Solution MC Quarrie Mit
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Problem Set 5 Solutions - McQuarrie Problems 3.20 MIT Dr. Anton Van Der Ven Fall Fa ll 2001 2001 Fall 2003 Problem 3-4
We have to derive the thermodynamic thermodynamic properties of an ideal monatomic gas from the following:
= eq 3 2mkT 2 V = e and q = 2
h
is the partition function for the grand canonical ensemble, where T, V, are fixed. The characteristic potenti potential al for the grand canonical ensemble is the ’’grand canonical potential’’
= E T S N = pV
(since E = T S pV pV + N )
d = SdT pdV pdV
N d
The thermodynamic properties for the grand canonical ensemble are:
S = ∂ ∂T N,V ∂ p = ∂V T, ∂ N = ∂ T ,V
The grand canonical potential is related to according to
= kT ln
but = e q so,
(see table 3-1)
3 23 5 2mkT 2 2 m 2 = kT q = kT 2 V e = (kT ) V exp h h2 kT
Starting with N,
2mkT 32
N = and p
p =
∂ ∂
T ,V
∂ ∂V
T ,
=
h2
= kT
2mkT h2
V exp exp
3 2
((3-8-1))
kT
exp kT exp
Putting those together we can get the ideal gas equation of state, namely
p = Now S,
S =
∂
5
3 2
= k ( kT ) ∂T N,V 2
N kT V
3 2m 2 h2
1
5 2
V e + (kT )
3 2m 2 h2
V
kT 2
e
3 3 5 2mkT 2 2mkT 2 S = k Ve Ve 2 h2 T h2 32 5 2mkT S = k V e 2 T h 2 N S = 5 k N 2 T
But from (3-8-1) we can get
e
= N V
((3-8-2))
3 2mkT 2 2
1
h
32 V 2mkT = kT ln 2 h N Putting this into (3-8-2) we get
3 5 2mkT 2 V S = k + k ln 2 h 2 N
N = N k
S = N k ln
ln e
32
2mkT 2
ln +
5 2
h
N
5
e 2
V
3 2mkT 2 V 2 h N
This is the same expression as that obtained obtained in the canonical ensemble (see (see Chapter 5). This is due to the equivalence of ensembles when N is very large. Problem 3-10
We are dealing with the isothermal-isobaric isothermal-isobaric ensemble this time, with the partition function for an ideal monatomic gas given to us in the problem as
3 5 (2m) 2 (kT ) 2
=
ph3
N
The isothermal-isobaric isothermal-isobaric is for fixed ( N, T, P ). The characteristic potential for this ensemble is the Gibbs free energy
G = E T S + pV dG = SdT + V dP + dN
The thermodynamic properties are:
∂G S = ∂T
V =
∂G ∂p
∂G = ∂N
p,N
T,N
p,T
G is related to the isothermal-isobaric partition function according to (see Table 3-1): 3 5 (2m) 2 (kT ) 2 G = kT ln = N kT ln 3
2
ph
3 5 G = N kT ln (2m) 2 + ln ln (kT kT ) ln ln h3 2
Starting with V,
V =
∂G ∂p
T ,N
lnln p p
T = Nk p
which is the ideal gas equation of state. Now for :
3 5 (2m) 2 (kT ) 2 = kT ln 3
=
∂G ∂N
p,T
ph
32 2mkT = kT ln 2 kT +kT ln p h
o
And now for S,
3 5 (2m) 2 (kT ) 2 5 = N k ln ln + N k 3 2 ph p,N 3 2mkT 2 V 5 2 S = N k ln 2 + N k lln n e h N 3 5 2mkT 2 V 2 e S = N k ln h2 N
∂G S = ∂T
These are the same expressions as obtained in the grand canonical ensemble (Problem 3-8) and in the canonical ensemble (see chapter 5). This is due to the equivalence of ensembles in the thermodynamic limit, i.e N is very large such that fluctuations are negligible. Problem 3-12
When looking at fluctuations, we derived
P (E ) ≈
1
√ 2kT 2C exp exp v
E E
2
2kT 2 C v
For an ideal monatomic gas (derived explicitly in chapter 5)
E = 3 N kT
2 ∂E 3 C v = = N k ∂T 2
We are now asked what is the probability th that at the N -particle system will sample an energy that differs by 10 10 4 % from the average energy, E = 32 N kT ? (We can let N = N A = 6 .022 022 1023M )
4 3 10 E = E = 10 6 E = 10 6 N kT 100 100 2 2 106 94 N 2 k2 T T 2 3 6 2 (E )2 N = = 10 10 2kT 2 C v 4 2kT 2 32 N k
So now we can go back to P (E ),
1 P (E ) ≈ exp exp kT 3N
√
3
3 6 10 10 4
2
N
P (E ) ≈
1 11 √ exp 4 .5 1011 exp kT 3N
This is an extremely small probability which validates our earlier assumptions. Problem 3-18
Derive an expression for the fluctuation of the pressure in the canonical ensemble.. We know the pressure in state i is
∂E i pi = ∂V
and by definition
p =
i
i
2
exp exp
∂E i ∂V
exp exp
Ei kT
Ei kT
Fluctuation is defined as p2 = ( p p) = p p2 p2 .
