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Substitute the Killing-flow four-acceleration and stationary pressure into the relativistic Euler equation. It givesFor 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,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.

A connected nonvacuum perfect fluid with equation of state $p=\rho /3$ obeys the stationary radiation density law $\rho =C/N^4$ with $C>0$. On a compact region of a regular spacetime, a smooth Killing vector field has bounded norm. Consequently $\rho$ cannot tend to zero at a finite regular free surface. A vanishing timelike norm instead makes the density diverge. A radiation-filled cavity can have a material wall with nonzero boundary pressure; this is a different boundary condition.