# 5.5: Photon Statistics


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$$\newcommand\Vpi ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[22], line 1, column 1 at template() at (Bookshelves/Thermodynamics_and_Statistical_Mechanics/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/05:_Noninteracting_Quantum_Systems/5.05:_Photon_Statistics), /content/body/p/span, line 1, column 23 $$
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$$\newcommand\Vsigma ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[26], line 1, column 1 at template() at (Bookshelves/Thermodynamics_and_Statistical_Mechanics/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/05:_Noninteracting_Quantum_Systems/5.05:_Photon_Statistics), /content/body/p/span, line 1, column 23 $$
$$\newcommand\Vvarsigma ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[27], line 1, column 1 at template() at (Bookshelves/Thermodynamics_and_Statistical_Mechanics/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/05:_Noninteracting_Quantum_Systems/5.05:_Photon_Statistics), /content/body/p/span, line 1, column 23 $$
$$\newcommand\Vtau ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[28], line 1, column 1 at template() at (Bookshelves/Thermodynamics_and_Statistical_Mechanics/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/05:_Noninteracting_Quantum_Systems/5.05:_Photon_Statistics), /content/body/p/span, line 1, column 23 $$
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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

$\mu=\pabc{F}{N}{T.V}=0\ .$

The number distribution, from Equation \ref{benum}, is then

$n(\ve)={1\over e^{\beta\ve}-1}\ .$

The grand partition function for a system of particles with $$\mu=0$$ is

$\Omega(T,V)= V\kT\!\!\int\limits_{-\infty}^\infty\!\!\!d\ve\,g(\ve)\,\ln\big(1-e^{-\ve/\kT}\big)\ ,$

where $$g(\ve)$$ is the density of states per unit volume.

Suppose the particle dispersion is $$\ve(\Bp)=A|\Bp|^\sigma$$. We can compute the density of states $$g(\ve)$$:

$\begin{split} g(\ve)&=\Sg\!\int\!\!{d^d\!p\over h^d}\>\delta\big(\ve-A|\Bp|^\sigma\big) ={\Sg\Omega\ns_d\over h^d}\!\int\limits_0^\infty\!\!dp\>p^{d-1}\,\delta(\ve-Ap^\sigma)\\ &={\Sg\Omega\ns_d\over \sigma h^d}\,A\nsub^{-{d\over \sigma}}\!\!\int\limits_0^\infty\!\!dx\> x^{d\over \sigma -1}\>\delta(\ve-x) ={2\,\Sg\over \sigma\,\RGamma(d/2)}\bigg({\sqrt{\pi}\over h A\nsub^{1/\sigma}}\bigg)^{\!\!d}\, \ve\nsub^{{d\over \sigma}-1}\,\RTheta(\ve)\quad , \end{split}$

where $$\Sg$$ is the internal degeneracy, due, for example, to different polarization states of the photon. We have used the result $$\Omega\ns_d=2\pi^{d/2}\big/\RGamma(d/2)$$ for the solid angle in $$d$$ dimensions. The step function $$\RTheta(\ve)$$ is perhaps overly formal, but it reminds us that the energy spectrum is bounded from below by $$\ve=0$$, there are no negative energy states.

For the photon, we have $$\ve(\Bp)=cp$$, hence $$\sigma=1$$ and

$g(\ve)={2\Sg\,\pi^{d/2}\over\RGamma(d/2)}\,{\ve^{d-1}\over (hc)^d}\>\RTheta(\ve)\ .$

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

$\begin{split} p(T)&=-\kT\!\!\int\limits_{-\infty}^\infty\!\!\!d\ve\,g(\ve)\,\ln\big(1-e^{-\ve/\kT}\big)\\ &=-{2\,\Sg\,\pi^{d/2}\over\RGamma(d/2)}\,(hc)^{-d}\,\kT\!\int\limits_0^\infty\!\!d\ve\,\ve\nsub^{d-1}\,\ln\big(1-e^{-\ve/\kT}\big)\\ &=-{2\,\Sg\,\pi^{d/2}\over\RGamma(d/2)}\,{(\kT)^{d+1}\over(hc)^{d}}\!\!\int\limits_0^\infty\!\!dt\> t^{d-1}\ln\big(1-e^{-t}\big)\ . \end{split}$

We can make some progress with the dimensionless integral:

$\begin{split} \CI\ns_d&\equiv -\int\limits_0^\infty\!\!dt\> t^{d-1}\ln\big(1-e^{-t}\big)\\ &=\sum_{n=1}^\infty{1\over n}\int\limits_0^\infty\!\!dt\>t^{d-1}\,e^{-nt}\\ &=\RGamma(d)\sum_{n=1}^\infty{1\over n^{d+1}}=\RGamma(d)\,\zeta(d+1)\ . \end{split}$

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

$\RGamma(z)={2^{z-1}\over\sqrt{\pi}}\,\RGamma\big(\frac{z}{2}\big)\,\RGamma\big(\frac{z+1}{2}\big)\ .$

Putting it all together, we find

$p(T)=\Sg\,\pi\nsub^{-{1\over 2}(d+1)}\,\RGamma\big(\frac{d+1}{2}\big)\,\zeta(d+1)\,{(\kT)^{d+1}\over (\hbar c)^d}\ . \label{photp}$

