Introduction to Statistical Physics
Solution Manual
Kerson Huang
ii
Chapter 1
1.1
Mass of water =106 g, temperature raised by 20◦ C.
Heat needed Q = 2 × 107 cal = 8.37×107 J.=23.2 kwh.
Work needed = mgh = 14×150×29000 = 6.09×107 ft-lb =22.9 kwh.
1.2
Work done along various paths are as follows
ab:
Z b
Z b
Vb
dV
P dV = N kB T1
= N kB T1 ln
V
V
a
a
a
cd:
µ
¶
Vb
Pd (Vd − Vb ) = N kB T3 1 −
Vd
de:
N kB T3
Z
e
d
dV
Va
= N kB T3 ln
V
Vd
No work is done along bc and ea. The total work done is the sum of the
above. Heat absorbed equals total work done, since internal energy is unchanged
in a closed cycle.
1.3
(a)
α=
1 ∂V
bV0 T b−1
=
V ∂T
T0b V
(b)
∆V
=
bV0 T b−1
∆T
T0b
N kB T
N kB T0b 1−b
T
=
V
V0
Work done = P ∆V = bN kB ∆T
P
=
1
2
CHAPTER 1.
1.4
Consider an element of the column of gas, of unit cross section, and height
between z and z+dz. The weight of the element is −gdM , where dM is the mass
of the element: dM = mndz, where m is the molecular mass, and n = P/kB T
is the local density, with P the pressure. For equilibrium, the weight must
equal the pressure differential: dP = −gdM .Thus, dP/P = −(mg/kB T )dz. At
constant T , we have dp/P = dn/n.Therefore
n(z) = n(0)e−mgz/kB T
1.5
No change in internal energy, and no work is done. Therefore total heat
absorbed ∆Q = ∆Q1 + ∆Q2 = 0. That is, heat just pass from one body to the
other. Suppose the final temperature is T . Then
∆Q1 = C1 (T − T1 ), ∆Q2 = C2 (T − T2 ). Therefore
T =
C1 T1 + C2 T2
C1 + C2
1.6
R
Work done by the system is − HdM . Thus the work on the system is
Z
κ
HdM =
T
Z
HdH =
κH 2
2T
1.7
Consider the hysteresis cycle in the sense indicated in Fig.1.6. Solve for the
magnetic field:
H = ±H0 + tanh−1 (M/M0 )
R
( + for lower branch, − for upper branch.). Using W = − HdM , we obtain
W =−
Z
M0
−M0
dM [H0 + tanh−1 (M/M0 )] −
= −4M0 H0
1.8
Z
−M0
M0
dM [−H0 + tanh−1 (M/M0 )]
3
log M
A log log plot of mass vs. A is shown in the following graph. The dashed
line is a straightline for reference.
1
10
100
1000
A
.
10000
4
CHAPTER 1.
Chapter 2
2.1
Use the dQ equation with P, T as independent variables:
dQ = CP dT + [(∂U/∂P )T + P (∂V /∂P )T ]dP
For an ideal gas (∂U/∂P )T = 0, P (∂V /∂P )T = −V. Thus
dQ = CP dT − V dP.
The heat capacity is given by
C = CP − V (∂P/∂T )path .
The path is P = aV b , or equivalently P b+1 = a(N kB T )b by the equation of
state. Hence
V (∂P/∂T )path = [ab/(b + 1)]V (N kB T )b T −1 = bN kB /(b + 1). Therefore
C = CP −
b
N kB
b+1
This correctly reduces to CP for b = 0.
2.2
Use a Carnot engine to extracted energy from 1 gram of water between 300
K and 290 K.
Max efficiency η = 1 − (290/300) = 1/30.
W = ηC∆T =
1
(4.164 J g−1 K−1 × 1 g × 10 K) = 1.39 J
30
Gravitational potential energy = 1 g × 9.8 kg s−2 × 110 m = 1.08 J
2.3
The highest and lowest available temperatures are, 600 F = 588.7 K and 70
F = 294.3 K.
The efficiency of the power plant is W/Q1 = 0.6[1 − (294.3/588.7)] = 0.3.
In one second: W = 106 J.
So Q2 = 2.33 × 106 J = CV ∆T . Use CV = 4.184 J g−1 K−1 ,
5
6
CHAPTER 2.
Flow rate = 6000 ×(0.305m)3
∆T =
2.33 × 106 J
3
(4.184 J g−1 K−2 )6000(0.305 m) 106 cm2 /m3
= 3.27 × 10−3 K
2.4
(a)
Since water is incompressible, a unit mass input gives a unit mass output.
The net heat supplied per unit mass is ∆Q = C(T1 − T ) − C(T − T2 ),
where C is the specific heat of water (per unit mass.) In steady state v 2 /2 =
∆Q. This gives
p
p
v = 2∆Q = 2C(T1 + T2 − 2T )
(b)
The entropy depends on the temperature like ln T . A unit volume of water
from each of the input streams has total entropy ln T1 + ln T2 This makes two
unit volumes in¡the output
2 ln T . Therefore the change
¢ stream, with entropy
√
in entropy is ln T 2 /T1 T2 ≥ 0. Thus T ≥ T1 T2 , and
p ¯¯
√ ¯¯p
vmax = 2C ¯ T1 − T2 ¯
2.5
(a)
P V1γ = 2P0 V0γ ,£ P V2γ = 2\P0 V0γ ¤
γ
(V1 /V2 )γ = 2. (L̄ + a)/(L − a) = 2.
a
21/γ − 1
= 1/γ
L
2
+1
(b)
∆U = ∆Q − W , ∆Q = 0.
CV ∆T = −W , ∆T = −W/CV .
T1 = 2T0 + ∆T = 2T0 − (W/CV ), T2 = T0 − ∆T = T0 + (W/CV ).
P =
R [2T0 − (W/CV )]
RT1
=
V1
A (L + a)
(c)
Ra
W = A 0 dx(P1 − P2 )
γ
γ
P1 = 2P0 V0γ / [A(L + x)] , P2 = P0 V0γ / [A(L − x)] .
µ
¶
2a
P0 V0 ³
a ´
1−
W =
1 − −γ
γ−1
L
L
7
2.6
(a)
P V = U/3, U = σV T 4 .
P = σT 4 /3.
dS = dQ/T = (dU + P dV )/T.
Integrate along paths with T =const, V =const.
S=
4
σV T 3.
3
(b)
S =Constant. ∴ T 3 ∼ V −1. Thus
T ∼ R−1
2.7
The heat absorbed by an ideal gas in an isothermal process is
∆Q = N kT ln(Vf /Vi )
where Vf and Vi are respectively the final and initial volume.The temperature
T in this formula is the ideal-gas temperature.
Draw a Carnot cycle on the P V diagram, and label the corners 1234 clockwise from the upper left.
The heat absorbed at the upper temperature T2 , and the heat rejected at
the lower temperature T1 , are
Q2 = N kT2 ln(V2 /V1 )
Q1 = N kT1 ln(V3 /V4 )
Because 23 and 12 lie on adiabatic lines, we have
V2 T2γ−1 = V3 T1γ−1
V1 T2γ−1 = V4 T1γ−1
Dividing one equation by the other yields V2 /V1 = V3 /V4 .
The efficiency of the cycle is therefore
η =1−
Q1
T1
=1−
Q2
T2
2.8
Diesel cycle:
Q2 = CP (T3 − T2 )
Q1 = CV (T4 − T1 )
η = 1 − (Q1 /Q2 ) = 1 − γ −1 [(T4 − T1 )/(T3 − T2 )]
We have P3 = P2 , hence
T3 /T2 = V3 /V2 = rc
The processes 12 and 34 are adiabatic, with T V γ−1 = constant. V4 = V1 .
Thus
8
CHAPTER 2.
T3 V3γ−1 = T4 V1γ−1
T2 V2γ−1 = T1 V1γ−1
Using the three relations derived, we obtain
η =1−
1
rcγ − 1
γ rγ−1 (rc − 1)
2.9
Otto cycle:
Q2 = CV (T3 − T2 )
Q1 = CV (T4 − T1 )
η = 1 − (Q1 /Q2 ) = 1 − [(T4 − T1 )/(T3 − T2 )]
The processes 12 and 34 are adiabatic, with T V γ−1 = constant. We have
V4 = V1 , V3 = V2 Thus
T1 V1γ−1 = T2 V2γ−1 .
T3 V2γ−1 = T4 V1γ−1 .
Taking the ratio of these equations, we have
T2 /T1 = T3 /T4 = rγ−1 .
Thus
η = 1 − r1−γ
2.10
First note Tb /Ta = Vb /Va = 2.
a→b
b→c
c→a
Work done
Pa (Vb − Va ) = Pa Va = N kTa
0 R
− P dV = −N kTa ln 2
Heat absorbed
CP ∆T = CP Ta
−CV Ta
−N kTa ln 2
W (Net work done) = N kTa (1 − ln 2)
Q2 (Heat absorbed) = CP Ta = 52 N kTa
η=
W
2
= (1 − ln 2) = 0.12
Q2
5
In comparison, η Carnot = 1 − (Tb /Ta ) = 0.5.
2.11
First note T2 = 4T1 . The P, V, T for the points A, B, C, D are as follows:
A
B
C
D
P
P1
2P1
2P1
P1
V
V1 = N kT1 /P1
2V1
V1
2V1
T
T1
4T1
2T1
2T1
9
(a)
Heat supplied along
¡
¢
ACB : CV T1 + CP (2T1 ) = ¡ 32 + 5 ¢ N kB T1 = 13
2 N kB T1 .
ADB : CP T1 + CV (2T1 ) = 52 + 3 N kB T1 = 11
2 N kB T1 .
AB : ∆U + ∆W = 32 N kB (2T1 ) + 32 P1 V1 = 6N kB T1 .
(b)
Heat capacity = ∆Q/∆T = 6N kB T1 /3T1 = 2N kB .
(c)
Work done = P1 V1 = N kB T1 . Heat absorbed = Heat absorbed along ACB
= (13/2)N kB T1 .
2
η=
13
2.12
(a)
Since no work is being done, and the temperatures diverge, heat must be
transferred from the colder body to the hotter body, with no other effect, and
this violates the Clausius statement of the second law.
(b)
The assertion is not true for physical black bodies, because they cannot be
point-like but have finite size. Even if the two bodies have identical shapes,
their optical images are not reciprocal. That is, the radiation from one body
may form an image that is larger than the other body, and thus not completely
absorbed by the other body.
10
CHAPTER 2.
Chapter 3
3.1
(a)
For a adiabatic process dS = 0, and the T dS equations give
CV dT = −(αT /κT )dV
CP dT = αT V dP
Dividing one by the other, we obtain
CP /CV = κT [−V (∂P/∂V )S ] = κT /κS
(b)
CV dT + (αT /κT )dV = CP dT − αT V dP T. Put
dT = (∂T /∂P )V dP + (∂T /∂V )P dV .
Equate the coefficients of dP and dV on both sides. One of them gives
CP − CV = (αT V /κT )(∂V /∂T )P = α2 T V /κT .
(c)
Using U = A + T S, H = G + T S (enthalpy), we have
CV
= (∂U/∂T )V = (∂A/∂T )V + S + T (∂S/∂T )V = T (∂S/∂T )V
= −T (∂ 2 A/∂T 2 )V
CP
= (∂H/∂T )P = (∂G/∂T )P + S + T (∂S/∂T )P = T (∂S/∂T )P
= −T (∂ 2 G/∂T 2 )P
3.2
The Sacker-Tetrode equation is
p
S = N kB [(5/2) − ln(nλ3 )], where n = N/V , and λ = 2π~2 /mkB T .
(a)
A = U − T S = (3/2)k
/ B T − T S = N kB T ln(nλ3 ) − N kB T.
G = A + P V = N kB T ln(nλ3 ).
(b)
Write ln(nλ3 ) = ln n + ln λ3 . The second term is a function of T only.
µ = (∂A/∂N )V,T = kB T ln(nλ3 )+N kB T (∂ ln n/∂N )V,T −kB T = kB T ln(nλ3 ).
µ = (∂G/∂N )P,T = kB T ln(nλ3 ) + N kB T (∂ ln n/∂N )P,T = kB T ln(nλ3 ).
11
12
CHAPTER 3.
3.3
The force on the bead is (P − Pa )A − mg, where
P = pressure in gas,
Pa = 1 atm.
The equation of motion for the displacement x is mẍ =(P − Pa )A − mg.
In equilibrium the pressure in the gas is P0 = Pa + (mg/A).
The volume is V0 = RT /P0 .
Assume adiabatic oscillations: P V γ = const.
This implies dP = −γ(P/V )dV ≈ −γ(P0 /V0 )Ax.
P = P0 + dP¡ ≈ P0 − γ(P¢0 /V0 )Ax.
Thus mẍ + γA2 P02 /RT x = 0.
The frequency of oscillations is
p
ω = AP0 γ/RT
3.4
Let the equilibrium pressure and temperature be P0 , T0 . Under an infinitesimal displacement x, suppose the pressure of compartment 1 changes by
dP . Since the process is adiabatic, we have P V γ = constant, or (dP/P ) +
γ(dV /V ) = 0. In terms of the temperature, we have T V γ−1 = constant, or
(dT /T ) + (γ − 1)(dV /V ) = 0.
(a)
For compartment 1, we have to first order
γP0 x
L
(γ − 1)T0 x
dT = −
L
dP = −
For compartment 2, replace x by −x.
(b)
The force acting on the piston is dF = AdP . The equation of motion for x
is dF = M ẍ, where M is the mass of the piston. Thus ẍ + (γAP0 /M L)x = 0,
and the frequency of small oscillations is
p
ω = γAP0 /M L
(c)
Due to the finite thermal conductivity of the piston, heat flows back and
forth between the two compartment, because of the oscillation in the temperature difference.Assume that the temperatures change so slowly that at any
moment we regard heat conduction as taking place between two heat reservoirs
of fixed temperatures. When an amount of heat dQ flows from 1 to 2, the
entropy increase is dS = (dQ/T2 ) − (dQ/T1 ). Thus
dS
=
dt
µ
1
1
−
T2 T1
¶
dQ
kB (∆T )2
≈ kB
=
dt
T1 T2
µ
∆T
T0
¶2
13
The temperature difference is
(∆T )2 = (T1 − T2 )2 = (2dT )2 =
Hence
4(γ − 1)2 T02 x2
L2
dS
= ax2
dt
where a = 4kB (γ − 1)2 /L2 .
