 Solution

In the non-scattering grey atmosphere approximation, $I$ is the frequency-integrated specific intensity along a ray, $s$ is distance along that ray, $\rho$ is the mass density, and $\kappa$ is the mass absorption coefficient. The local thermodynamic equilibrium source is the frequency-integrated Planck function $B={\sigma }_{\mathrm{SB}}{T}^4/\pi$. Across a thin element $ds$, absorption removes $\kappa \rho Ids$ and thermal emission adds $\kappa \rho Bds$. Their balance gives $dI/ds=-\kappa \rho (I-B)$.

Take the upward vertical as the polar axis, put $\mu =\cos \theta$, and define downward optical depth byAlong a ray, $dz/ds=\mu$ and $d\tau /ds=-\kappa \rho \mu$. The chain rule therefore gives $\boxed{\mu I_{\tau }=I-B}$. In a plane-parallel, azimuthally symmetric field, the solid angle element is $d\omega =d\phi d\mu$, soLocal radiative equilibrium requires zero net radiative heating of the material: absorption $4\pi \kappa \rho J$ balances emission $4\pi \kappa \rho B$. Where the absorption coefficient is nonzero this gives $B=J$, and substitution gives the required angular radiative transfer equation. It does not require the specific intensity itself to be isotropic.

At the top of an atmosphere with no external incoming long-wave radiation, impose $\boxed{I(0,\mu )=0(-1\le \mu <0)}$. These are inward directions; prescribing the same zero value for outward directions would remove the emergent radiation as well. The upward radiative flux and its derivative areThus $\boxed{\Phi \text{ is constant with height}}$: local radiative equilibrium expresses energy conservation through successive atmospheric layers.