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PHYS 449 - Statistical Mechanics I
Assignment 5 Solutions
1. (a) To estimate the number of occupied levels at room temperature (though any suitable temperature of yourchoosing is ok with me), we would use
h2π2n2max
2mL2= kBT.
This gives
nmax =
√2mL2kBT
h2π2.
The mass of N2 is m = 4.65 × 10−26 kg. The maximum occupied level is therefore
nmax =
√(2)(4.65 × 10−26)(1)2(1.38 × 10−23)(300)
(1.05459 × 10−34)2π2.
Plugging in numbers gives nmax ≈ 6 × 1010, which is a seriously large number of occupied levels. This is whytranslational motion is a degree of freedom at room temperature! We would need temperatures that were 1020
times lower than room temperature to start freezing out the translational motion of nitrogen!
(b) The mean energy is given by
U =∑i
niεi =
`max∑`=0
n`ε`.
Assuming that each energy level up to nmax is equally occupied, then the occupation of level ` is n` = N/nmax
for any value of `, where N is the total number of particles. The mean energy is therefore:
U =N
nmax
nmax∑n=0
h2π2n2
2mL2=
N
nmax
h2π2
2mL2
nmax∑n=0
n2 =N
nmax
h2π2
2mL2
(nmax)(1 + nmax)(1 + 2nmax)
6≈ h2π2n2max
6mL2,
where I have used the fact that nmax 1 in the last line, consistent with the result obtained in part (a).
Meanwhile, the number of arrangements is Ω = nNmax. Think of it this way: if there was only one particle thenthere would be nmax different energy levels to occupy; if the N particles are non-interacting then each has thismany levels to access. Putting this together with the result above gives
Ω =
(6mL2U
h2π2
)N/2
.
Using the thermodynamic identity 1/T = ∂S/∂U together with S = kB ln(Ω) gives
1
T=NkB2U
or U = NkBT/2, which is the result from the equipartition theorem. Note that this result doesn’t depend onthe value of nmax, only its existence. So the ‘classical’ result is recovered from the quantum mechanical valueof the energies only from the assumption that every accessible energy level is equally likely to be occupied (avery classical assumption).
For a three-dimensional box, one can imagine that the accessible states are again multiplicative (each dimensionis independent), so that Ω = n3Nmax. Everything goes through as before, but now one recovers the resultU = 3NkBT/2, as expected.
PHYS 449 - Assignment 5 Solutions 2
2. Most of the resulted for this question are found in the attached Mathematica notebook. But I can summarize.We have two equations of constraint: one for the total particle number and for the total (mean) energy:
N = n0 + n1 + n2 and U = [n0(0) + n1(1) + n2(4)] ε1,
where ε1 = h2π2/2mL2 is the overall energy scale. I can make my life simpler by assuming that the system isat equilibrium so that pi = ni/N and defining u ≡ U/Nε1. The two equations then become p0 + p1 + p2 = 1and p1 + 4p2 = u. We can solve these two equations for p0 and p1 to give p0 = 1 + 3p2 − u and p1 = u− 4p2.We can now write the total entropy of the system
S = −NkB [p0 ln(p0) + p1 ln(p1) + p2 ln(p2)]
= −NkB [(1 + 3p2 − u) ln(1 + 3p2 − u) + (u− 4p2) ln(u− 4p2) + p2 ln(p2)] ,
which is now solely a function of the (unknown) variable p2 and parameter u. To find out the value of p2 weneed to maximize the entropy, ∂S/∂p2 = 0, which (using Mathematica) gives
−NkB [ln(p2) + 3 ln(1 + 3p2 − u) − 4 ln(u− 4p2)] = 0,
which is equivalent to requiring that
p2(1 + 3p2 − u)3
(u− 4p2)4=p2p
30
p41= 1.
The rest of this work is done on the attached Mathematica notebook, which is heavily commented so I willstop this part here.