(d)
Energy dissipation, which has so far been ignored, occurs at the rate T0 dS/dt =
aT0 x2 . The time average of this rate is 12 aT0 x20 , where x0 is the amplitude of
oscillation. The energy of oscillation is E = 12 M ω 2 x20 . In one period of oscillation, the energy dissipated is ∆E = (2π/ω) 12 aT0 x20 . This gives a fractional
dissipation per cycle
∆E
2πT0
=
E
aM ω 3
3.5
(a)
P =−
µ
∂A
∂V
¶
T
= a0 (v0 − v)
(b)
κT = −v −1 (∂v/∂P )T = (a0 v)−1
α = v −1 (∂v/∂T )P = −v −1 (∂P/∂T )V (∂v/∂P )T , by chain rule.
α=
1 da0
a0 v dT
¶
= a0 (v02 − v 2 ) − f
(c)
µ=
µ
∂A
∂N
V,T
3.6
For this problem it is important to use the entropy expression with arbitrary
CV , instead of setting it to (3/2)kB . Write the adiabatic condition as
∆S = ∆S1 + ∆S2 = 0, or
(N1 + N2 )kB ln(Vf /Vi ) + (N1 CV 1 + N2 CV 2 ) ln(Tf /Ti ) = 0.
Thus, Tf /Ti = (Vi /Vf )ς ,where ζ = kB (N1 + N2 )/(N1 CV 1 + N2 CV 2 ).
This means T V ζ = constant. Putting T = P V /N kB T , where N = N1 +N2 T,
we have
P V ξ = constant
where
ξ =ζ +1=
n1 CP 1 + n2 CP 2
N1 (CV 1 + kB ) + N2 (CV 2 + kB )
=
N1 CV 1 + N2 CV 2
n1 CV 1 + n2 CV 2
14
CHAPTER 3.
3.7
(a)
Since the disks are thin, we can assume that their temperatures always
remain uniform.
Let the final temperature be T .
The changes in temperatures are respectively ∆T1 = T − T1 , ∆T2 = T − T2 .
For simplicity write CP 1 = C1 , CP 2 = C2 .
The amounts of heat absorbed are respectively ∆Q1 = C1 ∆T1 , ∆Q2 =
C2 ∆T2 .
Since the system is isolated ∆Q1 + ∆Q2 = 0. This gives
T =
C1 T1 + C2 T2
C1 + C2
(b)
Consider the instant when the two temperatures are T20 , T10 , (T20 > T10 ).
When an amount of heat dQ flows from 2 to 1, the entropy increase is
dS = (dQ/T10 ) − (dQ/T20 ).
We can express dQ in terms of the dT ’ through dQ = C1 dT10 = −C2 dT20 .
Thus we can rewrite dS = C1 (dT10 /T10 ) + C2 (dT20 /T20 ).
∆S = C1
Z
T
T1
dT10
+ C2
T10
Z
T
T2
T
T
dT20
= C1 ln
+ C2 ln
0
T2
T1
T2
3.8
The relations are straightforward mappings from a P V system to a magnetic
system.
3.9
(a)
The desired expression are straightforward mappings of those for a P V system.
(b)
The first relation is the condition that dA be an exact differential. The
second is obtained by using the equation of state M = κH/T .
(c)
The chain rule states (∂T /∂H)S (∂H/∂S)T (∂S/∂T )H = −1.
From (b) we have (∂H/∂S)T = −T 2 /(κH).
By definition, the heat absorbed at constant H is given by T dS = CH dT .
Thus (∂S/∂T )H = CH /T.
3.10
(a)
The important property to verify is that at constant T the entropy decreases
as the magnetic field H increases.
(b)
15
Isothermal magnetization: dT = 0.
The heat absorbed is
dQ = CM dT − HdM = −HdM . Therefore
∆Q = −
Z
0
H
HdM = −
κH 2
2T0
(c)
Adiabatic cooling: dQ = 0.
From dQ = CM dT − HdM
we¢ obtain
¡
dT = (H/CM ) dM = κ/aT 2 M dM . Multiply both sides by T 2 and integrate:
R0
R T1 2
T dT = (κ/a) M M dM.
T0
This gives T13 = T03 − (κ/2a) M 2 , or
T13 = T03 −
κ3 H 2
2aT02
This becomes negative when the magnetic field H is sufficiently large. However,
the equation becomes invalid long before that happens, for it is based on Curie’s
law, which is valid only for weak fields.
16
CHAPTER 3.
Chapter 4
4.1
The system is in contact with a heat reservoir, but initially not in equilibrium
with it. Let the stages of the process be labeled A,B,C:. We first calculate the
heat absorbed ∆Q, and the entropy change ∆S of the system.
(A) Water cools from 20◦ C to 0◦ C.
∆Q = RCP ∆T = −10 R× 4180 × 20 J = −8.36 × 105 J.
∆S = dQ/T = CP dT /T = CP ln(Tf /Ti ) = 41800 ln(273/293) = −2.96×
103 J/deg.
(B) Solidification at 0◦ C.
∆Q = −10 × 3.34 × 105 J =.−3.34 × 106 J.
∆S = ∆Q/T = −3.34 × 106 /273 = −1.22 × 104 J/deg.
(C) Ice cools from 0◦ C to -10◦ C.
∆Q = CP0 ∆T = −10 × 2090 × 10 J = −2.09 × 105 J.
∆S = CP0 ln(Tf /Ti ) = 20900 ln(263/273) = −7.80 × 102 J/deg.
Total heat absorbed by system: ∆Qsys = −4.39 × 106 J
Total entropy change of system: ∆Ssys = −1.39 × 104 J/deg.
The reservoir has a fixed temperature T0 = −10◦ C..
The total heat absorbed by reservoir equals that rejected by the system:
∆Qres = 4.39 × 106 J.
Entropy change of reservoir:
∆Sres = ∆Qres /T0 = 4.39 × 106 /263 = 1.67 × 104 J/deg.
∆Suniverse = ∆Sres + ∆Sres = 2.8 × 103 J/deg
4.2
Let P0 , T0 be the pressure and absolute temperature at the triple point. Let
L be the extensive latent heat (not specific latent heat.) Since the solid-gas
17
18
CHAPTER 4.
transition can be made either via a direct path or a solid-liquid-gas path, we
must have
Lsublimation = Lmelt + Lvap
Vaporization: dP/dT ≈ Lvap /T V = P Lvap /N kB T 2 .
·
µ
¶¸
Lvap
T0
1−
P = P0 exp
N kB T0
T
Melting: dP/dT = Lvap /T ∆V.
P = P0 +
Lmelt
T
ln
∆V
T0
Sublimation: dP/dT ≈ P (Lvap + Lmelt )/N kB T 2 .
·
µ
¶¸
Lvap + Lmelt
T0
P = P0 exp
1−
N kB T0
T
4.3
dP/dT = /T ∆v = [1.44 J/(18 − 20)cm3 ]T −1 .
∴ dT /dP = −c0 T , where c0 = 1.39 cm3 /J.
4.4
(a)
At a given v > v0 , the dashed line lies at a lower free energy than the
solid line. The latter represents a “stretched” that fills the whole volume. The
former represent a liquid drop at specific volume v0 that does not fill up the
entire volume. This is therefore the preferred state of the liquid. At v = v0 the
pressure is zero.
(b)
Now assume that the liquid coexists with its vapor, treated as an ideal gas.
We are in the transition region of a first-order phase transition. At the given
temperature, the liquid and gas have fixed densities, which must be consistent
with the requirement of equal pressure P and chemical potential µ. Denote
quantities for the liquid with subscript 1, and those for the vapor with subscript
2:
P1 = a0 (v0 − v),
µ1 = a0 (v02 − v 2 ) − f,
P2 = nkB T,
µ2 = kB T ln(nλ3 ).
where
p P1 , µ1 were obtained in Prob.3.5, and µ2 was given in Prob.3.2, with
λ = 2π~2 /mkB T . Thus, the conditions determining v and n are
a0 (v0 − v) = nkB T
a0 (v02 − v 2 ) − f = kB T ln(nλ3 )
19
From the first equation, we see that v0 − v > 0. It approaches zero as nT → 0.
(c)
Small n corresponds to (v0 − v) → 0. The second equation becomes −f ≈
kB T ln(nλ3 ). Thus
nλ3 ≈ exp(−f /kB T )
4.5
(a)
dP/dT = /[T (v2 − v1 )] ≈ /T v2 = /[T (kB T /P )]. Hence
T dP
=
P dT
kB T
(b)
T (K)
0.2
0.4
0.6
0.8
1.0
1.2
(ergs/g)
8.21×107
9.37
10.5
11.8
13.1
14.4
4.6
R
The accompanying sketch shows G = V dP . The system skips the closed
loop in the graph of G, because it is higher than need be
V
A
A
P
G
P
20
CHAPTER 4.
4.7
(a)
A(V, T ) = −RT ln(V − b) − (a/V ) + f (T )
As V → ∞, A(V, T ) → −RT ln V + f (T )
This should approach the ideal gas result (Prob.3.2) RT [ln(nλ3 ) − 1].
Therefore, up to an additive constant,
µ
¶
3
f (T ) = −RT 1 + ln T
2
(b)
CV = −T (d2 f /dt2 ) = (3/2)R, which is a constant.
4.8
T dS = CV dT + T (∂P/∂T )V dV = 0.
dT /dV = −(T /CV ) (∂P/∂T )V = −(RT /CV )(V − b)−1 .
Integrating this yields
ln T = −(R/CV ) ln(V − b)+ constant.
Thus the adiabatic condition is
T (V − b)R/CV = constant
When a = b = 0, the system reduces to an ideal gas, for which
R
CP − CV
=
=γ−1
CV
CV
Thus we recover T V γ−1 = constant.
4.9
The second virial coefficient for the van der Waal gas is given by c2 =
b − (a/RT ). A rough fit is
b ≈ 17 cm3 /mole
a ≈ 2100R deg cm3 / mole
4.10
Let ∆V = V1 − V2 be the difference in volume across the transition line.
Consider variations along the transition line, such as going from a to b, as
illustrated in the sketch. The chain rule says
µ
µ
¶ µ
¶
¶
∂∆V
∂T
∂P
= −1
∂T P ∂P ∆V ∂∆V T
This gives
µ
∂P
∂T
¶
∆V
=−
(∂∆V /∂T )P
α1 − α2
=
(∂∆V /∂P )T
κT 1 − κT 2
21
If the transition line refers to a second-order phase transition, then across this
line ∆V = 0, while the differences in α and κ are nonzero. Thus
∆α
dP
=
dT
∆κT
22
CHAPTER 4.
Chapter 5
5.1
n = 2.70 × 1019 atoms /cm3
v = 2 × 105 cm/s.
N = nv/6 ≈ 1024 s−1 cm−2 .
5.2
Let tV0 be the volume of the room, and V be the volume under consideration
The probability of finding an atom in V is V /V0 .
The probability of finding it elsewhere is 1 − (V /V0 ).
Since there are N independent atoms, the probability of finding none in V
is
µ
µ
¶N
¶¶
µ
V
V
= exp N ln 1 −
p= 1−
V0
V0
For small V /V0 we can use the expansion ln (1 − (V /V0 )) ≈ −V /V0 . Thus
p ≈ exp(−N V /V0 )
Under STP,
N=
V0
× (6.02 × 1023 mole−1 )
22.4 liter mole−1
For V0 = 27 × 103 liter,we have
1 cm3 :
for V
=
for V
= 1 A3 :
¡
¢
19
p ≈ exp −2.7 × 1019 ≈ 10−10
¡
¢
p ≈ exp −2.7 × 10−5 ≈ 1 − 2.7 × 10−5 = 0.99997
5.3
Let n = N/V
Probability of finding one atom in dV = ndV.
Probability of finding no atom in dV = 1 − ndV.
Probability of finding no atom in V = exp(−nV ).
23
24
CHAPTER 5.
p(r)dr = Prob.(one atom between r, r + dr)×Prob.(no atom in sphere of
radius r)
µ
¶
4
2
3
p(r) = 4πnr exp − πnr
3
5.4
For the beam to remain well-collimated, the atoms should suffer no scattering
by the air in the chamber along the flight path of length L. The condition is
therefore λ > L, where λ ≈ (nσ)−1 is the mean-free-path, where n is the density
of the air, and σ is the cross section for a collision between atoms in the beam
with an air molecule. Thus
1
n<
Lσ
For a rough estimate, take σ ≈ 10−16 cm2 . This gives n < 10−15 cm−3 . The
estimate can be refined by using a more precise value for σ.
5.5
(a)
The mass density of water is 1 g cm−3 . This corresponds to a number density
n = 2 × 1023 cm−3 . Thus λ = 5 × 1016 cm.
(b)
The rate of reaction is R = N Iσ, where N is the number of nucleons, I is
the neutrino flux, and σ is the reaction cross section. A person of mass 150,kg
contains N = 1029 nucleons. Thus R = 5 × 10−10 s−1 .
The collision time is τ = R−1 = 2 × 109 s ≈ 70 yrs.
Thus, one gets hit by a neutrino about once in a lifetime.
5.6
Following the hint, the answer is obtained straightforwardly:
Cn =
2πn/2
n n n
n
¡n
¢ −→
ln π − ln +
n→∞
2
2 2
2
Γ 2 +1
5.7
(a)
From (5.37), Γ(E, V ) = (∂Φ/∂E)∆, where
Z
N
Φ=V
dp1 · · · dpn
p21 +...+p2n <E
with n = 3N . Thus
Γ(E, V ) = K0 V n Σn
³√ ´
E = K0 V n nCn E (n−1)/2
25
(b) Using S = kB ln Γ, we have, up to an additive constant,
µ
¶
ln N
S(E)
ln C3N
3/2
= ln V +
+O
+ ln E
N kB
N
N
µ
¶
³
´
ln
N
= ln V E 3/2 + O
N
5.8
(a)
³√ ´
By the same reasoning as in the last problem, we obtain Γ(E, V ) = K0 Σn
E ,
where n = 6N . There is no volume dependence in the limit V → ∞, because
the particles are confined by the harmonic oscillator potential.
