# The Tolman-Ehrenfest effect for an ideal gas in a background of time-
independent electric, magnetic, and gravitational fields

H. Arthur Weldon<EMAIL_ADDRESS>Department of Physics and Astronomy, West
Virginia University , Morgantown, West Virginia, 26506-6315

###### Abstract

The statistical mechanics of an ideal gas of point particles moving in a time
independent background metric with $g_{0j}\neq 0$ is investigated. An explicit
calculation shows that when there is no background electrostatic or
magnetostatic field the thermodynamic pressure, energy density, and thermally
averaged energy-momentum tensor depend on temperature and chemical potential
only through the ratios $T_{0}/\sqrt{g_{00}}$ and $\mu_{0}/\sqrt{g_{00}}$. A
background magnetostatic field does not change this, however with a background
electrostatic field the previous results are multiplied by a factor
$\exp(-eA_{0}/T_{0})$, which is an exception to the strict Tolman-Ehrenfest
rule because the system is open.

##  I Introduction

The Tolman-Ehrenfest effect Tolman1 ; Tolman2 ; Tolman3  originated with the
observation that classical electromagnetic radiation in thermal equilibrium
with matter in a static background metric ($\partial_{0}g_{\mu\nu}=0$ and
$g_{0j}=0$) would have a thermally averaged energy-momentum tensor whose
temperature dependence always occurs in the ratio $T_{0}/\sqrt{g_{00}}$, where
$T_{0}$ is spacetime independent. There are a number of arguments Sanchez ;
Lima ; Cremaschini ; Kim ; Faraoni ; Kovtun ; Munoz  that support the Tolman-
Ehrenfest result for the static case $g_{0j}=0$. Ref. Munoz  specifically
treats the ideal gas.

There are arguments that the Tolman-Ehrenfest effects is valid when
$g_{0j}\neq 0$ Rovelli1 ; Haggard ; Rovelli2 ; Wald ; Santiago1 ; Santiago2 ;
Xia ; Santiago3 ; Chernodub . A special example in this category is that of a
rotating Minkowski reference frame Santiago3 ; Chernodub  in which $g_{t\phi}$
is non-zero but there is no curvature.

The Schwarzschild metric illustrates the issue. In the original
$(t,r,\theta,\phi)$ coordinates the metric is static:

$(ds)^{2}=\Big{(}1-{2GM\over r}\Big{)}(dt)^{2}-{(dr)^{2}\over
1-2GM/r}-r^{2}d\Omega^{2}$ (1)

with $d\Omega^{2}=d\theta^{2}+\sin^{2}\theta\,d\phi^{2}$. The thermodynamic
functions for an ideal gas in this background metric will depend on $T_{0}$
only through the ratio $T_{0}/\sqrt{1-2GM/r}$. A change to the outgoing
Eddington-Finkelstein time coordinate Eddington ; Finkelstein

$t^{\prime}=t-2GM\ln(r-2GM)$ (2)

changes the form of the line element to

$\displaystyle(ds)^{2}$ $\displaystyle=$ $\displaystyle\Big{(}1-{2GM\over
r}\Big{)}(dt^{\prime})^{2}+{4GM\over r}dt^{\prime}dr$ (3) $\displaystyle-$
$\displaystyle\Big{(}1+{2GM\over r}\Big{)}(dr)^{2}-r^{2}d\Omega^{2}.$

Since $g_{t^{\prime}r}\neq 0$ this metric is not static but stationary. It
will follow from the calculation described herein that an ideal gas of point
particles in this metric will be the same function of $T_{0}/\sqrt{1-2GM/r}$.

The more important situations are those in which the metric is stationary and
cannot be changed to static by a coordinate transformation. Though in this
paper the metric is not required to satisfy the Einstein field equations,
there are two familiar stationary metrics that do solve the field equations
and are not coordinate equivalent to a static metric: the Kerr metric
describing an uncharged, but rotating black hole; and the Kerr-Newmann metric
describing a charged, rotating black hole Newman . In Boyer-Lindquist
coordinates $(t,r,\theta,\phi)$ both metrics have $g_{t\phi}\neq 0$. The Kerr-
Newman black hole is surrounded by a static electric field and a static
magnetic field. There are many other stationary metrics that solve the field
equations Stephani . A limitation of the present analysis is that equilibrium
statistical mechanics, whether done in the canonical or the grand canonical
ensemble, applies to situations in which the total number of particles is
conserved. Thus if the metric has an event singularity statistical mechanics
is only applicable well outside the event horizon.

The Kerr-Newman example motivates extending the investigation to include
arbitrary background electrostatic and/or magnetostatic fields and this leads
to a specific exception to the Tolman-Ehrenfest effect. The exception can be
illustrated by a simple example. Consider an ideal gas of nonrelativistic
point particles with no gravitational field. In kinetic theory the
distribution function for such a gas in thermal equilibrium is

$f_{0}=n_{0}\Big{(}{2\pi\over mT_{0}}\Big{)}^{3/2}e^{-{\bf p}^{2}/2mT_{0}}.$
(4)

The particle density $n_{0}=\int[d^{3}p/(2\pi)^{3}]f_{0}$ is spatially
uniform. Suppose the particles have charge $e$ and a time-independent external
electric and magnetic field is imposed. After equilibrium is achieved the new
distribution function should satisfy the Vlasov equation Landau2

${\partial f\over\partial t}+v^{j}{\partial f\over\partial x^{j}}+e({\bf
E}+{\bf v}\times{\bf B})_{j}{\partial f\over\partial p_{j}}=0.$ (5)

In a complete analysis ${\bf E}$ and ${\bf B}$ would be the sum of the
external fields and the internal fields; the internal fields being determined
self-consistently by solving Maxwell’s equations with the charge density and
current density computed from the distribution function $f$. If the external
fields are much stronger than the internal fields produced by the charged
particles then ${\bf E}$ and ${\bf B}$ may taken as the external fields, with
time-independent scalar and vector potentials $A_{0}$ and ${\bf A}$. The
solution to the Vlasov equations is

$f({\bf x},{\bf p})=n_{0}\Big{(}{2\pi\over mT_{0}}\Big{)}^{3/2}e^{-[({\bf
p}-e{\bf A})^{2}/2m+eA_{0}]/T_{0}}$ (6)

where $A_{0}$ and ${\bf A}$ are evaluated at the position ${\bf x}(t)$ of the
particle and $f$ has no explicit time dependence. The particle density is

$n({\bf x})=\int{d^{3}p\over(2\pi)^{3}}f({\bf x},{\bf p})=n_{0}e^{-eA_{0}({\bf
x})/T_{0}}.$ (7)

The factor $\exp[-eA_{0}/T_{0}]$ will appear in other thermodynamic functions:
pressure, energy density, entropy density. (The gauge transformation that
eliminates $A_{0}$ would make ${\bf A}$ time-dependent and the distribution
function would no longer satisfy the Vlasov equation.) It will turn out that
in the presence of a stationary gravitational field the same factor
$\exp[-eA_{0}/T_{0}]$ will occur, whereas the Tolman-Ehrenfest expectation
would be a factor $\exp[-eA_{0}\sqrt{g_{00}}/T_{0}]$. This is not in conflict
with derivations of the Tolman-Ehrenfest effect Sanchez ; Lima ; Cremaschini ;
Kim ; Faraoni ; Kovtun ; Munoz ; Rovelli1 ; Haggard ; Rovelli2 ; Wald ;
Santiago1 ; Santiago2 ; Xia ; Santiago3  which rely on
$(T^{\mu\nu})_{;\mu}=0$.

