Stationary thermal geometric data

Differentiating $g(\xi ,\xi )$ with the metric-compatible connection gives $\xi (N_{\xi })=0$, since ${\xi }^a{\xi }^b{\mathrm{\nabla }}_a{\xi }_b=0$ by the Killing equation. The same equation gives ${\xi }^a{\mathrm{\nabla }}_a{\xi }_b=-\frac{1}{2}{\mathrm{\nabla }}_{b}g(\xi ,\xi )=N_{\xi }{\mathrm{\nabla }}_b{N}_{\xi }$. Therefore $a_b=u^a{\mathrm{\nabla }}_a{u}_b={\mathrm{\nabla }}_b\log N_{\xi }$. Also $\theta ={\mathrm{\nabla }}_a({\xi }^a\mathrm{/}{N}_{\xi })=0$ because ${\mathrm{\nabla }}_a{\xi }^a=0$ and $\xi (N_{\xi })=0$. The symmetric derivative is ${\mathrm{\nabla }}_{(a}{u}_{b)}=-u_{(a}{\mathrm{\nabla }}_{b)}\log N_{\xi }$; projecting both indices with $h$ gives zero, so ${\sigma }_{ab}=0$.

Put $f:=(k_B{N}_{\xi }T{)}^{-1}>0$, so $\beta =f\xi$. If $\beta$ is Killing, both $f\xi$ and $\xi$ have zero symmetric covariant derivative, and hence ${\xi }_{(b}{\mathrm{\nabla }}_{a)}f=0$. Contracting with $u^a{u}^b$ gives $u(f)=0$ because $\xi =N_{\xi }u$ and $N_{\xi }>0$; contracting with $u^a{h}^b{}_c$ gives $h^b{}_c{\mathrm{\nabla }}_{b}f=0$. The decomposition of a covector into its $u$ and rest-space parts now gives $df=0$.

In a convex coordinate ball in $O$, join any two coordinate points by a straight segment. The derivative of $f$ along that segment is zero by the covector decomposition just proved, so [F2] makes $f$ constant on the ball. Thus $f$ is locally constant on $O$. Each level set of $f$ and its complement are open; connectedness of $O$ therefore makes $f$ one positive constant. Equivalently $N_{\xi }T=T_K$ for a constant $T_K>0$, and then $\beta =\xi \mathrm{/}(k_B{T}_K)$.

Conversely, if $N_{\xi }T=T_K$ is constant, then $\beta =\xi \mathrm{/}(k_B{T}_K)$ is a constant multiple of the Killing field $\xi$ and is Killing. Under $\xi \mapsto b\xi$, $N_{\xi }\mapsto b{N}_{\xi }$ and $T_K\mapsto b{T}_K$, whereas $u=\xi \mathrm{/}{N}_{\xi }$ and $\beta =\xi \mathrm{/}(k_B{T}_K)$ remain fixed. These identities prove the converse and the normalization statement.

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 $\sigma =0$.

By the equivalent covariant characterization of a Killing field in Killing fields of a Lorentzian metric, and metric compatibility and torsion-freeness, $({\mathcal{L}}_{\xi }g)(X,Y)=g({\mathrm{\nabla }}_X\xi ,Y)+g(X,{\mathrm{\nabla }}_Y\xi )$: expand the Lie derivative from its flow definition, use $[\xi ,X]={\mathrm{\nabla }}_{\xi }X-{\mathrm{\nabla }}_X\xi$, and cancel the two ${\mathrm{\nabla }}_{\xi }$ terms with metric compatibility. Thus ${\mathrm{\nabla }}_{(a}{\xi }_{b)}=0$ implies ${\mathcal{L}}_{\xi }g=0$. The local flow identity ${\mathrm{\Phi }}_{s+h}={\mathrm{\Phi }}_h\circ {\mathrm{\Phi }}_s$ and the pullback order give $\frac{d}{ds}{\mathrm{\Phi }}_s^{\ast }g={\mathrm{\Phi }}_s^{\ast }({\mathcal{L}}_{\xi }g)=0.$ In local tensor components [F5] makes each component constant in $s$; since ${\mathrm{\Phi }}_0$ is the identity, ${\mathrm{\Phi }}_s^{\ast }g=g$. Hence ${\mathrm{\Phi }}_s$ is a local isometry.

