Tolman Redshift and Local Temperature

Thermal equilibrium in a gravitational field is not characterized by a spatially constant thermometer reading. In a stationary region with lapse $N=\sqrt{-\chi^2}$, the locally measured temperature obeys $N T_{\mathrm{loc}}=T_0$, where $T_0$ is conjugate to the normalized Killing time. The divergent static-frame temperature near a nonextremal horizon records infinite acceleration and redshift; by itself it does not mean that a freely falling observer encounters a divergent stress tensor.

Required background. Gravitational Actions and Boundary Terms supplies the boundary clock and energy used to define the thermal ensemble.

Helpful background. Quantum Fields in Curved Spacetime supplies stationary thermal states; Einstein Equations supplies gravitational redshift; and Black-Hole Geometry and Mechanics supplies Killing horizons and static observers.

Killing energy and local energy

Write a static metric in the region where the stationary Killing field $\chi=\partial_t$ is timelike as

$$ds^2=-N^2(\mathbf x)dt^2+h_{ij}(\mathbf x)dx^i dx^j.$$

A static observer has four-velocity $u^\mu=\chi^\mu/N$. If a particle has four-momentum $p^\mu$, its conserved Killing energy and locally measured energy are

$$E_0=-p_\mu\chi^\mu, \qquad E_{\mathrm{loc}}=-p_\mu u^\mu, \qquad E_0=N E_{\mathrm{loc}}.$$

The equilibrium Gibbs weight is defined relative to the boundary time: $\exp(-E_0/T_0)$. A local observer must describe the same weight as $\exp(-E_{\mathrm{loc}}/T_{\mathrm{loc}})$. Equality for every excitation immediately gives the Tolman relation

$$\boxed{N(\mathbf x)T_{\mathrm{loc}}(\mathbf x)=T_0}.$$

This derivation also fixes the normalization: rescaling $\chi$ changes both $E_0$ and $T_0$, while $u^\mu$, $E_{\mathrm{loc}}$, and $T_{\mathrm{loc}}$ are unchanged. The statement is therefore about a chosen stationary ensemble, not an invariant temperature field independent of its clock.

Hydrostatic interpretation

The same law follows from force balance. A static observer’s acceleration is $a_i=\partial_i\ln N$. For a perfect fluid at rest, stress-energy conservation gives

$$\partial_i p=-(\rho+p)\partial_i\ln N.$$

At zero chemical potential, the local thermodynamic identities $dp=s\,dT$ and $\rho+p=sT$ reduce this to

$$\partial_i\ln T_{\mathrm{loc}}=-\partial_i\ln N.$$

Integration yields $N T_{\mathrm{loc}}=T_0$. For a conserved neutral species, diffusive equilibrium also fixes $N\mu_{\mathrm{loc}}$, so $\mu_{\mathrm{loc}}/T_{\mathrm{loc}}$ is constant. In an electrostatic field, the conserved quantity includes electric work: with $\Phi(\mathbf x)=-\chi\cdot A(\mathbf x)+\chi\cdot A|_B$, it is $N\mu_{\mathrm{loc}}+q\Phi$ that is constant for particles of charge $q$. The chemical potential without that electric term need not obey a pure redshift law. In a gravitational cavity, the proper wall temperature—not a coordinate period alone—is the canonical datum. York’s two Schwarzschild branches arise after imposing that local boundary condition York 1986, §§II–V.

The conservation derivation clarifies what “equilibrium” means. There is no heat current in the static frame, the density operator is generated by the same Killing field everywhere, and the matter is in local thermodynamic equilibrium on scales short compared with the lapse-variation scale. If one instead prescribed a constant local temperature at two heights, the two Gibbs factors would assign different populations to a mode of fixed Killing energy. Heat would flow until the redshifted temperatures agreed.

The law is also operational. Two identical clocks at different lapse compare their transition frequencies by $\nu_0=N\nu_{\mathrm{loc}}$. A thermal spectrum whose local peak follows $\nu_{\mathrm{loc}}\sim T_{\mathrm{loc}}$ is reported at the common boundary frequency $\nu_0\sim T_0$. This frequency check is independent of the fluid derivation and shows that $NT$, rather than $T$ alone, is the equilibrium quantity.

Horizon limit and what diverges

Near a nonextremal Killing horizon, proper distance $\rho$ gives $N\simeq\kappa\rho$. If the equilibrium state has $T_0=T_{\mathrm H}=\kappa/(2\pi)$, then

$$T_{\mathrm{loc}}\simeq\frac{1}{2\pi\rho}.$$

Static observers become null in the limit and require acceleration $\lvert a\rvert\simeq1/\rho$. Their temperature is the corresponding local Unruh scale. The divergence is thus a property of an increasingly accelerated observer family. A horizon-regular Hartle–Hawking state can have finite stress tensor components in a freely falling orthonormal frame even though the static thermometer diverges. Conversely, the Boulware state is vacuum-like for static observers at infinity but its renormalized stress tensor is singular at the horizon. State regularity and detector temperature must not be conflated; see Wald 2001, §3, PDF.

The exact Gibbs-state argument requires a stationary equilibrium on the region being considered; it does not require local fluid thermodynamics to hold. The fluid argument above is a complementary approximation. The Unruh state of an evaporating black hole carries net flux and is not described by one spatially redshifted equilibrium temperature. Spacetimes with two horizons of unequal surface gravity likewise lack a single regular Gibbs state across the whole static patch. In slowly evolving geometries the relation can be used locally only with an explicit adiabatic error estimate.

An ergoregion supplies another boundary: the stationary Killing field becomes spacelike, so its norm cannot define the lapse of static observers. A rotating equilibrium must instead use the horizon generator $\chi=\partial_t+\Omega_{\mathrm H}\partial_\phi$ and remain inside a region where the corresponding thermal evolution is well defined. Simply inserting $\sqrt{-g_{tt}}$ into the Tolman formula would use an observer family that does not exist there.

Exercise

In the Rindler approximation $ds^2=-(\kappa\rho)^2dt^2+d\rho^2+dy^2+dz^2$, take the boundary-time temperature to be $T_0=\kappa/(2\pi)$. Find the local temperature and explain why its divergence does not establish a curvature or stress-tensor singularity.

Solution

Here $N=\kappa\rho$, so the Tolman law gives $T_{\mathrm{loc}}=T_0/N=1/(2\pi\rho)$. Rindler spacetime is exactly flat, so all curvature invariants vanish. The static worldlines have proper acceleration $1/\rho$ and become null at the horizon; the divergent reading is their Unruh temperature, not evidence that a freely falling frame or a regular state is singular.

References

• Wald, R. M. (2001), “The Thermodynamics of Black Holes,” Living Reviews in Relativity 4, 6. Open PDF.
• York, J. W. Jr. (1986), “Black-Hole Thermodynamics and the Euclidean Einstein Action,” Physical Review D 33, 2092–2099. DOI.