Throughout the discussion the particles are thermalized; the internal electric
and magnetic fields produced by the charged particles are neglected and
$A_{0}$, ${\bf A}$ are unthermalized, external potentials.

The central problem is that the non-vanishing of $g_{0\ell}$ makes the
Hamiltonian for a point particle moving on the geodesics of a time-independent
background metric rather complicated. It will be shown in Sec. II that the
particle Hamiltonian in the absence of a background electrostatic or
magnetostatic field is

$\overline{H}={1\over
g^{00}}\bigg{[}\sqrt{g^{00}(m^{2}\\!-\\!g^{ij}p_{i}p_{j})\\!+\\!(g^{0\ell}p_{\ell})^{2}}+g^{0\ell}p_{\ell}\bigg{]}.$
(8)

Here $p_{j}(t)$ are the canonical momenta and the metric components are
evaluated at the particle position $x^{i}(t)$. Though covariant metric
components $g_{\alpha\beta}$ do not appear in $\overline{H}$, the minimum
value of $\overline{H}$ occurs at $p_{j}=-mg_{j0}/\sqrt{g_{00}}$ and the
minimum energy is $m\sqrt{g_{00}}$.

##### Outline.

Sec. II derives the Hamiltonian $\overline{H}$ and incorporates a time-
independent background of static electric and magnetic fields. The Hamiltonian
leads to the partition function and the thermodynamic pressure in the grand
canonical ensemble in terms of two parameters $T_{0}$ and a chemical potential
$\mu_{0}$. The pressure due to the particles is of the form
$e^{\beta_{0}(\mu_{0}-eA_{0})}\overline{P}$.

Sec. III computes $\overline{P}$ explicitly and after using a particular
addition theorem for Bessel functions the final result is

$\overline{P}={m^{2}\over
2\pi^{2}}\Big{(}{T_{0}\over\sqrt{g_{00}}}\Big{)}^{2}K_{2}\Big{(}m{\sqrt{g_{00}}\over
T_{0}}\Big{)}$ (9)

with no dependence on $g_{0\ell}$ or $g_{jk}$.

Sec. IV contains the computation of the thermally averaged energy-momentum
tensor. The particle contribution has the perfect fluid form

$\langle T^{\mu\nu}_{\rm
part}\rangle=e^{\beta_{0}(\mu_{0}-eA_{0})}\big{[}U^{\mu}U^{\nu}(\overline{\rho}+\overline{P})-g^{\mu\nu}\overline{P}\big{]}.$
(10)

Sec. V discusses the low temperature limit and how to change from classical
Boltzman statistics to Bose or Fermi statistics.

Appendix A details the change of integration variables from the canonical
momenta to Euclidean momenta that makes possible the integration in Sec. III.

Appendix B proves a result used in Sec. IV, namely that the thermally averaged
energy-momentum tensor may be calculated from the variational derivative of
the partition function with respect to the time-independent metric:

${\delta\ln Z\over\delta g_{\mu\nu}}=-{\beta_{0}\sqrt{g}\over 2}\Big{[}\langle
T^{\mu\nu}_{\rm part}\rangle+T^{\mu\nu}_{\rm field}\Big{]}.$ (11)

Because $g_{0j}\neq 0$ it is not trivial to compute ${\cal H}_{\rm field}$ in
terms of the canonical momenta $\pi^{j}$ and the covariant fields $F_{jk}$.
The result is displayed in (34). The derivative of this Hamiltonian density
with respect to the metric, keeping $\pi^{j}$ and $F_{jk}$ fixed, yields the
usual Hilbert energy-momentum tensor $T^{\mu\nu}_{\rm field}$ when the momenta
are re-expressed in terms of field strengths.

Appendix C gives specific results for the energy density, number density, and
entropy density. At low temperature the entropy density is given by a local
form of the Sackur-Tetrode equation.

Greek letters run over 0,1,2,3; Latin letters over 1,2,3. The metric signature
is $(+---)$ and $g=|{\rm det}(g_{\mu\nu})|$; and $\hbar=c=1$.

##  II Hamiltonian and partition function

The first step is to derive the expression (42) for the thermodynamic
pressure. The starting point is the action for $N$ particles in a background
comprised of a time-independent metric plus electrostatic and magnetostatic
fields Weinberg

$\displaystyle S$ $\displaystyle=$
$\displaystyle-\sum_{n=1}^{N}\Big{\\{}m\int\sqrt{g_{\mu\nu}(x_{n})dx_{n}^{\mu}dx_{n}^{\nu}}+e\int
A_{\mu}(x_{n})dx_{n}^{\mu}\Big{\\}}$ (12) $\displaystyle-\int
d^{4}x{\sqrt{g}\over 4}F_{\mu\nu}(x)F^{\mu\nu}(x).$

The particle coordinates $x_{n}^{j}$ depend on the time coordinate $x_{n}^{0}$
or equivalently on the proper time
$d\tau_{n}=g_{\mu\nu}(x_{n})dx_{n}^{\mu}dx_{n}^{\nu}$. With $g_{0j}\neq 0$ the
proper time is not time reversal invariant and the particle motion is not
time-reversal invariance.

##### Hamiltonian for N particles.

The Lagrangian for $N$ particles is a sum of identical terms

$L_{N}=\sum_{n=1}^{N}L(x_{n}^{i},v^{j}_{n})$

where $v_{n}^{j}=dx^{j}/dx^{0}_{n}$ and each term has the form

$L(x^{i},v^{j})=-m\sqrt{g_{00}+2g_{j0}v^{j}+g_{jk}v^{j}v^{k}}-e(A_{0}+A_{j}v^{j}).$

The metric and the vector potential are evaluated at the position of the
particle $x^{k}(t)$. The canonical momentum is

$p_{j}={\partial L\over\partial
v^{j}}={-m(g_{j0}+g_{jk}v^{k})\over\sqrt{g_{00}+2g_{0i}v^{i}+g_{i\ell}v^{i}v^{\ell}}}-eA_{j}.$
(13)

In terms of the velocity $v^{j}=dx^{j}/dt$ this means that

$\displaystyle H_{\rm part}$ $\displaystyle=$ $\displaystyle p_{j}v^{j}-L$
(14) $\displaystyle=$
$\displaystyle{m(g_{00}+g_{0j}v^{j})\over\sqrt{g_{00}+2g_{0i}v^{i}+g_{i\ell}v^{i}v^{\ell}}}+eA_{0}.$

To express the velocity in terms of the momentum it is convenient to define

$p_{j}^{\prime}=p_{j}+eA_{j}.$ (15)

Note that $p_{j}$ depends only on $t$ and $A_{j}$ depends on $x^{i}(t)$. Eq.
(13) may be inverted to express the velocity in terms of momenta

$v^{j}={g^{j0}\over
g^{00}}-{c^{jk}p^{\prime}_{k}\over\sqrt{g^{00}[m^{2}-c^{i\ell}p^{\prime}_{i}p^{\prime}_{\ell}]}}.$
(16)