The number density is found to be

$\begin{split} n(T)&=\!\!\int\limits_{-\infty}^\infty\!\!\!d\ve\,{g(\ve)\over e^{\ve/\kT}-1}\\ &=\Sg\,\pi\nsub^{-{1\over 2}(d+1)}\,\RGamma\big(\frac{d+1}{2}\big)\,\zeta(d)\,\bigg({\kT\over\hbar c}\bigg)^{\!\!d}\ . \end{split}$

For photons in $$d=3$$ dimensions, we have $$\Sg=2$$ and thus

$n(T)={2\,\zeta(3)\over\pi^2}\,\bigg({\kT\over\hbar c}\bigg)^{\!\!3}\qquad,\qquad p(T)={2\,\zeta(4)\over\pi^2}\,{(\kT)^4\over(\hbar c)^3}\ .$

It turns out that $$\zeta(4)=\frac{\pi^4}{90}$$.

Note that $$\hbar c/\kB=0.22855\,{cm}\cdot\RK$$, so

${\kT\over\hbar c}=4.3755\,T[\RK]\,{cm}^{-1}\quad\Longrightarrow\quad n(T)=20.405\times T^3[\RK^3]\,{cm}^{-3}\ .$

To find the entropy, we use Gibbs-Duhem:

$d\mu=0=-s\,dT + v\,dp \quad\Longrightarrow\quad s=v\,{dp\over dT}\ ,$

where $$s$$ is the entropy per particle and $$v=n^{-1}$$ is the volume per particle. We then find

$s(T)=(d\!+\!1)\,{\zeta(d\!+\!1)\over\zeta(d)}\>\kB\ .$

The entropy per particle is constant. The internal energy is

$E=-{\pz\ln\Xi\over\pz\beta}=-{\pz\over\pz\beta}\big(\beta pV)= d\cdot p \,V\ , \label{photE}$

and hence the energy per particle is

$\ve={E\over N}=d \cdot p v = {d\cdot\zeta(d\!+\!1)\over\zeta(d)}\>\kT\ .$

## 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 $$L\ns_x\times L\ns_y\times L\ns_z$$. Suppose further that the dimensions are all expanded by a factor $$\lambda^{1/3}$$, the volume is isotropically expanded by a factor of $$\lambda$$. The cavity modes of the electromagnetic radiation have quantized wavevectors, even within classical electromagnetic theory, given by

$\Bk=\bigg({2\pi n\ns_x\over L\ns_x}\,,\,{2\pi n\ns_y\over L\ns_y}\,,\,{2\pi n\ns_z\over L\ns_z}\bigg)\ .$

Since the energy for a given mode is $$\ve(\Bk)=\hbar c |\Bk|$$, we see that the energy changes by a factor $$\lambda^{-1/3}$$ under an adiabatic volume expansion $$V\to\lambda V$$, where the distribution of different electromagnetic mode occupancies remains fixed. Thus,

$V\pabc{E}{V}{S}=\lambda\pabc{E}{\lambda}{S}=-\third E\ .$

Thus,

$p=-\pabc{E}{V}{S}={E\over 3V}\ ,$

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

$\begin{split} \pabc{E}{V}{T}&=\pabc{E}{V}{p}=3p\\ &=T\pabc{p}{T}{V}-p\ , \end{split}$

where the second line follows the Maxwell relation $$\big(\frac{\pz S}{\pz V}\big)\nd_p=\big(\frac{\pz p}{\pz T}\big)\nd_V$$, after invoking the First Law $$dE=T dS-p\,dV$$. Thus,

$T\,{dp\over dT}=4p \quad\Longrightarrow\quad p(T)=A\,T^4\ ,$

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 $$\theta$$ relative to a surface normal is

$d j\ns_\ve=c\cdot{E\over V}\cdot\cos\theta\cdot{d\Omega\over 4\pi}\ , \label{jephoton}$

where $$d\Omega$$ is the differential solid angle. Thus, the power emitted per unit area is

${dP\over dA}={c E\over 4\pi V}\!\int\limits_0^{\pi/2}\!\!\!d\theta\!\!\int\limits_0^{2\pi}\!\!d\phi\,\sin\theta\cdot\cos\theta= {c E\over 4V}=\frac{3}{4}\,c\,p(T)\equiv\sigma\,T^4\ ,$

where $$\sigma=\frac{3}{4} c A$$, with $$p(T)=A\,T^4$$ as we found above. From quantum statistical mechanical considerations, we have

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

is Stefan’s constant.

