Massive Jüttner stationarity and normalization

In coordinates put $G(x,P):=g_{ab}(x)P^a{P}^b$. Metric compatibility gives ${\mathrm{\partial }}_c{g}_{ab}={\mathrm{\Gamma }}^d{}_{ca}{g}_{db}+{\mathrm{\Gamma }}^d{}_{cb}{g}_{ad}$, while $\mathcal{L}G=P^c{\mathrm{\partial }}_c{g}_{ab}{P}^a{P}^b-2{\mathrm{\Gamma }}^a{}_{bc}{P}^b{P}^c{g}_{ad}{P}^d\mathrm{.}$ After substituting the metric-compatibility identity, each of its two terms matches one of the two connection contributions, by symmetry of the lower Christoffel indices and renaming dummy indices. Hence $\mathcal{L}G=0$. Therefore the geodesic spray is tangent to the total mass-shell level set over $O$. Along any local integral curve $(x(s),P(s))$ in that level set, the function $s\mapsto g_{x(s)}(P(s),N(x(s)))$ is continuous and never zero, where $N$ is the time-orienting future timelike field. Its negative sign at the initial point therefore persists, so the local flow preserves the future sheet (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model, Time orientation, future and past timelike/null/causal vectors, and the closed cones). Thus $\mathcal{L}$ is tangent to ${\mathcal{H}}_{M_{\ast },O}^{+}$, although its base component generally carries a point from one fiber to another.

If ${\beta }_x$ is future timelike, put $\alpha =\sqrt{-g({\beta }_x,{\beta }_x)}>0$, $u={\beta }_x\mathrm{/}\alpha$ and $T=1\mathrm{/}(k_B\alpha )$. Then $g(u,u)=-1$ and ${\beta }_x=u\mathrm{/}(k_{B}T)$. In the $u$-rest tetrad, $g({\beta }_x,P)=-\alpha E(q)$, so [F2] and [F4] give $I_{M_{\ast }}({\beta }_x)={\int }_{{\mathbb{R}}^3}{e}^{-\alpha E(q)}\frac{d^{3}q}{E(q)}=\frac{4\pi M_{\ast }{K}_1(\alpha M_{\ast })}{\alpha }<\mathrm{\infty }\mathrm{.}$

The other nonzero causal and spacelike types have these orthonormal normal forms. For a timelike vector choose $e_0$ to be the future unit vector in its direction or its negative, according to its time orientation. Its orthogonal complement is positive definite by [F7]; applying Gram–Schmidt to a basis of that complement gives $e_1,e_2,e_3$. For a spacelike vector set $e_3={\beta }_x\mathrm{/}\sqrt{g({\beta }_x,{\beta }_x)}$. Its orthogonal complement has inertia $(1,2)$ by [F7], so choose a unit timelike vector there and choose its sign to be future. The remaining orthogonal two-plane is positive definite, and Gram–Schmidt gives $e_1,e_2$. For a nonzero null vector start with a future tetrad and write ${\beta }_x={\beta }^0{e}_0+v$ with $v\perp e_0$. Nullity gives $\mathrm{\mid }v\mathrm{\mid }=\mathrm{\mid }{\beta }^0\mathrm{\mid }>0$. Set $e_3=v\mathrm{/}\mathrm{\mid }v\mathrm{\mid }$ and use Gram–Schmidt in the positive two-plane orthogonal to both $e_0$ and $e_3$ to choose $e_1,e_2$. Thus the normal forms are spacelike $(0,0,0,b)$, past timelike $(-\alpha ,0,0,0)$, past null $(-a,0,0,a)$, and future null $(a,0,0,a)$, with positive parameters.

When $\beta =u\mathrm{/}(k_{B}T)$, it is future timelike. The value of $\zeta$ in the Statement gives $e^{\zeta }{Z}_{M_{\ast }}\mathrm{/}(h_{\mathrm{r}\mathrm{e}\mathrm{f}}^3{c}^3)=n$. The complete moment calculation in [F4] therefore gives $J^a=n{u}^a$ and $T^{ab}=(\epsilon +p)u^a{u}^b+p{g}^{ab}$ with the displayed $\epsilon$ and $p$. This is a normalization of the specified mathematical ansatz.

