Assignment 3
Tyler Shendruk
February 26, 2010
1
Kadar Ch. 4 Problem 7
N diatomic molecules are stuck on metal. The particles have three states:
1. in plane and aligned with the x̂ axis
2. in plane but aligned with the ŷ axis
3. out of plane (aligned with the ẑ axis).
There is an energy cost associated with standing up but no energy cost associated with being in plane.
1.1
Part a)
1. How many microstates have the smallest value of energy?
The lowest energy state is when none of the molecules are paying the cost of standing up i.e. when all molecules line in plane:
Emin = 0
(1)
There are two orientations which each of the N molecules can take when
it’s in place so the number of states is
ΩEmin = 2N
(2)
2. What is the largest microstate energy?
E is maximized when every particle is out of plane:
Emax = N (3)
and only a single state (i.e. when every single particle is perpendicular to
the plane)can achieve that so
ΩEmax = 1
1
(4)
1.2
Part b)
1. For a microcanonical ensemble of energy E haw many states are there?
We have to ask how many ways can we get that energy E? That requires
E
(5)
molecules to be aligned perpendicular to the plane. So then the number
of ways for this to occur is
Nz =
Ω(E, N ) → Ω(Nz , N ) =
N!
Nz !(N − Nz )!
That is the number of ways to have Nz molecules out of the plane and
N − Nz molecules in the plane but the N − Nz molecules can be arranged
in 2N −Nz ways as we showed in the last part. So then the total number
of states to get E is
Ω(E, N ) =
N!
2N −Nz
Nz !(N − Nz )!
(6)
where Nz = E/.
2. What’s the entropy?
The entropy is
S = kB ln Ω
N!
N −Nz
2
= kB ln
Nz !(N − Nz )!
N!
= kB ln
+ (N − Nz ) ln (2)
Nz !(N − Nz )!
Break up the fraction and apply Stirling’s formula
S = kB [ln (N !) − ln (Nz !) − ln ((N − Nz )!) + (N − Nz ) ln (2)]
= kB [N ln (N ) − N − Nz ln (Nz ) + Nz − (N − Nz ) ln (N − Nz ) + (N − Nz ) + (N − Nz ) ln (2)]
= kB [N ln (N ) − Nz ln (Nz ) − (N − Nz ) ln (N − Nz ) + (N − Nz ) ln (2)]


= kB N ln (N ) − Nz ln (Nz ) + Nz ln (N ) − Nz ln (N ) − (N − Nz ) ln (N − Nz ) + (N − Nz ) ln (2)
|
{z
}
=0
= kB [(N − Nz ) ln (N ) + Nz (ln (N ) − ln (Nz )) − (N − Nz ) ln (N − Nz ) + (N − Nz ) ln (2)]
N − Nz
Nz
− (N − Nz ) ln
+ (N − Nz ) ln (2)
= kB −Nz ln
N
N
Nz
Nz
N − Nz
N − Nz
N − Nz
= −N kB
ln
+
ln
−
ln 2 .
N
N
N
N
N
If we write Nz = E/ we can say that the entropy is
E
E
E
E
E
S = −N kB
ln
+ 1−
ln 1 −
− 1−
ln 2 .
N
N
N
N
N
(7)
2
1.3
Part c)
Calculate the heat capacity.
We want to use C = dE/dT but we don’t know energy as a function of
temperature yet so first we must use
1
∂S =
T
∂E N
∂
E
E
E
E
E
= −N kB
ln
+ 1−
ln 1 −
− 1−
ln 2
∂E N N
N
N
N
"
#
1
E
1
1
E
E
1
1
1
−
= −N kB
ln
+
−
ln 1 −
1−
+
ln 2
N
N
N N
N
N 1 − NE
N
N
E
1
E
1
1
ln
− ln 1 −
+ ln 2
= −kB
N
N
kB
E
=−
ln
+ ln 2
N − E
Now we rearrange to find E(T )
E
= ln
+ ln 2
kB T
N − E
N − E
= exp
+ ln 2
E
kB T
N
− 1 = exp (ln 2) exp
E
kB T
N
= 2 exp
+1
E
kB T
N
E=
.