Using the methods developed in class:
Step 1: Multiply both si sides des by the partiti partition on function
E
pQ = =
i
pi exp exp
kT
i
Step 2: Get derivative with respect to m mechanical echanical variable’s conjugate. conjugate.
∂ ∂ ( pQ) = ∂V ∂V
p
i
E i exp exp kT
∂E i ∂V
p
eexp xp
i
i ∂E ∂V
1
kT
+
∂p Q ∂V
Q
p2 = p p2
∂p i ∂V
pi exp
i
=
i
∂p i ∂V
exp exp
Ei kT
E i kT
pi exp
+
i
+
Q
i
∂p i 2 p = kT ∂V
∂p ∂V
E i kT
E i pi exp exp kT
p exp i exp
p2 ∂p ∂ pi p2 + = + kT ∂ V ∂V kT
Rearranging a little...
Q
Note:
i
∂Q ∂p ∂ p + Q = ∂V ∂V ∂ V
Step 3: Divide through by the pa partition rtition functi function on Ei kT
pi exp
E i kT
E i ∂p ∂ exp exp + Q = kT ∂V ∂ V i i ∂p ∂p i E i 1 + Q = exp exp kT ∂V ∂V kT i
∂ p ∂V
Ei kT
∂E i ∂V
1
kT
1 i ∂E ∂V kT
Q
has no immediate macroscopic interpretation, it must be calculate in any specific case and depends
on the particular spectrum of
∂ Ei ∂V 2 2
i . This conclusion holds for all generalized forces in the form Ai = ∂E ∂a , where
a is an extensive displacement conjugate to A. Hence, we cannot make an unqualified assertion that fluctuations in all kinds of external forces will be small. 4
Compare this with fluctuations in extensive quantities such as E, H, or N which can be expressed in terms of thermodynamicc response variables suchs as heat capacities or compressibilities. thermodynami Specific calculations of the fluctuations in p of a perfect gas by Fowler is estimated as
p 2 ) ( p p p2
12 5 1012
≈
for a cubic centimeter of gas under standard standard conditions. This is approximatel approximately y
1
2
, where n is the number of
n 3
molecules in the gas.
(Source: The Principles of Statistical Mechanics, Richard C Tolman, Oxford University Press, first edition 1938)
Problem 3-24
2
k T 2 C p in an N, p, T ensemble. Show that H 2 H = kT
N, P, T fixed means we are working in the isothermal-isobar isothermal-isobaric ic ensemble.. The partition function in this ensemble is =
V j
E j pV
e
Where the E j ’s are the energies of the system when it has volume V. We also remember that H = E + pV. Using the methods developed in class: Step 1: Multiply both sides by the partition function
H =
Ej pV (E j + pV ) e
V,j
Step 2: Get the temperature derivative at constant (N, P ) (The conjugate variable to H in this case)
∂H 1 1 Ej pV = + H 2 (E j + pV ) e ∂T N,P kT kT 2 V,j
Step 3: Divide through by the partition function
∂H ∂T
1 + 2 H kT N,P
Ej pV V,j (E j + pV ) e
H
or
H 2
but we know
∂H ∂T
N,P
2
H H
2
= kT
1 = 2 kT
∂H ∂T N,P
= C p . So, 2
H 2
H H = kT 2 C p
5
2 (E j + pV ) e Ej pV
V,j
2 E j pV
V,j (E j + pV )
H 2
e
Problem 3-26
Show that
∂ ∂N
V,T
2
V = N 2
From Gibbs-Duhem we have
∂p ∂V
N,T
..
