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=(∂F∂N)∗T.V=0 .
The number distribution, from Equation ???, is then
n(ε)=1eβε−1 .
The grand partition function for a system of particles with μ=0 is
Ω(T,V)=VkBT∞∫−∞dεg(ε)ln(1−e−ε/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)=g∫ddphdδ(ε−A|p|σ)=gΩ∗dhd∞∫0dppd−1δ(ε−Apσ)=gΩ∗dσhdA−dσ†∞∫0dxxdσ−1δ(ε−x)=2gσΓ(d/2)(√πhA1/σ†)dεdσ−1†Θ(ε),
where g is the internal degeneracy, due, for example, to different polarization states of the photon. We have used the result Ω∗d=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)εd−1(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)=−kBT∞∫−∞dεg(ε)ln(1−e−ε/kBT)=−2gπd/2Γ(d/2)(hc)−dkBT∞∫0dεεd−1†ln(1−e−ε/kBT)=−2gπd/2Γ(d/2)(kBT)d+1(hc)d∞∫0dttd−1ln(1−e−t) .
We can make some progress with the dimensionless integral:
I∗d≡−∞∫0dttd−1ln(1−e−t)=∞∑n=11n∞∫0dttd−1e−nt=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)=2z−1√πΓ(z2)Γ(z+12) .
Putting it all together, we find
p(T)=gπ−12(d+1)†Γ(d+12)ζ(d+1)(kBT)d+1(ℏc)d .
The number density is found to be
n(T)=∞∫−∞dεg(ε)eε/kBT−1=gπ−12(d+1)†Γ(d+12)ζ(d)(kBTℏc)d .
For photons in d=3 dimensions, we have g=2 and thus
n(T)=2ζ(3)π2(kBTℏc)3,p(T)=2ζ(4)π2(kBT)4(ℏc)3 .
It turns out that ζ(4)=π490.
Note that ℏc/kB=0.22855cm⋅K, so
kBTℏc=4.3755T[K]cm−1⟹n(T)=20.405×T3[K3]cm−3 .
To find the entropy, we use Gibbs-Duhem:
dμ=0=−sdT+vdp⟹s=vdpdT ,
where s is the entropy per particle and v=n−1 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=−∂lnΞ∂β=−∂∂β(βpV)=d⋅pV ,
and hence the energy per particle is
ε=EN=d⋅pv=d⋅ζ(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 L∗x×L∗y×L∗z. 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=(2πn∗xL∗x,2πn∗yL∗y,2πn∗zL∗z) .
Since the energy for a given mode is ε(k)=ℏc|k|, we see that the energy changes by a factor λ−1/3 under an adiabatic volume expansion V→λV, where the distribution of different electromagnetic mode occupancies remains fixed. Thus,V(∂E∂V)∗S=λ(∂E∂λ)∗S=−13E .
Thus,p=−(∂E∂V)∗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
(∂E∂V)∗T=(∂E∂V)∗p=3p=T(∂p∂T)∗V−p ,
where the second line follows the Maxwell relation (∂S∂V)†p=(∂p∂T)†V, after invoking the First Law dE=TdS−pdV. Thus,TdpdT=4p⟹p(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
dj∗ε=c⋅EV⋅cosθ⋅dΩ4π ,
where dΩ is the differential solid angle. Thus, the power emitted per unit area isdPdA=cE4πVπ/2∫0dθ2π∫0dϕsinθ⋅cosθ=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=σT4⋅A where σ=5.67×10−8W/\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 R∗e=6.38×106\Rm, and the radius of the earth’s orbit a∗e=1.50×1011\Rm. Let’s assume that the earth has achieved a steady state temperature of T∗e. We balance the total power incident upon the earth with the power radiated by the earth. The power incident upon the earth is
P∗incident=πR2e4πa2e⋅σT4⊙⋅4πR2⊙=R2eR2⊙a2e⋅πσT4⊙ .
The power radiated by the earth is
P∗radiated=σT4e⋅4πR2e .
Setting P∗incident=P∗radiated, we obtain
T∗e=(R∗⊙2a∗e)1/2T∗⊙ .
Thus, we find T∗e=0.04817T∗⊙, and with T∗⊙=5780K, we obtain T∗e=278.4K. The mean surface temperature of the earth is ˉT∗e=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.
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 N∗T(ν,T) per unit frequency in thermodynamic equilibrium is (recall there are two polarization states)
N(ν,T)dν=2V8π3⋅d3keℏck/kBT−1=Vπ2⋅k2dkeℏck/kBT−1 .
We therefore have
N(ν,T)=8πVc3⋅ν2ehν/kBT−1 .
Since a photon of frequency ν carries energy hν, the energy per unit frequency E(ν) is
E(ν,T)=8πhVc3⋅ν3ehν/kBT−1 .
Note what happens if Planck’s constant h vanishes, as it does in the classical limit. The denominator can then be written
ehν/kBT−1=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=V⋅π215(kBT)4(ℏc)3 .
We can define the spectral density ρ∗ε(ν) of the radiation as
ρ∗ε(ν,T)≡E(ν,T)E(T)=15π4hkBT(hν/kBT)3ehν/kBT−1
so that ρ∗ε(ν,T)dν is the fraction of the electromagnetic energy, under equilibrium conditions, between frequencies ν and ν+dν, ∞∫0dνρ∗ε(ν,T)=1. In Figure [planck] we plot this in Figure [planck] for three different temperatures. The maximum occurs when s≡hν/kBT satisfies
dds(s3es−1)=0⟹s1−e−s=3⟹s=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=ℏck 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=dA⋅∫d3k(2π)3Θ(cosθ)⋅ε(k)⋅1ℏ∂ε(k)∂k∗z⋅1eε(k)/k∗BT−1 .
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+2αBC−2α8π2ℏ .
One can check that for g=2, C=ℏc, and α=1 that this result reduces to that of Equation [stefan].