MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
8.044 Statistical Physics I
Spring Term 2013
Solutions to Problem Set #8
Problem 1: Dust Grains in Space
a) H is separable: the 6 variables are statistically independent.
p(θ1 , θ2 , θ3 , L1 , L2 , L3 ) =
1
2π
b)
1
2π
1
2π
1
1
1
√
exp(−L23 /2I3 kT )
exp(−L21 /2I1 k T ) √
exp(−L22 /2I2 kT ) √
2πI1 kT
2πI2 kT
2πI3 kT
< L21 >=< L22 >= I1 kT < L23 >= I3 kT
~ is almost ⊥ to axis 3, the long axis.
⇒L
c)
h
iN
p
ZR = (Z1,R )N = (2π)9/2 I1 I2 I3 (kT )3/2
FR = −N kT ln Z1,R
SR = −
∂FR
∂T
= N k ln Z1,R + N kT
N
1 3 1
Z1,R
Z1,R 2 T
3
= N k ln Z1,R + N k
2
d) 3rd law is violated: limT →0 SR = N k ln(0) = −∞. At very low temperatures one must
switch to a quantum treatment of the rotational motion. Such a treatment will lead to a
result consistent with the 3rd law.
e) There is no energy gap behavior because there is no gap in the classically allowed rotational
energies. The quantum result, however, will show an energy gap.
1
Problem 2: Adsorption On a Stepped Surface
a) Z1 =
P
states exp(−state /kT ) = 0.01M + 0.14M exp(−∆/kT ) + 0.85M exp(−1.5∆/kT )
b)
nface
p
0.85M exp(−1.5∆/kT )
= face =
= 85 exp(−1.5∆/kT )
ncorner
pcorner
0.01M
c) Consider only the 2 lowest energy levels
E = N < one >
0.01M
0.14M exp(−∆/kT )
+ (∆)
= N (0)
0.01M + 0.14M exp(−∆/kT )
0.01M + 0.14M exp(−∆/kT )
≈ 14N ∆ exp(−∆/kT )
CA =
∂E
∂T
= 14N ∆
A
∆
kT 2
exp(−∆/kT ) = 14N k
∆
kT
2
exp(−∆/kT )
d) All states are equally likely ⇒ pface = 0.85 .
e) M possible states for each atom ⇒ limT →∞ S = N k ln M .
f) One expects energy gap behavior because there is an energy gap for the excitation of a
single atom.
2
Problem 3: Neutral Atom Trap
a) First write down the Hamiltonian for one atom.
p2y
p2x
p2
+
+ z + ar
2m 2m 2m
H1 =
Then compute the partition function
Z ∞
Z ∞
Z ∞
Z
p2
p2
p2
y
ar
1
x
z
− 2mkT
− 2mkT
− 2mkT
dpx
e
dpy
Z1 = 3
e
e
dpz
e− kT r2 sin θ drdθdφ
h −∞
{z
}
| √ {z
} | −∞√ {z
} | −∞√ {z
} |V R
2π mkT
2πmkT
2πmkT 3/2
= (
) 4π
h2
= 8πk 3 (
kT
a
3 Z
|0
2πmkT
4π
∞
0
exp[−ar/kT ] r2 dr
∞
y 2 e−y dy
{z
}
2
2πmk 3/2 9/2 −3
) T a
h2
In order to emphasize the dependence on the important variables, this can be written in the
form Z1 = AT α a−η where
A = 8πk 3 (
2πmk 3/2
)
h2
α = 9/2 and η = 3.
b) Remember to include correct Boltzmann counting.