Assignment 5 Solutions(1a) First define the fundamental constants:
hbar = 1.05459 * 10^H-34Lm = 4.65 * 10^H-26LL = 1
kB = 1.38 * 10^H-23LT = 300
nmax = Sqrt@2 * m * L^2 * kB * T hbar^2 Pi^2 ΑD
1.05459 ´ 10-34
4.65 ´ 10-26
1
1.38 ´ 10-23
300
5.92254 ´ 1010
1
Α
Here 1/Α is the fraction of room temperature, so room temperature has Α=1 and absolute zero would
have Α -> infinity. At room temperature there are something like 10^(10) energy levels occupied.
(1b) Sum over all energy levels up to nmax:
Clear@nmaxD; Simplify@Sum@n^2, 8n, 0, nmax<DD
1
6
nmax H1 + nmaxL H1 + 2 nmaxL
The rest is done on the analytical solutions pages.
(2a) Solve the two equations for the probabilities p0 and p1:
Clear@p0, p1, p2, uD;
ans = FullSimplify@Solve@8p0 + p1 + p2 1, p1 + 4 * p2 u<, 8p0, p1<DD
88p0 ® 1 + 3 p2 - u, p1 ® -4 p2 + u<<
8p0, p1< = 8p0, p1< . ans@@1DD
81 + 3 p2 - u, -4 p2 + u<
Check that they give the right equations:
FullSimplify@p1 + 4 * p2DFullSimplify@p0 + p1 + p2D
u
1
write the expression for the entropy at equilibrium:
Clear@kBD; S = -Num * kB * Hp0 * Log@p0D + p1 * Log@p1D + p2 * Log@p2DL
-kB Num Hp2 Log@p2D + H1 + 3 p2 - uL Log@1 + 3 p2 - uD + H-4 p2 + uL Log@-4 p2 + uDL
To find p2, need to maximize the entropy explicitly:
dS = FullSimplify@D@S, p2DD
-kB Num HLog@p2D + 3 Log@1 + 3 p2 - uD - 4 Log@-4 p2 + uDL
This implies that the following function is equal to unity:
func = p2 * H1 + 3 p2 - uL^3 H-4 p2 + uL^4
p2 H1 + 3 p2 - uL3
H-4 p2 + uL4
This is a fourth-order polynomial in p2 with a parameter u. Sounds ugly. We can try to solve this analyti-
cally for p2, but all we get is a mess:
Solve@func 1, p2D
::p2 ®
1
916
H27 + 229 uL -
1
2.
H-27 - 229 uL2
209 764
-
2
229
I-3 + 6 u + 29 u2M + I4 ´ 2
13 I-64 u + 48 u2
- 12 u3
+ u4MM
K229 J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27
u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
O +
1 I687 ´ 213M J6912 - 20 736 u + 23 328 u
2- 12 528 u
3+ 3483 u
4- 486 u
5+
27 u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
-
1
2.
H-27 - 229 uL2
104 882
+
1
229
I3 - 6 u - 29 u2M -
3
229
I-3 + 6 u + 29 u2M -
I4 ´ 213 I-64 u + 48 u
2- 12 u
3+ u
4MM
K229 J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27
u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
O -
1 I687 ´ 213M J6912 - 20 736 u + 23 328 u
2- 12 528 u
3+ 3483 u
4- 486 u
5+
-
2 ass5solb.nb
1 I687 ´ 2 M J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+
27 u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
-
-
H-27 - 229 uL3
12 008 989
+ I12 H-27 - 229 uL I-3 + 6 u + 29 u2MM 52 441 -
8
229
I-1 + 3 u - 3 u2
- 15 u3M
4 .H-27 - 229 uL2
209 764
-
2
229
I-3 + 6 u + 29 u2M + I4 ´ 2
13 I-64 u + 48 u2
- 12 u3
+ u4MM
K229 J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27 u6
-
-I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+
780 749 280 u7
- 197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+
222 588 u11
- 6183 u12MN
13
O + 1 I687 ´ 213M J6912 - 20 736 u +
23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27 u6
- -I47 775 744 -
286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
- 2 964 957 696 u4
+
3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
- 197 385 255 u8
+
33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
>,
:p2 ®
1
916
H27 + 229 uL -
1
2.
