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5.5: Photon Statistics

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Thermodynamics of the photon gas

There exists a certain class of particles, including photons and certain elementary excitations in solids such as phonons ( lattice vibrations) and magnons ( spin waves) which obey bosonic statistics but with zero chemical potential. This is because their overall number is not conserved (under typical conditions) – photons can be emitted and absorbed by the atoms in the wall of a container, phonon and magnon number is also not conserved due to various processes, In such cases, the free energy attains its minimum value with respect to particle number when

μ=(FN)T.V=0 .

The number distribution, from Equation ???, is then

n(ε)=1eβε1 .

The grand partition function for a system of particles with μ=0 is

Ω(T,V)=VkBTdεg(ε)ln(1eε/kBT) ,

where g(ε) is the density of states per unit volume.

Suppose the particle dispersion is ε(p)=A|p|σ. We can compute the density of states g(ε):

g(ε)=gddphdδ(εA|p|σ)=gΩdhd0dppd1δ(εApσ)=gΩdσhdAdσ0dxxdσ1δ(εx)=2gσΓ(d/2)(πhA1/σ)dεdσ1Θ(ε),

where g is the internal degeneracy, due, for example, to different polarization states of the photon. We have used the result Ωd=2πd/2/Γ(d/2) for the solid angle in d dimensions. The step function Θ(ε) is perhaps overly formal, but it reminds us that the energy spectrum is bounded from below by ε=0, there are no negative energy states.

For the photon, we have ε(p)=cp, hence σ=1 and

g(ε)=2gπd/2Γ(d/2)εd1(hc)dΘ(ε) .

In d=3 dimensions the degeneracy is g=2, the number of independent polarization states. The pressure p(T) is then obtained using Ω=pV. We have

p(T)=kBTdεg(ε)ln(1eε/kBT)=2gπd/2Γ(d/2)(hc)dkBT0dεεd1ln(1eε/kBT)=2gπd/2Γ(d/2)(kBT)d+1(hc)d0dttd1ln(1et) .

We can make some progress with the dimensionless integral:

Id0dttd1ln(1et)=n=11n0dttd1ent=Γ(d)n=11nd+1=Γ(d)ζ(d+1) .

Finally, we invoke a result from the mathematics of the gamma function known as the doubling formula,

Γ(z)=2z1πΓ(z2)Γ(z+12) .

Putting it all together, we find

p(T)=gπ12(d+1)Γ(d+12)ζ(d+1)(kBT)d+1(c)d .

The number density is found to be

n(T)=dεg(ε)eε/kBT1=gπ12(d+1)Γ(d+12)ζ(d)(kBTc)d .

For photons in d=3 dimensions, we have g=2 and thus

n(T)=2ζ(3)π2(kBTc)3,p(T)=2ζ(4)π2(kBT)4(c)3 .

It turns out that ζ(4)=π490.

Note that c/kB=0.22855cmK, so

kBTc=4.3755T[K]cm1n(T)=20.405×T3[K3]cm3 .

To find the entropy, we use Gibbs-Duhem:

dμ=0=sdT+vdps=vdpdT ,

where s is the entropy per particle and v=n1 is the volume per particle. We then find

s(T)=(d+1)ζ(d+1)ζ(d)kB .

The entropy per particle is constant. The internal energy is

E=lnΞβ=β(βpV)=dpV ,

and hence the energy per particle is

ε=EN=dpv=dζ(d+1)ζ(d)kBT .

Classical arguments for the photon gas

A number of thermodynamic properties of the photon gas can be determined from purely classical arguments. Here we recapitulate a few important ones.

  • Suppose our photon gas is confined to a rectangular box of dimensions Lx×Ly×Lz. Suppose further that the dimensions are all expanded by a factor λ1/3, the volume is isotropically expanded by a factor of λ. The cavity modes of the electromagnetic radiation have quantized wavevectors, even within classical electromagnetic theory, given by

    k=(2πnxLx,2πnyLy,2πnzLz) .

    Since the energy for a given mode is ε(k)=c|k|, we see that the energy changes by a factor λ1/3 under an adiabatic volume expansion VλV, where the distribution of different electromagnetic mode occupancies remains fixed. Thus,

    V(EV)S=λ(Eλ)S=13E .

    Thus,

    p=(EV)S=E3V ,

    as we found in Equation [photE]. Since E=E(T,V) is extensive, we must have p=p(T) alone.
  • Since p=p(T) alone, we have

    (EV)T=(EV)p=3p=T(pT)Vp ,

    where the second line follows the Maxwell relation (SV)p=(pT)V, after invoking the First Law dE=TdSpdV. Thus,

    TdpdT=4pp(T)=AT4 ,

    where A is a constant. Thus, we recover the temperature dependence found microscopically in Equation [photp].
  • Given an energy density E/V, the differential energy flux emitted in a direction θ relative to a surface normal is

    djε=cEVcosθdΩ4π ,

    where dΩ is the differential solid angle. Thus, the power emitted per unit area is

    dPdA=cE4πVπ/20dθ2π0dϕsinθcosθ=cE4V=34cp(T)σT4 ,

    where σ=34cA, with p(T)=AT4 as we found above. From quantum statistical mechanical considerations, we have

    \boldsymbol{\sigma={\pi^2 k_\ssr{B}^4\over 60\,c^2\,\hbar^3}=5.67\times 10^{-8}\,{\RW\over\Rm^2\,\RK^4} \label{stefan}}

    is Stefan’s constant.

