Jüttner partition, probability, and perfect-fluid moments

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 $⟨E⟩=\frac{3}{\alpha }+M_{\ast }\frac{K_1(\alpha M_{\ast })}{K_2(\alpha M_{\ast })}\mathrm{.}$

The bound $\mathrm{\mid }{q}_i\mathrm{\mid }\le E$ and the finite integrals of $e^{-\alpha E}$ and $E{e}^{-\alpha E}$ show that the current and stress integrals are absolutely convergent: in the rest tetrad, $\mathrm{\mid }{P}^a\mathrm{\mid }\le E$ and $\mathrm{\mid }{P}^a{P}^b\mathrm{\mid }\mathrm{/}E\le E$. Their common prefactor is $e^{\zeta }\mathrm{/}(h_{\mathrm{r}\mathrm{e}\mathrm{f}}^3{c}^3)=n\mathrm{/}{Z}_{M_{\ast }}$. Oddness in each $q_i$ makes the spatial current and off-diagonal stress entries zero; rotational invariance makes the three diagonal spatial entries equal. The time component of the current is $\frac{n}{Z_{M_{\ast }}}\int e^{-\alpha E}{d}^{3}q=n$, and the time-time stress entry is $\frac{n}{Z_{M_{\ast }}}\int E{e}^{-\alpha E}{d}^{3}q=n⟨E⟩$.

On $B_R\subset {\mathbb{R}}^3$, apply [F5] to $q\mapsto q{e}^{-\alpha E(q)}$. Its divergence is $(3-\alpha \mathrm{\mid }q{\mathrm{\mid }}^2\mathrm{/}E)e^{-\alpha E}$ and its outward flux is $4\pi R^3{e}^{-\alpha E(R)}$, which tends to zero as $R\to \mathrm{\infty }$ because $E(R)\ge R$. The divergence is absolutely integrable since its magnitude is at most $(3+\alpha E)e^{-\alpha E}$, whose integral is finite by the partition and energy calculations above. Dominated convergence [F7] on the increasing balls $B_m$ passes the bounded-ball identity to ${\mathbb{R}}^3$ and gives ${\int }_{{\mathbb{R}}^3}\frac{\mathrm{\mid }q{\mathrm{\mid }}^2}{E}{e}^{-\alpha E}{d}^{3}q=\frac{3}{\alpha }{Z}_{M_{\ast }}(\alpha )\mathrm{.}$ The trace of the spatial stress is $n\mathrm{/}{Z}_{M_{\ast }}$ times this integral, and its three diagonal entries are equal by step 4.1; therefore each is $p=n\mathrm{/}\alpha$.

The shell measure is independent of the future tetrad by [F1], and the integrals defining $J$ and $T$ are contractions of the geometric momentum vector with that measure. Thus their rest-tetrad components from step 4.1 and the pressure from step 5.1 determine the tensors intrinsically. In that tetrad $J=(n,0,0,0)$ and $T^{ab}=\mathrm{diag}(\epsilon ,p,p,p)$. Since $u=(1,0,0,0)$ and $g^{ab}=\mathrm{diag}(-1,1,1,1)$ there, these are exactly $J^a=n{u}^a$ and $T^{ab}=(\epsilon +p)u^a{u}^b+p{g}^{ab}$.