H-27 - 229 uL2
209 764
-
2
229
I-3 + 6 u + 29 u2M +
I4 ´ 213 I-64 u + 48 u
2- 12 u
3+ u
4MM
K229 J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27
u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
O +
1 I687 ´ 213M J6912 - 20 736 u + 23 328 u
2- 12 528 u
3+ 3483 u
4- 486 u
5+
27 u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
+
1
2.
H-27 - 229 uL2
104 882
+
1
229
I3 - 6 u - 29 u2M -
3
229
I-3 + 6 u + 29 u2M -
I4 ´ 213 I-64 u + 48 u
2- 12 u
3+ u
4MM
K229 J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27
u6
-
ass5solb.nb 3
u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
O -
1 I687 ´ 213M J6912 - 20 736 u + 23 328 u
2- 12 528 u
3+ 3483 u
4- 486 u
5+
27 u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
-
-
H-27 - 229 uL3
12 008 989
+ I12 H-27 - 229 uL I-3 + 6 u + 29 u2MM 52 441 -
8
229
I-1 + 3 u - 3 u2
- 15 u3M
4 .H-27 - 229 uL2
209 764
-
2
229
I-3 + 6 u + 29 u2M + I4 ´ 2
13 I-64 u + 48 u2
- 12 u3
+ u4MM
K229 J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27 u6
-
-I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+
780 749 280 u7
- 197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+
222 588 u11
- 6183 u12MN
13
O + 1 I687 ´ 213M J6912 - 20 736 u +
23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27 u6
- -I47 775 744 -
286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
- 2 964 957 696 u4
+
3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
- 197 385 255 u8
+
33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
>,
:p2 ®
1
916
H27 + 229 uL +
1
2.
H-27 - 229 uL2
209 764
-
2
229
I-3 + 6 u + 29 u2M +
I4 ´ 213 I-64 u + 48 u
2- 12 u
3+ u
4MM
K229 J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27
u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
O +
1 I687 ´ 213M J6912 - 20 736 u + 23 328 u
2- 12 528 u
3+ 3483 u
4- 486 u
5+
27 u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
-
4 ass5solb.nb
1
2.
H-27 - 229 uL2
104 882
+
1
229
I3 - 6 u - 29 u2M -
3
229
I-3 + 6 u + 29 u2M -
I4 ´ 213 I-64 u + 48 u
2- 12 u
3+ u
4MM
K229 J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27
u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
O -
1 I687 ´ 213M J6912 - 20 736 u + 23 328 u
2- 12 528 u
3+ 3483 u
4- 486 u
5+
27 u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
+
-
H-27 - 229 uL3
12 008 989
+ I12 H-27 - 229 uL I-3 + 6 u + 29 u2MM 52 441 -
8
229
I-1 + 3 u - 3 u2
- 15 u3M
4 .H-27 - 229 uL2
209 764
-
2
229
I-3 + 6 u + 29 u2M + I4 ´ 2
13 I-64 u + 48 u2
- 12 u3
+ u4MM
K229 J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27 u6
-
-I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+
780 749 280 u7
- 197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+
222 588 u11
- 6183 u12MN
13
O + 1 I687 ´ 213M J6912 - 20 736 u +
23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27 u6
- -I47 775 744 -
286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
- 2 964 957 696 u4
+
3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
- 197 385 255 u8
+
33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
>,
:p2 ®
1
916
H27 + 229 uL +
1
2.
H-27 - 229 uL2
209 764
-
2
229
I-3 + 6 u + 29 u2M +
I4 ´ 213 I-64 u + 48 u
2- 12 u
3+ u
4MM
K229 J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27
u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
O +
ass5solb.nb 5
1 I687 ´ 213M J6912 - 20 736 u + 23 328 u
2- 12 528 u
3+ 3483 u
4- 486 u
5+
27 u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
+
1
2.
