Surface Gravity and Hawking Temperature

A stationary Killing horizon has a temperature only after its time-translation generator has been normalized. If the future horizon generator is multiplied by a positive constant, its surface gravity, the Hamiltonian energy conjugate to it, and the Hawking temperature all acquire the same factor. The first law is therefore invariant: neither $\kappa$ nor $T_{\mathrm H}$ is an absolute number until the observer and boundary clock have been specified.

Required background. Gravitational Actions and Boundary Terms supplies the boundary Hamiltonian whose normalization is varied below.

Helpful background. Quantum Fields in Curved Spacetime supplies the fixed-background thermal-state construction; Einstein Equations supplies the classical field equation; and Black-Hole Geometry and Mechanics supplies Killing horizons and the classical first law.

Killing generators and surface gravity

Let $\mathcal H$ be a Killing horizon generated by a future-directed Killing field $\chi^\mu$. On the horizon, one equivalent definition of surface gravity is

$$\chi^\nu\nabla_\nu\chi^\mu=\kappa\chi^\mu.$$

The equation does not normalize $\chi^\mu$. For any constant $a\gt0$, the field $\chi'^\mu=a\chi^\mu$ generates the same null curves, but its parameter runs at a different rate. Substitution gives

$$\chi'^\nu\nabla_\nu\chi'^\mu =a^2\kappa\chi^\mu =(a\kappa)\chi'^\mu, \qquad \kappa'=a\kappa.$$

The normalization of the horizon generator is therefore part of the physical setup: in the asymptotically flat case the Killing field is normalized to one unit of time translation at infinity, while a cavity calculation normalizes it against the proper clock at the wall. It says which integral curves count as one unit of time for the boundary observer. Surface gravity measures the nonaffinity of that specifically normalized Killing generator; it is not a scalar assigned to the unparameterized null surface alone. This is why two authors can quote different numerical values for $\kappa$ while describing the same horizon geometry if their clocks have not been matched.

In an asymptotically flat stationary spacetime, the usual choice fixes the stationary Killing field to have norm $-1$ at infinity. In a cavity it is the proper boundary clock that is fixed instead. In de Sitter space there is no spatial infinity, so a chosen static observer or another geometric condition must supply the normalization. These are distinct physical boundary conditions, not cosmetic coordinate conventions. The normalization issue and the classical mechanics statement are reviewed explicitly by Wald 2001, §§2–3, PDF.

Temperature and energy scale together

For a nonextremal horizon, a regular equilibrium state, when it exists, has the Hawking temperature relative to the time generated by $\chi^\mu$,

$$T_{\chi}=\frac{\kappa_\chi}{2\pi}.$$

Regularity on the future horizon alone does not make a state thermal: the stationary Unruh state has outgoing Hawking radiation but no compensating incoming bath. Nor does every rotating spacetime admit a globally regular equilibrium state. The fixed-background derivation belongs to Hawking Radiation in Gravitational Collapse; here the point is the common normalization of thermal energy and horizon temperature.

The Hamiltonian charge is linear in its symmetry generator. Thus $H[a\chi]=aH[\chi]$, up to the same fixed reference subtraction, and $\delta H[a\chi]=a\delta H[\chi]$. For a nonrotating, uncharged stationary horizon the mechanical first law is

$$\delta H[\chi]=\frac{\kappa_\chi}{8\pi G}\,\delta A.$$

After the rescaling, both sides acquire $a$. With $S_{\mathrm{BH}}=A/(4G)$, the same equation reads

$$\delta H[\chi]=T_\chi\,\delta S_{\mathrm{BH}},$$

so the entropy is unchanged while energy and temperature are measured in the same rescaled time units. In a rotating or charged solution, the angular and electric work terms must be normalized with the same horizon generator and gauge choice; rescaling only $\kappa$ would destroy the first law.

The Euclidean route reaches the same conclusion. Near a nonextremal horizon, the Euclidean time angle is $\kappa\tau$; smoothness requires period $\beta_\chi=2\pi/\kappa_\chi$. Rescaling the Lorentzian generator rescales the conjugate time coordinate inversely, hence $\beta' = \beta/a$ and $T'=aT$. Gibbons and Hawking 1977, pp. 2752–2756 show how the boundary clock and Euclidean action enter the gravitational partition function.

Schwarzschild check and boundaries of the formula

For Schwarzschild with asymptotic time normalized by $g_{tt}\to-1$,

$$f(r)=1-\frac{2GM}{r}, \qquad r_h=2GM, \qquad \kappa=\frac12 f'(r_h)=\frac{1}{4GM}.$$

Therefore

$$T_{\mathrm H}=\frac{1}{8\pi GM}.$$

This check restores the physical mass $M$ rather than treating $GM$ as a mass, and it has the correct inverse-length dimension. If $\chi\mapsto a\chi$, then the energy called $M$ relative to that generator becomes $aM$ while the geometric horizon area is unchanged; the first law still matches term by term.

Three limitations matter. First, the formula fixes the Hawking scale of a stationary, nonextremal Killing horizon; interpreting it as a full equilibrium temperature requires the corresponding thermal state to exist. Second, the temperature seen by a local stationary detector is redshifted and need not equal $T_\chi$. Third, an evolving trapping horizon need not possess a unique Killing normalization or an exactly thermal state. Those cases require adiabatic or nonequilibrium reasoning rather than inserting a locally defined “surface gravity” into $T=\kappa/(2\pi)$.

There is also a useful extremal warning. Taking a smooth nonextremal family to $\kappa\to0$ gives a vanishing Hawking temperature in the normalized boundary time, but the exactly extremal Euclidean geometry has a different near-horizon topology and no ordinary conical tip that fixes a unique period. The limiting thermodynamic statement and the geometry at the endpoint must therefore be distinguished. Likewise, a negative rescaling would reverse the chosen future orientation; the temperature convention above restricts $a$ to be positive.

Exercise

Let an asymptotically normalized Schwarzschild generator be rescaled as $\chi' = 3\chi$. Determine $\kappa'$, $T'$, and $\delta H'$ in terms of the unprimed quantities, and verify the entropy form of the first law.

Solution

Linearity gives $\kappa'=3\kappa$, $T'=3T$, and $\delta H'=3\delta H$. Since the rescaling changes no horizon cross-section, $\delta S_{\mathrm{BH}}=\delta A/(4G)$ is unchanged. Hence $\delta H'=3T\,\delta S_{\mathrm{BH}}=T'\delta S_{\mathrm{BH}}$. Rescaling the temperature without the Hamiltonian energy would instead be inconsistent. This verifies the invariance directly.

References

• Gibbons, G. W., and S. W. Hawking (1977), “Action Integrals and Partition Functions in Quantum Gravity,” Physical Review D 15, 2752–2756. DOI.
• Wald, R. M. (2001), “The Thermodynamics of Black Holes,” Living Reviews in Relativity 4, 6. Open PDF.