Surface temperature of the earth

We derived the result P=σT4A where σ=5.67×108W/\Rm2K4 for the power emitted by an electromagnetic ‘black body’. Let’s apply this result to the earth-sun system. We’ll need three lengths: the radius of the sun R=6.96×108\Rm, the radius of the earth Re=6.38×106\Rm, and the radius of the earth’s orbit ae=1.50×1011\Rm. Let’s assume that the earth has achieved a steady state temperature of Te. We balance the total power incident upon the earth with the power radiated by the earth. The power incident upon the earth is

Pincident=πR2e4πa2eσT44πR2=R2eR2a2eπσT4 .

The power radiated by the earth is

Pradiated=σT4e4πR2e .

Setting Pincident=Pradiated, we obtain

Te=(R2ae)1/2T .

Thus, we find Te=0.04817T, and with T=5780K, we obtain Te=278.4K. The mean surface temperature of the earth is ˉTe=287K, which is only about 10K higher. The difference is due to the fact that the earth is not a perfect blackbody, an object which absorbs all incident radiation upon it and emits radiation according to Stefan’s law. As you know, the earth’s atmosphere retraps a fraction of the emitted radiation – a phenomenon known as the greenhouse effect.

[planck] Spectral density \rho_\ve(\nu,T) for blackbody radiation at three temperatures.
[planck] Spectral density ρε(ν,T) for blackbody radiation at three temperatures.

Distribution of blackbody radiation

Recall that the frequency of an electromagnetic wave of wavevector k is ν=c/λ=ck/2π. Therefore the number of photons NT(ν,T) per unit frequency in thermodynamic equilibrium is (recall there are two polarization states)

N(ν,T)dν=2V8π3d3keck/kBT1=Vπ2k2dkeck/kBT1 .

We therefore have

N(ν,T)=8πVc3ν2ehν/kBT1 .

Since a photon of frequency ν carries energy hν, the energy per unit frequency E(ν) is

E(ν,T)=8πhVc3ν3ehν/kBT1 .

Note what happens if Planck’s constant h vanishes, as it does in the classical limit. The denominator can then be written

ehν/kBT1=hνkBT+O(h2)

and

\boldsymbol{\CE\ns_\ssr{CL}(\nu,T)=\lim_{h\to 0} \CE(\nu)=V\cdot{8\pi\kT\over c^3}\,\nu^2\ .}

In classical electromagnetic theory, then, the total energy integrated over all frequencies diverges. This is known as the ultraviolet catastrophe, since the divergence comes from the large ν part of the integral, which in the optical spectrum is the ultraviolet portion. With quantization, the Bose-Einstein factor imposes an effective ultraviolet cutoff kBT/h on the frequency integral, and the total energy, as we found above, is finite:

E(T)=0dνE(ν)=3pV=Vπ215(kBT)4(c)3 .

We can define the spectral density ρε(ν) of the radiation as

ρε(ν,T)E(ν,T)E(T)=15π4hkBT(hν/kBT)3ehν/kBT1

so that ρε(ν,T)dν is the fraction of the electromagnetic energy, under equilibrium conditions, between frequencies ν and ν+dν, 0dνρε(ν,T)=1. In Figure [planck] we plot this in Figure [planck] for three different temperatures. The maximum occurs when shν/kBT satisfies

dds(s3es1)=0s1es=3s=2.82144 .

What if the sun emitted ferromagnetic spin waves?

We saw in Equation [jephoton] that the power emitted per unit surface area by a blackbody is σT4. The power law here follows from the ultrarelativistic dispersion ε=ck of the photons. Suppose that we replace this dispersion with the general form ε=ε(k). Now consider a large box in equilibrium at temperature T. The energy current incident on a differential area dA of surface normal to ˆz is

dP=dAd3k(2π)3Θ(cosθ)ε(k)1ε(k)kz1eε(k)/kBT1 .

Let us assume an isotropic power law dispersion of the form ε(k)=Ckα. Then after a straightforward calculation we obtain

dPdA=σT2+2α ,

where

σ=ζ(2+2α)Γ(2+2α)gk2+2αBC2α8π2 .

One can check that for g=2, C=c, and α=1 that this result reduces to that of Equation [stefan].


This page titled 5.5: Photon Statistics is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Daniel Arovas.

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