Define the pulled-back connection by $({\mathrm{\Phi }}_s^{\ast }\mathrm{\nabla }{)}_{X}Y:={\mathrm{\Phi }}_{s\ast }^{-1}({\mathrm{\nabla }}_{{\mathrm{\Phi }}_{s\ast }X}{\mathrm{\Phi }}_{s\ast }Y)$. A diffeomorphism preserves Lie brackets, so its pullback connection has zero torsion whenever $\mathrm{\nabla }$ does. Since ${\mathrm{\Phi }}_s^{\ast }g=g$, metric compatibility of $\mathrm{\nabla }$ also makes ${\mathrm{\Phi }}_s^{\ast }\mathrm{\nabla }$ compatible with $g$. The uniqueness in [F3] gives ${\mathrm{\Phi }}_s^{\ast }\mathrm{\nabla }=\mathrm{\nabla }$. The flow composition law also gives ${\mathrm{\Phi }}_{s\ast }\xi =\xi$.

Substituting ${\mathrm{\Phi }}_s^{\ast }\mathrm{\nabla }=\mathrm{\nabla }$ and ${\mathrm{\Phi }}_{s\ast }\xi =\xi$ into the curvature definition [F4] gives ${\mathrm{\Phi }}_s^{\ast }R=R$. Pullback commutes with tensor products and contractions by their slot definitions, and the preserved connection makes it commute with each covariant derivative. Repeating these identities proves preservation of every finite covariant derivative and contraction stated above; composing preserved scalar components with a smooth function preserves the resulting scalar invariant.

In the future massive shell write $P=(E,rn)$, where $E=\sqrt{M^2+r^2}$, $r>0$ and $n\in S^2$. Apply the assumed identity to the two shell vectors $(E,rn)$ and $(E,-rn)$ and subtract. Dividing the result by $2r>0$ gives $2E{A}_{0i}{n}_i+b_i{n}_i=0$ for every $E>M$ and every unit $n$. For fixed $n$, the left side is affine in $E$ and vanishes on an interval, so $A_{0i}{n}_i=b_i{n}_i=0$; varying $n$ yields $A_{0i}=b_i=0$.

On the future null cone write $P=\rho (1,n)$ with $\rho >0$ and $n\in S^2$. The identity becomes ${\rho }^{2}A((1,n),(1,n))+\rho b(1,n)=0$ for every $\rho >0$; comparing the two polynomial coefficients gives $A((1,n),(1,n))=b(1,n)=0$. Replacing $n$ by $-n$ in the covector equation gives $b_0=0$ and $b_i{n}_i=0$ for every $n$, so $b=0$.

The remaining massive-shell identity is $A_{00}(M^2+r^2)+r^2{n}^T{A}_{\mathrm{s}\mathrm{p}}n+b_0\sqrt{M^2+r^2}=0$, where $A_{\mathrm{s}\mathrm{p}}=(A_{ij})$. Varying $n$ shows $n^T{A}_{\mathrm{s}\mathrm{p}}n$ is a constant $a$ on $S^2$. Taking $n=e_i$ gives $A_{ii}=a$; taking $n=(e_i+e_j)\mathrm{/}\sqrt{2}$ gives $A_{ij}=0$ for $i\mathrm{\ne }j$. With $z=\sqrt{M^2+r^2}$ the remaining identity is $(A_{00}+a)z^2+b_{0}z-a{M}^2=0$ for every $z>M$. A polynomial vanishing on an interval has all coefficients zero; since $M>0$, this yields $a=A_{00}=b_0=0$, hence $A=b=0$.