H-27 - 229 uL2
104 882
+
1
229
I3 - 6 u - 29 u2M -
3
229
I-3 + 6 u + 29 u2M -
I4 ´ 213 I-64 u + 48 u
2- 12 u
3+ u
4MM
K229 J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27
u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
O -
1 I687 ´ 213M J6912 - 20 736 u + 23 328 u
2- 12 528 u
3+ 3483 u
4- 486 u
5+
27 u6
- -I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
-
197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
+
-
H-27 - 229 uL3
12 008 989
+ I12 H-27 - 229 uL I-3 + 6 u + 29 u2MM 52 441 -
8
229
I-1 + 3 u - 3 u2
- 15 u3M
4 .H-27 - 229 uL2
209 764
-
2
229
I-3 + 6 u + 29 u2M + I4 ´ 2
13 I-64 u + 48 u2
- 12 u3
+ u4MM
K229 J6912 - 20 736 u + 23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27 u6
-
-I47 775 744 - 286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
-
2 964 957 696 u4
+ 3 341 191 680 u5
- 2 038 188 096 u6
+
780 749 280 u7
- 197 385 255 u8
+ 33 096 924 u9
- 3 557 034 u10
+
222 588 u11
- 6183 u12MN
13
O + 1 I687 ´ 213M J6912 - 20 736 u +
23 328 u2
- 12 528 u3
+ 3483 u4
- 486 u5
+ 27 u6
- -I47 775 744 -
286 654 464 u + 752 467 968 u2
+ 671 293 440 u3
- 2 964 957 696 u4
+
3 341 191 680 u5
- 2 038 188 096 u6
+ 780 749 280 u7
- 197 385 255 u8
+
33 096 924 u9
- 3 557 034 u10
+ 222 588 u11
- 6183 u12MN
13
>>
Instead, let’s solve for p2 numerically for a range of values of u. The smallest u can be is zero, and the
largest is 4, so let’s loop over this range but not including the two limits. For each value of u, we can
numericaly solve for p2; because the function above is a fourth-order polynomial, we know that we will
obtain four roots. Not all of them need be real for all values of u. So the ‘If’ statement in the code below
checks if the imaginary part is zero. If so, then keep the solution for that value of u. Then dump the
values of u, p0, p1, p2, and p3 into arrays for plotting and manipulating later:
6 ass5solb.nb
Instead, let’s solve for p2 numerically for a range of values of u. The smallest u can be is zero, and the
largest is 4, so let’s loop over this range but not including the two limits. For each value of u, we can
numericaly solve for p2; because the function above is a fourth-order polynomial, we know that we will
obtain four roots. Not all of them need be real for all values of u. So the ‘If’ statement in the code below
checks if the imaginary part is zero. If so, then keep the solution for that value of u. Then dump the
values of u, p0, p1, p2, and p3 into arrays for plotting and manipulating later:
Clear@uD; grain = 100;
umax = 4.0;
For@i = 1, i < grain,
u = umax * i grain;
Clear@p2D; ans = Solve@func 1.0, p2D;
For@j = 4, j ³ 1,
If@Im@p2 . ans@@jDDD 0, p2 = p2 . ans@@jDDD;
j--;
D;
uu@iD = u;
q0@iD = p0;
q1@iD = p1;
q2@iD = p2;
Print@u, " ", p0, " ", p1, " ", p2D;
i++;
D;
0.04 0.960009 0.0399884 2.8901 ´ 10-6
0.08 0.920156 0.0797919 0.0000520296
0.12 0.880875 0.118833 0.000291746
0.16 0.842969 0.156041 0.000989738
0.2 0.807439 0.190081 0.00247983
0.24 0.775069 0.219909 0.00502284
0.28 0.746106 0.245192 0.00870205
0.32 0.720326 0.266233 0.0134418
0.36 0.697273 0.283635 0.0190912
0.4 0.676469 0.298041 0.0254896
0.44 0.657491 0.310012 0.032497
0.48 0.64 0.32 0.04
0.52 0.623729 0.328362 0.0479096
0.56 0.60847 0.335374 0.0561565
0.6 0.59406 0.341253 0.0646867
0.64 0.580373 0.34617 0.0734575
0.68 0.567305 0.35026 0.082435
0.72 0.554776 0.353633 0.0915919
0.76 0.542717 0.356377 0.100906