2 exp kBT + 1
−
(8)
We can now find the heat capacity
C=
dE
dT
2

= N (−1) 
1
2 exp
kB T
+1
N 2
C=2
kB T 2
1.4
 2 exp
kB T
exp kBT
2
2 exp kBT + 1
kB
(−1)
1
T2
(9)
Part d)
What’s the probability that a specific molecule is standing up at the energy
E?
3
We are looking for the probability that some specific molecule has an energy
and all the other molecules take care of the rest of the energy. This is the
probability of the configuration of N particles leading to energy E divided by
the probability of all the other N − 1 particles in a configuration resulting in an
energy of E − which was left over.
p(E, N )
p(E − , N − 1)
1
= p(E, N )
p(E − , N − 1)
1
=
Ω(E − , N − 1)
Ω(E, N )
Ω(E − , N − 1)
=
Ω(E, N )
(N − 1)!
Nz ! (N − Nz )! 1
=
2(N −1)−(Nz −1)
(Nz − 1)! ((N − 1) − (Nz − 1))!
N!
2N −Nz
(N − 1)!
Nz ! (N − Nz )! 2N −Nz
=
(Nz − 1)! (N − Nz )!
N!
2N −Nz
(N − 1)! Nz !
=
(Nz − 1)! N !
Nz
‘
=
N
p(n1 ) =
which of course if we had thought about the question before trying something
difficult, we knew was the answer all along.
If we want this as a function of temperature, we can just substitute the
values of Nz and E in to find
Nz
N
E
=
N
1
N
=
N 2 exp kB T + 1
p(n1 ) =
1
p(n1 ) =
2 exp
1.5
kB T
(10)
+1
Part e)
What’s the largest possible value
at any temperature?
of energy
Since T > 0, the term exp kB T is minimized when T → ∞. Therefore,
4
the maximum internal energy is
Emax = lim E(T )
T →∞
= lim
T →∞
=
2 exp
kB T
+1
N
2×1+1
Emax =
2
N
N
3
(11)
Sethna Ch. 3 Problem 7
Monatomic He gas mixed with diatomic H2 gas. Both are ideal-ish so they are
only weakly coupled. The question states as a fact:
3N/2
monatomic Ω1 (E1 ) ∝ E1
5N/2
diatomic Ω2 (E2 ) ∝ E2
2.1
.
Part a)
1. Calculate the probability of the monatomic gas being at an energy E1 .
The probability is the probability that the monatomic gas is at E1 and
the diatomic gas is at E − E1 when the total energy is E. So
Ω1 (E1 )Ω2 (E2 = E − E1 )
.
Ω(E)
p(E1 ) =
(12)
The key is determining what Ω(E) is.
Z
Ω(E) = dE1 Ω1 (E1 )Ω2 (E2 )
Z
3N/2 5N/2
∝ dE1 E1
E2
Z E
3N/2
=
dE1 E1
(E 0 − E1 )5N/2
0
Z
=
E
dE1 E1 (E 0 − E1 )N
3N/2
0
We know the integral
Z 1
Γ(m)Γ(N )
= a numerical value not dependant on x.
xm−1 (1−x)n−1 dx =
Γ(m + n)
0
(13)
5
So then the probability is
p(E1 ) =
Ω1 (E1 )Ω2 (E − E1 )
Ω(E)
3N/2
∝
E1
3N/2
∝ E1
(E − E1 )5N/2
Ω(E)
(E − E1 )5N/2
3N/2
p(E1 ) ∝ E1 (E − E1 )N
(14)
2. The maximum occurs when
1 dΩ1 1 dΩ2 =
Ω1 dE1 E ∗
Ω2 dE2 E ∗
1
(15)
2
where p(E1∗ ) is the maximum and E2∗ = E − E1∗ .
