .
N! "
12.9 The decomposition formula for the partition function for a model with “noninteracting” quantum numbers
Calculating the partition function for any given microscopic model is an extremely
challenging task and a great deal of efforts have been directed to develop various methods to
evaluate a partition function exactly or approximately. Fortunately, there are a handful of
physically important microscopic models whose partition functions we can calculate exactly.
Most of these models neglect interactions or forces among their constituent particles, and this is
exactly why they are analytically tractable. It is nevertheless remarkable that these drastically
simplified models can serve as minimal models for various universality classes of real
macroscopic systems.
What makes these models with non-interacting particles special is that their energy
eigenvalues can be written as a sum of the following form:
( )
( )
( )
Es1 ,s2 ,..sn = ! s11 + ! s22 + "" +! snn ,
where {s1 , s2 ,..sn } is a set of quantum numbers for a model. Note that each term on the righthand side of this equation depends only on a single quantum number. These quantum numbers
are then said to be “non-interacting.”
Using the following property of the exponential function,
exp (a1 + a2 + !! +an ) = exp (a1 ) exp( a2 ) !!exp( an ) ,
we can then show that the partition function for the model can be expressed as
Z=
# exp(!"E
s 1 ,s 2 ,..s n
%
s 1 ,s 2 ,..s n
(%
( %
(
) = ''# exp{!"$ ( ) }** ''# exp{!" $ ( ) }** + +''# exp{!"$ ( ) }**
&
1
s1
s1
= Z1 Z2 + +Zn
where ! = 1 (k BT ) and
{
}
()
Zi = $ exp !"# sii .
si
) & s2
2
s2
) & sn
n
sn
)
24
The main point of this result is that we can break up a “big” problem of calculating Z into small
problems of calculating Zi s. Very often, all the Zi s turn out be identical so that we only need to
calculate Z1 and obtain Z by
Z = Z1 n .
Derivation of the decomposition formula
Z=
# exp(!"E
s 1 ,s 2 ,..s n
s 1 ,s 2 ,..s n
=
# exp{!"$
) = # exp[ !" {$ ( ) + $ ( ) + %% +$ ( ) }]
1
s1
n
sn
s 1 ,s 2 ,..s n
(1 )
s1
! "$
( 2)
s2
! %% !"$ s(nn )
s1 ,s 2 ,..s n
=
2
s2
}
# exp{!"$ ( ) }exp {!"$ ( ) } %%exp{!" $ ( ) }
1
s1
2
s2
n
sn
s1 ,s 2 ,..s n
&
= ( # exp !" $ (s11 )
(' s1
{
}
)&
+ ( # exp !"$ (s2 )
2
+* (' s2
{
}
) &
+ %%( # exp !"$ (sn )
n
+* (' sn
{
}
)
+
+*
= Z1 Z2 % %Zn
Appendix#1: Derivation of the Gaussian approximation for g
In this appendix, we will show that the statistical weight g for energy value E is well
approximated by a Gaussian with a peak at E = U :
1
exp {S (E,V, N ) kB ! E (k B T )}
Z( T, V, N )E
&$
&(
( E ! U )2
1
"
exp
%
!
)
2
&' 2k BT CV ( T, V, N ) &*
2#k T 2 C (T, V ,N )
g( E,T, V, N ) =
B
V
According to the fundamental equation of thermodynamics, dU = TdS ! PdV , we find
dS =
1
P
dU + dV ,
T
T
which implies
" !S %
1 " !S %
.
=$
' =$ '
T # ! U & V ,n # !E & V , N E =U
25
By solving this equation for U, we can find U as a function of T, V, and N.: U = U(T ,V, N ) . We
then expand the entropy S around E = U :
S (E,V, N ) S(U, V, N ) 1 # " S &
1 # "2 S&
!
+ % (
E
)
U
+
%
(
(
)
kB
kB
kB $ "E ' V , N E = U
2k B $ " E 2 ' V ,N
=
( E ) U )2
E =U
S(U, V, N )
1
1
(E ) U ) )
( E ) U )2
+
2
kB
k BT
2kB T CV
=)
U ) TS(U, V, N ) E
1
( E ) U )2
+
)
2
kB T
k BT 2k BT CV
=)
F
E
1
( E ) U )2
+
)
k BT k B T 2k BT 2 CV
where we have used F = U ! TS(U, V, N ) and
" ! 2S %
$ 2'
# !E & V ,N
E =U
( ! " 1 %+
1 " !T %
1
1
1
=)
$ ', = . 2 $
'
=. 2
=. 2 .
"
%
* ! U # T & -V , N
T # !U & V ,N
T !U
T CV
$
'
# ! T & V, N
We then obtain
" S( E, V, N )
" F %1
"
%
1
E %'
1
2
' exp $ !
',
(
)
exp $
!
( exp $ !
E
!
U
$#
$# k BT '& U
$# 2kB T 2 CV
'&
E
kB
k B T '&
where we note that the function 1 E is relatively slowly varying inside the sharp peak of
2
exp ! ( E ! U) (2kB T 2 CV ) .so that 1 E can be replaced by 1 U . g now becomes
[
]
g( E,T, V, N ) =
(
" S(U ,V,N ) E %
1
'
exp $
!