The matrix $c^{jk}$ is given by

$c^{jk}=g^{jk}-{g^{j0}g^{k0}\over g^{00}}$ (17)

and satisfies

$g_{ij}c^{jk}=\delta_{i}^{k}.$ (18)

(For the special case in which $g_{ij}$ is diagonal, one can calculate
$g^{\mu\nu}$ by Cramer’s rule and show that $c^{jk}$ is diagonal and
$c^{jj}=1/g_{jj}$. ) The Hamiltonian for each particle is

$\displaystyle H_{\rm part}=\overline{H}+eA_{0}$
$\displaystyle\overline{H}={1\over
g^{00}}\bigg{[}\sqrt{g^{00}(m^{2}\\!-\\!c^{i\ell}p^{\prime}_{i}p^{\prime}_{\ell})}+g^{0\ell}p^{\prime}_{\ell}\bigg{]}.$
(19)

The minimum of $\overline{H}$ is $m\sqrt{g_{00}}$ and occurs at
$p^{\prime}_{j}=-mg_{j0}/\sqrt{g_{00}}$. For small velocity $\overline{H}$ may
be expanded as

$\overline{H}=m\sqrt{g_{00}}\bigg{\\{}1-{g_{ij}v^{i}v^{j}\over
2g_{00}}+2\Big{[}{g_{0\ell}v^{\ell}\over g_{00}}\Big{]}^{2}+{\cal
O}(v^{3})\bigg{\\}}$ (20)

which shows that for small velocity if $g_{0\ell}v^{\ell}\neq 0$ the energy is
higher than for a static metric.

##### The equation of motion.

Though equilibrium statistical mechanics does not employ the equation of
motion it is important to check that the Hamiltonian does give the correct
equation of motion for the particles. Hamilton’s first equation

$v^{j}={\partial H_{\rm part}\over\partial p_{j}}$ (21)

reproduces (16). Hamilton’s second equation

$-{dp_{j}\over dt}={\partial H_{\rm part}\over\partial x^{j}}$ (22)

will yield the equation of motion. It is convenient to employ the proper
velocity

$u^{\mu}={dx^{\mu}\over d\tau}$ (23)

with components

$\displaystyle u^{j}$ $\displaystyle=$
$\displaystyle{v^{j}\over\sqrt{g_{00}+2g_{0i}v^{i}+g_{i\ell}v^{i}v^{\ell}}}$
(24) $\displaystyle u^{0}$ $\displaystyle=$
$\displaystyle{1\over\sqrt{g_{00}+2g_{0i}v^{i}+g_{i\ell}v^{i}v^{\ell}}}.$ (25)

Since $u_{\alpha}=g_{\alpha\mu}u^{\mu}$, reference to (13) and (14) shows that

$\displaystyle p_{j}$ $\displaystyle=$ $\displaystyle-mu_{j}-eA_{j}$ (26)
$\displaystyle H_{\rm part}$ $\displaystyle=$ $\displaystyle mu_{0}+eA_{0}.$
(27)

Hamilton’s second equation (22) becomes

$m{du_{j}\over dt}+e{dA_{j}\over dt}=m{\partial u_{0}\over\partial
x^{j}}+e{\partial A_{0}\over\partial x^{j}}.$ (28)

A simple way to compute the spatial derivative of $u_{0}$ is to apply
$\partial/\partial x^{j}$ to $m^{2}=g^{\mu\nu}u_{\mu}u_{\nu}$, which results
in

$u^{0}{\partial u_{0}\over\partial x^{j}}=-{1\over
2}(\partial_{j}g^{\mu\nu})u_{\mu}u_{\nu}+eu^{k}\partial_{j}A_{k}.$ (29)

Multiplying (28) by $u^{0}$ and using $u^{0}(d/dt)=d/d\tau$ gives

$m\Big{[}{du_{j}\over d\tau}\\!-\\!{1\over
2}(\partial_{j}g_{\mu\nu})u^{\mu}u^{\nu}\Big{]}=eF_{j\alpha}u^{\alpha}.$ (30)

As yet there are only three equations. To obtain the fourth contract (30) with
$u^{j}$ and use $u^{j}(du_{j}/d\tau)=-u^{0}(du_{0}/d\tau)$ to obtain

$m{du_{0}\over d\tau}=eF_{0j}u^{j}.$ (31)

The four components of (30) and (31) are summarized by

$m\Big{(}{du_{\lambda}\over d\tau}-{1\over
2}(\partial_{\lambda}g_{\mu\nu})u^{\mu}u^{\nu}\Big{)}=eF_{\lambda\alpha}u^{\alpha}.$
(32)

The contravariant form of this equation is

$m\Big{(}{du^{\lambda}\over
d\tau}+\Gamma^{\lambda}_{\mu\nu}g_{\mu\nu}u^{\mu}u^{\nu}\Big{)}=eF^{\lambda\alpha}u_{\alpha},$
(33)

which is the correct equation of motion for a particle in a background
electric/magnetic field Landau1 . Solutions for the particle trajectories are
obtained in Hackmann .

##### Hamiltonian for the background electric/magnetic field.

The Hamiltonian for the background electromagnetic field is also needed. The
canonical Hamiltonian density given by Noether’s theorem in terms of the
canonical momenta $\pi^{j}=\partial{\cal L}_{\rm
field}/\partial(\partial_{0}A_{j})$ and canonical fields $F_{jk}$ is

$\displaystyle{\cal H}_{\rm field}$ $\displaystyle=$
$\displaystyle-{\pi^{j}\pi^{k}g_{jk}\over
2g^{00}\sqrt{g}}+{\pi^{j}F_{jk}g^{k0}\over g^{00}}$ (34) $\displaystyle+$
$\displaystyle{\sqrt{g}\over 4}F_{jk}F_{\ell
m}c^{j\ell}c^{km}+\pi^{j}\partial_{j}A_{0}$

as shown in Appendix C.

##### Partition function for the ideal gas.

In a gas of $N$ particles the Hamiltonian for each particle is of the form
(19) and is a function of each particle’s contravariant position coordinates
$x^{1},x^{2},x^{3}$ and its covariant momenta $p_{1},p_{2},p_{3}$. The
Hamiltonian for $N$ particles is the sum of the single particle Hamiltonians:

$H_{N}=\sum_{n=1}^{N}H_{\rm part}(x^{i}_{n},p^{n}_{j})$ (35)

and the total Hamiltonian is $H_{N}+\int d^{3}x{\cal H}_{\rm field}$. The
Boltzman factor $e^{-\beta_{0}H_{N}}$ is a product of N exponentials. The
integration of each factor $e^{-\beta_{0}H_{\rm part}}$ over its six-
dimensional phase space is the same and so the partition function for N
particles is

$Z_{N}={1\over N!}\Big{[}\int{d^{3}xd^{3}p\over(2\pi)^{3}}e^{-\beta_{0}H_{\rm
part}}\Big{]}^{N}e^{-\beta_{0}H_{\rm field}}.$ (36)

(Because of the convention $\hbar\to 1$ the denominator
$h^{3}=(2\pi\hbar)^{3}\to(2\pi)^{3}$.)