(b) Transcribing the result of the last problem, we have
µ
¶
S(E)
ln N
3
= ln E + O
N kB
N
5.9
Let the mean-free-path be λ ≈ 10−5 cm. To be away from the origin by a
distance L, a total of (L/λ)2 random steps would have to be taken. Since each
2
step lasts a collision time τ ≈ 10−10 s, the total time required is τ (L/λ) . For
L = 1 cm the time is:1 sec. For L = 1 m the time is 104 sec.
5.10
−1/2
For one coordinate, the probability of return after n collisions is (2πn)
,
according to (5.16). For the N -particle state to recur, all 6N coordinates have
to return at the same time. When this happens, different particles woud have
made different numbers of collisons n. For our order-of-magnitude estimate, we
can imagine that all particles have made an average numbers of collisions n̄,
each with probability p = (2πn̄)−1/2 which is a small but finite number. The
probability for gas as a whole to return to the initial state is then p6N . That is,
µ
¶
1
Recurrence time ≈ exp 6N ln
p
¡
¢
in units of the collision time. For N ∼ 1019 , this number is of order exp 1020 ,
which is so large that neither the value of p nor the units used makes any
significant difference.
26
CHAPTER 5.
Chapter 6
6.1
Let λ = (2mkB T )−1 .
R∞
R 3
d p ε f (p)
3
1 0 dpp4 exp(−λp2 )
R
h i= R 3
= kB T
=
2m 0∞ dpp2 exp(−λp2 )
2
d p f (p)
R∞
R 3 2
­ 2®
d p ε f (p)
15
1 0 dpp6 exp(−λp2 )
R∞
=
= R 3
=
(kB T )2
2
2
2
4m 0 dpp exp(−λp )
4
d p f (p)
­ 2®
3
− h i2 = (kB T )2
2
6.2
The energy distribution is defined through P (E)dE = f (p)4πp2 dp, where
f (p) is the Maxwell-Boltzmann distribution of momentum. Using E = p2 /2m,
we obtain
√
P (E) = c0 Ee−E/kB T
where c0 = nπ−1/2 (kB T )−3/2 .
6.3
The density is obtained by integrating the distribution function over the
momentum. The result is
n(z) = n(0)e−mgz/kB T
6.4
Using the equation of state of the ideal gas, we obtain P (1−γ)/γ T = C0 .After
some manipulation this leads to
dP
γ dT
mg
γ dT
=
−
dz =
P
γ−1 T
kB T
γ−1 T
27
28
CHAPTER 6.
Thus T changes with height z according to
kB
γ−1
dT
=−
mg
dz
γ
This can be integrated to yield
kB T (z) = kB T0 −
γ−1
mgz
γ
For T0 = 300 K and γ = 7/5, the temperatures becomes zero at z = 3.17×104 m.
(b)
From the above, we find
mg
dP
=−
dz
P
kB T
Using the expression for from the last part, we can integrate this to obtain
P
=
P0
µ
¶γ/(γ−1)
γ − 1 mgz
1−
γ kB T0
6.5
There is an effective temperature-dependent potential U (x), given through
exp(−U/kB T ) = c0 (1 + γx).
6.6
The answer is
¤
£
n1 (r)
= exp ω 2 r2 (m1 − m2 )/2kB T
n2 (r)
6.7
(a)
The most probable velocity is that at the maximum of the speed distribution.
This will be obtained in (c).
(b)
The pressure is given by
Z
1
p2
d p2px vx f (p) =
f (p)
d3 p p
P =
3
p2 + m2
px >0
p
where we have used vx = px / p2 + m2 . Write
Z
3
p
p2
m2
p
= p2 + m2 − p
p2 + m2
p2 + m2
29
The second can be neglected in theRultra-relativistic
limit p2 À m2 . Comparing
p
3
2
2
P with the energy density U/V = d p p + m f (p), we obtain
PV →
1
U
3
(c)
The velocity distribution f (v) is defined by
³
´
p
f (v)d3 v = C exp −βmv/ 1 − v 2 d3 p
√
Using p = mv/ 1 − v 2 , we obtain
µ
¶
Cm3
m
√
f (v) =
exp −
(1 − v 2 )5/2
kB T 1 − v 2
The non-relativistic limit corresponds to m(1 − v 2 )−1/2 ≈ 12 mv 2 + m + O(v 2 )
Now return to part (a). The most probable velocity v0 corresponds to the
maximum of the speed distribution 4πv 2 f (v). It is given by the root of the
equation
5
m 2
(1 − v 2 )3/2 + v 2 (1 − v 2 )1/2 −
v =0
2
2kB T
The non-relativistic and ultra-relativistic limits are (with c restored)
r
2kB T
v0
(kB T ¿ mc2 )
≈
c
mc2
µ
¶2
mc2
v0
(kB T À mc2 )
≈ 1−
c
5kB T
(d)
Relativistic effects become noticeable when kB T /mc2 is appreciable, say, at
10%. For H2 this corresponds to kB T = 0.1 × 2 GeV, or T = 2 × 1012 K.
6.8
(a)
¢
¡
The distribution is proportional to the velocity distribution exp −mvx2 /2kB T .
Substitute vx = c (f − f0 ) /f0 and then normalize the distribution. The result
is
¶−1/2
¶
µ
µ
mc2
2πkB T f0
2
exp
−
(f
−
f
)
P (f ) =
0
mc2
2kB T f02
(b)
The variance is
(f − f0 )2 =
R∞
0
³
´
2
2
df (f − f0 ) exp −λ (f − f0 )
³
´
R∞
2
df
exp
−λ
(f
−
f
)
0
0
30
CHAPTER 6.
√
of integration
to ν = λ(f −f0 ).
where λ = mc2 /(2kB T f02 ). Change the variable
√
p
The lower limit of integration becomes − λf0 = − mc2 /2kB T . This can be
replaced by −∞ when kB T ¿ mc2 , which is true in usual laboratory conditions.
We then obtain
kB T 2
(f − f0 )2 =
f
mc2 0
(c)
The line width
√ is given by the square root of the variance, and thus inversely
proportionalpto m. The H2 line width is therefore broader than that of O2
by a factor 32/2 = 4.
6.9
(a)
f (p) ∝ e−βcp ,
U/N = cp̄ = 3kB T,
CV = 3N kB .
(b)
P V = 13 U = N kB T.
6.10
Follow the hints and directions given in the problem.
6.11
W
=
Z
3
d pvx f (p) = C
vx >v0
r
= n
Z
∞
−λp2x
dpx vx e
mv0
µ
¶
v02
kB T
exp −
2πm
2mkB T
·Z
∞
−∞
¸
¡
¢ 2
2
dpy exp −λpy
6.12
p
≈ 104 m/s. This is to be compared
(a) The escape velocity is vc = 2GM/R p
with the most probable speed at STP v0 = 2kB T /m ≈ 2.2 × 103 m/s. The
fraction of gas that can escape is
Z ∞
Z
2
C ∞
4
2 −p2 /2mkB T
f=
dp4πp e
=√
dxx2 e−x
n mvc
π y
where C = n(2πmkB T )−3/2 and y = vc /v0 . Using the results of Prob.6.10(b),
we obtain
2
2y
f ≈ √ e−y
π
With y ≈ 4.5, we find f ≈ 5 × 10−8 .
(b)
31
The time it takes for an atom to go from sea level to the top of the atmosphere
through random collisions is
t≈
L2
= 3 × 1011 s ≈ 104 yr
λvc
where L = height of atmosphere ≈100 km, λ = mean-free-path ≈3×10−7 m.
6.13
(a)
The number of atoms with momentum magnitude between p and p + dp
is V 4πp2 f (p)dp, where V is the volume, and f (p) is the Maxwell-Boltzmann
distribution. Thus
∆N
∆E
= 4πV
= 4πV
Z
∞
p0
Z ∞
dpp2 f (p)
dpp2
p0
p2
f (p)
2m
Using the results of Prob 6.10(b), and .N = nV , and E = 32 N kB T , we obtain
the fractional changes
∆N
N
∆E
E
= 2πy 1/2 e−y
=
4π 3/2 −y
y e
3
for y = 0 /kB T À 1..
(b)
From kB T = 23 E/N , we obtain kB ∆T = 23 [(∆E/N ) − E(∆N/N 2 )], hence
µ
¶
2
∆E
∆N
∆T
1/2
=
−
≈ 2πy
y − 1 e−y
T
E
N
3
Taking the logarithm of the equation for ∆N/N,we have ln(N/∆N ) = y −
ln(2πy 1/2 ), which gives
³
´
y ≈ ln(N/∆N ) + ln 2π ln(N/∆N )1/2 = ln(πN/∆N ) + ln ln(N/∆N )
We then find, to leading order,
∆T
4 ∆N
≈
T
3 N
r
ln
∆N
N
32
CHAPTER 6.
6.14
(a)
Let the axis along the needle be labeled 1, and a perpendicular axis 2.
The moments of inertia about these axes are I1 , I2 , with I2 À I1 . By the
equipartition of energy we have
J12
J2
kB T
= 2 =
2I1
2I2
2
where Ji is the components of angular momentum along the axis i. Thus
r
I2
J2
=
À1
J1
I1
That is, the angular momentum is nearly parallel to the axis of the needle.
(b)
The equipartition of energy states
1
1
CV 2 = kB T
2
2
This gives
p
V 2 = 6.5 µV .
6.15
(a)
Take 1 mole of N2 . The mass is 28 g.
For v = 7 km/s, the kinetic energy is
K.E. = 12 M v 2 = 0.5(0.028)(7000)2 = 686 kJ.
When this energy is converted into heat, the temperature rise is of the order
of
∆T =K.E./kB = 5500 K.
Thus, the astronauts would be fried.
(b)
A constant deceleration a is equivalent to the application of a potential mxa,
where x is the distance, and m is the mass of an air molecule. The Boltzmann
factor gives a relative density distribution
µ
¶
∆n
mxa
= exp −
n
kB T
which equals the fractional change in pressure ∆P/P at constant temperature.
The difference in pressure between the points x = 0 and x = L is therefore
∆P = P0 [(1 − exp(−mxL/kB T )].
(c)
Let the total mass of air be M = N a, where N is the total number of air
molecules. The force is
F = A∆P =
AP0 maL
P0 V
=
N ma = N ma = M a
kB T
N kB T
33
The stopping time is t = v/a, which corresponds to a distance
The work done is therefore
W =
1 2
2 at
= v 2 /2a.
1
F v2
= M v2
2a
2
Thus the translational kinetic energy is completely converted to mechanical
work, and ∆T = 0..
(d) The translational velocity of the air must be, at all times, much smaller
than sound velocity, relative to the walls of the container.
34
CHAPTER 6.
Chapter 7
7.1
R
Particle flux : IN = vx >0 d3 pvx f (p)
R
Energy flux: IE = vx >0 d3 pvx f (p)mv 2 /2
Average energy of an escaped particle
¡
¢
R∞
IE
m 0 dvv 5 exp −mv 2 /2kB T
R
=
= 2kB T
IN
2 0∞ dvv 3 exp (−mv 2 /2kB T )
Thus, the escaped atoms come to thermal equilibrium at a temperature T1 , with
3
4
2 kB T1 = 2kB Ṫ . Hence T1 = 3 T . This assumes that the total amount of gas
escaped is so small that the temperature of the original system is unchanged.
7.2
Let
p n1 , n2 denote the densities of U-238 and U-235 respectively. Flux
= n kB T /2πm.
After one stage of effusion,
µ ¶ r
n1
n1
m2
=
n2
n2 0 m1
After k stages,
µ
n1
n2
¶
Find k such that n1 = n2 .
k=
=
k
µ
n1
n2
¶ µ
0
m2
m1
¶k/2
2 ln (n1 /n2 )0
2 ln (99.27/0.75)
=
= 775
ln (m1 /m2 )
ln (238/235)
7.3
In an adiabatic process P V γ = constant. Using the equation of state, we
find that P (N kB T /P )γ = constant, or P 1−γ T γ = constant. Differentiating this
relation with respect to P we obtain
µ
¶
∂T
γ−1T
=
∂P S
γ P
35
36
CHAPTER 7.
The particle density is n = P/kB T . Thus
µ
¶
∂n
1
=
∂P S
kB T γ
Hence
1
nkB T γ
r
kB T γ
c =
m
κS
=
7.4
Test the condition K/c ¿ CP . Using the data given, we find K/c = 1.44 ×
10−5 , CP = 0.24T, in the mixed unit given. Thus the condition is well-fulfilled,
and shows that sound propagated adiabatically.
7.5
From (7.19) ∂ 2 ρ/∂t2 + ρ∇ · ∂u/∂t = 0. Instead of the Euler equation
ρ∂u/∂t = −∇P , use the Navier-Stokes equation.(7.48). Then in first-order
approximation (7.21) is replaced by
.∂ 2 ρ
4ν 2
− ∇2 P +
∇ (∇ · u) =0
2
∂t
3
where ν is the viscosity. Use the continuity equation ρ∇ · u = − ∂ρ/∂t, and
convert ∇2 P to ∇2 ρ as in (7.22). The result is
1
4ν 2 ∂ρ
.∂ 2 ρ
− 2 ∇2 ρ −
∇
=0
∂t2
c
3ρ
∂t
For a sinusoidal wave ρ = ρ0 + ρ1 exp(ikx − iωt), the last term is i(4νkω/3ρ0 )ρ1 .
Thus the damping coefficient is 4νkω/3ρ0 .
7.6
The one-dimensional diffusion equation has solution
µ
¶
x2
N
exp −
n(x, t) = √
4πDt
4πDt
The gas is characterized by the diffusion constant D. Suppose the detector has
spatial resolution ∆x. We want to find the time t at which n(L, t)∆x = 1. That
leads to the implicit equation
t=
1
L2
³
´
4πD ln N ∆x/√4πDt
37
In the first approximation, we put t = L2 /4πD on the right side. This gives
t=
1
L2
4πD ln (N ∆x/L)
The logarithm is not very sensitive to ∆x.
7.7
The insulating power
η is the inverse of the coefficient of thermal conduc√
tivity. Thus η ∝ σ m, where σ is the collision cross section, and m the mass
of the molecule of the gas. Assuming that the molecular diameter increases like
m1/3 we have
η ∝ m5/6
To double η, we need to increase m by a factor 26/5 = 2.3. To double the
insulating power of air, we would need a gas of molecular weight 69.