We find
3N 3N/2−1 1 5N 5N/2−1 E1
E
= 5N/2
3N/2 2
∗
2 2
E1
E2
E1
E−E1∗
1 3N 1 5N =
E1 2 E ∗
E2 2 E−E ∗
1
1
1
5
3
=
E1∗
E − E1∗
5
E − E1∗ = E1∗
3
E1∗ =
3
E
8
(16)
That’s a little theoretical for me. We could have found the exact same
solution from the more physically transparent strategy of saying the state
with the maximum probability is the equilibrium state and entropy must
be maximized at equalibrium. Therefore
dS1 dS2 =
dE1 E ∗
dE2 E−E ∗
1
1
d
d
kB ln Ω1 =
kB Ω2 dE1
dE2
E∗
E−E1∗
1
d
d
3N/2 5N/2 kB
ln E1
∗ = kB dE2 ln E2
dE1
E
E−E1∗
1
3N d
5N d
ln E1 = kB
ln E2 kB
2 dE1
2 dE2
∗
E1
E−E1∗
3N 1 5N 1 kB
= kB
2 E1 E ∗
2 E2 E−E ∗
1
5N
1
3N 1
= kB
kB
2 E1∗
2 E − E1∗
6
1
which is rearranged to find
3
E
8
in accordance with our previous result.
E1∗ =
The third useful way to find it is to just take the derivative of probability
(Eq. 14) with respect to E1 and set it equal to zero. That energy is an
extrema.
2.2
Part b)
Take the logarithm of the probability, do a Taylor expansion and then reexponentiate. What is the approximation to the probability distribution?
ln p (E1 ) =
ln p (E1∗ )
1 ∂ 2 ln p (E1 ) ∂ ln p (E1 ) 2
∗
(E1 − E1∗ ) + . . .
(E1 − E1 ) +
+
∂E1 E1 =E ∗
2
∂E12
∗
E1 =E1
1
(17)
Evaluate the derivatives as
∂ ln p (E1 )
5N
∂
3N
ln E1 +
ln (E − E1 )
=
∂E1
∂E1 2
2
3N 1
1
5N
=
+
(−1)
2 E1
2 E − E1
N 3
5
−
=
.
2 E1
E − E1
Therefore when we remember Eq. 16, we have
∂ ln p (E1 ) 5
N 3
−
=
∂E1 E1 =E ∗
2 E1∗
E − E1∗
1
N
3
5
=
−
2 38 E
E − 38 E
N 8
5
=
− 5
2 E
8E
= 0.
On the other hand, the second derivative is
∂
N 3
5
∂ 2 ln p (E1 )
=
−
∂E12
∂E1 2 E1
E − E1
"
#
N 3
5
=−
+
2
2 E12
(E − E1 )
7
which we again evaluate at E1 = E1∗ and find
"
#
N
∂ 2 ln p (E1 ) 3
5
=−
+
2
∂E12
2 (E1∗ )2
(E − E1∗ )
E1 =E1∗
"
#
5
N
3
=−
2 +
2
3
2
E − 38 E
8E
"
#
N
5
64
=−
+
2
5
2 3 (E)2
8E
N 64
64
=−
+
2 3E 2
5E 2
64 N 1 1
=−
+
2 E2 3 5
256 N
=−
15 E 2
N
∝ 2.
E
Now we can put all of this back into our logarithm and the take the exponent
of both sides.
1 256 N
2
ln p (E1 ) ≈ ln p (E1∗ ) + 0 −
(E1 − E1∗ )
2
2
15
E
1 256 N
∗
∗ 2
exp (ln p (E1 )) = exp ln p (E1 ) −
(E1 − E1 )
2 15 E 2
!
2
1 (E1 − E1∗ )
∗
p (E1 ) ≈ p (E1 ) exp −
15 E 2
2 256
N
(18)
Comparing this to a gaussian
p(x) = √
1 (x − µ)2
exp −
2
σ2
2πσ 2
1
we identify the mean to be
< E1 >≈ E1∗ ,
(19)
meaning the distribution is symmetric i.e. has approximately zero skew and
that the standard deviation is
√
15 E
√ .