Z =
1 N
Z
N! 1
F = −kT ln Z = −kT (N ln Z1 − N ln N + N )
= −N kT ln(Z1 /N ) − N kT
S = −
∂F
∂T
N
= N k ln(Z1 /N ) + N k + N kT
= N k ln(Z1 /N ) + (11/2)N k
3
1
Z1 /N
(9/2)
Z1 /N
T
c)
dQ = 0
no heat is exchanged with surroundings
dQ = dS/T
process is said to be reversible
⇒ dS = 0,
S is constant
⇒ Z1 is constant, using the result from b)
9/2
⇒ T 9/2 /a3 is constant and = T0 /a30
T
T0
9/2
T
3
a
=
a0
2/3
a
= T0
a0
Problem 4 Two-Dimensional H2 Gas
a)
±3
9~2 /2I
2m
±2
4~2 /2I
2m
±1
~2 /2I
2m
0
0
m
1m
degeneracy
m = (~2 /2I) m2
Zrot,1 =
X
exp[−(state)/kT ] =
m=−∞
states
= 1+2
∞
X
∞
X
exp[−(~2 /2IkT ) j 2 ]
j=1
4
exp[−(~2 /2IkT ) m2 ]
b)
p(m = 3)
Z −1 exp[−9~2 /2IkT ]
= −1
= exp[−(5/2)~2 /IkT ]
p(m = 2)
Z exp[−4~2 /2IkT ]
c)
p(m = 3 | = 9~2 /2I) =
Z −1 exp[−9~2 /2IkT ]
= 1/2
2(Z −1 exp[−9~2 /2IkT ])
Z −1 exp[−~2 /2IkT ]
1
=
p(m = 1 | ≤ ~ /2I) =
Z −1 + 2(Z −1 exp[−~2 /2IkT ])
2 + exp[~2 /2IkT ])
2
d)
Zrot,1 =
∞
X
2
Z
∞
exp[−(~2 /2IkT ) m2 ] dm
exp[−(~ /2IkT ) m ] →
m=−∞
−∞
∞
m2
exp[−
] dm =
2(IkT /~2 )
−∞
Z
=
2
2πIkT
~2
|
{z
1/2
}
Gaussian normalization
1 ∂Z1
Erot = N < >= N −
Z1 ∂β
1
Z1
1
= N −
−
= (1/2)N = (1/2)N kT
Z1
2β
β
5
∝ β −1/2
Problem 5: Why Stars Shine
a) The electostatic potential outside the charged sphere depends only on r, the magnitude
of the distance from the center of the sphere.
|e|
φ(r) =
r≥R
r
The potential energy of another proton, considered to be a point particle, in this field is
e2
V (r) = qφ(r) =
r
Then the minimum energy that the second proton must have to get within a radial distance
R of the first is
e2
(4.8 × 10−10 )2
Emin = V (R) =
=
= 1.92 × 10−6 ergs
R
1.2 × 10−13
b) In problem 4 of problem set 2 we found the following expression for the kinetic energy of
a particle in a three dimensional classical gas.
r
2 1
E
p(E) = √
exp[−E/kT ]
π kT kT
Now find the probability p+ that a given proton in the stellar plasma has an energy greater
than Emin .
Z ∞
p+ ≡ prob(E > Emin ) =
p(E) dE
Emin
2 1
= √
π kT
Z
∞
Emin
r
E
2
exp[−E/kT ] dE = √
kT
π
Z
∞
√
y exp[−y] dy
ymin =Emin /kT
2 √
≈ √
ymin exp[−ymin ]
π
This is going to turn out to be a very small number, probably too small to be represented
on a hand calculator. Therefore, let’s work toward getting its logarithm.
#
"
r
Emin
2
+ log10 exp[−Emin /kT ]
log10 (p+ ) = log10 √
kT
π
E
E
log10 exp[−Emin /kT ] = − min log10 (e) = −(.4343) min
kT
kT
Emin
1.920 × 10−6
= 3.476 × 102
=
kT
1.381 × 10−16 × 4 × 107
log10 (p+ ) = 1.323 − 1.510 × 102 = −149.6
p+ = 0.2 × 10−149
6
c)
r
<v> =
=
8kT
πm
8 × 1.381 × 10−16 × 4 × 107
π × 1.67 × 10−24
1/2
= 9.18 × 107 cm/sec
σ = π(2R)2 = π(2.4 × 10−13 )2 = 1.81 × 10−25 cm2
n =
ρ
Mproton
=
100
= 5.99 × 1025 protons/cm3
1.67 × 10−24
L = (nσ)−1 = 9.22 × 10−2 cm
τcollision = L/ < v >= 1.01 × 10−9 sec
d) The fusion rate per proton is p+ times the collision rate per proton. But in general a rate
equals the reciprocal of the characteristic time between events, so
τfusion = τcollision /p+ =
1.01 × 10−9
= 5 × 10140 sec
0.2 × 10−149
The universe is about 15 billion years old, corresponding to a time
Tuniverse = 15 × 109 × 365 × 24 × 60 × 60 = 4.7 × 1017 sec
If the mass of the sun is 2 × 1033 grams then the number of protons it contains is given by
Nprotons =
2 × 1033
= 1.2 × 1057
1.67 × 10−24
Then for the entire sun, the total number of fusions per second is found as follows.
number of fusions per second = Nprotons × fusion rate per proton
= Nprotons /τfusion
= 1.2 × 1057 / 5 × 10140 = 2 × 10−84 sec−1
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8.044 Statistical Physics I
Spring 2013
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