7.8
This is a hypothetical exercise, since we are ignoring an important heat
source, the radiation from the sun. (See Prob.17.3). The total rate of heat
generated is 43 πR3 W , and this must equal the rate of heat radiated 4πR2 σTi4 ,
where T1 is the surface temperature, and σ is Stefan’s constant. This give the
surface temperature
µ
¶1/4
RW
T1 =
3σ
In the interior, the rate of heat generation per unit volume is ρW , and this
equals ∇ · q, where q is the heat flux vector. Using q = −κ∇T , we have the
equation for the temperature distribution ∇2 T = −ρW/κ. Assuming that T is
spherically symmetric, and using spherical coordinates, we obtain
d 2 dT (r)
ρW 2
r
=−
r
dr
dr
κ
Integration of this equation, observing that T cannot be singular at r = 0, gives
T (r) = T0 −
ρW 2
r
6
where T0 is the temperature at r = 0. The surface temperature is
T1 = T0 −
ρW 2
R
6
Thus
1
T0 = ρW R2 +
6
µ
RW
3σ
¶1/4
38
CHAPTER 7.
7.9
(a)
The heat absorbed by per unit volume is dQ = −∇ · qdt, which defines the
heat flux vector q. Putting dQ = T ds, we have
∂s
1
+ ∇·q=0
∂t
T
(b)
Consider the heat flux due to heat conduction q = −κ∇T. Write (∇·q)/T =
∇ · (q/T ) − q · ∇(1/T ). The last equation can be rewritten
q
∂s
= −∇ · + κ
∂t
T
µ
∇T
T
¶2
The second term, which is always positive, is the rate of irreversible entropy
production.
7.10
Suppose the thickness of the ice sheet is x. Consider a unit square of the ice
sheet. The mass of the sheet increases at the rate ρdx/dt, and generates heat
at the rate ρdx/dt. This must equal the heat flux, which we can represent as
qx =
κ∆T
x
for a small thickness x. Thus
dx
κ∆T
=
dt
ρx
Chapter 8
8.1
(a)
The number of ways to choose the n atoms to remove from N sites is
N !/ [n! (N − n)!]
(b)
The number of ways to choose the n interstitials out of M is
M !/ [n! (M − n)!]
(c)
The total energy is E = n∆.The phase space volume is
Γ(n) =
N !M !
n! (N − n)!n! (M − n)!
Using the Stirling approximation, we obtain the entropy
³
³
N
S
n´
M
n´
= ln Γ(n) = n ln
− (N − n) ln 1 −
+ n ln
− (M − n) ln 1 −
kB
n
N
n
M
The temperature is defined through
1 ∂S
1 ∂ ln Γ(n)
1
=
=
kB T
kB ∂E
∆
∂n
This gives
∂
∆
=
ln Γ(n) = ln
kB T
∂n
µ
¶
µ
¶
N
M
− 1 + ln
−1
n
n
(d)
The previous equation can be rewritten as
µ
¶
n2
∆
= exp −
(N − n) (M − n)
kB T
39
40
CHAPTER 8.
The low- and high-temperature limits are
√
n ≈
N M exp (−∆/2kB T ) (kB T ¿ ∆)
1
1
1
≈
+
(kB T À ∆)
n
N
M
(e)
n
≈ exp (−∆/2kB T )
N
For T = 300 K: n/N ≈ e−20 = 2 × 10−9 .
For T = 1000 K: n/N ≈ e−6 = 2.5 × 10−3 .
8.2
(a)
Since each link can be pointed left or right independently, the number of
ways to choose N+ links to point right is Γ = N !/[N+ !(N − N+ )!]. We must
have N− = N − N+ . The entropy is S = kB ln Γ, which leads to
S
= −r ln r − (1 − r) ln(1 − r)
N kB
where r = N+ /N is the fraction of right-pointing links.
(b)
The internal energy is independent of N+ , and we can set it to zero. Thus
the free energy is A = −T S, where T is just a constant scale factor.
(c)
The tension τ can be obtained from dU = 0 = T dS + τ dL, where L is the
length of the chain:
L = a(N+ − N− ) = a (2N+ − N ) = aN (2r − 1)
We obtain
τ
1
1
1
∂S
=−
=
ln
kB T
2aN kB ∂r
2a r(1 − r)
where kB T is just a scale factor. The tension is never zero. It is minimum when
r = 1/2, and goes to infinity when the chain is fully stretched to the right
(r = 1) or to the left (r = 0).
In this model, “temperature” is not a relevant concept, since energy is irrelevant. The factor T in T dS is an arbitrary scale factor.
If we give each left-pointing link an energy , then the total energy would be
E = N− = N (1−r). The temperature would be given by T −1 = − ln [r(1 − r]],
apart from a scale factor
8.3
(a)
Assume that a link can be up or down independently. The partition function
is the product of the partition functions of the individual links. The possible
41
N
energies are 0 and mga. Thus QN = [1 + exp (−βmga)] . We have ignored the
fact that the energy of the nth link depends on its height, and therefore on the
states of the preceding links. We have also ignored is the restriction that the
links cannot go above the ceiling.
(b)
∂ ln QN
N mga
=
∂β
exp (βmga) + 1
The length of the chain is L = (N − N 0 )a, where N 0 = U/(mga) is the number
of up links. Thus
Na
L=
1 + exp (−βmga)
(c)
Since U = mga[N − (L/a)], the force constant is mg.
U =−
8.4
(a)
The possible states are labeled by the number of open links n = 0, 1, 2, · · · , N .
The energy with n open links is En = n∆. The partition function is
QN =
N
X
e−βn∆ =
n=0
1 − e−β(N̄+1)∆
1 − e−β∆
(b)
The average number of open links is
(N + 1) e−β(N̄ +1)∆
1 ∂ ln QN
e−β∆
−
=
∆ ∂β
1 − e−β∆
1 − e−β(N̄ +1)∆
The second term is negligible for large N. At low temperatures β∆ À 1 we
have
n̄ ≈ e−β∆
n̄ = −
8.5
(a)
There are 6 sites in each hexagon, but each site is shared by 3 hexagons.
Thus we can assign 2 sites to a hexagon. On the other hand, each hexagon is
associated with one interstitial site. Thus, in an infinite lattice, there are half
as many interstitial sites as lattice sites.
(b)
The entropy is given by
S
kB
= ln Γ(E) = ln Γvacancy + ln Γinterstitial
Γvacancy
=
Γinterstitial
=
N!
M !(N − M )!
(N/2)!
M !(N/2 − M )!
42
CHAPTER 8.
The energy is E = M ∆, and the volume fixed, and proportional to N . Thus,
this gives S(E, V ).Using the Stirling approximation, we obtain
S
kB
= {N ln N − M ln M − (N − M ) ln(N − M )} + {N → N/2}
µ
¶ µ
¶ µ
¶
M
N
2M
N
− (N − M ) ln 1 −
−
− M ln 1 −
= 2M ln √
N
2
N
2M
(c)
From T −1 = ∂S//∂E we obtain the relation
µ
E
N∆
¶2
1
=
2
µ
¶µ
¶
E
2E
1−
1−
e−∆/kB T
N∆
N∆
This can be easily solved, but we only give the high- and low-temperature limits:
½ 1/2
E
2 exp(−∆/2kB T ) (kB T À ∆)
≈
1/3
(kB T ¿ ∆)
N∆
The above is equal to the average interstitial fraction M/N at a given temperature.
8.6
(a)
The partition function for N non-interacting particles is QN = QN
1 , where
Q1 is that for a single particle:
Q1 =
3
X
exp (−β n ) = 2e−β(bx
2
−cx/2)
2
+ e−β(bx
+cx)
n=1
The free energy per particle is a(x, T ) = −kB T ln Q1 .
(b)
We find the equilibrium value of x̄ by minimizing a(x, T ) with respect to x,
or maximizing Q1 .Assume that x̄ is small, and expand the exponential to order
x2 . The condition Q01 = 0 gives two roots:
½
£ 0
¤
x̄ =
(4kB T /c) 1 − (4bkB T /c2 )
Since x̄ cannot be negative, the nontrivial root is acceptable only when T < Tc ,
where
c2
kB Tc =
4b
Examining the sign of Q001 shows that when T < Tc the nontrivial roots corresponds to a maximum, while x̄ = 0 corresponds to a minimum. For T > Tc , the
only solution is x̄ = 0, which corresponds to a maximum. Thus there is a phase
transition at T = Tc .
43
8.7
For a classical relativistic gas,
QN (V, T ) =
Z
"
#
N p
X
d3N p d3N q
VN 1 N
exp −β
(cpi )2 + (mc2 )2 =
I (β)
3N
N !h
N ! h3N
i=1
where
I(β) =
Z
h p
i
d3 p exp −β (cp)2 + (mc2 )2
Using the Sterling approximation to write N ! ≈ N N , we obtain
· µ
¶
¸
V
AN (V, T ) = −N kB T ln
+
1
+
ln
I(β)
N h3
In the nonrelativistic limit kB T ¿ mc2 we have
·
µ
¶¸
Z
2
p2
3/2
I(β) ≈
d3 p exp −β mc2 +
= e−mc /kB T (2πmkB T )
2m
·
µ 3 ¶
¸
λ N
2
AN (V, T ) ≈ N mc + kB T ln
− kB T
V
¡
¢
µ ≈ mc2 + kB T ln nλ3
In ultra-relativistic situations kB T À mc2 we can neglect the rest energy, and
take
µ
¶3
Z
Z ∞
kB T
3
−βcp
2 −βcp
I(β) ≈
d pe
= 4π
dp p e
= 8π
c
0
" Ã
!
#
µ
¶
3
π2N
~c
AN (V, T ) ≈ N kB T ln
−1
V
kB T
¡
¢
µ ≈ kB T ln nL3
where
L = π2/3
~c
kB T
8.8
The partition function is QN = ξ N , where ξ is the partition function for one
particle:
Z
X
2 2
2 2
V
V
ξ=
e−β~ k /2m =
d3 ke−β~ k /2m = 3
3
λ
(2π)
k
p
¡
¢
where λ = 2π~2 /mkB T . The free energy is A = −kB T ln QN = −N kB T ln V /λ3 ,
The equation of state is P = −∂A/∂V = N kB T /V.
44
CHAPTER 8.
8.9
(a)
The partition is QN = ξ N , where
Z ∞
Z
2 2
1 ∞
2πkB T
−βp2 /2m
ξ=
dpe
dqe−βmω q /2 =
τ −∞
τω
−∞
(b)
The free energy is
A = −kB T ln QN = −N kB T ln
µ
2πkB T
τω
¶
Thus
S
U
CV
·
µ
¶¸
∂A
2πkB T
= −
= N kB 1 + ln
∂T
τω
= A + T S = N kB T
∂U
=
+ N kB
∂T
8.10
(a)
¡
¢N
QN = eβµ0 B + e−βµ0 B
(b)
hM i = −
eβµ0 B − e−βµ0 B
1 ∂
ln QN = µ0 N βµ B
β ∂B
e 0 + e−βµ0 B
(c)
­ 2®
1 ∂2
4µ20 N
2
ln
Q
=
M − hM i = 2
N
2
β ∂B 2
(eβµ0 B + e−βµ0 B )
Chapter 9
9.1
(a)
Q(z, T ) =
¶
N0 µ
X
N0
N=0
N
z N e−βN =
¶
N0 µ
X
¢N0
N0 ¡ −β ¢N ¡
= 1 + ze−β
ze
N
N =0
(b)
z ∂
hN i
1
=
ln Q = −1 β
N0
N0 ∂z
z e +1
(c)
U
= −
C
=
∂
N0
ln Q = −1 β
∂β
z e +1
2
N0 kB (β ) eβ
∂U
=
2
∂T
z (z −1 eβ + 1)
9.2
(a)
The grand partition function for the O2 lattice gas is
¶
N µ
X
¢N
N ¡ −β 1 ¢N1 ¡
z1 e
= 1 + z1 e−β 1
Q1 (z, T ) =
N1
N1 =0
The fraction of occupied sites is
z1 ∂
1
hN1 i
=
ln Q1 = −1 β
N0
N0 ∂z1
z1 e 1 + 1
45
46
CHAPTER 9.
Setting the above to f = 0.9, with z1 = 10−5 and T = 310 K, we find,
1
= kB T ln
z1 (1 − f )
≈ −0.37 eV
f
(b)
The grand partition function is now given by
Q(z, T ) =
N N−N
X
X 1 µ N ¶µN − N1 ¶ ¡
¢N1 ¡ −β 2 ¢N2
z1 e−β 1
z2 e
N2
N1
N1 =0 N2 =0
¶
N µ
X
¢N −N1
N ¡ −β 1 ¢N1 ¡
=
z1 e
1 + z2 e−β 2
N1
N1 =0
¡
¢N
= 1 + z1 e−β 1 + z2 e−β 2
The fraction of sites occupied by O2 is
z1 e−β 1
z1 ∂
hN1 i
ln Q =
=
−β
N
N ∂z1
! + z1 e 1 + z1 e−β
Set this to 0.1 and solve for
2.
With
2
1
2
from (a), we obtain
= −0.55 eV
9.3
(a)
E(M ) = − M
µ ¶
N
Γ(M ) =
M
(b)
The grand partition function of the adsorbed gas is
¶
N µ
X
¢N
N ¡ β ¢M ¡
= 1 + zeβ
ze
Q(z, T ) =
M
M=0
where z = eβµ .The average fraction of occupied sites can be obtained either by
maximizing the summand using the Stirling approximation:
·µ ¶
¸
¡
¢
N ¡ β ¢M
ln
ze
≈ M ln zeβ + N ln N − M ln M − (N − M ) ln (N − M )
M
or by calculating the grand canonical average:
z ∂
1
M̄
=
ln Q = −1 −β
N
N ∂z
z e
+1
47
(c)
¡
¢
The chemical potential for an ideal gas is given in Prob.3.2: µ = kB T ln nλ3 ,
p
where λ = 2π~2 β/m, and n = βP . In equilibrium, the chemical potntail of
the adsorbed gas must equal that of the surrounding gas. Thus
M̄
λ3 βP
= −β
N
e
+ λ3 βP
(d)
M 2 − M̄ 2 = z
∂ ∂
ze−β
z
ln Q =
2
∂z ∂z
(e−β + z)
9.4
(a)
The equation of state is
µ
¶
¶µ
1
8
3
V −
P+ 2 = T
3
V
3
Differentiating both sides with respect to P at constant T , we find
∂V
=
−
∂P
P+
3
V2
V − 13
¡
¢=
− V63 V − 13
¢2
¡
V − 13
¡
¢
8
6
1 2
3T − V 3 V − 3
Near the critical point we put V = 1 and let T → 1+ . Thus
κT = −
1 ∂V
1
≈
V ∂P
6 (T − 1)
(b) The fractional density fluctuation near the critical point diverges:
T
1
n2 − n̄2
= κT ≈
n̄2
V
6 (T − 1)
9.5
¡
¢
¡
¢
The condition for equilibrium is ln n+ L3 +ln n− L3 = 0, or n+ n− = L−6 .