σ E1 ≈
(20)
16 N
We know the definition of the standard deviation is
q
σE1 = < (E1 − E1∗ )2 >
which is the energy fluctuation. So we see that the energy fluctuation is
1
∆E ∝ √
N
8
(21)
3
Harden Problem 3
A polymer is made up of N segments which are either in a
• relaxed state α of length a
• contracted state β of length a/2
β’s energy is larger than α’s (we can say the α state energy is zero).
3.1
Part a)
Calculate the number of states for a polymer of energy 0 < E < N i.e. not all
compressed.
The energy is fixed so this is microcanoncial again the length is
L = Nα a + Nβ
a
2
(22)
where Nα is the number of segments in the α state, etc which add up to the
total number of segments
N = Nα + Nβ .
(23)
The number of segments in the β state determine the energy to be
H = E = Nβ .
(24)
We’ll start by finding the number of states Ω as a function of Nβ which
means we want to rearrange Eqs 22 and 23 in terms of Nβ .
a
L = Nα a + Nβ
2
Nβ
= a Nα +
2
Nβ
= a N − Nβ +
2
Nβ
=a N−
nn
2
L Nβ
N= +
a 2 L
Nβ = 2 N −
a
(25)
(26)
(27)
Using Eq. 24, we find the number of states
Ω(N, Nβ ) =
Ω(N, E) =
N!
Nβ ! (N − Nβ )!
E
9
N!
! N−
E
!
(28)
We’re going to need it later so we also calculated the number of states as a
function of length:
N!
Nβ ! (N − Nβ )!
N!
Ω(N, L) =
L
2 N − a ! N −2 N −
Ω(N, Nβ ) =
Ω(N, L) =
3.2
2 N−
L
a
L
a
!
N!
! 2 La − N !
(29)
Part b)
Calculate the entropy as a function of energy and as a function of length.
S = kB ln Ω
"
S(N, E) = kB ln
"
S(N, L) = kB ln
N!
E
! N −
#
E
N!
L
2 N − a ! 2 La − N !
10
(30)
!
#
(31)
3.3
Part c)
We use the Stirling approximation and the very important formula
1
∂S =
T
∂E N
"
#
∂
N!
= kB
ln E E
∂E
! N − !
E
E
∂
ln N ! − ln
! − ln N −
!
= kB
∂E
∂
E
E
= −kB
ln
! + ln N −
!
∂E
∂ E
E
E
E
E
E
= −kB
ln
− + N−
ln N −
− N−
∂E E
E
E
∂ E
ln
+ N−
ln N −
−N
= −kB
∂E ∂ E
E
E
E
= −kB
ln
+ N−
ln N −
∂E E
E 1 1
1
E
E
1
1
1
= −kB
ln
+
+ −
ln N −
+ N−
−
E/ N − E/
1
E
E
1 1
1
= −kB
ln
+ − ln N −
−
E
E
kB
ln
− ln N −
=−
"
!#
E
N− kB
− ln
=−
E
kB
E
=−
ln
.
N − E
We can now rearrange for E as a function of T :
E
−
= ln
kB T
N − E
E
= exp −
N − E
kB T
E = N e−/kB T − Ee−/kB T
E = N
e−/kB T
1 + e−/kB T
E(T ) =
N
1 + e/kB T
(32)
The length should be directly related to E. From Eq. 27 and 24, we find the
11
length to be
L
Nβ = 2 N −
a
E
L
=2 N−
a
L
E
=N−
a
2
Ea
L = Na −
2
a
1
= Na − N
2
1 + e/kB T
Na
1
=
2−
2
1 + e/kB T
N a 2 + 2e/kB T − 1
=
2
1 + e/kB T
N a 1 + 2e/kB T
L=
2
1 + e/kB T
12
(33)