Exact thermal Killing kinematic equivalence

Differentiating $u^b{u}_b=-1$ gives $u^b{\mathrm{\nabla }}_a{u}_b=0$. Decompose the first derivative slot into its component parallel to $u$ and its rest-space projection: ${\mathrm{\nabla }}_a{u}_b=-u_a{a}_b+h_a{}^c{h}_b{}^d{\mathrm{\nabla }}_c{u}_d$. Symmetrizing the rest-space term and using the definitions of expansion and shear yields ${\mathrm{\nabla }}_{(a}{u}_{b)}=-u_{(a}{a}_{b)}+{\sigma }_{ab}+\theta h_{ab}\mathrm{/}3$. Differentiating $\beta =u\mathrm{/}(k_{B}T)$ then gives ${\mathrm{\nabla }}_{(a}{\beta }_{b)}=(k_{B}T{)}^{-1}[-u_{(a}(a_{b)}+{\mathrm{\nabla }}_{b)}\log T)+{\sigma }_{ab}+\frac{1}{3}\theta h_{ab}]\mathrm{.}$

Contracting this identity with $u^a{u}^b$ shows that ${\mathrm{\nabla }}_{(a}{\beta }_{b)}=0$ implies $D\log T=0$, hence $DT=0$ because $T>0$.

Projecting both indices with $h$ gives ${\sigma }_{ab}+\frac{1}{3}\theta h_{ab}=0$. Taking its $h$-trace gives $\theta =0$, since $h$ has rank three and $\sigma$ is trace-free; substituting back gives ${\sigma }_{ab}=0$.

Contracting one index with $u$ and projecting the other with $h$ gives $h_a{}^b(a_b+{\mathrm{\nabla }}_b\log T)=0$. Thus the displayed temperature-gradient condition also follows from the Killing equation.

Conversely, the four stated conditions make the parallel, mixed and rest-space projections in step 1.1 all zero. Since $T_{x}M=\mathrm{span}\{u_x\}\oplus u_x^{\perp }$ at every $x\in O$, these projections determine the full symmetric tensor ${\mathrm{\nabla }}_{(a}{\beta }_{b)}$; it is zero, so $\beta$ is Killing. In particular, the acceleration/temperature-gradient relation alone omits the necessary conditions $DT=0$, $\theta =0$ and $\sigma =0$.