Homework #5
1. Kittel, Ch4-2. Surface temperature of the Sun. The value of the total radiant energy
flux density at the earth from the Sun normal to the incident rays is called the solar
constant of the Earth. The observed value integrated over all emission wavelengths
and refereed to the mean Earth-Sun distance is:
solar constant = 0.136 J s-1 cm-2.
(49)
(a) Show that the total rate of energy generation of the Sun is 4x1026 Js-1. (b) From
this result and the Stefan-Bolzmann constant σ B = 5.67 × 10 −12 Js −1cm −2 K −4 , show
that the effective temperature of the surface of the Sun treated as a black body is
T ≈ 6000 K . Take the distance of the Earth from the Sun as 1.5x1013 cm and the
radius of the Sun as 7x1010 cm.
(a) The total rate of energy generated by the Sun should equal the total rate of energy
received by 4πr 2 , where r = 1.5 × 1013 cm is the distance between the Sun and the
Earth. Then with the solar constant, we can calculate the total rate of energy
generation of the Sun as
W = 4πr 2 × 0.136 Js −1cm −2 = 4π (1.5 × 1013 ) 2 × 0.136 Js −1 = 3.8 × 10 26 Js −1 .
(b) The total rate of energy generated by the Sun should equal the total energy emitted
by the surface of the Sun. Thus W = 4πR 2 × σ B T 4 , where R and T are the radius and
surface temperature of the Sun, respectively.
1
1
1
 W
4 
4
3.8 × 10 26
15 4



(
)
T = 
=
=
1
.
09
×
10
= 5.7 × 10 3 K .
2 
 4π × 5.67 × 10 −12 × 7 2 × 10 20 
πσ
4
R
B




2. Kittel, Ch4-5. Surface temperature of the Earth. Calculate the temperature of the
surface of the Earth on the assumption that as a black body in thermal equilibrium it
reradiates as much thermal radiation as it receives from the Sun. Assume also that the
surface of the Earth is at a constant temperature over the day-night cycle. Use
T⊕ = 5800 K ; R⊕ = 7 × 1010 cm ; and the Earth-Sun distance of 1.5x1013 cm.
The total rate of energy emitted by the Sun is 4πR⊕2 × σ B T⊕4 . At the position of the
2
4πR⊕2 × σ B T⊕4  R⊕ 
 σ B T⊕4 . Here
Earth, the rate of energy received per unit area is
= 
2
4πr
r 
r is the Sun-Earth distance. Thus the rate of energy received by the earth from the
2
R 
Sun is πR  ⊕  σ B T⊕4 . This energy is eventually balanced by the rate of energy
r 
emitted by the Earth which equals 4πRe2 × σ B Te4 . Therefore we have
2
e
2
R 
πR  ⊕  σ B T⊕4 = 4πRe2 × σ B Te4 , or
r 
2
e
Te = T⊕
3.
2
 R⊕  4
 T⊕ = 4Te4 .