In the grand canonical ensemble the partition function depends on the chemical
potential $\mu_{0}$:

$Z=\sum_{N=0}^{\infty}(e^{\beta_{0}\mu_{0}})^{N}Z_{N},$ (37)

and therefore

$\ln Z=\int{d^{3}xd^{3}p\over(2\pi)^{3}}e^{-\beta_{0}(H_{\rm
part}-\mu_{0})}-\beta_{0}H_{\rm field}.$ (38)

The momentum integration variable may be shifted from the canonical $p_{j}$ to
$p_{j}^{\prime}$

$p_{j}^{\prime}=p_{j}+eA_{j};$ (39)

this removes $A_{j}$ from the partition function as expected from the Bohr-Van
Leeuwen theorem Huang .

The partition function is directly related to the thermodynamic pressure
Dowker

$\ln Z=\beta_{0}\int d^{3}x\sqrt{g}\,P.$ (40)

Therefore

$P=P_{\rm part}-{1\over\sqrt{g}}{\cal H}_{\rm field},$ (41)

where the pressure exerted by the gas of particles is

$P_{\rm
part}={T_{0}\over\sqrt{g}}\int{d^{3}p^{\prime}\over(2\pi)^{3}}e^{-\beta_{0}(H_{\rm
part}-\mu_{0})}.$ (42)

The next step is to perform this integration.

##  III Calculation of the particle pressure

The terms in $H_{\rm part}$ involving $\mu_{0}$ and $eA_{0}$ have no momentum
dependence and so

$\displaystyle P_{\rm part}$ $\displaystyle=$ $\displaystyle
e^{\beta_{0}(\mu_{0}-eA_{0})}\overline{P}$ (43) $\displaystyle\overline{P}$
$\displaystyle=$
$\displaystyle{T_{0}\over\sqrt{g}}\int{d^{3}p^{\prime}\over(2\pi)^{3}}e^{-\beta_{0}\overline{H}}.$
(44)

The dependence on $\mu_{0}$ may be written

$\beta_{0}\mu_{0}={\sqrt{g_{00}}\over T_{0}}{\mu_{0}\over\sqrt{g_{00}}}.$ (45)

and is in agreement with Oscar Klein’s argument Klein  that both $T_{0}$ and
$\mu_{0}$ will always occur divided by $\sqrt{g_{00}}$. The factor
$\exp[-eA_{0}/T_{0}]$ in (43) is an exception to the Tolman-Ehrenfest rule and
will remain in the final result. This section will show that $\overline{P}$ is
given by (67) and is a function only of the ratio $T_{0}/\sqrt{g_{00}}$.

##### Euclidean momenta.

To compute $\overline{P}$ it is necessary to change the integration variables
from $p_{j}^{\prime}$ with metric $g_{jk}$ into Euclidean momenta $k_{a}$. The
details of this transformation are given in Appendix A. The Hamiltonian
becomes

$\overline{H}={1\over\sqrt{g^{00}}}\Big{[}\sqrt{m^{2}+{\bf k}^{2}}+{\bf
s}\cdot{\bf k}\Big{]},$ (46)

where ${\bf s}$ is a Euclidean vector with length squared

${\bf s}^{2}=g_{0j}g^{j0}=1-g_{00}g^{00}.$ (47)

The minimum of $\overline{H}$ occurs at $k_{a}=-m{\rm
s}_{a}\big{/}\sqrt{1-{\rm s}^{2}}$ and the value of the minimum is still
$m\sqrt{g_{00}}$. The change in integration variables requires

$d^{3}p^{\prime}=\sqrt{gg^{00}}\,d^{3}k,$ (48)

which leads to

$\overline{P}=T_{0}\sqrt{g^{00}}\int{d^{3}k\over(2\pi)^{3}}e^{-\beta_{0}\overline{H}}.$
(49)

##### Dimensionless variables.

The change to a dimensionless integration variable ${\bf u}={\bf k}/m$ and
introduction of a dimensionless parameter

$z={m\over T_{0}\sqrt{g^{00}}}.$ (50)

converts the exponent in the integrand of (44) to

$\beta_{0}\overline{H}=z\big{[}\sqrt{1+{\bf u}^{2}}+{\bf s}\cdot{\bf
u}\big{]}.$ (51)

The pressure integral is

$\overline{P}(s,z)={m^{4}\over
z}\int{d^{3}u\over(2\pi)^{3}}e^{-\beta_{0}\overline{H}}.$ (52)

It appears that $\overline{P}$ is a function of the two variables $s$ and $z$.
To show that it actually depends only on a particular combination of these two
variables requires performing the integration. The angular integration gives

$\overline{P}(s,z)={m^{4}\over
2\pi^{2}sz^{2}}\int_{0}^{\infty}\\!\\!udu\,e^{-z\sqrt{1+u^{2}}}\sinh[szu].$
(53)

Next expand the $\sinh$ in an infinite series:

$\overline{P}(s,z)={m^{4}\over
2\pi^{2}z}\sum_{\ell=0}^{\infty}{(sz)^{2\ell}\over(2\ell\\!+\\!1)!}\int_{0}^{\infty}du\,e^{-z\sqrt{1+u^{2}}}u^{2\ell+2}.$
(54)

The necessary integrals are modified Bessel functions of the second kind
Gradshteyn :

$\int_{0}^{\infty}du\,e^{-z\sqrt{1+u^{2}}}u^{2\ell+2}={\Gamma(\ell+{3\over
2})\over\Gamma({1\over 2})}\Big{(}{2\over z}\Big{)}^{\ell+1}K_{\ell+2}(z)$
(55)

and the pressure becomes

$\overline{P}(s,z)={m^{4}\over
8\pi^{2}}\sum_{\ell=0}^{\infty}{s^{2\ell}\over\ell!}\Big{(}{z\over
2}\Big{)}^{\ell-2}K_{\ell+2}(z).$ (56)

By using the asymptotic value of $K_{\ell+2}(z)$ either for low temperature
($z\gg 1$) or for high temperature ($z\ll 1$) one can perform the sum on
$\ell$ and obtain the leading term and the sub-leading term in either regime
and confirm the Tolman-Ehrenfest effect to that order. This suggests that the
infinite series representation can be simplified.

##### Simpler expression for $\overline{P}$.

It will turn out that the entire series is only a function of the single
variable

$z\sqrt{1-s^{2}}=m{\sqrt{g_{00}}\over T_{0}}.$ (57)

To show this define the derivative combination

${\cal D}=z{\partial\over\partial z}+({1\over s}-s){\partial\over\partial s},$
(58)

with the property

${\cal D}(z\sqrt{1-s^{2}})=0.$ (59)

Application of ${\cal D}$ to (56) gives

$\displaystyle{\cal D}\overline{P}(s,z)$ $\displaystyle=$
$\displaystyle{m^{4}\over
8\pi^{2}}\sum_{\ell=0}^{\infty}\bigg{[}{s^{2\ell}\over\ell!}2\ell({1\over
s^{2}}-1)\Big{(}{z\over 2}\Big{)}^{\ell-2}K_{\ell+2}(z)$
$\displaystyle+{s^{2\ell}\over\ell!}(\ell-2)\Big{(}{z\over
2}\Big{)}^{\ell-2}K_{\ell+2}(z)$
$\displaystyle+{s^{2\ell}\over\ell!}\Big{(}{z\over 2}\Big{)}^{\ell-2}z{d\over
dz}K_{\ell+2}(z)\bigg{]}.$

