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

Specify the radiation normalization first. Let $u_{\nu }$ be radiation energy density per ordinary frequency, with SI units $\mathrm{J}{\mathrm{m}}^{-3}\mathrm{H}{\mathrm{z}}^{-1}$. Then $A_{21}$ is the spontaneous emission rate per upper-state chemical atom, while $B_{12}{u}_{\nu }$ and $B_{21}{u}_{\nu }$ are the induced absorption and stimulated emission rates, evaluated at a sharp transition frequency. For a finite spectral line profile, replace $u_{\nu }$ by its profile-weighted average.

The Einstein coefficients obeyHere $g_i$ are the statistical weights of the two energy levels, and the field is unpolarized and isotropic. The units and physical dimensions areThe apparently time-independent dimension of $B$ follows because a per-hertz density includes an extra unit of time. If instead $B$ multiplies the mean intensity $J_{\nu }$, use $B^{(J)}=(4\pi /c)B$ and $A/B^{(J)}=2h{\nu }_0^3/c^2$. These distinct normalizations must not be mixed in the radiative transfer equation.

Use the energy-density convention of part (a), a normalized common spectral line profile $\int \varphi (\nu )d\nu =1$, and isotropic spontaneous emission. For a narrow spectral line centered at ${\nu }_0$,The factor $1/c$ follows from $u_{\nu }=4\pi J_{\nu }/c$: angle-integrated absorption removes $c{\kappa }_{\nu }{u}_{\nu }$ of energy per volume and time. The emission coefficient is per solid angle, whereas the absorption coefficient has units of inverse length.

The Boltzmann distribution in thermal equilibrium gives $n_2/n_1=(g_2/g_1)e^{-h{\nu }_0/(k_{B}T)}$. Dividing the two expressions and using the relations between Einstein coefficients,Thus the source function equals the Planck function at the transition frequency, and in the usual narrow-line approximationWith an exactly common finite-width $\varphi$ and constant line-center Einstein coefficients, their ratio is constant across the profile; exact equality to a varying $B_{\nu }$ at every frequency is only a narrow-line approximation. Exact thermal radiative transfer requires frequency-resolved detailed balance, which supplies the corresponding small profile corrections. This distinction does not alter the intended equilibrium result.

For constant nonzero absorption coefficient, multiply the radiative transfer equation by the integrating factor $e^{{\kappa }_{\nu }z}$:Integrating from the entrance to $z$ givesThe boundary condition is immediate. Differentiation gives $I_{\nu }^{\prime }=-{\kappa }_{\nu }[I_{\nu }(0)-S_{\nu }]e^{-{\kappa }_{\nu }z}=-{\kappa }_{\nu }{I}_{\nu }+j_{\nu }$, checking the equation. For positive optical depth ${\kappa }_{\nu }z\gg 1$, the incident specific intensity is forgotten and $I_{\nu }\to S_{\nu }$. Negative absorption coefficient instead describes gain and does not have this relaxation limit. At ${\kappa }_{\nu }=0$ use the nonsingular form directly: $I_{\nu }(z)=I_{\nu }(0)+j_{\nu }z$.

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.

To derive fermionic Einstein-coefficient relations, first consider an idealized emitter that can exchange one ultrarelativistic fermion with the bath, with zero chemical potential. Let ${\mathcal{D}}_{\nu }=d4\pi {\nu }^2/c^3$ be its density of states per volume and ordinary frequency, with $d$ populated helicity states; for one neutrino helicity $d=1$, whereas the corresponding photon count is two. The Fermi-Dirac distribution givesPauli blocking changes the final-state factor in downward emission from the bosonic $1+f$ to $1-f$. Define positive coefficients $B_{12},C_{21}$ and $A_{21}=C_{21}h\nu {\mathcal{D}}_{\nu }$. The upward rate per lower emitter is $B_{12}{u}_{\nu }$, and the downward rate per upper emitter isThermal detailed balance, $n_1{B}_{12}{u}_{\nu }=n_2(A_{21}-C_{21}{u}_{\nu })$ with $n_2/n_1=(g_2/g_1)e^{-h\nu /(k_{B}T)}$, yieldsFor example, substituting $e^{h\nu /(k_{B}T)}=x$ reduces the balance equation to $B_{12}=(g_2/g_1)C_{21}$ because $x{f}_{\nu }=1-f_{\nu }$.

If one insists on the original additive notation $A_{21}+B_{21}{u}_{\nu }$, its formal coefficient is negative:A negative correction means inhibited emission into occupied modes, not a negative total decay rate; $0\le f_{\nu }\le 1$ ensures positivity. The dimensions are those of part (a), and the factor of two relative to the photon density is a helicity count, independent of the statistical sign.

There is a physical qualification: ordinary electronic energy levels of the same chemical atom have the same fermion parity, and cannot interconvert by emission or absorption of a single neutrino. A realizable single-fermion emitter must change the relevant particle-number and parity sector, with matching conservation laws. For ordinary atomic levels, neutrino-pair processes would require a different calculation. A bath with nonzero chemical potential would also require the corresponding chemical-equilibrium population ratio. The calculation above is the intended single-mode statistical analogy under explicit assumptions.