r 
R⊕
7 × 1010
= 5800 ×
= 280 K .
2r
2 × 1.5 × 1013
Kittel, Ch4-7. Free energy of photon gas. (a) Show that the partition function of a
photon gas is given by
Z = ∏ [1 − exp( − ω n / τ )]−1 .
(53)
n
where the product is over the modes n. (b) The Helmholtz free energy is found
directly from (53) as
F = τ ∑ log[1 − exp( − ω n / τ )] .
(54)
n
Transform of the sum to an integral; integrate by parts to find
F =−
π 2Vτ 4
.
45 3c 3
(a) For a single mode of ω, the corresponding partition function is
∞
Z ω = ∑ exp( −nω / τ ) =
n =0
1
.
1 − exp( − ω / τ )
(55)
1
.
1 − exp( − ω n / τ )
Then Z = ∏ Z ω = ∏
ω
n
(b) F = −τ log Z = −τ log ∏
n
1
= τ ∑ log[1 − exp( − ω n / τ )].
1 − exp( − ω n / τ )
n
For small separation between ω from adjacent modes, the sum can be replaced by
2V
integration, ∑ → 3 ∫ 4πp 2 dp . The factor 2 comes from the degeneracy of the
h
n
left- and right-polarized modes for each ω. Note ω = pc for photon, and ω runs
from zero to infinity.
Then
2V
→ 3
h
∑
n
∞
∫
0
2V
4πp dp = 3
h
2
F = τ ∑ log[1 − exp( − ω n / τ )] =
n
Let x =
4.
∞
∫
0
V
4π 3ω 2 dω
= 2 3
3
c
π c
Vτ
π 2c3
∞
∫
∞
∫
ω 2 dω .
0
dω ⋅ ω 2 log[1 − exp( − ω / τ )] .
0
ω
.
τ
F=
Vτ 4
π 2 3c 3
=−
Vτ
3π 2 3c 3
4
∞
∫
dx ⋅ x 2 log[1 − exp( − x )] =
0
∞
∫
0
Vτ 4
3π 2 3c 3
∞
∫
d ( x 3 ) log[1 − exp( − x )]
0
x dx
Vτ
π
π Vτ 4
=
−
=
−
−
ex −1
3π 2 3c 3 15
45 3c 3
3
4
4
2
Kittel, Ch4-11. Heat capacity of solids in high temperature limit. Show that in the
limit of T >> θ the heat capacity of a solid goes towards the limit of CV → 3Nk B , in
conventional units. To obtain higher accuracy when T is only moderately larger than
θ, the heat capacity can be expanded as a power series in 1/T, of the form
a 

CV = 3Nk B 1 − ∑ nn  .
T 
n

(56)
Determine the first nonvanishing term in the sum. Check your result by inserting
T = θ and comparing with Table 4.2.
The partition function for mode ω, is
∞
Z ω = ∑ exp( −nω / τ ) =
n =0
Then Z = ∏ Z ω = ∏
ω
n
1
.
1 − exp( − ω / τ )
1
.
1 − exp( − ω n / τ )
F = −τ log Z = τ ∑ log[1 − exp( −ω n / τ )].
n
U =τ2
ω n
∂ log Z
.
=∑
∂τ
exp( ω n / τ ) − 1
n
CV = k B
ω n exp( ω n / τ ) ω n
exp( ω n / τ )
∂U
= kB ∑
⋅ 2 = kB ∑
2
∂τ
[exp( ω n / τ ) − 1] τ
[exp( ω n / τ ) − 1]2
n
n
In the high T limit,
CV ≈ k B ∑
n
 ω 
⋅ n  .
 τ 
2
ω n
<< 1, exp( ω n / τ ) ≈ 1 + ω n / τ .
τ
1
( ω n / τ ) 2
 ω 
⋅  n  = k B ∑ 1 . Since there are 3N modes,
 τ 
n
2
Then CV ≈ 3Nk B .
If we keep higher order terms in the exp( ω n / τ ) expansion,
exp( ω n / τ )
1  ω 
 ω n 
⋅
 ≈ 1−  n  −m.
2 
[exp( ω n / τ ) − 1]  τ 
12  τ 
2
2
2


a 
1  ω 

CV ≈ k B ∑ 1 −  n  − m = 3Nk B 1 − ∑ nn  .
T 
n
n

 12  τ 

The first non-vanishing term in the expansion is
2

1  ω  
CV = k B ∑ 1 −  n   .
n
 12  τ  
Replace the sum with integration
∑
n
→
9N
ω D3
ωD
∫
0
ω 2 dω , we have
∑ 1 = 3N .
n
ωD
2

1  ω   9 Nk B
2
d
ω
ω
1
−

 =

∫0
ω D3
 12  τ  
2

1  ω D2  

 
= 3Nk B 1 −
 20  τ  
CV =
9 Nk B
ω D3
2
 ω D3
1   ω D5 
−
 


 3 12  τ  5 
With the relation of ω D = k Bθ , where θ is the Debye temperature, we have
2
2