The last line is simplified by using the identity Gradshteyn

$z{d\over dz}K_{\ell+2}(z)=(\ell+2)K_{\ell+2}(z)-zK_{\ell+3}(z).$ (61)

All terms proportional to $K_{\ell+2}$ combine:

$\displaystyle{\cal D}\overline{P}(s,z)$ $\displaystyle=$
$\displaystyle{m^{4}\over
8\pi^{2}}\sum_{\ell=1}^{\infty}{s^{2\ell-2}\over(\ell\\!-\\!1)!}2\Big{(}{z\over
2}\Big{)}^{\ell-2}K_{\ell+2}(z)$ (62) $\displaystyle-$
$\displaystyle{m^{4}\over
8\pi^{2}}\sum_{\ell=0}^{\infty}{s^{2\ell}\over\ell!}\Big{(}{z\over
2}\Big{)}^{\ell-2}zK_{\ell+3}(z).$

The two series cancel and thus ${\cal D}\overline{P}(s,z)=0$ and so
$\overline{P}(s,z)$ is a function only of the single variable (57):

$\overline{P}(s,z)={m^{4}\over 8\pi^{2}}\Psi(z\sqrt{1-s^{2}}\,),$ (63)

or equivalently

$\sum_{\ell=0}^{\infty}{s^{2\ell}\over\ell!}\Big{(}{z\over
2}\Big{)}^{\ell-2}K_{\ell+2}(z)=\Psi(z\sqrt{1-s^{2}}).$ (64)

To obtain an explicit form for $\Psi$, set $s=0$

$\Big{(}{2\over z}\Big{)}^{2}K_{2}(z)=\Psi(z).$ (65)

This determines the function $\Psi$ and so for $s\neq 0$

$\overline{P}(s,z)={m^{4}\over
2\pi^{2}}{K_{2}(z\sqrt{1-s^{2}}\,)\over[z\sqrt{1-s^{2}}]^{2}}.$ (66)

In more physical variables

$\overline{P}={m^{2}\over
2\pi^{2}}\Big{(}{T_{0}\over\sqrt{g_{00}}}\Big{)}^{2}K_{2}\Big{(}m{\sqrt{g_{00}}\over
T_{0}}\Big{)}.$ (67)

(That the two expressions (56) and (67) for $\overline{P}$ are equal amounts
to

${1\over 4}\sum_{\ell=0}^{\infty}{s^{2\ell}\over\ell!}\Big{(}{z\over
2}\Big{)}^{\ell-2}K_{\ell+2}(z)={K_{2}(z\sqrt{1-s^{2}}\,)\over[z\sqrt{1-s^{2}}]^{2}},$
(68)

which is a known addition theorem for Bessel functions Watson .)

##### Energy density.

For later purposes it is convenient to compute the quantity

$\overline{\rho}={T_{0}\over\sqrt{g}}\int{d^{3}p^{\prime}\over(2\pi)^{3}}\overline{H}e^{-\beta_{0}\overline{H}}=T_{0}{\partial\overline{P}\over\partial
T_{0}}-\overline{P}.$ (69)

Using (61) gives

$\overline{\rho}+\overline{P}={m^{3}\over
2\pi^{2}}\Big{(}{T_{0}\over\sqrt{g_{00}}}\Big{)}K_{3}\Big{(}m{\sqrt{g_{00}}\over
T_{0}}\Big{)}.$ (70)

##  IV The energy-momentum tensor

Appendix B shows that the thermally averaged energy-momentum tensor may be
computed from the variation derivative of the partition function. Eqs. (109)
and (126) give

${\delta\ln Z\over\delta g_{\mu\nu}}=-{\beta_{0}\sqrt{g}\over 2}\Big{[}\langle
T^{\mu\nu}_{\rm part}\rangle+T^{\mu\nu}_{\rm field}\Big{]}.$ (71)

Since $\ln Z=\beta_{0}\int d^{3}x\sqrt{g}P$ this is equivalent to

$\langle T^{\mu\nu}_{\rm part}\rangle+T^{\mu\nu}_{\rm
field}=-{2\over\sqrt{g}}{\partial(\sqrt{g}P)\over\partial g_{\mu\nu}}.$ (72)

Using $\sqrt{g}P$ from (41) gives

$\displaystyle\langle T^{\mu\nu}_{\rm part}\rangle$ $\displaystyle=$
$\displaystyle-{2\over\sqrt{g}}{\partial(\sqrt{g}P_{\rm part})\over\partial
g_{\mu\nu}}$ (73) $\displaystyle T^{\mu\nu}_{\rm field}$ $\displaystyle=$
$\displaystyle{2\over\sqrt{g}}{\partial{\cal H}_{\rm field}\over\partial
g_{\mu\nu}}.$ (74)

##### Field EMT.

In Appendix B, ${\cal H}_{\rm field}$ is expressed in terms of canonical
momenta $\pi^{j}$ and canonical fields $F_{jk}$ with the result (121). The
partial derivative (74) is computed keeping $\pi^{j}$ and $F_{jk}$ fixed; when
the result is re-expressed in terms of the fields the result is the usual
Hilbert energy-momentum tensor:

$T^{\mu\nu}_{\rm
field}=-F^{\mu\alpha}F^{\nu\beta}g_{\alpha\beta}+{g^{\mu\nu}\over
4}F_{\alpha\beta}F^{\alpha\beta}.$ (75)

##### Particle EMT.

To implement Eq. (73) begin with $P_{\rm
part}=e^{\beta_{0}(\mu_{0}-eA_{0})}\overline{P}$. Since $\overline{P}$ depends
only on the $g_{00}$ component of the metric the derivative in (73) is

$\langle T^{\mu\nu}_{\rm
part}\rangle=e^{\beta_{0}(\mu_{0}-eA_{0})}\Big{[}-2\delta^{\mu}_{0}\delta^{\nu}_{0}{\partial\overline{P}\over\partial
g_{00}}-g^{\mu\nu}\overline{P}\Big{]}$ (76)

As shown in (66) the $g_{00}$ dependence of $\overline{P}$ occurs only through
the ratio $T_{0}/\sqrt{g_{00}}$ and so a $g_{00}$ derivative is equivalent to
a $T_{0}$ derivative:

$-2g_{00}{\partial\overline{P}\over\partial
g_{00}}=T_{0}{\partial\overline{P}\over\partial
T_{0}}=\overline{\rho}+\overline{P}.$ (77)

Therefore

$\langle T^{\mu\nu}_{\rm
part}\rangle=e^{\beta_{0}(\mu_{0}-eA_{0})}\Big{\\{}\delta^{\mu}_{0}\delta^{\nu}_{0}{(\overline{\rho}+\overline{P})\over
g_{00}}-g^{\mu\nu}\overline{P}\Big{\\}}.$ (78)

The thermal average of the particle velocity is zero because $v^{j}=\partial
H_{\rm part}/\partial p_{j}$:

$\int{d^{3}p^{\prime}\over(2\pi)^{3}}\,v^{j}e^{-\beta_{0}(H_{\rm
part}-\mu_{0})}=0.$ (79)