Given n − n+ = n0 , we find#
"s
4
n+
1
=
+1−1
n0
2
(n0 L3 )2
#
"s
1
n−
4
=
+1+1
n0
2
(n0 L3 )2
48
CHAPTER 9.
9.6
(a)
Let Ni be the number of molecules of type Xi present. The reaction consumes ν i molecules of type Xi . Thus the change in Ni is proportional to ν i , with
the same proportionality constant for all i. Hence δN = δNi /ν i is independent
of i.
(b)
Minimizing the free energy, we have
0 = δA =
X ∂A
X
δNi =
µi ν i δN = 0
∂Ni
i
i
P
Since δN is arbitrary, we have i µi ν i = 0.
9.7
(a)
In a fixed volume, the densities obey the relations
δn3
δn1
= δn2 = −
2
2
Hence A = n1 − 2n2 and B = n1 + n3 remain constant.
(b)
The chemical
¡
¢ potential forpa classical ideal gas is, according to Prob.3.2,
µ = kB T ln nλ3 , where λ = 2π~2 /mkB T . The condition for chemical equi¡
¢
¡
¢
¡
¢
librium is 2 ln n1 λ31 + ln n2 λ32 − 2 ln n3 λ33 = 0 where the λi are independent
of thus densities. Thus
n21 n2
= K0
n23
¡
¢3/2
where K0 = (4/9)3 mkB T /π~2
. Two other conditions are
n1 + n3
n1
= n0
= 2n2
These imply
n31 = 2K0 (n0 − n1 )
High-temperature limit K0 → ∞ :
2
µ
¶
r
n0
n1 ≈ n0 1 −
2K0
Low-temperature limit K0 → 0
1/3
n1 ≈ (2K0 n0 )
9.8
49
The density depends on the power series
y=
∞
X
bz
=1
We seek an expansion of the equation state in the form
P∞
bz
P
= 1 + a2 y 2 + a3 y 3 + · · ·
= P∞=1
nkB T
b
z
=1
To
the lowest two orders, we have
̰
!2
̰
!3
P∞
X
X
bz
=1
P∞
= 1 + a2
bz
+ a3
bz
+ ···
=1 b z
=1
=1
We expand both sides to order z 3 , obtaining
z + b2 z 2 + b3 z 3 + · · ·
i
¡
¢h
¢2
¡
= z + 2b2 z 2 + 3b3 z 3 + · · · 1 + a2 z + 2b2 z 2 + · · · + a3 (z + · · · )3 + · · ·
= z + (2b2 + a2 ) z 2 + (4b2 a2 + a3 + 3b3 ) z 3 + · · ·
Equating the coefficients of z 2 and z 3 on both sides, we obtain
a2
a3
= −b2
= 4b22 − 2b3
50
CHAPTER 9.
Chapter 10
10.1
The energy residing in a mode of frequency ω of the transmission line is
µ
¶
~ω
~ω
~ω
T
1
−
E = β~ω
≈
k
=
B
e
−1
2kB T
β~ω + 1 (β~ω)2 + · · ·
2
The second term above gives the first quantum correction. As a estimate use
the fundamental mode ω = πc/L, where c is the velocity of light, and L the
length of the transmission line. The Nyquist theorem becomes
µ
¶
π~c
V 2 = 4kB T R∆ν 1 −
2LkB T
For L = 1 mm, the correction amounts to approximately 1% at T = 300 K.
10.2
The accompanying sketch illustrates the construction that would lead to a
fractal of dimension 2.
(a) Start with a straight line of unit length.
(b) Halve the step size, and double the path length by taking more steps.
The way to do this is not unique. Pick one of the ways.
(c) In the next iteration, each previous segment is independently replaced
by a path of twice the length with half the step size.The path length L depends
on the step size τ according to L ∝ τ 1−D , with D = 2.
51
52
CHAPTER 10.
10.3
Ignoring the possibility that two suspended particles collide with each another, we can regard the suspension as an ideal gas in equilibrium with. the
medium, which acts as a heat reservoir. Therefore its partial pressure obeys
the ideal gas law.
10.4
It is straightforward to show that
¡
¢
1
n(x, t) = √
exp −x2 /4Dt
4πDt
satisfies the diffusion equation. To show the initial condition, note
½
0 (x 6= 0)
n(x, t) −→
∞ (x − 0)
t→0
and, for all t 6= 0,
Therefore
Z
∞
dxn(x, t) = 1
−∞
n(x, t) −→ δ(x)
t→0
10.5
(a)
For the Brownian particles: D = 4 × 10−9 cm2 /s.
For O2 : D ≈ 0.1cm2 /s
q
Thus an O2 molecule will travel 14 × 1010 ρ ≈ 10 cm.
(b)
From Einstein’s relation η = D/kB T.
F = u/η = kB T u/D ≈ 10−5 dyne.
10.6
Perrin obtained A0 = 7.05 × 1023 , which would have led to
kB
=
e =
8.32 × 106
= 1.18 × 10−16 cgs (Modern value:1.381 × 10−16 )
7.05 × 1023
2.9 × 1014
= 4.14 × 10−10 cgs (Modern value: 4.803 × 10−10 )
7.05 × 1023
10.7
(a)
Substitute j = −D∇n, into the continuity equation ∇· j + ∂n/∂t = 0 to
obtain
∂n
−D∇2 n +
=0
∂t
53
(b)
With a drift current produced by a uniform constant external force Fext ,
The total particle current is
j = −D∇n +
n
Fext
η
where η is the mobility. Thus the diffusion equation generalized to
1
∂n
−D∇2 n + Fext · ∇n +
=0
η
∂t
(c)
The absorption,contributes a term −V (r)n to the rate of change of the particle density. From this point of view, the Schrödinger equation describes a
diffusion in imaginary time, with absorption, of the wave function ψ. What
makes quantum mechanics distinctive is that ψ is a complex probability amplitude, and not a probability.
54
CHAPTER 10.
Chapter 11
11.1
If the showers are distributed at random, the probability that one occurred
on Tuesday would be 1/7, and the probability that it did not occur would be
6/7. The probability that none of the 12 showers occur on Tuesday would be
(6/7)12 = 0.157. Better bring the umbrella.
11.2
If parking tickets were issued at random, the probability of getting 12 tickets
on two days of the week would be (2/7)12 = 3 × 10−7 . This is so small that we
must reject the assumption that tickets were given out at random, and advise
the student to use a parking lot on those days. Of course, this assumes that the
police maintains the same tactic.
11.3
What determines whether the man goes north or south is the correlation
between northbound and southbound trains, as illustrated in the sketch. If he
enters the station during the interval x, he goes north. Otherwise he goes south.
Since he went north 70% of the time, we conclude x = 0.7.
11.4
Generate a sequence of random number between 0 an 1. Divide the interval
(0,1) into say 10 equal bins, and keep a running score of the number of random
numbers in each bin, as they are being generated. At the end of the run, plot
55
56
CHAPTER 11.
a histogram of the numbers in each bin. If the sequence is truly random, the
histogram should fit a Poisson distribution..
11.5
The current-voltage characteristic of the device is shown in the accompanying sketch.
Let the probability of finding the voltage to have a value between V and
V + dV be P (V )dV.
Let the probability of finding the current to have a value between I and
I + dI be Q(I)dI.
The current is never negative. So Q(I) = 0 for I < 0. For I > 0, we have
Q(I) = P (V )
where
dV
V0
= P (V )
dI
I + I0
µ
I
V = V0 ln 1 +
I0
In the range V ≤ 0, we must have
Z
0
dIQ(I) =
−∞
Z
(I > 0)
¶
0
−∞
dV P (V ) ≡ α
Thus Q(I) should contain a term αδ(I). The complete result is
½
V0
P (V ) I+I
+ αδ(I) (I ≥ 0)
0
.
Q(I) =
0
(I < 0)
11.6
Let the probability density for y be Q(y). We have
Z x
Z y
dy 0 Q(y 0 ) =
dx0 P (x0 )
−∞
−∞
√
where x = y/b. The integrands on both sides are zero for negative arguments.
Thus
µ 02 ¶
Z x
Z y
³ y ´
0
x
0
0
0x
dy Q(y ) =
dx
exp −
= exp −
−1
a
2a
2ab
0
0
Differentiating this gives
57
Q(y) =
11.7
Solution provided in text.
³ y ´
1
exp −
2ab
2ab
(y ≥ 0)
58
CHAPTER 11.
Chapter 12
12.1
(a)
Z
∞
dτ eiωτ G(τ )
Z ∞
iωτ
= ν
dτ e
dtf (t)f (t + τ ) + 2πδ(ω)I 2
−∞
−∞
¯ ¯
= ν ¯fω2 ¯ + 2πδ(ω)I 2
R∞
R∞
where I = ν −∞ dtf (t)I = ν −∞ dtf (t) by Campbell’s theorem (11.29)..
(b) R
∞
fω = 0 dteiωτ −λt = (iω − λ)−1 .
R∞
I = ν 0 dte−λt = ν/λ.
S(ω) =
−∞
Z ∞
S(ω) =
ν
2πν 2
+
δ(ω)
ω 2 + λ2
λ2
The white-noise component is ν/λ2 , the first term in the limit λ À ω.
12.2
We can
with
R ∞use the result of RProb.12.1(a),
∞
fω = −∞ dteiωt ϕ(t) −→ −∞ dtϕ(t) ≡ q.
ω→0
R∞
I = ν −∞ dtϕ(t) = νq.
S(ω) −→ νq + 2πν 2 q 2 δ(ω)
ω→0
12.3
(a)
I(t) and I(t + τ ) are the same if there are an even number of sign changes
during τ and equal but opposite if there are an odd number of sign changes.
Thus
hI(t)I(t + τ )i = a2 (Peven − Podd )
59
60
CHAPTER 12.
(b)
The probability that there are k crossing in the time interval τ > 0 is given
by the Poisson distribution
P (k; ν) =
(ντ )k −ντ
e
k!
The probability there are an even and odd number of crossings are given respectively by
Ã
!
(ντ )2
(ντ )4
−ντ
Peven = e
1+
+
+ ···
2!
4!
Ã
!
3
5
(ντ )
(ντ )
−ντ
ντ +
Podd = e
+
+ ···
3!
5!
Thus for τ > 0
2 −ντ
hI(t)I(t + τ )i = a e
Ã
(ντ )2 (ντ )3
1 − ντ +
−
+ ···
2!
3!
!
= a2 e−2ντ
If τ < 0, then hI(t)I(t + τ )i = hI(t − |τ |)I(t)i = hI(t)I(t + |τ |)i by invariance
in time translation. Thus the general answer is obtained by replacing τ by |τ |.
(c)
S(ω) =
Z
∞
−∞
dτ hI(t)I(t + τ )i = a2
= 2a2 Re
Z
∞
dτ eiωτ −ντ =
0
12.4
From (12.2)
W3 (3, 1, 2) =
From (12.24) and (12.29)
Z
Z
∞
dτ eiωτ −ντ + a2
0
2νa2
ω2 + ν 2
Z
0
dτ eiωτ +ντ
−∞
dx4 W4 (3, 1, 4, 2)
W3 (3, 1, 2) = W2 (3, 1)P (3, 1|2) = W2 (3, 1)P (1|2)
W4 (3, 1, 4, 2) = W3 (3, 1, 4)P (3, 1, 4|2) = W3 (3, 1, 4)P (4|2)
= W2 (3, 1)P (1|4)P (4|2)
Substituting these into the last equaton we obtain
Z
W2 (3, 1)P (1|2) =
dx4 W2 (3, 1)P (1|4)P (4|2)
Z
P (1|2) =
dx4 P (1|4)P (4|2)
61
12.5
The problem one faces in a computer simulation of a phase transition is
“critical slowing down”. It is easy to obtain a rough value for Tc , but very
difficult to attain precision. This is because on a finite lattice the transition
will not be sharp, but increasing the lattice size also increases the time to reach
thermal equilibrium. It becomes increasingly difficult for large blocks of spins
to flip, since all spins have to flip at the same time, by chance. Indeed, this
is why the largest block, namely the whole lattice, does not flip at all, leading
to spontaneous magnetication. To speed up the simulation, one has to improve
the algorithm by making trial flips of not just single spins, but blocks of spins
of random sizes.
12.6
(a)
Since the diffusion equation is invariant under translations in space and time,
the solution (10.28) can be generalized to a starting position x0 and starting
time t0 by by replacing x, t by x − x0 , t − t0 , respectively.
(b)
Denote the transition probability from step i to step j by
Ã
!
1
(xi − xj )2
P (i|j) = p
exp −
4D(ti − tj )
4πD(ti − tj )
To begin, show that for n = 2,
P (2|0) =
The right side is
1
p
X=
4πD (t2 − t1 ) (t1 − t0 )
Z
∞
Z
dx1 P (2|1)P (1|0)
Ã
1
dx1 exp −
4D
−∞
Ã
(x2 − x1 )2 (x1 − x0 )2
−
t2 − t1
t1 − t0
!!
The exponent can be wrtiten as
¶
¶
µ
µ
1
x20
2B
A
x22
2
−
+
−
x1 +
x1
4D t2 − t1 t1 − t0
4D
A
where
A =
B
=
1
1
+
t2 − t1
t1 − t0
x2
x0
+
t2 − t1
t1 − t0
Performing the integral, we obtain
Ã
Ã
!!
p
2
2
4πD/A
1
(x1 − x0 )
B2
(x2 − x1 )
p
exp −
X =
−
−
4D
t2 − t1
t1 − t0
A
4πD (t2 − t1 ) (t1 − t0 )
Ã
!