In the quadratic null-cone equation, comparing $n$ and $-n$ gives $A_{0i}=0$. Its value on $e_i$ yields $A_{ii}=-A_{00}$, and its value on $(e_i+e_j)\mathrm{/}\sqrt{2}$ yields $A_{ij}=0$ for $i\mathrm{\ne }j$. Thus $A=\mathrm{diag}(A_{00},-A_{00},-A_{00},-A_{00})=\lambda g$ with $\lambda =-A_{00}$. This proves the null clause and completes the lemma.

Write $P(q)=(E(q),q)$ in the first tetrad and let $q^{\mathrm{\prime }}=Q(q)$ be the spatial coordinates of the same future-shell vector in the second tetrad. The tetrad change $\mathrm{\Lambda }$ maps the entire future shell bijectively to itself; the spatial coordinate and its inverse are smooth, so $Q$ is a global $C^1$ diffeomorphism of ${\mathbb{R}}^3$. Form the $4\times 4$ matrix $C(q)=[P(q),{\mathrm{\partial }}_{1}P(q),{\mathrm{\partial }}_{2}P(q),{\mathrm{\partial }}_{3}P(q)]$. Block elimination gives $\det C(q)=E(q)-\mathrm{\mid }q{\mathrm{\mid }}^2\mathrm{/}E(q)=M_{\ast }^2\mathrm{/}E(q)$.

The chain rule on the shell gives $\mathrm{\Lambda }C(q)=C(q^{\mathrm{\prime }})\mathrm{diag}(1,DQ(q))$. Taking absolute determinants and using [F2] and step 1.1 yields $\frac{M_{\ast }^2}{E(q)}=\frac{M_{\ast }^2}{E(q^{\mathrm{\prime }})}\mathrm{\mid }\det DQ(q)\mathrm{\mid },\mathrm{\mid }\det DQ(q)\mathrm{\mid }=\frac{E(q^{\mathrm{\prime }})}{E(q)}\mathrm{.}$

For any Borel $A$ on the shell, apply [F3] to the nonnegative Borel function $h(q^{\mathrm{\prime }})={\mathbf{1}}_A(P_{e^{\mathrm{\prime }}}(q^{\mathrm{\prime }}))\mathrm{/}E(q^{\mathrm{\prime }})$. Then ${\mu }_{M_{\ast },x,e^{\mathrm{\prime }}}(A)={\int }_{{\mathbb{R}}^3}\frac{{\mathbf{1}}_A(P_e(q))}{E(q^{\mathrm{\prime }})}\mathrm{\mid }\det DQ(q)\mathrm{\mid }{d}^{3}q={\int }_{{\mathbb{R}}^3}\frac{{\mathbf{1}}_A(P_e(q))}{E(q)}{d}^{3}q={\mu }_{M_{\ast },x,e}(A),$ where the first equality uses the common geometric vector and the second uses step 2.1. This is equality of positive Borel measures, completing the proof.

For $z>0$, the defining integrand of $K_j(z)$ is positive. Near $t=0$ it is bounded. For large $t$, $\cosh t\ge e^t\mathrm{/}2$ and $\cosh (jt)\le e^{jt}$; after $y=e^t$, the bound $y^{j-1}{e}^{-zy\mathrm{/}2}$ is integrable because $y^{j-1}\le e^{zy\mathrm{/}4}$ for all sufficiently large $y$. Hence each $K_j(z)$ is finite and positive. By [F1]–[F2], writing $r=\mathrm{\mid }q\mathrm{\mid }$ and then $r=M_{\ast }\sinh t$ gives $I_{M_{\ast }}=4\pi M_{\ast }^2{\int }_0^{\mathrm{\infty }}{\sinh }^{2}t{e}^{-z\cosh t}dt=2\pi M_{\ast }^2(K_2(z)-K_0(z)),z=\alpha M_{\ast }\mathrm{.}$ The endpoint terms in the integral of $\frac{d}{dt}(\sinh t{e}^{-z\cosh t})$ vanish: the value is zero at $t=0$, and its absolute value tends to zero at infinity since the exponential $e^{-z\cosh t}$ dominates $\sinh t$. Thus $K_2(z)-K_0(z)=2K_1(z)\mathrm{/}z$, and $I_{M_{\ast }}=4\pi M_{\ast }{K}_1(z)\mathrm{/}\alpha$.