The normalized velocity vector of the ideal gas is therefore
$U^{\mu}=\delta^{\mu}_{0}/\sqrt{g_{00}}$, which allows (78) to be expressed in
the perfect fluid form

$\displaystyle\langle T^{\mu\nu}_{\rm part}\rangle$ $\displaystyle=$
$\displaystyle
e^{\beta_{0}(\mu_{0}-eA_{0})}\Big{\\{}U^{\mu}U^{\nu}(\overline{\rho}+\overline{P})-g^{\mu\nu}\overline{P}\Big{\\}}$
(80) $\displaystyle=$ $\displaystyle\Big{\\{}U^{\mu}U^{\nu}(\rho_{\rm
part}+P_{\rm part})-g^{\mu\nu}P_{\rm part}\Big{\\}}.$

Because of the external potential $A_{0}({\bf x})$ the covariant divergence of
$\langle T^{\mu\nu}_{\rm part}\rangle$ is not zero:

$\langle T^{\mu\nu}_{\rm part}\rangle_{;\mu}=g^{\nu j}{e\over
T_{0}}(\partial_{j}A_{0}).$ (81)

##  V Discussion

##### Low temperature example.

The spatial dependence of $\overline{P}$ comes entirely from $g_{00}$ and so
$\overline{P}$ is constant on surfaces of constant $g_{00}$. The full particle
pressure $P_{\rm part}=e^{\beta_{0}(\mu_{0}-eA_{0})}\overline{P}$ has somewhat
different isobaric surfaces if $A_{0}\neq 0$. The result (67) for
$\overline{P}$ may be evaluated when $Z=m\sqrt{g_{00}}/T_{0}\gg 1$ using the
asymptotic behavior $K_{2}(Z)\to e^{-Z}\sqrt{\pi/2Z}$:

$P_{\rm part}={T_{0}\over\sqrt{g_{00}}}\Big{(}{mT_{0}\over
2\pi\sqrt{g_{00}}}\Big{)}^{3/2}e^{\beta_{0}(\mu_{0}-eA_{0}-m\sqrt{g_{00}})}+\dots.$
(82)

The resulting thermodynamic functions (number density, entropy density, and
energy density) are displayed in Appendix C. The gradient of the pressure is

${\partial_{j}P_{\rm part}\over P_{\rm
part}}=-\beta_{0}\Big{[}m{\partial_{j}g_{00}\over
2g_{00}}+e\partial_{j}A_{0}\Big{]}-{5\over
4}{\partial_{j}g_{00}\over(g_{00})^{3/2}}.$ (83)

For the Kerr-Newman metric in Boyer-Lindquist coordinates $t,r,\theta,\phi$
both $g_{00}$ and $A_{0}$ depend on $r$ and $\theta$. At large distances the
$\theta$ dependence is negligible

$\displaystyle g_{00}$ $\displaystyle\approx$ $\displaystyle 1-{2GM\over r}$
(84) $\displaystyle A_{0}$ $\displaystyle\approx$ $\displaystyle{Q\over r}$
(85)

and the pressure gradient is radial:

${dP_{\rm part}\over dr}\approx n\Big{(}-{GMm\over r^{2}}+{eQ\over
r^{2}}\Big{)}$ (86)

after using $\beta_{0}P_{\rm part}\approx n$ and neglecting the ${\cal
O}(T_{0}/m)$ correction. The first term in the parenthesis is the
gravitational force exerted on the particle by the central mass; the second
term is the electrostatic force on the particle.

##### Quantum statistics.

The calculations presented are for distinguishable point particles obeying
Boltzman statistics. If instead the particles are indistinguishable and obey
either Bose-Einstein or Fermi-Dirac statistics the particle pressure becomes

$P_{\rm
part}=-{T_{0}\over\xi\sqrt{g}}\int{d^{3}p\over(2\pi)^{3}}\ln\big{[}1-\xi
e^{-\beta_{0}(H_{\rm part}-\mu_{0})}\big{]}$ (87)

where $\xi=1$ for Bose statistics and $\xi=-1$ for Fermi statistics. When
expanded in a series

$P_{\rm part}={T_{0}\over\xi\sqrt{g}}\sum_{b=1}^{\infty}{(\xi
e^{\beta_{0}(\mu_{0}-eA_{0})})^{b}\over
b}\int{d^{3}p\over(2\pi)^{3}}e^{-b\beta_{0}\overline{H}}$ (88)

the integral is the same as (44) except that $\beta_{0}$ is replaced by
$b\beta_{0}$. The value of the integral may be read off from (67)

$P_{\rm part}={m^{2}\over 2\pi^{2}\xi}\sum_{b=1}^{\infty}(\xi
e^{\beta_{0}(\mu_{0}-eA_{0})})^{b}\Big{(}{T_{0}\over
b\sqrt{g_{00}}}\Big{)}^{2}K_{2}\Big{(}mb{\sqrt{g_{00}}\over T_{0}}\Big{)}$
(89)

and satisfies the Tolman-Ehrenfest rule except for the $eA_{0}$ dependence. If
the ensemble contains particles and antiparticles the total particle pressure
requires adding to (89) another such series with $\mu_{0}$ replaced by
$-\mu_{0}$. A simple check of (89) is that of massless bosons with
$\mu_{0}=A_{0}=0$ in which case

$P_{\rm part}={\pi^{2}\over 90}\bigg{(}{T_{0}\over\sqrt{g_{00}}}\bigg{)}^{4}.$
(90)

##  Appendix A Change to Euclidean momenta

The calculation in Sec. III of the partition function requires integrating
$\exp[-\beta_{0}\overline{H}]$ with respect to the canonical momenta
$dp_{1}dp_{2}dp_{3}$, where

$\overline{H}={1\over
g^{00}}\bigg{[}\sqrt{g^{00}(m^{2}-c^{ij}p^{\prime}_{i}p^{\prime}_{j})}+g^{0j}p^{\prime}_{j}\bigg{]},$
(91)

with $c^{ij}$ as defined in (17), and $p_{j}^{\prime}=p_{j}+eA_{j}$. The
spatial metric $g_{jk}$ may be expanded at any point $x^{i}$ in terms of three
Euclidean frame vectors

$g_{jk}=-\sum_{a=1}^{3}f_{(a)j}f_{(a)k},$ (92)

which are the analogs of vierbeins in three-dimensional space. (The minus sign
arises because the eigenvalues of $g_{jk}$ are negative.) The contravariant
form of the frame vectors is

$f_{(a)}^{i}=-c^{ij}f_{(a)j}.$ (93)

Equations (92), and (93) imply

$\displaystyle\sum_{a=1}^{3}f_{(a)}^{i}f_{(a)j}$ $\displaystyle=$
$\displaystyle\delta^{i}_{\;\;j}$ $\displaystyle f_{(a)}^{j}f_{(b)j}$
$\displaystyle=$ $\displaystyle\delta_{ab}$ (94)
$\displaystyle\sum_{a=1}^{3}f_{(a)}^{i}f_{(a)}^{j}$ $\displaystyle=$
$\displaystyle-c^{ij}.$