CV
1 θ  
1 θ 
= 1−   .
CV = 3Nk B 1 −    , or
CV (T = ∞)
20  T 
 20  T  
θ
T
Table 4.2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.5
2
3
4
5
6
7
8
9
10
15
5.
24.943
24.93
24.89
24.83
24.75
24.63
24.5
24.34
24.16
23.96
23.74
22.35
20.59
16.53
12.55
9.2
6.23
4.76
3.45
2.53
1.891
0.576
2

1 θ  
CV = 24.9431 −   
 20  T  
24.943
24.93
24.89
24.83
24.74
24.63
24.49
24.33
24.15
23.93
23.70
22.14
19.95
13.72
4.99
-6.24
-19.95
-36.17
-54.88
-76.07
-99.77
-255.67
Kittel, Ch4-14, Heat capacity of liquid 4He at low temperature. The velocity of
longitudinal sound waves in liquid 4He at temperatures below 0.6K is
2.383 × 10 4 cms −1 . There are no transverse sound waves in the liquid. The density is
0.145gcm −3 . (a) Calculate the Debye temperature. (b) Calculate the heat capacity per
gram on the debye theory and compare with the experimental value
CV = 0.0204 × T 3 , in J g-1K-1. The T3 dependence of the experimental value suggests
that phonons are the most important excitations in liquid 4He below 0.6K. Note that
the experimental value has been expressed per gram of liquid. The experiments are
due to J. Wiebes, C. G. Niels-Hakkenberg, and H. C. Krammers, Physca 32, 625
(1957).
(a) Since there are no transverse wave, the degeneracy is one. With the relation of
ω = pv , where v = 2.38 × 10 4 cm / s is the speed of sound, we have
V
3N = 3
h
∫
V
4πp dp = 3
h
2
ωD
∫
0
Vω D3
4π 3ω 2 dω
=
.
v3
6π 2 v 3
1
1
 18 Nπ 2  3
ω D v  18 Nπ 2  3
 or θ =
 .
ω D = v ⋅ 
=
⋅
kB
k B  V 
 V 
The density N/V can be obtained from the mass density ρ and the atomic mass m of
He.
N ρ
0.145gcm −3
= =
= 2.17 × 10 22 cm −3 .
− 24
V m 4 × 1.67 × 10 g
Then
1
1
v  18 Nπ 2  3 1.05 × 10 −27 × 2.383 × 10 4
22
2 3
 =
(
)
⋅ 
⋅
18
×
2
.
17
×
10
×
3
.
14
= 28.4 K .
θ=
kB  V 
1.38 × 10 −16
(b) The heat capacity is CV =
N=
12π 4 Nk B
× T 3 . For each gram, the number of He is
5θ 3
1
= 1.5 × 10 23 .
− 24
4 × 1.67 × 10
12π 4 1.5 × 10 23 × 1.38 × 10 −23
× T 3 = 0.021 × T 3 Jg −1 K −1 , which agrees very
5 × 28.4 3
well with the experimental result.
Then CV =
6. Kittel, Ch4-15, Angular distribution of radiant energy flux. (a) Show that the
spectral density of radiant energy flux that arrives in the solid angle dΩ is
cuω cos θ ⋅ dΩ / 4π , where θ is the angle the normal to the unit area makes with the
incident ray, and uω is the energy density per unit frequency range. (b) Show that the
1
sum of this quantity over all incident rays is cuω .
4
(a) Noting that only the normal component contributes to the emission, we have
WdΩ / 4π = cu cos θ ⋅ dΩ / 4π .
θ
(b) The averaged emission power is
W =
π /2
∫
0
cu
=
2
π /2
∫
0
dθ
2π
∫
dφ ⋅ cu cos θ ⋅ sin θ / 4π
0
dθ cos θ ⋅ sin θ =
cu
4