Stationary thermal geometric data — Examples

Put $a=s\mathrm{/}{L}_0$. The Christoffel formula in [F1] gives as the only nonzero coefficients (up to symmetry of the lower indices) ${\mathrm{\Gamma }}^s{}_{ij}=a{a}^{\mathrm{\prime }}{\delta }_{ij},{\mathrm{\Gamma }}^i{}_{sj}={\mathrm{\Gamma }}^i{}_{js}=\frac{a^{\mathrm{\prime }}}{a}{\delta }_j^i\mathrm{.}$ With $a^{\mathrm{\prime }}=1\mathrm{/}{L}_0$ and $a^{\mathrm{\prime }\mathrm{\prime }}=0$, the curvature formula in [F1] gives ${\mathrm{Ric}}_{ss}=-3a^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}a=0$ and ${\mathrm{Ric}}_{ij}=(a{a}^{\mathrm{\prime }\mathrm{\prime }}+2(a^{\mathrm{\prime }}{)}^2){\delta }_{ij}=2L_0^{-2}{\delta }_{ij}$. Contracting with $g^{ss}=-1$ and $g^{ij}=a^{-2}{\delta }^{ij}$ yields $S=6(a^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}a+(a^{\mathrm{\prime }}\mathrm{/}a{)}^2)=6\mathrm{/}{s}^2$.

Suppose a timelike Killing field $X$ existed on a nonempty open $U\subseteq O$. It is nowhere zero, so [F2] applies and gives $X(S)=0$. Since $dS=-12s^{-3}ds$, this forces $X^s=0$ at every point of $U$. Therefore $g(X,X)=a(s{)}^2{\sum }_i(X^i{)}^2\ge 0$, contradicting timelikeness. This proves (a), including every smaller open region.

Suppose a normalizable massive stationary exponential existed on a nonempty open $U\subseteq O$. Choose any point of $U$ and a connected open coordinate ball $V\subseteq U$ around it. Restricting the ansatz to $V$ preserves smoothness, stationarity, and the finite shell normalizer. The theorem [F3] would then supply a future timelike Killing field on $V$, which step 2.1 rules out. Thus no such massive state exists on $U$. This conclusion concerns only that ansatz and exact stationarity.

Metric compatibility makes the geodesic spray tangent to the null shell, since, with $G=g_{ab}{P}^a{P}^b$, $\mathcal{L}G=P^c({\mathrm{\partial }}_c{g}_{ab})P^a{P}^b-2g_{ad}{P}^d{\mathrm{\Gamma }}^a{}_{bc}{P}^b{P}^c=0$ by the metric-compatibility identity in [F5]. On $G=0$ away from $P=0$, the fiber derivative $2g(P,\cdot )$ is nonzero, so this is a smooth level hypersurface; its nonzero and future conditions are open within it. Thus restriction of $\mathcal{L}$ to ${\mathcal{H}}_{0,O}^{+}$ is defined. For a future null vector $P$, the null-shell Liouville derivative of $f=e^{\zeta +g(\beta ,P)}$ is $\frac{\mathcal{L}f}{f}=P^a{\mathrm{\nabla }}_a\zeta +P^a{P}^b{\mathrm{\nabla }}_{(a}{\beta }_{b)}\mathrm{.}$ By [F4], its vanishing on the full future null shell is equivalent to $d\zeta =0$ and ${\mathrm{\nabla }}_{(a}{\beta }_{b)}=\lambda g_{ab}$; conversely those conditions make it zero because $g(P,P)=0$. The factor is smooth since $\lambda =\frac{1}{4}{g}^{ab}{\mathrm{\nabla }}_{(a}{\beta }_{b)}$. For $K:=(s\mathrm{/}{L}_0){\mathrm{\partial }}_s$, direct differentiation gives $({\mathcal{L}}_{K}g{)}_{ss}=-2\mathrm{/}{L}_0$ and $({\mathcal{L}}_{K}g{)}_{ij}=2a^2{\delta }_{ij}\mathrm{/}{L}_0$, with mixed components zero; hence ${\mathcal{L}}_{K}g=2g\mathrm{/}{L}_0$. Also $g(K,K)=-(s\mathrm{/}{L}_0{)}^2<0$, and $K$ is future-directed since $s>0$. Thus $\beta =bK$ is future timelike and conformal Killing, with ${\mathrm{\nabla }}_{(a}{\beta }_{b)}=(b\mathrm{/}{L}_0)g_{ab}$; for constant $\zeta$ its null-shell Liouville derivative vanishes. This proves the explicitly constructed massless conformal branch without asserting a massive Killing equilibrium.