= Perfect-fluid alignment in a static spacetime
{title2=$u^a=k^a/\sqrt{k^bk_b}$}
For <metric signature> $(+---)$, a <perfect fluid> has $T_{ab}=(\rho+p)u_au_b-pg_{ab}$. The <Einstein field equations> and <static Killing field is a Ricci eigenvector> imply $(\rho+p)(k\cdot u)u_{[a}k_{b]}=0$. Two nonzero timelike vectors cannot be orthogonal, so $k\cdot u\ne0$. If $\rho+p\ne0$, the <four-velocity> is parallel to the static <Killing vector>. Choosing both future-directed gives the displayed normalization. For $\rho+p=0$, the <stress-energy tensor> is proportional to the <metric tensor> and does not determine a preferred fluid velocity; the strict enthalpy hypothesis is therefore material.
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