Homework 4
Duygu Can
01 Dec 2010
homework 4 solutions
1.1
Problem 4.1
Question
Show that the number of photons
in a cavity of volume V is
P
< sn > in equilibrium at temperature τ
N = 2.404π 2 V (τ /h̄c)3
2
(1.1)
3
From (23) the entropy is σ = (4π V /45)(τ /h̄c) , whence σ/N ' 3.602. It is
believed that the total number of photons in the universe is 108 larger than the
total number of nucleons(protons, neutrons). Because both entropies are of the
order of the respective number of particles (see Eq. 3.76), the photons provide
the dominant contribution to the universe, although the particles dominate the
total energy. We believe that the entropy of the photons is essentially constant,
so that the entropy of the universe is approximately constant with time.
Solution
N
=
X
< sn >
(1.2)
n
=
X
1
eh̄ωn /τ
n
−1
(1.3)
q
where ωn = nπc/L and n = n2x + n2y + n2z . The sum over nx ,ny ,nz is replaced
by an integral over the volume element dnx dny dnz as follows,
Z
X
1 ∞
(...) = 2 ×
4πn2 dn(...)
(1.4)
8 0
Here the factor 2 arises since there are two independent polarizations of the
electromagnetic field, and there is also a factor of 18 = 213 since only the positive
octant of the space is involved. Then,
X
1
N =
(1.5)
h̄ω
/τ
n
e
−1
n
Z ∞
n2
≈ π
dn h̄nπc/Lτ
(1.6)
e
−1
0
Setting u = h̄nπc/Lτ we get,
N
=
=
≈
Z
Lτ 3 ∞
u2
π (
)
du u
h̄πc
e −1
0
τ
π −2 V ( )3 Γ(3)ζ(3)
h̄c
τ
−2
2.404π V ( )3
h̄c
4
2
(1.7)
(1.8)
(1.9)
Problem 4.5
While taking the integral wwe have utilized,
Z
Γ(s)ζ(s) =
0
∞
xs−1 dx
ex − 1
(1.10)
where Γ is the Gamma function and ζ is the Riemann ζ-function (http :
//en.wikipedia.org/wiki/Riemann zeta f unction).
1.2
Problem 4.5
Question
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 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 = 5800K; R = 7 × 1010 cm; and
the Earth-Sun distance of DES = 1.5 × 1013 cm.
Solution
Total energy that sun produces,
U
= J S
(1.11)
2
= σT4 4πR
(1.12)
Flux on Earth,
JE
U
2
4πDES
R 2
= σT4 (
)
DES
=
(1.13)
(1.14)
Energy received by Earth,
2
UE = πRE
JE
(1.15)
2
UE = 4πRE
σTE4
(1.16)
Energy re-radiated from Earth,
Equating these two energies and solving for TE we get,
r
TE
R
T
2DES
≈ 280K
=
(1.17)
(1.18)
3
1. Homework 4 Solutions
1.3
Problem 4.9
Question
Consider a transmission line of length L on which electromagnetic waves satisfy
the one-dimensional wave equation v 2 ∂ 2 E/∂x2 = ∂ 2 E/∂t2 , where E is an
electric field component. Find the heat capacity of the photons on the line,
when in thermal equilibrium at temperature τ . The enumaration of modes
proceeds in the usual way for one dimension: take the solutions as standing
waves with zero amplitude at each end of the line.
Solution
Wave equation in one dimension is,
v2
∂2E
∂2E
=
2
∂x
∂t2
(1.19)
Assume solution in the seperable form, E = X(x)T (t). Subtitute this form of
the solution into the wave equaintion.
v2
∂2T
∂2X
T
=
X
∂x2
∂t2
(1.20)
Dividing both parts by X(x)T (t) we get,
v2 ∂ 2 X
1 ∂2T
=
2
X ∂x
T ∂t2
(1.21)
Notice that LHS of the equation above depends only on x and the RHS of the
equation depends only on t. Then they must both equal to a constant, say
−ω 2 (The negative sign is chosen to satisfy the BC’s on both ends). Then the
solution of the wave equation in one dimension is,
E = E0 sin(ωt) sin(ωx/v)
(1.22)
Since the wave is confined to (0, L), ωn = nπv/L.
The total energy is,
U
=
X
< sn > h̄ωn
(1.23)
h̄ωn
h̄ω
e n /τ −
(1.24)
n
=
X
n
=
≈
4
1
h̄nπv/L
h̄nπv/Lτ − 1
e
n
Z
Lτ 2 ∞ duu
h̄πv 0 eu − 1
X
(1.25)
(1.26)
Problem 4.9
=
=
=
Lτ 2
Γ(2)ζ(2)
h̄πv
Lτ 2 π 2
h̄πv 6
Lπτ 2
h̄6v
(1.27)
(1.28)
(1.29)
The heat capacity is,
CL
=
=
∂U
)L
∂τ
Lπτ
3h̄v
(1.30)
(1.31)
5