SdT V dp + N d = 0
At constant T , we then get
d
dp
We can use the chain rule and get:
d d = dp dN
Using partial derivative manipulation
dN dp
But for a single component system
=
V N
V,T
V,T
dN = dV V,T
dV
P,T
dV dp
V = N
= the molar volume dN N dV =
dN p,T
V,T
dp
Putting this all together we get
dN dp
V
V 2 = 2 N V,T
d dN
dp
N,T
and we get
N,T
dp dV
N,T
Problem 4-2
Show that
6N V
h2 12mkT
23
given in table 4-1 is very large for electrons in metals at T = 300 300 K. K.
Take Na-metal having the following properties
10 m - stable in the bcc crystal structure structure with lattice constant a = 4 .23 23 10 10 10 - two Na atoms per bcc unit ce cell ll - number of valance valance electrons per Na atom = 1
- valance electrons electrons in Na can be considered ne nearly arly free So we can get the following values to substitute into the original equations:
N 2 = 3 V (4 (4 . 23 23 1010)
34 h = 6 .6262 6262 1034 J s 23 J k = 1 . 3807 3807 1023 K 31 1095 1031 kg m = 9 .1095 e
Putting those all together we get
6N V
h2 12mkT
32
= 1524 1524 >> 1
Therefore Boltzmann statistics cannot be applied to electrons in metals. Must use Fermi-Dirac statistics. 6
Problem 4-6
∞
S = =
n 2 1 xn 1 x2
N =0 {nj }
with n1 and n2 = 0, 1, and 2. The Let’s consider
means {nj }
n1 n 2
with the restriction that n1 + n2 = N. N .
2 for several values of N. x1n1 x2n
{nj }
N = 0
→ possible combinations of n1 and n2 are 0 and 0. =⇒ =1
{nj }
N = 1
→ possible combinations of n1 and n2 are n1
n2
0 1
1 0
=⇒ N = 2
{nj }
n1
n2
2 1 0
0 1 2
⇒
N = 3
1 1
n 2 1 xn 1 x2 = x 1 + x2
→ possible combinations of n1 and n2 are
=
N
N
2 2 2
2 2 1 n x12 xn 1 x2 = x 2 + x 1 x2 +
{nj }
→ possible combinations of n1 and n2 are n1
n2
2 1
1 2
N
3 3
n 2
n1
⇒
2
2
x1 x 2 x1 x2 = x 1x2 + = {n } j
N = 4
→ possible combinations of n1 and n2 are n1
n2
2
2
=
⇒
For N > 4
→
{nj }
N
4
n 2 2 2 1 xn 1 x2 = x 1 x2
{nj }
2 x1n1 x n 2 = 0 because n1 + n2 � 4
Putting everything together we get
2 x1n1 x 2n x2 + x22 + x1 x2 + x 12 x 2 + x 12 x22 21 + x1 x22 + x = 1 + x1 +
N {nj }
7
Now lets consider 2
S =
1 + xk + x 2k
k =1 =1
S = 1 + x1 + x 21 1 + x2 + x22 = 1 + x1 + x2 + x22 + x1 x2 + x21 + x1 x22 + x 21 x 2 + x21 x22
This expression contains the exact same terms as that obtained with
N
n 2 1 xn 1 x2 .
{nj }
Problem 4-8
We need to show this ( remember the upper (lower) signs is for Fermi-Dirac (Bose-Einstei (Bose-Einstein) n) :
S = k
[nj ln nj (1 nj )ln(1 )ln(1 nj )] )]
j
Start with the partition function and go from there. Becuase of the equivalence of ensembles in the thermodynamic limit, we can calculate the entropy using the ensemble that offers the most mathematical convenience. For Fermions
∂ ln ln ∂T V,
or Bosons, this is the grand canonical ensemble. S = k ln + kT
� εj 1 + e e kT
=
j
ln =
1
� εj ln ln 1 + + e kT
j
∂ ln = ∂T
j
∂ e
� ε j
kT
∂T
� ε j kT
1 e
=
εj kT 2
( εj ) exp kT � εj
1 e kT
j
ε eε j jkT � ε j ln 1 e kT + ε j 1 e
S = k
j
To make this a little easier to manipulate we can write this shorthand by making the following substitutions. Let:
u = 1
ε j
e and v = e u = 1 v and u v = 1 v ε j
nj =
So we now have
{
S = k
j
{
S = k
u
{ }
ln v = k ln ln u nj ln
}
j
j
Remembering that u v = 1
}
j
{
S = k
j
S = k
( 1
j
ln u nj ln nj nj ln u
ln u nj ln nj = k ( 1 nj ) ln
1 ln nj ln nj ( 1 nj ) ln u
ln nj nj )ln(1 )ln(1 nj ) nj ln
}
nj ln ln nj (1 (1 nj )ln(1 )ln(1 nj )
{
8
}
Problem 4-12
N-distinguishable independent independent particles, each of which can be in state +εo or -εo .
