MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
8.044 Statistical Physics I
Spring Term 2013
Solutions to Problem Set #3
Problem 1: Clearing Impurities
Since we are asked for an approximate answer we will resort to the central limit theorem.
For this we need < x > and < x2 > for a single sweep of the laser beam.
Z
∞
<x> =
∞
2
Z
∞
2
x p(x) dx = a
3
|0
2
x p(x) dx = a2
3
2
<x > =
∞
2
2
Var(x) = < x > − < x > =
∞
Z
Z
|0
2
ξ exp(−ξ) dξ = a
3
{z
}
1
∞
4
ξ 2 exp(−ξ) dξ = a2
3
{z
}
2
4 4
−
3 9
a2 =
8 2
a
9
The general form of the central limit theorem is
p(d) ≈ √
1
2πσ 2
exp[−(d− < d >)2 /2σ 2 ]
with
< d > = 36 × < x > = 24 a
σ 2 = 36 × Var(x) = 32 a2
Although it was not asked for, here is a sketch of the resulting probability density.
p(d)
0.1a
d
10a
30a
20a
1
40a
50a
Problem 2: Probability Densities of Macroscopic verses Microscopic Variables
a) Let E1 be the kinetic energy of a single atom in the gas. We can begin with the expression
for p(E1 ) found in problem 4 on Problem Set 2.
r
2 1
E1
p(E1 ) = √
exp(−E1 /kT )
π kT kT
Z ∞ r
2 1
E1
< E1 > = √
E1
exp(−E1 /kT ) dE1
kT
π kT 0
Z ∞ p
2
= √ kT
ξ ξ exp(−ξ) dξ
π
0
|
{z
√ }
Γ(5/2) = (3/4) π
= (3/2)kT
<
E12
2 1
> = √
π kT
r
∞
Z
E12
0
2
= √ (kT )2
π
Z
E1
exp(−E1 /kT ) dE1
kT
∞
ξ2
p
ξ exp(−ξ) dξ
|
{z
√ }
Γ(7/2) = (15/8) π
0
= (15/4)(kT )2
Var(E1 ) = < E12 > − < E1 >2 = (3/2)(kT )2
r
σE1 / < E1 > =
2
= 0.82
3
b) For the sum of N statistically independent variables, the mean is the sum of the means
and the variance is the sum of the variances. Thus if EN is the total kinetic energy of the
gas
r
√ p
N 3/2 kT
1
2
σEN / < EN >=
= √
= 1.6 × 10−10
N (3/2) kT
3
N
2
Problem 3: Temperature
a) Solve each equation for V.
nR
c0 H 0
V =(
)(Θ +
)
P
M0
nR cH
V =(
)(
)
P
M
Equate these two and factor out nR/P .
cH
c0 H 0
some constant,
=Θ+
=
call it h
M
M0
Substitution into the first equation gives P V /nR = h, so at equilibrium
PV
cH
c0 H 0
=
=Θ+
nR
M
M0
b) P V /nR = h looks like the ideal gas law with h → T , so call h ≡ T and thus find the
following equations of state.
PV
= nRT
H
T
M = c
M
0
for an ideal gas
for a Curie Law Paramagnet
Paramagnet with ordering
phase transition to a
ferromagnet at t = Θ
H0
= c
T −Θ
0
Problem 4: Work in a Simple Solid
Substitute the given model expression relating volume changes to changes in the pressure and
the temperature, dV = −V KT dP + V α dT , into the differential for work. As a simplification
we are told to replace the actual volume V by its value at the starting point V1 in the
coefficients entering the differential for the work. Of course the volume itself can not really
remain constant, for in that case d/ W = −P dV = 0.
d/ W = −P dV = KT P V1 dP − αP V1 dT
Along path “a”
Z
Z
W1→2 =
d
/W +
where dP=0
Z
= −αP1 V1
d
/W
where dT=0
2
Z
dT + KT V1
1
2
P dP
1
= −αP1 V1 (T2 − T1 ) +
3
1
KT (P22 − P12 )V1
2
Along path “b”
Z
Z
W1→2 =
d
/W +
where dT=0
Z
d
/W
where dP=0
2
Z
dT
1
1
=
2
P dP − αP2 V1
= K T V1
1
KT (P22 − P12 )V1 − αP2 V1 (T2 − T1 )
2
Along “c” dT and dP are related at every point along the path,
dT =
T2 − T1
dP,
P2 − P1
so the expression for the differential of work can be written as
T2 − T1
) dP
P2 − P1
T2 − T1
=
KT V1 − αV1
P dP
P 2 − P1
d/ W = KT P V1 dP − αP V1 (
Now we can carry out the integral along the path.
