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Physics LibreTexts

5.5: Photon Statistics

  • Page ID
    18810
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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    \)
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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.

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

    Distribution of blackbody radiation

    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].