(x2 − x0 )2
1
exp −
= p
4D(t2 − t0 )
4πD(t2 − t0 )
62
CHAPTER 12.
Next show that the result for n − 1 implies that for n, where n > 2. The
integrals one has do is similar to the one above. This will complete the proof
by induction.
12.7
The instructions are fairly explicit.
Chapter 13
Solutions are given in the text.
63
64
CHAPTER 13.
Chapter 14
14.l
p
The relativistic energy is E = (pc)2 + (mc2 )2 . In the ultra-relativistic domain we can neglect the mass term and thus E = pc. The deBroglie wavelength
is h/p = hc/E. The thermal wavelength is therefore proportional to hc/kB T .
14.2
14.3
U
=
CV
=
³
´
3
3
P V ≈ N kB T 1 ± 2−5/2 nλ3
2
2
³
´
∂U
3
≈ N kB 1 ∓ 2−7/2 nλ3
∂T
2
The upper sign is for fermions, lower sign is for bosons.
14.4
(a)
N=
X
λ
nλ = z
X
λ
65
exp (−β λ ) = zQ
66
CHAPTER 14.
(b)
The internal energy per particle is defined by
P
U
λ λ exp (−β λ )
= P
N
λ exp (−β λ )
Thus,
−
∂ ln Q
1 X
=
∂β
Q
λ
exp (−β λ ) = U
λ
(c)
Q =
X
αβγ
¡
¡
exp −β
trans
α
+
rot
β
+
vib
γ
¢¢
= Qtrans Qrot Qvib
∂
(ln Qtrans + ln Qrot + ln Qvib )
∂β
1 ∂U
= ctrans + crot + cvib
N ∂T
U
N
= −
cV
=
14.5
Qtrans
= V (2π)−3 4π
Z
0
U
N
∞
¡
¢
dkk 2 exp −β~2 k2 /2m = V /λ3
= −∂ ln λ−3/2 /∂β =
3
kB T
2
3
ctrans
=
kB
2
.
14.6
(a)
ln Qrot
U
N
¡
¡
¢¢
¡
¢
≈ ln 1 + exp −β~2 /I ≈ exp −β~2 /I
µ
¶
∂ ln Qrot
~2
β~2
= −
=
exp −
∂β
I
I
crot
≈3
kB
µ
β~2
I
¶2
µ
¶
β~2
exp −
I
(b)
Z
∞
Qrot
≈
U/N
≈ kB T
0
¡
¢
d 2 exp −β~2 2 /2I ∝ .β −1
67
crot
≈1
kB
(c)
The internal energy rises exponentially from T = 0 to approach a linear
behavior. The qualitative behaviors are as shown in the accompanying sketch.
C
Urot
rot
k
kT
kT
2
h /I
14.7
(a)
Qvib
=
∞
X
n=0
U
N
= −
¢−1
¡
exp (−β~ω(n + 1/2)) = eλ/2 eλ − 1
∂ ln Qvib
~ω eλ + 1
=
∂β
2 eλ − 1
cvib
= e−β~ω
kB
µ
(λ=β~ω)
β~ω
1 − e−β~ω
¶2
Cvib
k
kT
hω
(b)
¿
À
1
1 eλ + 1
∂
n+
=−
ln Qvib =
2
∂λ
2 eλ − 1
*µ
¶2 + ¿
À2
1
1
∂2
eλ
− n+
n+
=
ln Qvib =
2
−2
2
2
∂λ
(eλ − 1)
where λ=β~ω.
68
CHAPTER 14.
14.8
kB Tvib
kB Trot
≈ ~ω
~2
≈
I
For H2 , Tvib = 6100 K, Trot = 85.4 K.
From T = 0, the specific heat rises to 3kB /2 before it reaches Trot , then
increases by kB around T = Trot , and increases by kB again around T = Tvib .
14.9
= γ n + bσ n
µ
¶
1
= ~ω n +
2
µ
¶2
1
= ~ω n +
2
n
γn
σn
h i=
P
n
(γ n + bσ n ) e−β(γ n +bσn )
P −β(γ +bσ )
n
n
ne
Expanding this to first order in b, we have
³
´
h i
¯
≈
+ bν 2 − bβ~ω ν 3 − ν 2 ν̄
~ω
~ω
where ν = n̄ + 1/2, and a bar denotes average with respect to the unperturbed
system with b = 0.
∂
− ν 2 = ν 3 − ν 2 ν̄
∂λ
From Prob.14.7(b) we have
ν2 =
e2λ + 6eλ + 1
2
4 (eλ − 1)
from which we obtain
ν3
−
ν 2 ν̄
¢
¡
∂ 2 eλ 3eλ + 4
=− ν =
3
∂λ
2 (eλ − 1)
where λ=β~ω. Thus
¢
¡
h i−¯
e2λ + 6eλ + 1 λeλ 3eλ + 4
≈
2 −
3
b~ω
4 (eλ − 1)
2 (eλ − 1)
The specific heat is obtained by differentiating the above with respect to T.
Chapter 15
15.1
The cross section for a partially polarized beam is
σ pol = |α|2 σ 1 + |β|2 σ2 = Tr (ρσ)
2
2
where |α| + |β| = 1. With respect to the present basis (i.e., the spin states 1
and 2) we have
µ
¶
µ
¶
2
|α|
σ1 0
0
ρ=
σ=
2
0 σ2
0
|β|
The matrix trace is indepedent of the basis.
15.2
(a)
Qclassical =
Z
dpdq −βH
1
=
e
τ
τ
Z
−∞
(b)
Qquantum =
∞
X
∞
e−βEn =
n=0
¡
¢
dp exp −βp2 /2m
Z
∞
¡
¢
2π
dq exp −βmω 2 q 2 /2 =
βτ
ω
−∞
·
µ
¶¸
1
e−β~ω/2
exp −β~ω n +
=
2
1 − e−β~ω
n=0
∞
X
(c)
Qquantum −→
β→0
1
2π
=
β~ω
βhω
Comparison with Qclassical gives
τ =h
which is Planck’s constant.
69
70
CHAPTER 15.
15.3
Postulate the form
1
e−β(K+V ) = e−βK e−βV e− 2 β
2
X
to second order in β. Expand both sides to second order:
1
e−β(K+V ) ≈ 1 − β (K + V ) + β 2 (K + V )2
2
e
−βK −βV −β 2 X
e
Therefore
e
·
¸
¸·
¸·
1 2 2
1 2 2
1 2
≈
1 − βK + β K
1 − βV + β V
1− β X
2
2
2
¢
1 2¡ 2
≈ 1 − β (K + V ) + β K + V 2 + 2KV − X
2
i
1 2h
2
≈ 1 − β (K + V ) + β (K + V ) + KV − V K − X
2
X = KV − V K = [K, V ]
15.4
(a)
¡
¢N
QN = eβµ0 B + e−βµ0 B
(b)
hM i = −
eβµ0 B − e−βµ0 B
1 ∂
ln QN = µ0 N βµ B
β ∂B
e 0 + e−βµ0 B
(c)
­ 2®
1 ∂2
4µ20 N
2
ln QN =
M − hM i = 2
2
2
β ∂B
(eβµ0 B + e−βµ0 B )
15.5
(a)
Nb
Nf
N
z −1 e−β + 1
X
i
=
−1
β
z e k +1
=
k
where k = ~2 k2 /2m.
(b)
The condition is Nb + Nf = N , or
1
1
f3/2 (z) = 1
+
z −1 e−β + 1 nλ3
71
p
where λ = 2π~2 /mkT .
(c)
For small z, the condition becomes
zeβ + z/(nλ3 ) = 1
Thus
¡
¢
z = nλ3 1 − nλ3 eβ
This is valid for nλ3 ¿ 1.
(d)
For high temperatures β → 0. Thus nλ3 ¿ 1, and z ¿ 1.
From Nf = (V /λ3 )f3/2 (z), or nf λ3 = f3/2 (z), we obtain
nf λ3
nf
n
¡
¢
≈ z = nλ3 1 − nλ3 eβ
≈ 1 − nλ3 eβ
For low temperatures we expect most particles to be in one of the bound states,
and thus nf /n → 0.For β → ∞, the condition for z becomes
zf3/2 (z) = nλ3 e−β
This means that z is small, so the condition reduces to z 2 = nλ3 e−β .Thus
1
nf
≈ √ 3 e−β
n
nλ
/2
15.6
hnk i =
1
Q
X
nk exp [−β ( 1 n1 +
{n1 ,n2 ,··· }
Differentiating both sides with respect to
−
2 n2
1 ∂
hnk i =
β∂ p
1
Q
−
X
{n1 ,n2 ,··· }
1
β
µ
∂ 1
∂ pQ
p
+ · · · ) + βµ] = −
with p 6= k, we obtain
nk np exp [−β ( 1 n1 +
¶
X
1 ∂
ln Q
β∂ k
2 n2
+ · · · ) + βµ]
nk exp [−β ( 1 n1 +
{n1 ,n2 ,··· }
2 n2
+ · · · ) + βµ]
= hnp nk i − hnp i hnk i
¡
¢−1
does not depend on
We know that hnk i = z −1 eβ k ± 1
is zero, or
hnp nk i = hnp i hnk i (p 6= k)
p.
Thus the above
72
CHAPTER 15.
15.7
hσ 2 i − hσi2 =
XX
k∈G p∈G
[hnk np i − hnk i hnp i] =
Xh
k∈G
hn2k i − hnk i
2
i
The last relation follows from the fact that terms with k 6= p do not contribute,
2
2
as shown in the last problem. By (15.35), hn2k i − hnk i = hnk i ∓ hnk i . This
directly leads to the answer desired.
Chapter 16
16.1
The fraction of electrons that can excited is of the order of kB T / F . Hence
the effective density is nkB T / F , where n is the electron density. The meanfree-path iswhere σ is the collision cross section.
λ≈
16.2
n = 4.35 × 1027 cm−3
F = 24.6 MeV
Av. energy per nucleon =
3
5 F
F
nσkB T
= 14.8 MeV
16.3
(a)
The Fermi wave number kF is given through (2s + 1)V (4π/3)kF3 = N .
Thus, kF = [3n/4π(2s + 1)]1/3.
pF
F
= ~kF
q
=
p2F c2 + m2 c4
(b)
where
(c)
R
p
=
R
|p|<pF
Z hp
i
(cp)2 + (mc2 )2 − mc2
U = 2V
p
Z
Z
2
P = 2 px vx =
(p · v)
3 p
p
d3 p/h3 .
73
74
CHAPTER 16.
For n1/3 << mc/~, particles near the Fermi surface are non-relativistic:
p
p2
(cp)2 + (mc2 )2 ≈ mc2 +
2m
We have v = p/m. Hence
U ≈V
Z
2
PV ≈ V
3
p
Z
p2
2m
p
2
p2
= U
m
3
For n1/3 >> mc/~, particles near Fermi surface are ultra-relativistic:
p
(cp)2 + (mc2 )2 ≈ cp
We have v =cp/p. Hence
U
PV
≈ 2V
≈
2
V
3
Z
cp
p
Z
cp =
p
1
U
3
(d)
F
= 6 × 10−5 eV.
16.4
(a)
Let p± be the Fermi momenta of the spin-up and spin-down gases. The
energy of an atom of up(down) spin is
(H) =
Thus
N± =
p2±
∓ µH
2m
V 4π 3
4πV
(2m)3/2 [ (H) ± µH]3/2
p =
h3 3 ±
3h3
(b)
For complete polarization, we have N− = 0, hence (H) = µH, and
N+ =
4πV
(4mµH)3/2
3h3
The total density is now n = N+ /V . The minimum field is
Hmin =
µ
3π 2
4
¶2/3
~2 n2/3
µm
75
16.5
(a)
Consider a shell of thickness dr in the gas. Let the pressure differential be
dP . The inward force acting on a patch of the shell of area dA is −dAdP . In
hydrostatic equilibrium this must equal the gravitational attraction due to the
mass at the center. Thus
−P dA = γM ρ(r)r−2 dAdr
dP
γM ρ(r)
= −
dr
r2
(b)
P =
Thus
2
n
5
∝ n5/3 ∝ ρ5/3
F
dρ
dr
= −K 2
r
ρ1/3
Assuming ρ(∞) = 0, we have
ρ(r) =
C0
r3/2
16.6
(a)
Nb
=
=
Nf
N
z −1 e−β
X
k
+1
i
z −1 eβ k + 1
where k = ~2 k 2 /2m.
(b)
The condition is Nb + Nf = N , or
1
z −1 e−β
+
+1
p
where λ = 2π~2 /mkB T .
(c)
For small z, the condition becomes
1
f3/2 (z) = 1
nλ3
zeβ + z/(nλ3 ) = 1
Thus
This is valid for nλ3 ¿ 1.
¡
¢
z = nλ3 1 − nλ3 eβ
76
CHAPTER 16.
(d)
For high temperatures β → 0. Thus nλ3 ¿ 1, and z ¿ 1.
From Nf = (V /λ3 )f3/2 (z), or nf λ3 = f3/2 (z), we obtain
nf λ3
nf
n
¡
¢
≈ z = nλ3 1 − nλ3 eβ
≈ 1 − nλ3 eβ
For low temperatures we expect most particles to be in one of the bound states,
and thus nf /n → 0.For β → ∞, the condition for z becomes
zf3/2 (z) = nλ3 e−β
This means that z is small, so the condition reduces to z 2 = nλ3 e−β .Thus
nf
1
≈ √ 3 e−β
n
nλ
/2
16.7
The probability of finding an electron with energy ∆ above the Fermi level
¢−1
¡
.The probability for finding an electron with energy ∆
is P (∆) = eβ∆ + 1
below the Fermi level is P (−∆). Therefore
Q(∆) = 1 − P (−∆) =
1
eβ∆ + 1
16.8
(a)
The number of states in a volume element in momentum space is 2L2 dpx dpy /h2 .
The density of states is
2A
D(p) =
(2π~)2
(b)
Obtain D( ) through D( )d = D(p)πpdp:
D( ) =
(c)
¤
£
N = 2 A/(2π)2 πkF2
kF
=
F
=
mA
2π~2
r
2πN
A
π~2 N
mA
77
(d)
E=
Z
F
d D( ) =
0
(e)
σ=
∂E
π~2
=−
∂A
2m
π~2 N 2
2m A
µ
N
A
¶2
(f)
√
In 3D D( ) ∝
, while in 2D D( ) is independent of . When the temperature increases from T = 0, the average energy increases like 0 + kB T , and hence
the density of states in 3D increases like kB T , whereas it remains constant in
2D. Thus we expect the chemical potential to be less sensitive to temperature
in the 2D case. That is, the temperature dependence is weaker.