T2 − T1
W1→2 =
KT V1 − αV1
P2 − P1
=
Z
2
P dP
| {z }
1
2
(P2 − P12 ) =
2
1
(P1 + P2 )(P2 − P1 )
2
1
1
1
KT (P22 − P12 )V1 − α(P1 + P2 )(T2 − T1 )V1
2
2
Note that the work done along each path is different due to the different contributions from
the α (thermal expansion) term. Path “b” requires the least work; path “a” requires the
most.
4
Problem 5: Work and the Radiation Field
The differential of work is d/ W = −P dV and one immediately thinks about trying to express
P in terms of V in order to simplify the integral. However, along path “a” this is not
necessary: along one part dV = 0 and along the other the temperature, and hence the
pressure, is a constant.
Z 2
Z 2
W1→2 =
P dV −
P dV
1 T constant
1 V constant
}
{z
|
0
1
= − σT14
3
=
Z
2
1
1
dV = − σT14 (V2 − V1 )
3
1 4
V1
σT1 V2 ( − 1)
V2
3
Since the figure in the problem indicates that V1 > V2 , the underlined result is positive.
Along path “b” there are no shortcuts and we must prepare to carry out the integral. Since
d
/ W is expressed in terms of dV , we convert the T dependence of P into a function of V .
V T 3 = a constant ≡ V1 T13
T = (
V1 1/3
) T1
V
T 4 = T14 (
V1 4/3
V
) = T14 ( )−4/3
V
V1
Now we can carry out the integral over the path to obtain the total work done.
Z 2
Z 2
1
T 4 (V ) dV
W1→2 = −
P dV = − σ
3
1
1
Z 2
1
V
= − σT14
( )−4/3 dV
3
1 V1
h2
1
1 V
= − σT14 V1 − ( )−1/3
3
1/3 V1
1
V2 −1/3
4
= σT1 V1 ( )
−1
V1
V1 1/3
4
= σT1 V1 ( ) − 1
V2
Since V1 > V2 , this quantity is also positive.
5
Problem 6: Equation of State for a Ferromagnet
a) We are looking for the magnetization as a function of the field and the temperature,
M (H, T ), so we form the differential of M as follows.
∂M
∂M
dM =
dH +
dT
∂H T
∂T H
We are given the two coefficients in the expansion, but must make sure that their cross
derivatives are equal as is required for an exact differential.
∂
∂M
1
a
=
∂T ∂H T
Tc (1 − T /Tc )2
∂
∂M
1
f 0 (H)
=
Tc (1 − T /Tc )2
∂H ∂T H
The equality of these two expressions requires that f 0 (H) = a. Integration gives f (H) =
aH + c but we are told that f (0) = 0 so we know that c = 0. Thus
f (H) = aH
b) Now we must integrate the exact differential to find the state function M (H, T ).
Z ∂M
M (H, T ) =
dT + g(H)
∂T H
=
∂M
∂H
f (H)
+ M0 (1 − T /Tc )1/2 + g(H)
(1 − T /Tc )
=
f 0 (H)
+ g 0 (H)
(1 − T /Tc )
by calculation from above
=
a
+ 3bH 2
1 − T /Tc
as given
T
Now we set about finding an expression for g(H).
g 0 (H) = 3bH 2
g(H) = bH 3 + K
M (H = 0, T = Tc ) = 0 ⇒ K = 0
Now putting all the pieces together gives
M (H, T ) = M0 (1 − T /Tc )1/2 +
6
aH
+ bH 3
(1 − T /Tc )
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8.044 Statistical Physics I
Spring 2013
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