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For metric signature $(+---)$, a stationary perfect fluid follows a timelike Killing vector field $K$ with four-velocity $U=K/N$. Its Killing-flow four-acceleration is $U^b{\nabla }_b{U}_a=-{\nabla }_a\log N$. The relativistic Euler equation and stationarity of the pressure give the displayed hydrostatic equilibrium equation. This uses stationarity along $K$ and does not require hypersurface orthogonality.

Let $N=\sqrt{K^a{K}_a}>0$ on the fluid region and choose the Killing vector field $K$ to be future-directed. The normalized four-velocity isThe Killing equation implies both $K^b{\nabla }_{b}N=0$ andConsequently its Killing-flow four-acceleration isFor a scalar, the Lie derivative is ${\mathcal{L}}_{K}p=K^b{\nabla }_{b}p$. Thus the assumed stationarity of the pressure gives $U^b{\nabla }_{b}p=0$. These are the ingredients needed for hydrostatic equilibrium along a Killing vector; hypersurface orthogonality of $K$ is not needed.

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.