$#
ZE
kB
k BT '&
" F %1
"
%
1
1
' exp $ !
exp $ !
( E ! U )2 '
2
$# kB T '& U
$# 2kB T CV
'&
Z
We can also calculate the partition function using the function D:
26
#
Z = " exp[! Es (k BT )] =
s
#
= $ dE
E0
$ dED( E,V, N ) exp[! E (k T)]
B
E0
% S(U, V, N )
1
E (*
exp '
!
'&
E
kB
k BT *)
, F /1 #
%
(
1
2
11 $ dE exp ' !
*
+ exp.. !
E
!
U
(
)
'& 2k B T 2 CV
*)
- k B T 0 U E0
, F / 2kB T 2 CV
11
= exp.. !
U
- kBT 0
, F / 2kB T 2 CV
11
+ exp.. !
U
- kBT 0
,
E !U
$ 2 d ..- 2k T 2C
B
V
2k B T C V
#
( E0 !U )
% ,
/
1 exp ' ! . E ! U
' .
1
2
0
'& - 2kB T CV
/2(
1 *
1 *
0 *)
#
$ dx exp (! x )
2
!#
, F / 2 2kB T CV
11
= exp.. !
U
- kBT 0
2
where we have used
"
# dx exp( !x ) =
2
$.
!"
Finally, we find
g( E,T, V, N ) !
# S(U, V, N )
1
E &(
exp %
"
!
%$
ZU
kB
kB T ('
# (E " U )2 &
1
%"
(.
exp
%$ 2k BT 2 CV ('
2 )kB T 2 CV
The above result for Z is consistent with F = !k B T ln Z as
# 2 "k T 2 C
F
B
V
ln Z = !
+ ln%%
k BT
U
$
where we have neglected ln
(
&
()! F ,
(
k BT
'
)
2 !kB T 2 CV U since F ~ O( N ) while
" 2 !k T 2 C
B
V
ln $$
U
#
%
( "
%+
' ~ O* ln $ 1 ' - << O( N ) .
'
) # N &,
&
27
Appendix#2:
Isothermal compressibility ! T is related to the variance for a system’s pressure Ps .
*,
$ #P '
V
! T = + "V & s ) "
% #V ( N
k BT
-,
(P
s
" Ps
)
2
.,"1
/
0,
so that
!Ps2 "
( Ps # Ps
)
2
=
% $P (
kB T ,.
1 0.
- #V ' s * # 1
& $V ) N
V ./
+ T .2
Derivation
!
,. # $ 1
' 0.
$ #P '
1
=&
)
= - && + Pse ! *Es )) 1
V " T % # V ( T , N ./ #V % Z s
( 2.T, N
=
$ # ! *Es '
$ #P ' !*E
1
1 $ #Z '
1
! *E
Ps &
e )
! 2&
) + Ps e s + + & s ) e s
+
( T, N Z % #V ( T ,N s
Z s % #V
Z s % #V ( N
P
1
Ps2 !*E s
= +
e
! s
Z s k BT
Z
Ps2
P
=
! s
kB T
Z
=
=
Ps
2
(P
s
Ps
+k T e
s
2
! Ps
kB T
! Ps
k BT
)
$ #
+ &% #V e
s
!* Es
B
!*E s
'
$ #P '
) + & s)
( T ,N
% #V ( N
$ #P '
+ & s)
% #V ( N
$ #P '
+ & s)
% #V ( N
2
$ #P '
+ & s)
% #V ( N
As the volume of the system is extensive or V ! N , we obtain
!Ps2 "
1
1
"
.
V
N
28
SUMMARY FOR Ch.12
The overall scheme of the canonical ensemble approach
N
! !
!
1. Construct a microscopic model with a Hamiltonian: H = " Ti + #( r1, r2 ,..., rN ) .
i=1
2. By solving the Schrödinger equation for the model,
!
! !
!
! !
!
H"s ( r1, r2 ,..., rN ) = E s (V,N )"s ( r1, r2 ,..., rN ) ,
calculate its energy eigenvalues: {Es (V, N )}s .
!
3. Calculate the partition function Z(T ,V, N ) = " exp {! Es (V, N ) / kB T} .
s
4. Calculate the Helmholtz free energy: F( T, V,n ) = !k BT ln Z(T ,V,nN Avogadro) .
5
Calculate the state variables from the Helmholtz free energy:
# "F &
P = !%
( ,
$ "V ' T, n
) " # F& ,
U = !T 2 * % ( - ,
+ "T $ T ' .V ,n
( )
and
( )
# " F&
S = !% (
$ " T ' V ,n
( )
6. The decomposition formula: for Es1 ,s2 ,..sn = ! s11 + ! s22 + "" +! snn ,
%
' # exp !"$ s(1 )
exp
!
"
E
=
#
s 1 ,s 2 ,..s n
1
'& s1
s 1 ,s 2 ,..s n
= Z1 Z2 + +Zn
(
Z=
)
{
}
where Zi = $ exp !"# s(ii) .
si
{
(%
* ' # exp !" $ (s2 )
2
*) '& s2
}
{
}
( %
* + +'# exp !"$ (sn )
n
*) '& sn
{
}
(
*
*)