78
CHAPTER 16.
Chapter 17
17.1
(a)
The operators p and q are hermitian, and defined by [p, q] = −i~. Solving
for a and a† , we have
a = (2~mω)−1/2 p − i (mω/2~)1/2 q
a†
= (2~mω)
which give
−1/2
p + i (mω/2~)
1/2
q
£ †¤
a, a = 1
We can write
p2
2m
=
1
mω 2 q 2
2
=
¡
¢¤
1 £ †
~ω a a + aa† + a2 + a†2
4
¡
¢¤
1 £ †
~ω a a + aa† − a2 + a†2
4
Hence
H
¢
p2
1 ¡
1
+ mω 2 q 2 = ~ω a† a + aa†
2mµ 2
2
¶
1
= ~ω a† a +
2
=
(b)
Define the eigenstate |ni by
a† a|ni = n|ni
hn|ni = 1
The number hn|a† a|ni is the norm of the state vector a|ni, and therefore
non-negative.
This means that the eigenvalues of a† a cannot be negative.
79
80
CHAPTER 17.
Multiplying both sides of the first relation by a, we have
aa† a|ni = na|ni
Using aa† = a† a + 1, we have
(a† a + 1)a|ni = na|ni
(a† a)a|ni = (n − 1)a|ni
This means that a|ni has eigenvalue n − 1.
If n were not an integer, then successive application of a would eventually
make n negative.
Since this cannot happen, n must be an integer.
(c)
We have shown that a|ni has eigenvalue n − 1.
By the same method we can show a† |ni has eigenvalue n + 1. Thus
a|ni = C|n − 1i
where C is a constant. The norm of this state is C ∗ C, which should be n.
We can choose the real solution
√
C= n
Similarly we can show
a† |ni =
√
n + 1|n + 1i
17.2
(a)
The star is completed enclosed by the shell of dust cloud, which absorbs all
the radiation from the star. The dust cloud has two surfaces, an outer one and
an inner one, and we assume that they have approximately the same area.
Let the temperature of the star by T , and that of the dust cloud T 0 .
Let the power radiated by the star be R, and that of the dust cloud be R0
from each surface.
The net power outflow from the star is R − R0 . The net influx is R0 . In
equilibrium R − R0 = R0 . Hence the power radiated to the outside world is
R0 =
1
R
2
(b)
Since R ∝ T 4 , R0 ∝ T,4 we have T 0 /T = (R0 /R)1/4 = 2−1/4 .
17.3
We are ignoring heat generated due to radioactivity in the Earth’s interior.
(See Prob. 7.8.)
81
Let the subscripts S and E identify quantities relating respectively to the
Sun and Earth, which are separated by a distance L. The radiation per unit
surface area is σT 4 . From the viewpoint of the Sun, the fractional solid angle
2
subtended by the Earth is πRE
/(4πL2 ). Thus the power received by the Earth
is
2
πRE
σTS4 (4πRS2 )
4πL2
2
This must equal the power radiated by the Earth σTE4 (4πRE
). Thus the radius
of the Earth drops out and we obtain
r
TE
RS
=
TS
2L
17.4
Energy flows into the house due to transmission of radiation originating
outside, and reflection of radiation originating inside. Thus the rate of inflow
is (1 − r)T04 + rT 4 . The loss rate is (1 − r)T 4 . In equilibrium we must have
(1 − r)T04 + rT 4 = (1 − r)T 4 .Thus
T
=
T0
µ
1−r
1 − 2r
¶1/4
17.5
From (8.61) S = T −1 (U + P V − N µ).
For photons µ = 0, P V = U/3. Thus
S=
4U
4
= V σT 3
3T
3
.
K.
17.6
(a)
From (10.5) the photon density is n = κ(kT /~c)3 ≈ 4000 cm−3 for T = 2.73
(b)
From Prob.10.5, the entropy density at T = 2.73 K is S/V = k(4π/45)(kT /~c)3 .
The entropy per photon is independent of the temperature:
µ
¶
S
4π
=
k = 1.21k
N
45κ
where k is Boltzmann’s constant.
(c)
When S = constant, T ∝ V −1/3 . .When the radius of the universe doubles,
the temperature drops by a factor 2−1/3 = 0.793.
82
CHAPTER 17.
17.7
(a)
There are 3N modes, each with energy ~ω 0 .
U=
3N ~ω 0
exp (β~ω 0 − 1)
(b)
2
CV = ∂U/∂T → 3N (~ω 0 ) kT −2 exp (−β~ω 0 )
(c)
¡
¢
A = 3N kT ln 1 − e−β~ω0
17.8
(a)
£
¡
£
¡
¢¤
¢¤
A = Nsolid − + 3kT ln 1 − e−β~ω + Ngas kT 1 − ln nλ3
(b)
The chemical potentials must be equal:
¡
¢
¡
¢
ln nλ3 = − + 3kT ln 1 − e−β~ω
(c)
P (T ) = nkT
17.9
(a)
The free energy is
¡
¢
A = φ + 3N kT ln 1 − e−β~ω0
The condition for equilibrium is P = −∂A/∂V = 0, which gives
∂φ
3γn~ω 0
=
∂V
exp (β~ω 0 ) − 1
where n = N/V
(b)
V
= V0 +
α =
where n0 = N/V0 .
3γn~ω 0
K [exp (β~ω 0 ) − 1]
3kγn~ω 0
KT 2 [exp (β~ω 0 ) − 1]2
83
17.10
(a)
The heat capacity of an electron gas is
CV
Nk
≈
kTF
=
Thus
π2 T
2 TF
~2 ¡ 2 ¢2/3
3π n
2m
a=
π2
2TF
(b)
The Debye heat capacity for T ¿ TD is given by
CV
12π 4
≈
Nk
5
Thus
b=
µ
T
TD
¶3
12π 4
3
5TD
.
(c)
The plot is CV /nR vs. T 2 . Thus, the intercept at T 2 = 0 gives Ra, and the
slope of the line gives Rb. From the plot, a rough reading gives
Intercept
Slope
= Ra = 1.2 × 10−3 cal mole−1 K−2
= Rb = 4 × 10−6 cal mole−1 K−4
Thus
a = 6 × 10−3 K−1
b = 2 × 10−6 K−3
which lead to
TF
TD
= 820 K
= 490 K
The electron density is n ≈ 1020 cm−3 .
.
17.11
The energy is
p
= ak 2/3 , where a = ~ σ/ρ.
Z
d2 k
U
=
A
(2π)2 exp (β ) − 1
84
CHAPTER 17.
Changing the variable of integration to x = β = βak 2/3 , we find
where
U
C0 ³ ρ ´2/3
(kT )7/3
=
A
3π~4/3 σ
C0 =
Z
0
∞
dx
x4/3
= 1.68
ex − 1
Chapter 18
18.1
In the transition P = kB T λ−3 g5/2 (1).Since this is proportional to T 5/2 , we
have dP/dT = 52 kB λ−3 g5/2 (1) The specific volume of the gas phase at transition
¤−1
£
This represents the change
is given by its inverse density v0 (T ) = λ3 g3/2 (1)
in specific volume in the first-order transition, since the other phase has v0 = 0.
Thus
5kB g5/2 (1) 1
dP
=
dT
2 g3/2 (1) v0 (T )
We can read off the change in specific entropy:
s0 =
The latent heat of transition is
5kB g5/2 (1)
2 g3/2 (1)
= T s0 .
18.2
Just above the transition z is slightly less than 1. Using the expansion
given in the problem we have
√
nλ3 = g3/2 (z) = g3/2 (1) − b ν + aν
where ν = − ln z. Thus the equation for the fugavity is
√
aν − b ν + c = 0
where
"
3
c = g3/2 (1) − nλ = g3/2 (1) 1 −
≈
3
T − Tc
g3/2 (1)
2
Tc
85
µ
Tc
T
¶3/2 #
86
CHAPTER 18.
Choose the solution that goes to zero when c → 0. To order c2 we have
³a´
√
ν =c+
c2
b
This leads to the expansion
z = 1 − c2 −
2a 3
c + ···
b
18.3
The parametric equation of state is
λ3 P
kB T
z3
z2
+
+ ···
25/2 35/2
z2
z3
λ3 n = z + 3/2 + 3/2 + · · ·
2
3
= z+
We want to obtain P as a power series in λ3 n. To do this, invert the second
¢2
¢2
¡
¡
equation by writing z = λ3 n + k2 λ3 n + k3 λ3 n + · · · . The coefficients k
are determined by the second equation. Substitute this into the first equation
to obtain an expansion of P.
The details are as follows: Let x = λ3 n. We have
x = (x + k2 x2 + k3 x3 ) + 2−3/2 (x + 2k2 x2 ) + 3−3/2 x3 + O(x4 )
which gives
λ3 P
kB T
2
= 2−3/2 , k3 =
1
4
− 3−3/2 . Thus
= (x + k2 x2 + k3 x3 ) + 2−5/2 (x + 2k2 x2 ) + 3−5/2 x3 + O(x4 )
µ
¶
1
2
x3 + O(x4 )
− √
= x − 2−5/2 x2 +
8 9 3
which gives
a2
a3
= −2−5/2
1
2
=
− √
8 9 3
18.4
Putting λ3 = cT −3/2 ,we have
½ 5/2
3
3
T g5/2 (z) (T > Tc )
U =
P V = V kB
T 5/2 g5/2 (1) (T < Tc )
2
2c
½ 5 3/2
dz
∂U
3
g5/2 (z) + T 5/2 g3/2 (z) z1 dT
2T
CV =
= V kB
5 3/2
g5/2 (1)
∂T
2c
2T
(T > Tc )
(T < Tc )
87
We differentiate the relation nλ3 = g3/2 (z) to obtain
3 nλ3
1 dz
=−
z dT
2 T g1/2 (z)
which approaches zero when z → 1, because g1/2 (z) → ∞. Thus CV is continuous at z = 1.
We use g1/2 (z) = z(d/dz)g3/2 (z) = −dg3/2 (z)/dν, where ν = − ln z . From
Prob.11.2, we have the expansion near z = 1 :
g1/2 (z) = b0 ν −1/2 + b1 + · · ·
where b0 = 1.7275, b1 = 1.460. Thus near z = 1, or ν = 0,we have
1 dz
dν
3 nc −5/2 √
T
ν
=−
≈
z dT
dT
2 b0
Differentiate CV , and setting ν = 0, we obtain
·
µ
¶¸
3
N kB
d 1 dz
0
0
C+ − C− = N kB Tc
= −ξ
2
dT z dT
Tc
ν=0
where + and − indicate approaching the limit from above and below, respectively, and
¸2
·
µ
¶2
27 g3/2 (1)
27 2.612
ξ=
=
= 3.86
16
b0
16 1.7275
18.5
(a)
Let = ~2 k2 /2m. The total number of particles is
N
N
V
=
=
1
z −1 e−β∆
−1
+
X
k6=0
1
z −1 eβ
−1
1
1
1
+ 3 g3/2 (z)
−1
−β∆
V z e
−1 λ
The first term is the condensate density. When V → ∞, it is negligible except
when z = e−β∆ .
(b)
The unperturbed transition temperature T0 is determined by the condition
nλ30 = g3/2 (1). The perturbed transition temperature Tc is determined by the
condition nλ3c = g3/2 (e−β c ∆ ). We can use the approximation
p
g3/2 (e−β c ∆ ) ≈ g3/2 (1) − a ∆/kB T0
88
CHAPTER 18.
where a = 3.455. Let Tc = T0 + δT To lowest order in ∆/kB T0 we obtain
r
∆
δT
= c0
T0
kB T0
2 a
c0 =
= 0.881
3 g3/2 (1)
18.6
(a)
The internal energy due to phonons is
Z ∞
π 2 V (kB T )4
~ck
4πV
dkk2
=
U=
3
exp (β~ck) − 1
30 (~c)3
(2π) 0
The heat capacity per unit mass is given by
Cphonon =
1 ∂U
2π 2 k 4
T3
=
mN ∂T
15 ρ (~c)3
where ρ is the mass density. Using data for liquid helium
c = 2.39 × 104 cm s−1
ρ = 0.144 g cm−3
we have
Cphonon = 0.021 T 3 J g−1 K−1
where T is the absolute temperate in K.
(b)
The internal energy per unit volume due to rotons is
Z ∞
U
4π
∆ + ~2 (k − k0 )2 /2σ
2
dkk
=
V
exp (β∆) exp (β~2 (k − k0 )2 /2σ) − 1
(2π)3 0
We are interested in temperatures below 1 K. Since ∆ ≈ 9 K, we can ignore
the term −1 in the denominator. The main contributions will come from the
neighborhood of the roton minimum, i.e., at k = k0 . Thus we only need to keep
the term ∆ in the numerator. Now change variable of integration to q = k − k0 .
The lower limit of can be replaced by −∞ for low temperatures. Thus
Z ∞
¡
¢
1
U
−β∆
dq(q + k0 )2 exp −β~2 q 2 /2σ
≈ 2 ∆e
V
2π
−∞
We can replace (q + k0 )2 by k02 . Thus we obtain
r
U
k02 ∆ σkB T −∆/kB T
e
≈
V
π
2π~2
r
k02 ∆2 σkT e−∆/kB T
Croton ≈
πρ
2π~2 kB T 2
89
Using data for liquid helium
= 8.65 K
= 1.92 × 108 cm−1
= 1.07 × 10−24 g
∆/kB
k0
σ
we obtain
Croton
µ
¶
387
8.65
≈ 3/2 exp −
J g−1 K−1
T
T
where T is the absolute temperature in K.
18.7
The specific heat is
C = 0.021 T 3 +
µ
¶
387
8.65
exp
−
J g−1 K−1
T
T 3/2
where T is the absolute temperate in K. The numbers in this formula are based
on neutron scattering data independent of the specific heat. The following
plot compares this formula to data on specific heat shown as dots. It is
an absolute comparison with no adjustible parameters. At T =1 K, there is
already noticeable discrepancy, the main source of which probably comes the
approximations we used in the roton specific heat.