## Surface temperature of the earth

We derived the result $$P=\sigma T^4\cdot A$$ where $$\sigma=5.67\times 10^{-8}\,\RW/\Rm^2\,\RK^4$$ 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 $$R\ns_\odot=6.96\times 10^8\,\Rm$$, the radius of the earth $$R\ns_\Re=6.38\times 10^6\,\Rm$$, and the radius of the earth’s orbit $$a\ns_\Re=1.50\times 10^{11}\,\Rm$$. Let’s assume that the earth has achieved a steady state temperature of $$T\ns_\Re$$. We balance the total power incident upon the earth with the power radiated by the earth. The power incident upon the earth is

$P\ns_{incident}={\pi R_\Re^2\over 4\pi a_\Re^2}\cdot\sigma T_\odot^4\cdot 4\pi R_\odot^2= {R_\Re^2\,R_\odot^2\over a_\Re^2}\cdot \pi\sigma T_\odot^4\ .$

The power radiated by the earth is

$P\ns_{radiated}=\sigma T_\Re^4\cdot 4\pi R_\Re^2\ .$

Setting $$P\ns_{incident}=P\ns_{radiated}$$, we obtain

$T\ns_\Re=\bigg({R\ns_\odot\over 2\,a\ns_\Re}\bigg)^{\!\!1/2}\,T\ns_\odot\ .$

Thus, we find $$T\ns_\Re=0.04817\,T\ns_\odot$$, and with $$T\ns_\odot=5780\,\RK$$, we obtain $$T\ns_\Re=278.4\,\RK$$. The mean surface temperature of the earth is $${\bar T}\ns_\Re=287\,\RK$$, which is only about $$10\,\RK$$ 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.

Recall that the frequency of an electromagnetic wave of wavevector $$\Bk$$ is $$\nu=c/\lambda=ck/2\pi$$. Therefore the number of photons $$\CN\ns_T(\nu,T)$$ per unit frequency in thermodynamic equilibrium is (recall there are two polarization states)

$\CN(\nu,T)\,d\nu={2\, V \over 8\pi^3}\cdot{d^3\!k\over e^{\hbar ck/\kT}-1} = {V\over\pi^2}\cdot{ k^2\,dk\over e^{\hbar ck/\kT}-1} \ .$

We therefore have

$\CN(\nu,T)={8\pi V\over c^3}\cdot{\nu^2\over e^{h\nu/\kT}-1}\ .$

Since a photon of frequency $$\nu$$ carries energy $$h\nu$$, the energy per unit frequency $$\CE(\nu)$$ is

$\CE(\nu,T)={8\pi hV\over c^3}\cdot{\nu^3\over e^{h\nu/\kT}-1}\ .$

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

$e^{h\nu/\kT}-1={h\nu\over \kT} + \CO(h^2)$

and

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

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 $$\nu$$ part of the integral, which in the optical spectrum is the ultraviolet portion. With quantization, the Bose-Einstein factor imposes an effective ultraviolet cutoff $$\kT/h$$ on the frequency integral, and the total energy, as we found above, is finite:

$E(T)=\int\limits_0^\infty\!\!d\nu\>\CE(\nu)=3pV=V\cdot{\pi^2\over 15}\,{(\kT)^4\over (\hbar c)^3}\ .$

We can define the spectral density $$\rho\ns_\ve(\nu)$$ of the radiation as

$\rho\ns_\ve(\nu,T)\equiv {\CE(\nu,T)\over E(T)}={15\over\pi^4}\,{h\over\kT}\,{(h\nu/\kT)^3\over e^{h\nu/\kT}-1}$

so that $$\rho\ns_\ve(\nu,T)\,d\nu$$ is the fraction of the electromagnetic energy, under equilibrium conditions, between frequencies $$\nu$$ and $$\nu+d\nu$$, $$\int\limits_0^\infty \!d\nu\,\rho\ns_\ve(\nu,T)=1$$. In Figure [planck] we plot this in Figure [planck] for three different temperatures. The maximum occurs when $$s\equiv h\nu/\kT$$ satisfies

${d\over ds}\bigg( {s^3\over e^s-1} \bigg)= 0 \qquad\Longrightarrow\qquad {s\over 1-e^{-s}}=3 \qquad\Longrightarrow\qquad s=2.82144\ .$

## 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 $$\sigma T^4$$. The power law here follows from the ultrarelativistic dispersion $$\ve=\hbar ck$$ of the photons. Suppose that we replace this dispersion with the general form $$\ve=\ve(\Bk)$$. Now consider a large box in equilibrium at temperature $$T$$. The energy current incident on a differential area $$dA$$ of surface normal to $$\zhat$$ is

$dP=dA\cdot\!\int\!\!{d^3\!k\over (2\pi)^3}\>\RTheta(\cos\theta)\cdot\ve(\Bk)\cdot {1\over\hbar} {\pz \ve(\Bk)\over\pz k\ns_z}\cdot{1\over e^{\ve(\Bk)/k\ns_\RB T}-1}\ .$

Let us assume an isotropic power law dispersion of the form $$\ve(\Bk)=C k^\alpha$$. Then after a straightforward calculation we obtain

${dP\over dA}=\sigma\,T^{2+{2\over\alpha}}\ ,$

where

$\sigma=\zeta\big(2+\frac{2}{\alpha}\big)\,\RGamma\big(2+\frac{2}{\alpha}\big)\cdot {\Sg \, k_\RB^{2+{2\over\alpha}} \, C^{-{2\over\alpha}}\over 8\pi^2\hbar}\ .$

One can check that for $$\Sg=2$$, $$C=\hbar c$$, and $$\alpha=1$$ 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.