Since $f>0$, differentiating its logarithm along the geodesic spray gives $\frac{\mathcal{L}f}{f}=\mathcal{L}\log f=P^a{\mathrm{\nabla }}_a\zeta +P^a{P}^b{\mathrm{\nabla }}_a{\beta }_b=P^a{\mathrm{\nabla }}_a\zeta +P^a{P}^b{\mathrm{\nabla }}_{(a}{\beta }_{b)}\mathrm{.}$ The last equality holds because $P^a{P}^b$ is symmetric; the covariant expression follows by differentiating $g(\beta ,P)$ along a geodesic, where $P$ is parallel.

For spacelike ${\beta }_x=(0,0,0,b)$, $b>0$, let ${\theta }_0\in (0,\pi )$ be the unique angle with $\cos {\theta }_0=1\mathrm{/}2$. By [F6] the cap $n_3\ge 1\mathrm{/}2$ has area ${\int }_0^{2\pi }{\int }_0^{{\theta }_0}\sin \theta d\theta d\phi =2\pi (1-\cos {\theta }_0)=\pi$. On it $g({\beta }_x,P)=br{n}_3\ge br\mathrm{/}2$. For $r\ge M_{\ast }$, $E\le \sqrt{2}r$, so the shell integral dominates a positive multiple of $\int r{e}^{br\mathrm{/}2}dr=\mathrm{\infty }$. For past timelike ${\beta }_x=(-\alpha ,0,0,0)$, the exponent is $\alpha E\ge \alpha r$ and the full-sphere radial measure is $4\pi r^2\mathrm{/}E\ge cr$ for all sufficiently large $r$, so its integral diverges. For past null ${\beta }_x=(-a,0,0,a)$, the same cap gives $g({\beta }_x,P)=a(E+r{n}_3)\ge ar\mathrm{/}2$, again forcing divergence.

For future null ${\beta }_x=(a,0,0,a)$, $a>0$, set $\delta =M_{\ast }\mathrm{/}r\in (0,1\mathrm{/}2]$ for $r\ge 2M_{\ast }$. Let ${\theta }_0\in (0,\pi )$ be the unique angle with $\cos {\theta }_0=1-\delta$. By [F6] the cap $n_3\ge 1-M_{\ast }\mathrm{/}r$ has area ${\int }_0^{2\pi }{\int }_0^{{\theta }_0}\sin \theta d\theta d\phi =2\pi (1-\cos {\theta }_0)=2\pi M_{\ast }\mathrm{/}r$. On this cap, $E-r{n}_3=(E-r)+r(1-n_3)\le \frac{M_{\ast }^2}{2r}+M_{\ast },$ so $e^{g({\beta }_x,P)}=e^{-a(E-r{n}_3)}$ is bounded below by a fixed positive constant. Since $r^2\mathrm{/}E\ge r\mathrm{/}\sqrt{2}$, the radial integral dominates a positive constant times $\int dr=\mathrm{\infty }$. For ${\beta }_x=0$, the normalizer is $4\pi {\int }_0^{\mathrm{\infty }}{r}^2\mathrm{/}Edr$, which diverges since $r^2\mathrm{/}E\ge r\mathrm{/}\sqrt{2}$ for $r\ge M_{\ast }$.

The causal classification exhausts all vectors by zero, timelike, spacelike and null type. Step 1.2 proves finiteness for future timelike vectors; steps 2.2–2.3 prove divergence for every other type. Therefore $I_{M_{\ast }}({\beta }_x)<\mathrm{\infty }$ exactly when ${\beta }_x$ is future timelike.

At a fixed event, apply [F3] to $A_{ab}={\mathrm{\nabla }}_{(a}{\beta }_{b)}$ and $b_a={\mathrm{\nabla }}_a\zeta$ in the last expression of step 2.1. Its vanishing on every future massive-shell vector is equivalent to $A=0$ and $b=0$. By [F8], this says that $\beta$ is Killing and $d\zeta =0$; the same identity proves the converse.

On each convex coordinate ball in $O$, the condition $d\zeta =0$ makes $\zeta$ have zero derivative along every straight coordinate segment. The mean value theorem [F5] makes $\zeta$ constant on that ball. Thus every level set of $\zeta$ and its complement are open; connectedness of $O$ makes $\zeta$ one constant on all of $O$.