Homework 4

Duygu Can 01 Dec 2010

homework 4 solutions

1.1 Problem 4.1

Question

Show that the number of photons∑

< sn > in equilibrium at temperature τin a cavity of volume V is

N = 2.404π2V (τ/h̄c)3 (1.1)

From (23) the entropy is σ = (4π2V/45)(τ/h̄c)3, whence σ/N ' 3.602. It isbelieved that the total number of photons in the universe is 108 larger than thetotal number of nucleons(protons, neutrons). Because both entropies are of theorder of the respective number of particles (see Eq. 3.76), the photons providethe dominant contribution to the universe, although the particles dominate thetotal 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 =∑n

< sn > (1.2)

=∑n

1eh̄ωn/τ − 1

(1.3)

where ωn = nπc/L and n =√n2x + n2

y + n2z. The sum over nx,ny,nz is replaced

by an integral over the volume element dnxdnydnz as follows,∑(...) = 2× 1

8

∫ ∞0

4πn2dn(...) (1.4)

Here the factor 2 arises since there are two independent polarizations of theelectromagnetic field, and there is also a factor of 1

8 = 123 since only the positive

octant of the space is involved. Then,

N =∑n

1eh̄ωn/τ − 1

(1.5)

≈ π

∫ ∞0

dnn2

eh̄nπc/Lτ − 1(1.6)

Setting u = h̄nπc/Lτ we get,

N = π4(Lτ

h̄πc)3

∫ ∞0

duu2

eu − 1(1.7)

= π−2V (τ

h̄c)3Γ(3)ζ(3) (1.8)

≈ 2.404π−2V (τ

h̄c)3 (1.9)

2

Problem 4.5

While taking the integral wwe have utilized,

Γ(s)ζ(s) =∫ ∞

0

xs−1dx

ex − 1(1.10)

where Γ is the Gamma function and ζ is the Riemann ζ-function (http ://en.wikipedia.org/wiki/Riemann zeta function).

1.2 Problem 4.5

Question

Calculate the temperature of the surface of the Earth on the assumption that asa black body in thermal equilibrium it reradiates as much thermal radiation itreceives from the Sun. Assume also that the surface of the Earth is at a constanttemperature over the day-night cycle. Use T� = 5800K; R� = 7×1010cm; andthe Earth-Sun distance of DES = 1.5× 1013cm.

Solution

Total energy that sun produces,

U� = J�S� (1.11)= σT 4

�4πR2� (1.12)

Flux on Earth,

JE =U

4πD2ES

(1.13)

= σT 4�(

R�DES

)2 (1.14)

Energy received by Earth,

UE = πR2EJE (1.15)

Energy re-radiated from Earth,

UE = 4πR2EσT

4E (1.16)

Equating these two energies and solving for TE we get,

TE =√

R�2DES

T� (1.17)

≈ 280K (1.18)

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1. Homework 4 Solutions

1.3 Problem 4.9

Question

Consider a transmission line of length L on which electromagnetic waves satisfythe one-dimensional wave equation v2∂2E/∂x2 = ∂2E/∂t2, where E is anelectric field component. Find the heat capacity of the photons on the line,when in thermal equilibrium at temperature τ . The enumaration of modesproceeds in the usual way for one dimension: take the solutions as standingwaves with zero amplitude at each end of the line.

Solution

Wave equation in one dimension is,

v2 ∂2E

∂x2=∂2E

∂t2(1.19)

Assume solution in the seperable form, E = X(x)T (t). Subtitute this form ofthe solution into the wave equaintion.

v2 ∂2X

∂x2T = X

∂2T

∂t2(1.20)

Dividing both parts by X(x)T (t) we get,

v2

X

∂2X

∂x2=

1T

∂2T

∂t2(1.21)

Notice that LHS of the equation above depends only on x and the RHS of theequation depends only on t. Then they must both equal to a constant, say−ω2 (The negative sign is chosen to satisfy the BC’s on both ends). Then thesolution 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 =∑n

< sn > h̄ωn (1.23)

=∑n

h̄ωneh̄ωn/τ − 1

(1.24)

=∑n

h̄nπv/L

eh̄nπv/Lτ − 1(1.25)

≈ Lτ2

h̄πv

∫ ∞0

duu

eu − 1(1.26)

4

Problem 4.9

=Lτ2

h̄πvΓ(2)ζ(2) (1.27)

=Lτ2

h̄πv

π2

6(1.28)

=Lπτ2

h̄6v(1.29)

The heat capacity is,

CL =∂U

∂τ)L (1.30)

=Lπτ

3h̄v(1.31)

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