N + = number of particles with energy +εo N = number of particles with energy εo with N + + N = N
The total energy is given as:
E = N + εo N εo = 2N + εo N εo E We now have to evaluate the partition function Q by summing exp over levels and compare it to the result exp kT N Q = q ..
We know Q =
E j exp exp kT
j
where j labels a state in which the system can reside.
Instead of summing over the states that the system can be in, we can also sum over the possible energy levels,
making sure we take account of the degeneracy of each energy level, i.e.
Q =
E
E ( ( E E ) exp exp kT
where (E ) is the number of states with energy E.
For this system we have already stated the allowed energy levels are E = 2 N + ε N ε where N + can vary from 0 → N. For each allowed energy level E , there are (E ) possible states compatible wit with h this energy. o
o
Since the N
particles are distinguishable,
(E ) =
N ! N ! = N+ ! N ! N + (N N + )!
This represents the number of ways that N + particles out of N can be in the + state.
So now we can write Q =
E ) exp ( (E exp ( E )
E
=
⇒
Since E is uniquely a function of N +
N +
N ! exp [ (2 exp (2 N + ε o N εo )] )] N + (N N + )!
This can be rewritten in a nicer form if we remember the binomial expansion n n! xn1 n (1 (1 + x) = n1 !(n n1 )!
n1 =0 =0
So we have
Q = eN εo
N +
2εo N + N ! e N + (N N + )!
if we let x = e2εo and rewrite using the binomial expansion we have
N
2ε o Q = eN εo 1 + e or
Q = eεo + eεo
N
Now we need to compare this to Q = q , where q is the single particle partition function defined as N
9
q =
e ε l
l
where l labels the single particle states. In this example there are two single particle states, with energy - εo and +εo .
⇒ q = e
ε o
=
ε o +e
and
N
Q = q N = e ε o + eεo
The last part of this problem asks us to calculate and plot the heat capacity for this system. We know C = v
and E is given as
∂ ln ln eεo + e εo ∂ ln ln Q E = kT 2 = kT 2 N ∂T ∂T ε o εo ε o εo εo kT 2 e e eε o 2 kT 2 e = N εo ε o E = kT N ε e o + eε o e + eε o
ε o ) E = N εo tanh ( ε
and C v is then
ε ∂E o = N k C v = ∂T kT 2 Plotting
C v Nk vs.
kT εo
2
ε
10
2
sech sech
o
kT
∂E ∂T
Problem 5-4
Calculate the entropy of Ne at 300K and 1 atm.
The entropy of an ideal gas (eqn 5-20):
3 5 2mkT 2 V e 2 S = N k ln 2 h N
Note this is neglecting electroni electronicc excitations (see Chapter 5). Some data: 26 kg m = 3 .351 351 1026 27 27 J k = 1 .3807 3807 10 k 34 h = 6 .6262 J s 6262 1034 p = 1 atm = 1 . 013 013 105 P a
Putting that together we can get
V kT 26 = = 4 .0889 0889 1026 N p
and
3 2mkT 2 h2
31 = 8 . 852 852 1031
7 S = N k ln 4.41 10 N S = k ln ln 4 . 41 41 107
where N = the number of Ne atoms.
We are now asked to estimate the translational degeneracy
-From our study of fluctuation theory, we found that the fluctuation in energy of a thermodynamic system (with N very large) is exceedingly small. (see discussion on page 63 of -Therefore, the energy of the gas is essentially always very close to E McQuarrie) and we can use the expression for the entropy in the microcanonica microcanonicall ensemble.
