Blundell Thermal Physics · Problem 23.7

Photons

Physics — Statistical Mechanics · photon-number-entropy-energy

Problem

Show that the total number NN of photons in black-body radiation contained in a volume VV is N=∫0∞g(ω) dωeℏω/kBT−1=2ζ(3)π2(kBTℏc)3V,(23.74)N = \int_0^\infty \frac{g(\omega)\,\mathrm{d}\omega}{\mathrm{e}^{\hbar\omega/k_\mathrm{B}T}-1} = \frac{2\zeta(3)}{\pi^2}\left(\frac{k_\mathrm{B}T}{\hbar c}\right)^3 V,\tag{23.74} where ζ(3)=1.20206\zeta(3) = 1.20206 is a Riemann zeta function (see Appendix C.4). Hence show that the average energy per photon is UN=π430ζ(3)kBT=2.701 kBT,(23.75)\frac{U}{N} = \frac{\pi^4}{30\zeta(3)}k_\mathrm{B}T = 2.701\,k_\mathrm{B}T,\tag{23.75} and that the average entropy per photon is SN=2π445ζ(3)kB=3.602 kB.(23.76)\frac{S}{N} = \frac{2\pi^4}{45\zeta(3)}k_\mathrm{B} = 3.602\,k_\mathrm{B}.\tag{23.76} The result for the internal energy of a photon gas is therefore U=2.701NkBTU = 2.701 Nk_\mathrm{B}T, whereas for a classical ideal gas one obtains U=32NkBTU = \frac{3}{2}Nk_\mathrm{B}T. Why should the two results be different? Compare the expression for the entropy of a photon gas with that for an ideal gas (the Sackur–Tetrode equation); what is the physical reason for the difference?

Related theory

Browse the problem library