Past exam of the atomic and laser physics course of the University of Oxford / 2021 / 3 / d

In local thermodynamic equilibrium, write the one-ray radiative transfer equation asLarge total optical depth removes the boundary contribution, but an additional local condition is essential: the temperature and opacity vary little over an absorption length ${\kappa }_{\nu }^{-1}$. Iterating once in that small gradient givesThe integrated Planck function and its derivative areUsing the specified Rosseland mean opacity, with ${\kappa }_R$ measured here in inverse length, the source's definition $F_{\mathrm{ray}}=\int I_{\nu }d\nu$ therefore givesThe printed leading term $\sigma T^4$ lacks a factor $1/\pi$ under that definition. An isothermal optically thick medium is a concrete counterexample: $I_{\nu }=B_{\nu }$ gives $F_{\mathrm{ray}}=\sigma T^4/\pi$ exactly, while the printed formula gives $\sigma T^4$.

A physical radiative flux requires an angular integral. For direction cosine $\mu$, the same gradient expansion gives $I_{\nu }(\mu )=B_{\nu }-\mu {\kappa }_{\nu }^{-1}{\partial }_z{B}_{\nu }$. Integration over all directions gives the net radiative diffusion fluxThe isotropic equilibrium contribution cancels between opposite directions. For comparison the outward-hemisphere flux alone is $\sigma T^4-8\sigma T^3(dT/dz)/(3{\kappa }_R)$ at this order. None of these consistent definitions yields the printed mixture of coefficients.