0.8 0.531076 0.358566 0.110359
0.84 0.519805 0.36026 0.119935
0.88 0.508866 0.361512 0.129622
0.92 0.498228 0.362363 0.139409
ass5solb.nb 7
0.96 0.487862 0.362851 0.149287
1. 0.477745 0.363007 0.159248
1.04 0.467855 0.362859 0.169285
1.08 0.458177 0.362431 0.179392
1.12 0.448692 0.361743 0.189564
1.16 0.439389 0.360814 0.199796
1.2 0.430255 0.359661 0.210085
1.24 0.421278 0.358296 0.220426
1.28 0.412449 0.356734 0.230816
1.32 0.40376 0.354986 0.241253
1.36 0.395203 0.353062 0.251734
1.4 0.386771 0.350972 0.262257
1.44 0.378457 0.348724 0.272819
1.48 0.370256 0.346325 0.283419
1.52 0.362163 0.343783 0.294054
1.56 0.354172 0.341104 0.304724
1.6 0.346279 0.338294 0.315426
1.64 0.338481 0.335358 0.32616
1.68 0.330774 0.332301 0.336925
1.72 0.323154 0.329128 0.347718
1.76 0.315619 0.325842 0.35854
1.8 0.308165 0.322447 0.369388
1.84 0.30079 0.318947 0.380263
1.88 0.293491 0.315345 0.391164
1.92 0.286267 0.311644 0.402089
1.96 0.279115 0.307847 0.413038
2. 0.272033 0.303956 0.424011
2.04 0.26502 0.299973 0.435007
2.08 0.258074 0.295902 0.446025
2.12 0.251193 0.291743 0.457064
2.16 0.244376 0.287498 0.468125
2.2 0.237623 0.28317 0.479208
2.24 0.230931 0.278759 0.49031
2.28 0.224299 0.274267 0.501433
2.32 0.217728 0.269696 0.512576
2.36 0.211215 0.265047 0.523738
8 ass5solb.nb
2.4 0.20476 0.260319 0.53492
2.44 0.198363 0.255516 0.546121
2.48 0.192023 0.250637 0.557341
2.52 0.185738 0.245682 0.568579
2.56 0.17951 0.240654 0.579837
2.6 0.173337 0.235551 0.591112
2.64 0.167219 0.230375 0.602406
2.68 0.161155 0.225126 0.613718
2.72 0.155147 0.219804 0.625049
2.76 0.149193 0.214409 0.636398
2.8 0.143294 0.208942 0.647765
2.84 0.137449 0.203401 0.65915
2.88 0.13166 0.197787 0.670553
2.92 0.125925 0.1921 0.681975
2.96 0.120246 0.186338 0.693415
3. 0.114623 0.180502 0.704874
3.04 0.109057 0.174591 0.716352
3.08 0.103548 0.168603 0.727849
3.12 0.0980965 0.162538 0.739365
3.16 0.0927043 0.156394 0.750901
3.2 0.0873723 0.15017 0.762457
3.24 0.0821016 0.143865 0.774034
3.28 0.0768938 0.137475 0.785631
3.32 0.0717507 0.130999 0.79725
3.36 0.0666742 0.124434 0.808891
3.4 0.0616665 0.117778 0.820556
3.44 0.0567305 0.111026 0.832243
3.48 0.0518691 0.104175 0.843956
3.52 0.0470861 0.0972186 0.855695
3.56 0.0423857 0.0901524 0.867462
3.6 0.0377732 0.0829691 0.879258
3.64 0.0332546 0.0756605 0.891085
3.68 0.0288378 0.0682162 0.902946
3.72 0.0245324 0.0606235 0.914844
3.76 0.0203505 0.052866 0.926784
3.8 0.0163088 0.0449216 0.93877
ass5solb.nb 9
3.84 0.01243 0.03676 0.95081
3.88 0.00874817 0.0283358 0.962916
3.92 0.00531965 0.0195738 0.975107
3.96 0.00225737 0.0103235 0.987419
OK, everything seems to be working. Now let’s plot the three probabilities as a function of u. I encode
p0 as red, p1 as blue, and p2 as green:
plotp0 = Table@8uu@iD, q0@iD<, 8i, 1, grain<D;
plotp1 = Table@8uu@iD, q1@iD<, 8i, 1, grain<D;
plotp2 = Table@8uu@iD, q2@iD<, 8i, 1, grain<D;
Show@8ListPlot@plotp0, PlotStyle ® RedD,
ListPlot@plotp1, PlotStyle ® BlueD, ListPlot@plotp2, PlotStyle ® GreenD<D
1 2 3 4
0.2
0.4
0.6
0.8
For very small u, all of the probability is in p0, which makes sense: for low energies all of the particles
are going to be in the lowest energy state. As u increases, p0 drops and p1 and p2 increase until at
u=5/3 they are all exactly 1/3 each. Beyond this u, the value of p2 continues to increase toward unity,
signifying that all particles are in the highest energy state.