The same polar substitution in $Z_{M_{\ast }}$ gives $Z_{M_{\ast }}=4\pi M_{\ast }^3{\int }_0^{\mathrm{\infty }}{\sinh }^{2}t\cosh t{e}^{-z\cosh t}dt=\pi M_{\ast }^3(K_3(z)-K_1(z))\mathrm{.}$ The endpoint terms in the integral of $\frac{d}{dt}(\sinh (2t)e^{-z\cosh t})$ vanish by the same exponential bound. Using $\cosh (2t)-1=2{\sinh }^{2}t$ and $\cosh (3t)-\cosh t=4{\sinh }^{2}t\cosh t$ gives $K_3(z)-K_1(z)=4K_2(z)\mathrm{/}z$. Hence $Z_{M_{\ast }}=4\pi M_{\ast }^2{K}_2(z)\mathrm{/}\alpha$, which is finite and positive.

Interpret $d^{3}q$ as $d{\lambda }_3(q)$ on the measure space $({\mathbb{R}}^3,\mathcal{L}({\mathbb{R}}^3),{\lambda }_3)$ of [F3]. The density $Z_{M_{\ast }}^{-1}{e}^{-\alpha E(q)}$ is positive and Borel, hence $\mathcal{L}({\mathbb{R}}^3)$-measurable by [F3]. Thus ${\nu }_u$ is the measure on $\mathcal{L}({\mathbb{R}}^3)$ obtained by integrating this density against ${\lambda }_3$; its total mass is $Z_{M_{\ast }}^{-1}{Z}_{M_{\ast }}=1$ by the definition of $Z_{M_{\ast }}$, so it is a probability measure on $({\mathbb{R}}^3,\mathcal{L}({\mathbb{R}}^3))$. Also $h_{\mathrm{r}\mathrm{e}\mathrm{f}}^3{c}^3$ has units ${\mathrm{J}}^3{\mathrm{m}}^3$, so $n{h}_{\mathrm{r}\mathrm{e}\mathrm{f}}^3{c}^3$ and $Z_{M_{\ast }}$ both have units ${\mathrm{J}}^3$; hence $\zeta$ is dimensionless.

Differentiation under the integral sign in $Z_{M_{\ast }}(\alpha )$ is valid near any ${\alpha }_0>0$: for $\alpha \ge {\alpha }_0\mathrm{/}2$, $E{e}^{-\alpha E}\le C{e}^{-{\alpha }_{0}E\mathrm{/}4}$, where $C={\sup }_{E\ge M_{\ast }}E{e}^{-{\alpha }_{0}E\mathrm{/}4}<\mathrm{\infty }$ and the majorant is integrable by $Z_{M_{\ast }}({\alpha }_0\mathrm{/}4)<\mathrm{\infty }$. Thus $-{\mathrm{\partial }}_{\alpha }\log Z_{M_{\ast }}=⟨E⟩$. For the defining $K_2$ integral, on $z\in [z_0\mathrm{/}2,3z_0\mathrm{/}2]$ the derivative integrand is bounded in absolute value by $e^{3t}{e}^{-(z_0\mathrm{/}4)e^t}$, an integrable function after $y=e^t$. [F4] therefore permits differentiation and gives $K_2^{\mathrm{\prime }}(z)=-\frac{1}{2}(K_1(z)+K_3(z))$, since $\cosh t\cosh (2t)=(\cosh (3t)+\cosh t)\mathrm{/}2$. Combining this with the recurrence in the preceding calculation gives $K_2^{\mathrm{\prime }}(z)=-K_1(z)-2K_2(z)\mathrm{/}z$. Differentiating the formula for $Z_{M_{\ast }}$ now shows $⟨E⟩=\frac{3}{\alpha }+M_{\ast }\frac{K_1(\alpha M_{\ast })}{K_2(\alpha M_{\ast })}\mathrm{.}$