The quadratic term in $\overline{H}$ is

$-c^{ij}p_{i}p_{j}=\sum_{a=1}^{3}(f^{i}_{(a)}p^{\prime}_{i})(f^{j}_{(a)}p^{\prime}_{j}),$
(95)

which suggests introducing Euclidean momenta $k_{a}$

$k_{a}=f^{j}_{(a)}\,p^{\prime}_{j}.$ (96)

(This is not a canonical transformation: the defining Poisson bracket
$\\{x^{i},p_{j}\\}=\delta^{i}_{j}$ implies $\\{x^{i},k_{a}\\}=f^{i}_{(a)}$.)
$\overline{H}$ is now

$\overline{H}={1\over g^{00}}\bigg{[}\sqrt{g^{00}(m^{2}+{\bf
k}^{2})}+g^{0j}\sum_{a=1}^{3}f_{(a)j}k_{a}\bigg{]}.$ (97)

Define a Euclidean vector

$s_{a}={g^{0j}f_{(a)j}\over\sqrt{g^{00}}}$ (98)

with length

$s^{2}=\sum_{a=1}^{3}[s_{a}]^{2}=g_{0j}g^{0j}=1-g_{00}g^{00}.$ (99)

The Hamiltonian has the simple form

$\overline{H}={1\over\sqrt{g^{00}}}\Big{[}\sqrt{m^{2\\!}+\\!{\bf k}^{2}}+{\bf
s}\cdot{\bf k}\Big{]}.$ (100)

The inverse of relation (96) is

$p^{\prime}_{j}=\sum_{a=1}^{3}f_{(a)j}k_{a},$ (101)

which gives for the Jacobian of the change from $p^{\prime}_{j}$ to Euclidean
$k_{a}$

$\displaystyle d^{3}p^{\prime}$ $\displaystyle=$ $\displaystyle\sqrt{|{\rm
det}(g_{ij})}|\,d^{3}k=\sqrt{{g^{00}\over|{\rm det}(g^{\mu\nu})|}}d^{3}k$
(102) $\displaystyle=$ $\displaystyle\sqrt{gg^{00}}\,d^{3}k$

where $g=|{\rm det}(g_{\mu\nu})|$. These results yield (49).

##  Appendix B $T^{\mu\nu}$ for particles and background field

###  B.1 Calculation of $T^{\mu\nu}_{\rm part}$ from $H_{\rm part}$

The energy-momentum tensor for the particles may be calculated using the
Hilbert variational principle:

$\displaystyle-{\sqrt{g}\over 2}T^{\mu\nu}_{\rm part}(x)$ $\displaystyle=$
$\displaystyle{\delta\over\delta g_{\mu\nu}(x)}\int dx^{\prime 0}L_{\rm
part}(x^{\prime})$ (103) $\displaystyle=$ $\displaystyle\delta^{3}({\bf
x}-{\bf x}^{\prime})\bigg{[}{\partial L_{\rm part}\over\partial
g_{\mu\nu}}\bigg{]}_{\partial_{0}x^{i},x^{i}}$

where the metric variation is performed before the equations of motion are
imposed on the particle velocity and position. The following steps depend
three facts: (i) $L_{\rm part}$ does not depend on the momentum $p_{j}$, (ii)
$L_{\rm part}=p_{j}\partial_{0}x^{j}-H_{\rm part}$ and
$p_{j}\partial_{0}x^{j}$ does not depend on the metric, (iii) $H_{\rm part}$
does not depend on the velocity $\partial_{0}x^{i}$:

$\displaystyle\bigg{[}{\partial L_{\rm part}\over\partial
g_{\mu\nu}}\bigg{]}_{\partial_{0}x^{i},x^{i}}$ $\displaystyle=$
$\displaystyle\bigg{[}{\partial L_{\rm part}\over\partial
g_{\mu\nu}}\bigg{]}_{\partial_{0}x^{i},x^{i},p_{j}}$ (104) $\displaystyle=$
$\displaystyle-\bigg{[}{\partial H_{\rm part}\over\partial
g_{\mu\nu}}\bigg{]}_{\partial_{0}x^{i},x^{i},p_{j}}$ (105) $\displaystyle=$
$\displaystyle-\bigg{[}{\partial H_{\rm part}\over\partial
g_{\mu\nu}}\bigg{]}_{x^{i},p_{j}}.$ (106)

Therefore

${\sqrt{g}\over 2}T^{\mu\nu}_{\rm part}(x)=\delta^{3}({\bf x}-{\bf
x}^{\prime})\bigg{[}{\partial H_{\rm part}\over\partial
g_{\mu\nu}}\bigg{]}_{x^{i},p_{j}}.$ (107)

The partition function for the particles

$\ln Z_{\rm part}=\int{d^{3}xd^{3}p\over(2\pi)^{3}}e^{-\beta_{0}(H_{\rm
part}-\mu_{0})}$ (108)

has a variational derivative

${\delta\ln Z_{\rm part}\over\delta
g_{\mu\nu}(x^{\prime})}=-\beta_{0}{\sqrt{g}\over
2}\int{d^{3}pd^{3}x\over(2\pi)^{3}}T^{\mu\nu}_{\rm part}e^{-\beta_{0}(H_{\rm
part}-\mu_{0})}$ (109)

This is the starting point of Sec. IV.

##### Explicit form for $T^{\mu\nu}_{\rm part}$.

It is not difficult to explicitly differentiate $H_{\rm part}$ with respect to
$g_{\mu\nu}$ by considering the metric dependence of the covariant particle
velocity $u_{\alpha}$. From (26) $u_{j}$ is independent of the metric because
$p_{j}$ and $A_{j}$ are. The derivative of
$g^{\alpha\beta}u_{\alpha}u_{\beta}=1$ with respect to $g^{\mu\nu}$ yields

${\partial u_{0}\over\partial g^{\mu\nu}}=-{u_{\mu}u_{\nu}\over 2u^{0}}.$
(110)

This gives the derivative of $H_{\rm part}=mu_{0}+eA_{0}$. A change from the
contravariant metric to the covariant metric gives

$\sqrt{g}\,T^{\mu\nu}_{\rm part}=\delta^{3}({\bf x}-{\bf
x}^{\prime})m{u^{\mu}u^{\nu}\over u^{0}}$ (111)

as expected.

###  B.2 Calculation of $T^{\mu\nu}_{\rm field}$ from ${\cal H}_{\rm field}$

####  B.2.1 The canonical ${\cal H}_{\rm field}$

Because $g_{0j}\neq 0$ it is not trivial to compute ${\cal H}_{\rm field}$ in
terms of the canonical momenta $\pi^{j}$ and the magnetic field $F_{jk}$. The
starting point is the Lagrange density

${\cal L}_{\rm field}=-{\sqrt{g}\over
4}F_{\mu\nu}F_{\alpha\beta}g^{\mu\alpha}g^{\nu\beta},$ (112)

which implies a canonical momentum

$\pi^{j}={\partial{\cal L}_{\rm
field}\over\partial(\partial_{0}A_{j})}=-\sqrt{g}F^{0j}$ (113)

and the Hamiltonian density

$\displaystyle{\cal H}_{\rm field}$ $\displaystyle=$
$\displaystyle\pi^{j}\partial_{0}A_{j}+{\sqrt{g}\over
4}F_{\alpha\beta}F^{\alpha\beta}$ (114) $\displaystyle=$
$\displaystyle\pi^{j}F_{0j}+{\sqrt{g}\over
4}F_{\alpha\beta}F^{\alpha\beta}+\pi^{j}\partial_{j}A_{0}.$