(2b) Let’s return again to the entropy S.
Clear@u, p2D; S
-kB Num Hp2 Log@p2D + H1 + 3 p2 - uL Log@1 + 3 p2 - uD + H-4 p2 + uL Log@-4 p2 + uDL
To obtain the temperature (or rather 1/T), we need to take the derivative of the entropy with respect to
u. The problem is that we don’t know the explicit u-dependence of p2, only its numerical values at
various values of u. But we do know that p2 is some function of u, so let’s rewrite the entropy making
this u-dependence explicit. Then we can formally take the derivative of the entropy with respect to u:
S2 = -kB Num Hp2@uD Log@p2@uDD +
H1 + 3 p2@uD - uL Log@1 + 3 p2@uD - uD + H-4 p2@uD + uL Log@-4 p2@uD + uDLdS2 = FullSimplify@D@S2, uDD Num Ε1
-kB Num
HLog@u - 4 p2@uDD Hu - 4 p2@uDL + Log@p2@uDD p2@uD + Log@1 - u + 3 p2@uDD H1 - u + 3 p2@uDLL
-
1
Ε1
kB HLog@u - 4 p2@uDD H1 - 4 p2¢@uDL + Log@p2@uDD p2
¢@uD + Log@1 - u + 3 p2@uDD H-1 + 3 p2¢@uDLL
By the way, we really want the derivative with respect to U. Since u = U NΕ1 this means that d/dU =
H1 NΕ1)d/du. That’s why I divided by NΕ1 above. In any case, the expression now depends on p2¢@uD,
which we don’t know. One possibility would be to plot p2 as a function of u, and to try to extract the
numercal derivative. Thankfully, there is a nicer way. We know that the constitutive equation for p2 is:
10 ass5solb.nb
By the way, we really want the derivative with respect to U. Since u = U NΕ1 this means that d/dU =
H1 NΕ1)d/du. That’s why I divided by NΕ1 above. In any case, the expression now depends on p2¢@uD,
which we don’t know. One possibility would be to plot p2 as a function of u, and to try to extract the
numercal derivative. Thankfully, there is a nicer way. We know that the constitutive equation for p2 is:
func
p2 H1 + 3 p2 - uL3
H-4 p2 + uL4
Again we can explicitly write this in terms of p2[u]:
func2 =
p2@uD H1 + 3 p2@uD - uL3
H-4 p2@uD + uL4