S = k ln
Compare this with where is the degeneracy at fixed energy E.
S = k ln ln 4 . 41 41 107
and we get
N
N = 4. 41 41 107
. which makes sense since it should be large because N is on the order of 1023
Problem 5-9
What is the DeBroglie wavelength of Arÿ at 298K ?
h2 = 2mkT Use:
26 kg m = 6 .634 634 1026 27 J k = 1 .3807 3807 1027 k
h = 6 .6262 1034 J s 11
12
and we get = 1. 6 10 11 m
Now compare this with the inter-atomic distance The volume per Ar atom is
V kT = N p with p = 1.013 013 10 10 5 Pa. So
V N
= 4.06 06 10 10 26 m3 . The interatomic distance ∼
V N
(See page 83 on the relevance of this result.)
12
1 3
= 3 .0 10 10 9 >>
Problem Set 5 Additional Problems Solutions 3.20 MIT Dr. Anton Van Der Ven Fall Fa ll 2001 2001 Fall 2002
Problem 1
(a) A system of N non-interacting particles with two possible states either 0 or ε. A good rule is too assume ’’particles’’ (e.g. atoms, electrons, etc) are indistinguishable unless they are localized in a crystal or on a surface. The ’’particles’’ number of atoms in the excited state can be determined using Boltzmann statistics under the assumption that we are working at high temperature and/or low density::
N ε = N ε
where ε is the probability an atom will be in state ε. This probability is determined using the single particle partition function and can be written as
ε exp kT exp ε = ε; exp exp kT
(McQuarrie 4-14)
i
But our system can be in only two states, so the sum in the denominator can be found explicitly: ε exp kT ε = exp [0] exp [0] + exp exp
So N ε can be written as
ε kT
ε exp kT exp N N ε = N ε = N = ε 1 + exp exp kT 1 + exp exp N ε =
(b) The total energy is simply U = N ε = N
U = N
i
N ε 1+exp[ kT ]
ε kT
4-12 and 4-13) i εi (McQuarrie
N [(1 2 ) 0 + 2 ε] i εi = N ( 1 ε1 + 2 ε2 ) =
i
ε exp kT U = N ( 2 ε) = N ε ε 1 + exp exp kT U =
Nε ε 1+exp[ kT ]
Note: Since many particles will occupy the same state (either 0 or ε) these particles must must be Bosons. At lower temperatures we would have to use Bose-Einstein statistics (McQuarrie 4-26) which would lead to a much more complicated problem since we would have to determine the chemical potential . 1
Problem 2
The magnetization is given as N
M =
ni o
i=1 =1
This is basically saying that we have atoms localized in a crystal and the magnetic moment at each site can be
either up or down. The problem to determine the thermodynam thermodynamic ic properties as a function of are Let T, N, H. V, also assume for simplicity that weasks canus work at constant volume. Therefore, our controlling variables T, N, H.us We need to make the appropriate Legendre Legendre transform to the entropy. Remember the entropy can be writte written n starting from E :
E = T S pV pV H + HM M + N
rearranging to get things in terms of S and we get
S = E + pV H M N k Legendre transform such that our controlling variables variables are V, N, T, H
S E + H M = ( T S E + H M ) = = ln k where is the characteristic potential for this ensemble with V , N , T , H constant and is the partition function. can be written as
=
exp[ (E )]] state exp[ state M state state H )
states
where we sum over all possible energy states and magnetizations M state state .
