Statistical Mechanics Spring 2012 Homework 5Due Friday April 6

1. The occupancy of the first excited state for temperatures below the con-densation temperature is given by

〈N1〉 =1

eβ~ω0 − 1≈ kT

~ω0∼ N1/3

(T

Tc

)since kTc � ~ω0.

〈N1〉N∼ 1

N2/3

(T

Tc

)→ 0 in the thermodynamic limit.

2. Pathria and Beale 7.19. The density of states for a two-dimensional har-monic oscillator isa(ε) = ε/(~ω0)2 so the number particles in the trap is given by

N(T, µ) =

∫dε

ε

(~ω)21

eβ(ε−µ) − 1.

As T → Tc, µ→ 0 so

N =

∫dε

ε

(~ω)21

eβc(ε) − 1=

(kTc~ω0

)2 ∫xdx

ex − 1= ζ(2)

(kTc~ω0

)2

=π2

6

(kTc~ω0

)2

.

so kTc = ~ω√

6N/π2. The condensate fraction for T ≤ Tc isN0/N = 1 − (T/Tc)

2. For this two-dimensional theory to be valid, theoccupancy of the first excited z-state must be negligible which requires~ωz � kTc ∼

√N~ω0, i.e. ωz �

√Nω0.

3. Solution: for power-law density of states

a(ε) = αεσ

(d− 1)!(ε1)σ+1

〈N〉 =

(kT

ε1

)σ+1 ∞∑j=1

zj

jσ+1

〈N〉 = α

(kT

~ω0

)σ+1

gσ+1(z)

N = α

(kT

~ω0

)σ+1

ζσ+1

ζ(σ + 1) =

∞∑j=1

1

jσ+1≈∫ ∞1

ds

sσ+1=

1

σfor σ � 1.

kTcε1

=

(N

αζσ+1

)1/(σ+1)

≈(σN

α

)1/(σ+1)

→ 0 as σ → 0.

1

For d-dimensional harmonic oscillator,

a(ε) =εd−1

(d− 1)!(~ω0)d

〈N〉 =

∫ ∞0

a(ε)dε

eβ(ε−µ) − 1=

(kT

~ω0

)d ∞∑j=1

zd

jd

〈N〉 =

(kT

~ω0

)dgd(z)

N =

(kTc~ω0

)dζ(d)

ζ(d) =

∞∑j=1

1

jd≈∫ ∞1

ds

sd=

1

d− 1for d− 1� 1.

kTc~ω0

=

(N

ζ(d)

)1/d

≈ ((d− 1)N)1/d → 0 as d→ 1.

The ground state occupancy and internal energy below the critical tem-perature are given by

N0

N= 1−

(T

Tc

)d,

U(T ) =d!ζ(d+ 1)

(d− 1)!

(kT )d+1

(~ω)d, i.e.

U(T )

NkTc=dζ(d+ 1)

ζ(d)

(T

Tc

)d+1

≈ (d− 1)ζ(2)

(T

Tc

)for d− 1� 1.

4. Pathria and Beale 7.21 Using expressions (7.2.12) and (7.2.23), we readilyget

U

N̄=

π4

30ζ(3)kT ' 2.7 kT .

Note that the numerical factor appearing here is actually Γ(4)ζ(4)/Γ(3)ζ(3).

5. Pathria and Beale 7.23 The total power radiated by the sun (radius Rsand temperature Ts is

Ps = 4πσT 4sR

2s

where σ = π2k4T 4/60~3c5, so the intensity is the earth’s distance fromthe sun (Re)is

Ie =4πσT 4

sR2s

4πR2e

= σ

(RsRe

)2

.

2

The radiation pressure acting on a blackbody is P = I/c. The equilibriumtemperature of a flat plate blackbody absorber (one-sided) is determinedby the power absorbed being equal to the power emitted, i.e. σT 4

e =Ie so Te = Ts

√Rs/Re ' 396 K. This, not coincidentally, is about the

equilibrium temperature of the earth. A better estimate would include theemissivity of the sun and earth, the albedo of the earth, and the sphericalshape of the the earth and its rotation which averages the temperatureover the whole earth.

6. Pathria and Beale 7.24 The number density of photons in the cosmicmicrowave background (CMB) follows from equation (7.3.23)

n =2ζ(3)

π2

(kT

~c

)3

' 4.10× 108 m−3 ' 410. cm−3

The energy density is

u =π2

15

(kT )4

(~c)3' 4.17× 10−14 J/m3.

The entropy density is

s =4π2k

45

(kT

~c

)3

' 1.48× 109km−3 ' 2.04× 10−14 J/m3K.

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