p2@uD H1 - u + 3 p2@uDL3
Hu - 4 p2@uDL4
Because the function equals unity, the derivative of this function with respect to u must be equal to zero:
ans = FullSimplify@D@func2, uDD
IH-1 + u - 3 p2@uDL2 IH-4 + uL p2@uD + Iu - u2
+ 12 p2@uDM p2¢@uDMM Hu - 4 p2@uDL5
ans2 = Solve@ans 0, p2¢@uDD
::p2¢@uD ®
H-4 + uL p2@uD
-u + u2
- 12 p2@uD>>
p2¢@uD = p2
¢@uD . ans2@@1DD
H-4 + uL p2@uD
-u + u2
- 12 p2@uD
So we have a nice analytical formula for the derivative of p2 with respect to u, even though we don’t
have an explicit form for p2[u]. We can susbstitute this back into the expression for dS/dU:
dS2b = FullSimplify@dS2D
-
1
Ε1 HH-1 + uL u - 12 p2@uDLkB HH-1 + uL u HLog@u - 4 p2@uDD - Log@1 - u + 3 p2@uDDL +
H-4 H-1 + uL Log@u - 4 p2@uDD + H-4 + uL Log@p2@uDD + 3 u Log@1 - u + 3 p2@uDDL p2@uDL
We can copy this, now again replacing p2[u] with p2:
dS3 = -
1
Ε1 HH-1 + uL u - 12 p2LkB HH-1 + uL u HLog@u - 4 p2D - Log@1 - u + 3 p2DL +
H-4 H-1 + uL Log@u - 4 p2D + H-4 + uL Log@p2D + 3 u Log@1 - u + 3 p2DL p2L
-
1
H-12 p2 + H-1 + uL uL Ε1
kB HH-1 + uL u H-Log@1 + 3 p2 - uD + Log@-4 p2 + uDL +
p2 HH-4 + uL Log@p2D + 3 u Log@1 + 3 p2 - uD - 4 H-1 + uL Log@-4 p2 + uDLL
Finally, we can generate a list of temperatures for each value of u:
ass5solb.nb 11
Clear@uD; grain = 100;
umax = 4.0;
For@i = 1, i < grain,
u = umax * i grain;
p0 = q0@iD;
p1 = q1@iD;
p2 = q2@iD;
tt@iD = 1 dS3;
Print@u, " ", tt@iDD;
i++;
D;
0.04
0.314628 Ε1
kB
0.08
0.408978 Ε1
kB
0.12
0.499202 Ε1
kB
0.16
0.592835 Ε1
kB
0.2
0.691362 Ε1
kB
0.24
0.793815 Ε1
kB
0.28
0.898612 Ε1
kB
0.32
1.00469 Ε1
kB
0.36
1.11174 Ε1
kB
0.4
1.22003 Ε1
kB
0.44
1.3301 Ε1
kB
0.48
1.4427 Ε1
kB
0.52
1.5586 Ε1
kB
0.56
1.67869 Ε1
kB
0.6
1.80389 Ε1
kB
0.64
1.93521 Ε1
kB
0.68
2.07373 Ε1
kB
12 ass5solb.nb
0.72
2.22072 Ε1
kB
0.76
2.37755 Ε1
kB
0.8
2.54587 Ε1
kB
0.84
2.72758 Ε1
kB
0.88
2.92491 Ε1
kB
0.92
3.1406 Ε1
kB
0.96
3.37792 Ε1
kB
1.