Since the particles are non-interacting, the energy at N, = constant and H = 0 is constant. E is independent
of the number or arrangement arrangement of up versus down spins. Since the absolute scale of energy energy is not important for thermodynamics, we can arbitrarily set the constant energy equal to zero giving us thermodynamics,
=
M state exp[ M exp[ state H ] =
states
N
exp exp
n1 ,,n n 2 ,...n N N
=
i=1 =1
N
exp[ exp[ n ni o H ] =
n1 ,,n n 2 ,...n N i=1 =1
We can evaluate the sum since ni =
N
ni o H = +1 +1
exp exp [n n i o H ]
n1 ,n 2 ,...n N i=1 =1
exp[ n exp[ ni o H ]
i=1 =1 ni =1
1 so,
o H ] + exp = (exp[ exp [ o H ]])) N
We know the characteristic potential of an ensemble is related to the partition function for that ensemble according according to
= ln → = kT ln Furthermore, we know from thermo that
d = SdT pdV pdV + dN M dH
which gives us the following relationships for the properties of the system:
S = M =
∂ ∂T
∂
V,N,H
∂H
2
V,N,T
= p =
∂ ∂N
V,T,H
∂ ∂V
T,N,H
and from stat mech we have from above
= kT N ln ln [exp [exp [ o H ] + exp exp [ o H ]] ]]
Lets get the entropy, S
∂ S = ∂T
V,N,H
S = kN ln [ o H ] + exp + kT N ln (exp (exp exp [ o H ]) ])
o H kT 2
(exp [ (exp o H ] exp exp [ o H ]]) )
o H ] + exp exp [ exp exp [ o H ]
S = kN { ln ln (exp (exp [ H ] + exp exp [ H H ]) ]) tanh tanh ( H ) } o
Now for the magnetizati magnetization on ∂ M = ∂H
V,N,T
= kTNo
o
o
exp [ exp o H ] exp exp [ o H ] exp [ exp o H ] + exp exp [ o H ]
M = N o tanh tanh ( o H )
The energy E
o
= E T S H M
= N o tanh tanh ( o H )
E = + T S + H M = 0
However if you define a quantity called the internal magnetic energy (which is a quantity analogous to the
enthalpy in the T, p, p, N ensemble)
E H H = E H M
you can get
E H o H ) tanh ( H = N o H tanh
The last part of this problem asks you to determine the behavior of the energy and entropy as T → 0. E (T → 0) = N H H H
lim →∞ kN →∞ S = lim
o
H ] + exp [ H ]) tanh ( H )} = 0 { ln (exp [ o
o
o
o
S (T 0) = 0 is in accordance with the third law of thermodynami thermodynamics. cs.
→
Problem 3
(a) N
M =
i=1 =1
(see solution to Problem 2 - Method 1)
∂ M = ∂H
V,N,T
= kTNo
o ni
exp [ exp o H ] exp exp [ o H ] o H ] + exp exp [ exp exp [ o H ]
o M = N o tanh ( H )
3
= N o tanh tanh ( o H )
(b) The partition function for the N, V, H, T fixed ensemble is
s exp [ (E exp
=
s
M H )])] s
where s = states
We want to determine the fluctuations in the extensive quantity M. Use the 3-step procedure developed in class. Step 1: Multiply both sides by the partition function
M =
M s exp exp [ (E s
states
M H )])]
s
Because E s is always constant (see comment in Problem 2) we can therefore arbitrarily set it to zero and write
N
N
N
M =
exp o H ni exp
o ni
n1 ,n 2,...nN
i=1 =1
i=1 =1
Step 2: Get derivative with respect to mechanical variable’s conjugate.
2
∂M +M ∂H
∂M ∂ + M M = ∂H ∂H n
N
n1 ,,n n 2 ,...n ,...nN
N
N
N
o
N
H exp o H exp
o ni
i=1 =1
N
exp o H exp
ni
i=1 =1
,n n2,...nN 1 ,
N
=
ni
ni
i=1 =1
o
n1 ,,n n 2 ,...n ,...nN
i=1 =1
N
2
ni
i=1 =1
N
exp o H exp
ni
i=1 =1
Step 3: Divide through by the partition function
∂M 2 ∂H + M = M 2 or we can write it like this
2 (M ) = M 2
with:
∂M ∂H
(c) As
→ ∞ , tanh tanh (H H )
1 ∂M ∂H
→
,N
2
M =
= N 2 1 tanh2 (H )
2 1. Therefore M T ( M ) T →0 = N and
words, the ground state with all the spins aligned has no fluctuations. 4
= 2 1 1 = 0. In other
T →0
{ }
Problem 4
From the first and second law we have
dE = T dS pdV pdV + H dM + dN
E = T S pV pV
H M + N
(Euler Form)
We are working with fixed N, T, V, M in this problem. We need to make a Legendre transform because T is a controlling variable instead of E.
S E = F = ln Q k
Q =
exp [ E state ] exp
F = kT ln Q
From the differential form of F, namely
states
dF = SdT pdV pdV + H dM + dN
we get that
∂F H = ∂M
T,V,N
We are working under constant magnetization so N
M = o
ni = o (n+ n)
i=1 =1
where n+ = number of up spins and n = number of down spins. That means that the sum in the expression of Q must be performed over only those states with fixed M (i.e. fixed n+ and n)
Also the atoms in the system do not interact, meaning that the energy is independent of the number and
arrangement of up/down spins and is therefore constant = E o. o.
