= Solution
Substitute the <Killing-flow four-acceleration> and stationary <pressure> into the <relativistic Euler equation>. It gives
$$
\boxed{\nabla_ap=-(\rho+p)\nabla_a\log N,
\qquad N=\sqrt{K^cK_c}.}
$$
For the <equation of state> of <blackbody radiation>, $p=\rho/3$, this becomes $d\rho=-4\rho\,d\log N$. The <barotropic hydrostatic first integral> therefore gives, on each connected nonvacuum fluid region,
$$
\boxed{\rho N^4=C>0,\qquad \rho=C(K^cK_c)^{-2}.}
$$
This is the <stationary radiation density> law. Combining it with the <Stefan–Boltzmann law> $\rho\propto T^4$ also gives the <Tolman–Ehrenfest relation> $TN=\mathrm{constant}$.
At a regular finite <free surface>, a smooth <Killing vector field> has finite norm. Hence $C/N^4$ cannot approach zero there. More globally, its norm has a finite upper bound on a <compact> regular fluid closure, which bounds the density away from zero whenever $C>0$. If the timelike norm instead tends to zero, the density diverges. Thus a <radiation fluid has no regular free surface>: a finite radiation-filled cavity needs a material wall supporting nonzero <pressure>, rather than an unsupported radiation-vacuum interface.
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