3.64096 Ε1
kB
1.04
3.93478 Ε1
kB
1.08
4.26585 Ε1
kB
1.12
4.64247 Ε1
kB
1.16
5.07558 Ε1
kB
1.2
5.57985 Ε1
kB
1.24
6.1754 Ε1
kB
1.28
6.89072 Ε1
kB
1.32
7.76739 Ε1
kB
1.36
8.86867 Ε1
kB
1.4
10.2958 Ε1
kB
1.44
12.2214 Ε1
kB
1.48
14.9662 Ε1
kB
1.52
19.2004 Ε1
kB
ass5solb.nb 13
1.56
26.6 Ε1
kB
1.6
42.8631 Ε1
kB
1.64
107.876 Ε1
kB
1.68 -
217.111 Ε1
kB
1.72 -
54.5984 Ε1
kB
1.76 -
31.3716 Ε1
kB
1.8 -
22.0734 Ε1
kB
1.84 -
17.0611 Ε1
kB
1.88 -
13.9238 Ε1
kB
1.92 -
11.7734 Ε1
kB
1.96 -
10.2062 Ε1
kB
2. -
9.01234 Ε1
kB
2.04 -
8.07174 Ε1
kB
2.08 -
7.31089 Ε1
kB
2.12 -
6.68223 Ε1
kB
2.16 -
6.15359 Ε1
kB
2.2 -
5.70248 Ε1
kB
2.24 -
5.31264 Ε1
kB
2.28 -
4.9721 Ε1
kB
2.32 -
4.67179 Ε1
kB
2.36 -
4.40473 Ε1
kB
14 ass5solb.nb
2.4 -
4.16546 Ε1
kB
2.44 -
3.94968 Ε1
kB
2.48 -
3.75388 Ε1
kB
2.52 -
3.57525 Ε1
kB
2.56 -
3.41147 Ε1
kB
2.6 -
3.26059 Ε1
kB
2.64 -
3.12103 Ε1
kB
2.68 -
2.9914 Ε1
kB
2.72 -
2.87056 Ε1
kB
2.76 -
2.75751 Ε1
kB
2.8 -
2.65141 Ε1
kB
2.84 -
2.55152 Ε1
kB
2.88 -
2.45718 Ε1
kB
2.92 -
2.36784 Ε1
kB
2.96 -
2.28299 Ε1
kB
3. -
2.2022 Ε1
kB
3.04 -
2.12506 Ε1
kB
3.08 -
2.05122 Ε1
kB
3.12 -
1.98035 Ε1
kB
3.16 -
1.91217 Ε1
kB
3.2 -
1.84641 Ε1
kB
ass5solb.nb 15
3.24 -
1.7828 Ε1
kB
3.28 -
1.72112 Ε1
kB
3.32 -
1.66115 Ε1
kB
3.36 -
1.60266 Ε1
kB
3.4 -
1.54545 Ε1
kB
3.44 -
1.48931 Ε1
kB
3.48 -
1.43401 Ε1
kB
3.52 -
1.37934 Ε1
kB
3.56 -
1.32505 Ε1
kB
3.6 -
1.27086 Ε1
kB
3.64 -
1.21645 Ε1
kB
3.68 -
1.16145 Ε1
kB
3.72 -
1.10535 Ε1
kB
3.76 -
1.0475 Ε1
kB
3.8 -
0.986956 Ε1
kB
3.84 -
0.922253 Ε1
kB
3.88 -
0.850861 Ε1
kB
3.92 -
0.767586 Ε1
kB
3.96 -
0.657798 Ε1
kB
This is very interesting! The numerics clearly show that temperature initially increases monotonically
with u, but sharply rises as u approaches 5/3, the point at which all of the probabilities are equal.
Beyond this energy, the temperature goes negative!! This is unphysical, which means that the maxi-
mum (mean) energy occurs when the temperature goes to (positive) infinity, corresponding to equal
probabilities in all energy levels. The inference is that at high temperatures, the probabilities of occupy-
ing energy levels are all equal. This is precisely the ‘classical result’ found in the previous assignment
when we maximized the entropy for a d-dimensional coin. The assignment asked us to plot the mean
energy as a function of temperature, so here goes:
16 ass5solb.nb
This is very interesting! The numerics clearly show that temperature initially increases monotonically
with u, but sharply rises as u approaches 5/3, the point at which all of the probabilities are equal.
Beyond this energy, the temperature goes negative!! This is unphysical, which means that the maxi-
mum (mean) energy occurs when the temperature goes to (positive) infinity, corresponding to equal
probabilities in all energy levels. The inference is that at high temperatures, the probabilities of occupy-
ing energy levels are all equal. This is precisely the ‘classical result’ found in the previous assignment
when we maximized the entropy for a d-dimensional coin. The assignment asked us to plot the mean
energy as a function of temperature, so here goes:
lastplot = Table@8tt@iD * kB Ε1, uu@iD<, 8i, 1, 41<D;
ListPlot@lastplotD
5 10 15 20
0.5
1.0
1.5
The mean energy (u) clearly asymptotes at 5/3 at high temperatures.
ass5solb.nb 17