Q =
exp [ E states] = exp
states with magnetization M
exp exp [ E o ] = exp exp [ E o ]
states with magnetization M
where is the number of states that are consistent with a magnetization M = o (n+ n) .
N !
=
n+ ! (N n+ )! N ! Q = exp [ E o ] exp n+ ! (N n+ )!
F = kT ln
N ! exp [ E ] exp n+ ! (N n+ )! o
using Stirling’s approximatio approximation, n,
= E o
N ! kT ln ln n+ ! (N n+)!
F = E o kT (N ln N n+ ) + ( N N n+ ) ln N N ) n+ ln ln n+ + n+ (N n+ ) ln ln ( N
{
}
but we know from above that
M = o (n+ n) = o (2n + N ) = n+ =
⇒
M + o N 2o
we can substitute this back into our expression for F to get F as a function of M
F = E o
kT
N ln N
M + o N
2o
M + o N
2 o
ln
5
N M o N M o ln 2 o 2o
Remembering that H =
∂M T,V,N
1 M + o N 1 1 o N M 1 H = = kT ln ln + ln ln + 2 o 2o 2 o 2 o 2 o 2o T,V,N kT M + o N N M H = ln ln ln ln o 2 o 2o 2o M + o N 2o H = ln o N M and it is easy to show that M = o N tanh tanh ( o H ) which is the same result as in the other ensemble of Problems
∂F ∂M
∂F
2 and 3. This is due to the equivalence of ensembles in the thermodynamic thermodynamic limit.
Problem 5
(a) Grand canonical ensemble:
=
∞
N =0
j
exp [ E j ] exp
exp [N exp ] =
∞
Q (N, V, T ) exp exp [N ]
N =0
This sum extends
over allstates with total # of particles = N
j
q N j ] = Q( N exp [ E exp , V , T ) = N !
for a one component gas of non-interacting particles. particles. We can write q in closed form as we have done before as
q = So now can be written as: ∞
=
q N
N =0 =0
N !
3 2mkT 2 h2
exp [N exp N ] =
V
∞
qe
N
N !
N =0 =0
But if we let a = qe then we can write this sum in a form for which the solution is know, namely ∞ N
N =0 =0
a = e a N !
If we now substitute in for q we can write as
= ezV
e z = 2 h
with
3 2m 2
(b) The characteristic potential for the grand canonical ensemble is
pV = = kT lnln = E T S N d = SdT pdV N d 6
p =
∂ ∂V
∂ (zV ) = kT k T = kT k T z ∂V T,
∂ ∂
N =
∂z = kT k T V = zV z V ∂ T ,V
N N putting those two parts together and eliminating z gives us the familiar ideal gas equation p = kT . V
(c)
2 (N )
2 () = 2
2
(because V fixed)
N 2
2
We need to get (N ) = N 2 N . Follow the 3-step procedure.
Step 1: Multiply both sides by the partition function ∞
N
N =0 =0
∞
N
q N
N =0 =0
N
e
N !
q N N e N !
N = =
∞
=
q N N N e N !
N =0 =0
Step 2: Get derivative with respect to mechanical variable’s conjugate.
∞
∂N + N ∂
N =0 =0
q N N N N e N !
∞
=
N 2
N = =0 0
∂N + N N = N N 2 ∂
Step 3: Divide through by the partition function
∂N 2 + N = N 2 ∂
2
N 2 N = kT k T
32
from (b) we have that N = zV with
2m
z = eh2
∂N ∂
∂N ∂
.
∂z V = zV = ∂ 2
z V N 2 N = zV
divide by V to get the densities
2 2 =
z N = 2 V V
2 2 N = 2 2 N
2 2
1 = =
√ N
7
kT pV
q N N e N !
(d) To second order in N
2 ) 1 2 ∂ ln P (N ln P (N ) = ln P ( N ) + + (N ) ∂N 2 2 N =N
assuming a Gaussian distribution around N with variance N
P (N )
= exp exp P (N )
101212 N
2
7
10 = e 10
8
2
we get
which is very unlikely.
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