Spedic Heat ( J / g-deg )
0.1
0.08
0.06
0.04
0.02
0.5
0.6
0.7
0.8
0.9
1
T (K)
18.8
(a)
¢ 5/2
¡
P1 = −∂E0 /∂V = − (3/5) N ∂ F /∂V = c1 ~2 /m n1 ,
¡
¢2/3
/5.
c1 = 6π 2
(b)
P2 = kB T λ−3 g5/2 (1) = c2 (m2 /~)3/2 (kB T )5/2 ,
=3/2
c2 = (2π)
g5/2 (1).
(c)
90
CHAPTER 18.
kB T << F ,
¡
¢
2/3
kB T << ~2 /2m (3n1 /4π) .
(d)
m1
=
m2
µ
c2
c1
¶2/3 µ
3
4π
¶10/9 µ
kB T
F
¶5/3
(e)
For Fermi degeneracy m1 /m2 << 1. For Bose degeneracy T < Tc . To fulfill
both conditions we must have
µ ¶5/3
n1
m1
K
<
<< 1
n2
m2
where K is a numerical constant.
18.9
In 2D the fugacity z is determined by
Z
d2 k
1
N =A
2 z −1 exp(β~k 2 /2m) − 1
(2π)
where A is the area of the system. By expending the right side in a power
series in z we obtain
1
mkB T
N
ln
=
2
A
2π~
1−z
For 0 ≤ z ≤ 1, there is no upper bound to N, except when T = 0. This means
that, the states of nonzero momentum can accommodate any N , and there is
no Bose-Einstein condensation. The except occurs at T = 0, when all particles
go into the zero-momentum state.
18.10
For a gas of N photons with number conservation, the fugacity z is determined by
Z
d3 k
1
N = 2V
3 z −1 exp(β~ck) − 1
(2π)
The zero-momentum state is ignored in the continuum approximation used. By
expanding the right side in a power series in z we obtain
n=
2
π2
µ
kB T
~c
¶3
g3 (z)
P
where n = N/V , and gn (z) = ∞=1 −n . The function g3 (z) is finite at z = 1,
d
d
but has an infinite second derivative. We can see by noting that z dz
z dz
g3 (z) =
g1 (z) diverges at z = 1. Thus n is finite at z = 1, but cannot be continued
91
beyond z = 1. There is Bose-Einstein condensation to the zero-momentum state
when n exceeds the critical density
nc =
2
π2
µ
kB T
~c
¶3
g3 (1)
At constant density n the condensation occurs below the critical temperature
~c
Tc =
kB
·
π2 n
2g3 (1)
¸1/3
92
CHAPTER 18.
Chapter 19
19.1
Z
Z
Z ∞
eikr cos θ
1
eikr − e−ikr
=
dkk 2
3
2
2
k + 2r0
4π ir 0
k + 2r0
(2π) 0
−1
¡ √ ¢
¡ √ ¢
Z ∞
ikr
exp −r 2r0
1 ∂
e
1 ∂ exp −r 2r0
√
= − 2
dk
=−
=
4π r ∂r −∞ k2 + 2r0
4πr ∂r
4πr
2r0
m(x) =
1
∞
dkk 2 2π
1
d(cos θ)
The dk integration was carried out over a contour in the complex plane.
19.2
(a)
Minimize E(η, ε) with respect to ε:
∂E
= 2cε + gη 2
∂ε
gη2
ε̄ = −
2c
0 =
(b)
Eeff (η) = (at − g) η 2 + b̃η 4
g2
b̃ = b −
4c
(c)
Minimize Eeff (η) with respect to η:
i
h
∂Eeff
= η 2 (at − g) + 4b̃η 2
∂η
(
0
(t > g/a)
q
=
(g − at) /2b̃ (t < g/a)
0 =
η̄
93
94
CHAPTER 19.
The transition temperature corresponds to t = g/a, or Tc +g/a. Thus, the effect
of the coupling g is to raise the transition temperature. The equilibrium order
parameter in the low-temperature phase is, more explicitly,
r³
r
g´
a
Tc +
η̄ =
−T
2
2 (b − g /4c)
a
√
which increase as g increases from 0, and becomes ∞ at g = 4c. The system
becomes unstable for larger values of g, for Eeff has no lower bound.
19.3
(a)
If b = 0, we have the usual quartic curve. Just add a cubic term to get the
graphs shown. We see that the transition temperature is greater than T0 . As
we show in more detail later, S̄ jumps abruptly from a finite value to 0, making
a first-order phase transition. This is typical of a Landau free energy with a
cubic term.
E
S
S
T<T0
T=T0
T>T0
T>>T0
(b)
We see from the sketch above that the conditions for E(S) to be at a
minimum are
(i) ∂E/∂S = 0,
(ii) E ≤ 0,
(iii) S̄ ≤ 0.
The first leads to the condition
¡
¢
S 2at + 3bS + 4cS 2 = 0
The possible roots are
S
S
= 0 Ã
!
r
32cat
3b
−1 ± 1 −
=
8c
9b2
For the nontrivial root to correspond to the minimum, it is necessary that
32cat/9b2 < 1. We see from the graphs that when t = 0 the solution is not
S = 0. Thus we must choose the − sign. Thus
!
Ã
r
3b
32cat
S̄ = −
1+ 1−
8c
9b2
95
if 32cat < 9b2 and E(S̄) < 0. Otherwise S̄ = 0.
(c)
We note that at T = Tc we have ∂E/∂S = 0 and E = 0. Excluding the
trivial root, we can write these two conditions as
2at + 3bS + 4cS 2
at + bS + cS 2
= 0
= 0
Multiplying the second equation by 4 and subtracting it from the first, we find
bS = −2at. Substittuting this into the second equation gives t = b2 /4ac, or
Tc = T0 +
b2
4ac
(d)
When the temperature is increase from below to T = Tc , or t = b2 /4ac, we
have
b
S̄ = −
2c
and E(S̄) = 0. When T is further decreased it jumps to S̄ = 0, where E = 0.
Thus the transition is first-order. This is illustrated in the accompanying sketch.
E
T<Tc
S
S
First-order
transition
T=Tc
T>Tc
Since E is the free energy, the entropy of the system is
−
∂E
= −aS 2
∂T
(It was unthoughtful of us to have used the symbol S for the order parameter!)
Thus, the latent heat is
µ
¶ µ ¶2
b
b2
2
L = aTc S̄ = a T0 +
4ac
2c
(e)
96
CHAPTER 19.
According to the model S̄ decreases steadily as the temperature decreases
below Tc . But the model breaks down when the magnitude of S̄ becomes too
large, for it must saturate when molecular alignment becomes perfect.
19.4
(a)
E = E0 + atφ2 + b̄φ4
¢
¡
b̄ = b + c cos4 θ + sin4 θ
Minimize E with respect to φ:
¢
¡
∂E
= φ 2at + 4b̄φ2
∂φ
at
2
φ̄ = −
2b̄
a2 t2
E(φ̄, θ) = E0 −
2b̄
0 =
Thus the minimum of E occurs at the minimum of b̄.
(b)
c < 0 : θ = 0, or θ = π/2.
b̄min
c = 0 : θ irrelevant.
= b − |c|
s
a|t|
¡
¢
φ̄ =
2 b̄ − |c|
b̄min
= b
r
φ̄ =
a|t|
2b̄
c > 0 : θ = π/4.
c
= b+
s 2
a|t|
φ̄ =
2b̄ + c
b̄min
In all cases, the critical exponent for the order parameter is β = 1/2.
(c)
To find the susceptibility, we turn on an infinitesimally small external field
h = (h1 , h2 ), so that
¡
¡
¢
¡
¢2
¢
E = E0 + at φ21 + φ22 + b φ21 + φ22 − |c| φ41 + φ42 − h1 φ1 − h2 φ2
97
First consider the case h = (h1 , 0).We can take φ =(φ1 , 0). The condition
∂E/∂φ1 = 0 gives
2atφ1 + 4 (b − |c|) φ31 − h1 = 0
Differentiating with respect to h1 yields the longitudinal susceptibility:
χ1 =
∂φ1
1
1
=
2 = − 3at
∂h1
2at + 12 (b − |c|) φ1
−1
The last step is obtained by using φ21 = −at [2 (b − |c|)] . The relevant critical
exponent is γ = 1.
Next consider the case h = (0, h2 ). We must consider φ =(φ1 , φ2 ) with a
small φ2 induced by the transverse field. The field causes the order parameter
to deviate from the x axis by an infinitesimal angle θ, while the magnitude φ is
unchanged, as illustrated in the accompanying sketch.
y
h2
θ
x
We write E in the form
£
¡
¢¤
E = E0 + atφ2 + b − |c| sin4 θ + cos4 θ φ4 − h2 φ sin θ
The condition ∂E/∂φ = 0 gives
£
¡
¢¤
2atφ + 4 b − |c| sin4 θ + cos4 θ φ3 − h2 sin θ = 0
Differentiating with respect to h2 and setting h2 = 0, θ = 0, we obtain
1
∂θ
=
∂h2
16|c|φ3
The transverse susceptibility is
χ2 =
∂φ2
∂θ
1
|c| − b 1
=φ
=
2 = 16a|c| t
∂h2
∂h2
16|c|φ
Thus, the associated critical exponent is γ = 1.
19.5
For given a, let b be the lengths of a diagonal line, and θ the angle between
a diagonal line and the horizontal. Then b cos θ = 12 , and a + 2b sin θ = 1.
p
These give the relation b = 12 1 + (1 − a)2 . The total length of the highway is
p
L = 4b + a = 2 1 + (1 − a)2 + a. The minimum occurs at a = 1 − √13 = 0.306.
√
The minumum length of the highway is L = 1 + 3 = 2.732.
98
CHAPTER 19.
Chapter 20
20.1
The wave-number difference between the condensates is given by k1 − k2 =
2π
−3
,
cm.
λ where λ is the deBroglie wavelength. From the data we have λ/2 =1.5×10
The relative velocity is thus given by
~
(k1 − k2 ) = 0.05 cm/s
m
.
20.2
¡
¢
2
From (20.16) |ψ| = g −1 µ − 12 mω 2 r2 . With (20.11), we put µ = gn0 ,
where n0 = N0 /V . Thus
µ
¶
r2
2
|ψ| = n0 1 −
16πan0 r04
√
The half-width is r02 8πan0 . With the data given, we obtain
a = 5.7 × 10−7 cm
20.3
(a)
From Prob.(17.1), the coordinate of a 1D harmonic oscillator
in
p is expressed
¢
¡
terms of creation and annihilation operators through q = i ~/2mω a − a† .
Thus
¸
·
¢
~
1 1¡ 2
x2 =
a†x ax + −
ax + a†2
x
mω
2 2
and similarly for y and z.The last two terms have no diagonal elements. Therefore
µ
¶
~
3
hn|x2 + y 2 + z 2 |ni =
nx + ny + nz +
mω
2
(b)
99
100
CHAPTER 20.
We know that
¯ À
¿ ¯ 2
¯p
mω 2 r2 ¯¯
¯
n = En
n¯
+
2m
2 ¯
From part (a) we have
1
mω 2
hn|r2 |ni = En
2
2
Subtracting the two equations yields the desired result.
(c)
In the N -boson system, the probability of finding a boson in state n is given
by the fraction of particles in that state:
P (n) =
1
1
N z −1 exp (En /kT ) − 1
Therefore
¡
¢
X
r02 X nx + ny + nz + 32
2
hri =
hn|r |niP (n) =
N n z −1 exp (En /kT ) − 1
n
2
20.4
Estimate the transition temperature T0 by putting the chemical potential
equal to the zero-point energy: µ = 32 ~ω. Thus
N
Z
≈
∞
dnx dny dnz
0
µ
=
kT0
~ω
¶3 Z
∞
1
exp (~ω (nx + ny + nz ) /kT0 ) − 1
dxdydz
0
and we obtain
1
exp (x + y + z) − 1
kT0
= bN 1/3
~ω
where b is given by
b−3 =
Z
∞
dxdydz
0
1
exp (x + y + z) − 1
20.5
(a)
In the semiclassical approximation
N
µ
µ
¶¶
3
dnx dny dnz exp −β~ω nx + ny + nz +
2
0
µ
¶
3
= z (β~ω)−3 exp − β~ω
2
≈ z
Z
∞
101
Thus
3
z ≈ N (β~ω) exp
µ
3
β~ω
2
¶
=
µ
T0
bT
¶3
exp
µ
3
β~ω
2
¶
where we have used T0 defined in Prob.20.4. The chemical potential is
µ ¶
3
T0
µ ≈ kT ln z = ~ω + 3kT ln
2
bT
We should put b ≈ 1 in the spirit of this approximation, so µ → 32 ~ω at T = T0 .
(b)
2
hri
r02
µ
¶
3
dnx dny dnz nx + ny + nz +
2
µ
µ
¶¶
3
× exp −β~ω nx + ny + nz +
2
¸
µ
¶·
3
z
3
−4
−3
=
exp − β~ω 3 (β~ω) + (β~ω)
N
2
2
≈
Z
z
N
We neglected the second term, and obtain
hri2
3kT
≈
= 3b
2
r0
~ω
µ
T
T0
¶
N 1/3
(c)
According to (b), hri2 is a linear function of T . However, the formula becomes invalid below T ≈ T0 , where a condensate begins to form. The condensate wave function is that for nx = ny = 0, and occupies a central region in
the harmonic potential of radius r0 . Thus below T0 the linear plot flattens to a
constant value r02 . (See sketch)
<r 2 >
r 02
T
T0
102
CHAPTER 20.
Chapter 21
21.1
The equation for the magnetic field is the same as that for the vector potential:
¶
µ 2
∂
−
λ
B(x) = 0
∂x2
where λ = 0 outside the medium (x < 0), and λ = 16πe2 n/mc2 inside (x > 0).
Since B = B0 outside, and B(x) must be continuous, the solution inside is
³ √ ´
B(x) = B0 exp − λx
(x > 0)
The penetration depth is λ−1/2 .
21.2
(a)
From Ohm’s law U0 − V = IR. Thus
U0 −
~ dϕ
= I0 R sin ϕ
2e dt
(b)
~ dϕ
= U0 (1 − κ sin ϕ)
2e dt
The left side is the voltage. When it approaches zero, we must have sin ϕ = κ−1 .
This is possible if κ > 1. Thus there is a nonzero current I = κ−1 I0 .
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