To express this in terms of canonical variables requires the identity

$g_{jk}\pi^{k}=-\sqrt{g}F^{0k}g_{kj}=-\sqrt{g}\,g^{0\mu}F_{\mu j}$ (115)

which may be solved for $F_{0j}$:

$F_{0j}={1\over g^{00}}\bigg{[}-{g_{jk}\pi^{k}\over\sqrt{g}}+F_{j\ell}g^{\ell
0}\bigg{]}.$ (116)

This allows the first term in (114) to be expressed in terms of momentum and
covariant field strength. For the second term in (114) use the quantity

$c^{\alpha\mu}=g^{\alpha\mu}-{g^{\alpha 0}g^{\mu 0}\over g^{00}}$ (117)

to obtain

$F_{\alpha\beta}F^{\alpha\beta}=F_{\alpha\beta}c^{\alpha\mu}c^{\beta\nu}F_{\mu\nu}+2{F^{0\alpha}g_{\alpha\mu}F^{0\mu}\over
g^{00}}.$ (118)

Because $c^{\alpha\mu}$ is zero if either $\alpha=0$ or $\mu=0$ and
$F^{0\alpha}$ is zero if $\alpha=0$ the identity simplifies to

$F_{\alpha\beta}F^{\alpha\beta}=F_{jk}c^{j\ell}c^{km}F_{\ell
m}+2{F^{0j}g_{jk}F^{0k}\over g^{00}}.$ (119)

This allows the second term in (114) to be expressed as

${\sqrt{g}\over 4}F_{\alpha\beta}F^{\alpha\beta}={\pi^{j}\pi^{k}g_{jk}\over
2g^{00}\sqrt{g}}+{\sqrt{g}\over 4}F_{jk}c^{j\ell}c^{km}F_{\ell m}.$ (120)

The resulting Hamiltonian density in terms of canonical variables is therefore

$\displaystyle{\cal H}_{\rm field}$ $\displaystyle=$
$\displaystyle-{\pi^{j}\pi^{k}g_{jk}\over
2g^{00}\sqrt{g}}+{\pi^{j}F_{jk}g^{k0}\over g^{00}}$ (121) $\displaystyle+$
$\displaystyle{\sqrt{g}\over 4}F_{jk}F_{\ell
m}c^{j\ell}c^{km}+\pi^{j}\partial_{j}A_{0}.$

A simple check is that $\partial_{0}A{{}_{j}}=\partial{\cal H}_{\rm
field}/\partial\pi_{j}$.

####  B.2.2 The metric derivative of ${\cal H}_{\rm field}$

The canonical momenta $\pi^{j}$ and the magnetic field $F_{jk}$ do not depend
on the metric. The metric dependence of ${\cal H}_{\rm field}$ due to
$\sqrt{g}$ is easy to compute from (121) and gives

$\bigg{[}{\partial\sqrt{g}\over\partial g^{\mu\nu}}\bigg{]}{\partial{\cal
H}_{\rm field}\over\partial\sqrt{g}}=-{\sqrt{g}\over 2}\Big{(}{g_{\mu\nu}\over
4}F_{\alpha\beta}F^{\alpha\beta}\Big{)}$ (122)

after using (120) to convert canonical momentum back into fields. To compute
the metric derivative with $\sqrt{g}$ fixed

${\partial{\cal H}_{\rm field}\over\partial g^{\mu\nu}}\bigg{|}_{\sqrt{g}}$
(123)

is more difficult and must be calculated for the separate cases
$\mu\nu=00,j0,jk$. After computing the metric derivatives it is necessary to
use the relation

$F^{0k}=g^{0\mu}F_{\mu j}c^{jk}$ (124)

which follows from (115). The final result is

${\partial{\cal H}_{\rm field}\over\partial g^{\mu\nu}}=-{\sqrt{g}\over
2}\bigg{[}-F_{\mu\alpha}F_{\nu\beta}g^{\alpha\beta}+{g_{\mu\nu}\over
4}F_{\alpha\beta}F^{\alpha\beta}\bigg{]}.$ (125)

A change from contravariant $g^{\mu\nu}$ to covariant $g_{\mu\nu}$ and the
relation $\ln Z_{\rm field}=-\beta_{0}H_{\rm field}$ give

${\delta\ln Z_{\rm field}\over\delta g_{\mu\nu}}=-\beta_{0}{\sqrt{g}\over
2}T^{\mu\nu}_{\rm field}.$ (126)

##  Appendix C Thermodynamics functions

###  C.1 Exact relations

##### Thermodynamic quantities.

The thermodynamic pressure due to the particles is

$P_{\rm
part}={T_{0}\over\sqrt{g}}\int{d^{3}p^{\prime}\over(2\pi)^{3}}e^{-\beta_{0}(H_{\rm
part}-\mu_{0})}.$ (127)

Differentiation gives

$\mu_{0}{\partial P_{\rm part}\over\partial\mu_{0}}+T_{0}{\partial P_{\rm
part}\over\partial T_{0}}=P_{\rm part}+\rho_{\rm
part}+n{eA_{0}\over\sqrt{g_{00}}}.$ (128)

where

$\rho_{\rm
part}={1\over\sqrt{g}}\int{d^{3}p^{\prime}\over(2\pi)^{3}}\overline{H}\,e^{-\beta_{0}(H_{\rm
part}-\mu_{0})}.$ (129)

The first term in (128) is related to the number density:

${n\over\sqrt{g_{00}}}={\partial P_{\rm
part}\over\partial\mu_{0}}=\beta_{0}P_{\rm part}.$ (130)

This is the ideal gas law

$P_{\rm part}=n{T_{0}\over\sqrt{g_{00}}}$ (131)

in local form. The second term in (128) is related to the local entropy
density

$\displaystyle s$ $\displaystyle=$ $\displaystyle\sqrt{g_{00}}\,{\partial
P_{\rm part}\over\partial T_{0}}$ (132) $\displaystyle=$ $\displaystyle{1\over
T}\big{[}P_{\rm part}+\rho_{\rm part}+n{eA_{0}\over\sqrt{g_{00}}}-\mu
n\big{]}.$

###  C.2 Low temperature regime

Using the approximation (82) the number density is

$n({\bf x})=\Big{(}{mT_{0}\over
2\pi\sqrt{g_{00}}}\Big{)}^{3/2}e^{\beta_{0}(\mu_{0}-eA_{0}-m\sqrt{g_{00}})}.$
(133)

The entropy density $s({\bf x})$ can be computed from (132) and afterwards if
the chemical potential is expressed in terms of the density the result is

$s({\bf x})=n\bigg{[}{5\over 2}+\ln\bigg{\\{}{1\over n}\bigg{(}{mT_{0}\over
2\pi\sqrt{g_{00}}}\bigg{)}^{3/2}\bigg{\\}}\bigg{]},$ (134)

which is the local form of the Sackur-Tetrode equation Pereira . The energy
density of the particles is

$\rho_{\rm part}({\bf x})=n\bigg{[}m+{3\over
2}{T_{0}\over\sqrt{g_{00}}}\bigg{]}.$ (135)

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