research38 min studyLesson 825 in Physics

Surface gravity and stationary horizon laws

Normalize the horizon generator and apply stationary first-law variations without mistaking them for general horizon dynamics.

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What you’ll understand

  • Fix surface-gravity normalization through the asymptotic time translation.
  • Compute Kerr surface gravity and horizon angular velocity.
  • Separate equilibrium variations and controlled physical-process laws from arbitrary evolution.

The phrase surface gravity can mislead if it suggests the finite acceleration of someone hovering on a horizon. For the nonextremal static-horizon comparison, the exterior hovering-acceleration limit diverges, while its suitably redshifted limit remains finite. More fundamentally, surface gravity measures the nonaffinity of a horizon-generating Killing vector. Its numerical value depends on that vector's normalization. The stationary laws become physically interpretable only after the normalization, asymptotic charges, units, and class of variations have been fixed.

A Killing horizon needs a specified generator

Keep signature (−,+,+,+) and G=c=1, so surface gravity has inverse-length units. In an asymptotically flat stationary axisymmetric black hole, normalize the stationary Killing vector t to have norm minus one at infinity and the axial generator phi to have closed orbits of parameter period two pi. The horizon Killing generator is their combination with the horizon angular velocity. It is null on the horizon, even though t alone need not be null there for a rotating hole.

Along the horizon, the Killing generator satisfies a nonaffine geodesic equation whose coefficient is kappa. Multiplying the generator by a positive constant multiplies kappa by the same constant. This changes a time normalization, not the geometry or a physical temperature defined with a fixed asymptotic clock. A dynamical event horizon need not be a Killing horizon, and alternative dynamical surface-gravity definitions require extra choices. The stationary definition must not be treated as a unique number for every evolving horizon.

χa=ta+ΩHφa,χb∇bχa=κχaon H,χ↦bχ⟹κ↦bκ,κSchw=14M\chi^a=t^a+\Omega_H\varphi^a,\qquad \chi^b\nabla_b\chi^a=\kappa\chi^a\quad\text{on }\mathcal H,\qquad \chi\mapsto b\chi\Longrightarrow\kappa\mapsto b\kappa,\qquad \kappa_{\mathrm{Schw}}=\frac1{4M}
b is a positive constant. The Schwarzschild value uses the unit time translation at infinity. In SI units, this geometric κ corresponds to acceleration c²κ, while its Hawking temperature is ħcκ/(2πkB).↔ Scroll wide equations horizontally.

The zeroth and first laws have equilibrium hypotheses

The zeroth law states that kappa is constant across an appropriate connected stationary Killing horizon. One sufficient derivation uses the Einstein equation with the dominant energy condition; other geometric versions use additional symmetry or bifurcation assumptions. Constancy is not obtained merely from assigning the name horizon to a surface. For the uncharged Kerr family, the outer and inner radii and rotation parameter determine kappa, angular velocity, and area explicitly.

The first law relates infinitesimal changes between nearby stationary solutions, with mass and angular momentum defined using the normalized asymptotic symmetries. Controlled physical-process versions also exist, using perturbative assumptions and suitable initial and final conditions. Neither form is a general nonlinear equation for an arbitrary instantaneous area and mass. A finite change must be integrated with the changing coefficients or computed from the full solutions, not obtained by multiplying one initial kappa by a large area jump.

r±=M±M2−a2,a=JM,κ=r+−r−2(r+2+a2),ΩH=ar+2+a2,δM=κ8πδA+ΩHδJr_\pm=M\pm\sqrt{M^2-a^2},\quad a=\frac JM,\qquad \kappa=\frac{r_+-r_-}{2(r_+^2+a^2)},\quad \Omega_H=\frac a{r_+^2+a^2},\qquad \delta M=\frac{\kappa}{8\pi}\delta A+\Omega_H\delta J
These formulas are for an uncharged subextremal Kerr hole in geometric units. M,a,r± have length units, J and A have length-squared units, and ΩH has inverse-length units. The displayed first law has no charge variation; charged laws need an additional potential-charge term with a specified gauge convention.↔ Scroll wide equations horizontally.

Redshift distinguishes kappa from a hovering acceleration

In the static Schwarzschild exterior, let f equal one minus two M over r. The proper acceleration needed to remain at fixed r is astat=M/(r2f)a_{\mathrm{stat}}=M/(r^2\sqrt f). Its redshifted value is M divided by r squared, tending to kappa only in the horizon limit. It is neither constant at all radii nor the acceleration of a freely falling observer. The stationary Killing time and the observer's proper time are different clocks.

The connection to temperature adds quantum assumptions and a state choice. A Schwarzschild equilibrium state has a redshifted thermal interpretation, while a collapse state with outgoing Hawking flux is not the same global equilibrium bath. Extremal limits, where kappa tends to zero while area can remain nonzero, require particular care and are not captured by casually extrapolating every nonextremal derivation. Stationary mechanics supplies exact geometric relations within its domain; it does not settle the microscopic entropy or the endpoint of evaporation.

A normalized Kerr first-law variation

An uncharged Kerr hole has M=5L and a=3L. Compute r±, kappa, and OmegaH. For infinitesimal variations δA=18πεL² and δJ=3εL² with epsilon much smaller than one, find δM to first order.

  1. The square root is M2−a2=4L\sqrt{M^2-a^2}=4L, so r+=9Lr_+=9L and r−=1Lr_-=1L. The common factor r+2+a2=90L2r_+^2+a^2=90L^2.

  2. The surface gravity is κ=8/(180L)=2/(45L)\kappa=8/(180L)=2/(45L), and ΩH=3/(90L)=1/(30L)\Omega_H=3/(90L)=1/(30L). These use the fixed asymptotic time normalization.

  3. The area contribution is (2/45)(18/8)ϵL=0.10ϵL(2/45)(18/8)\epsilon L=0.10\epsilon L. The rotation contribution is (1/30)(3)ϵL=0.10ϵL(1/30)(3)\epsilon L=0.10\epsilon L. Hence δM=0.20ϵL+O(ϵ2)\delta M=0.20\epsilon L+O(\epsilon^2), a differential relation rather than a finite jump formula.

The result

The radii are 9L and 1L, kappa=2/(45L), OmegaH=1/(30L), and the stated small variation gives δM=0.20εL to first order.

Explore the model

Interactive lab

Interactive lab · vectors

A generator normalization changes components

For each fresh comparison, Reset and set BOTH Angle=30° and Magnitude. Compare magnitudes 1 and 2, predicting projections (√3/2,1/2) and (√3,1), with the same direction. This illustrates positive vector rescaling only. The Euclidean vector is not a null Killing generator, and the instrument has no metric, surface gravity, first law, or asymptotic clock.

A generator normalization changes componentsPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forNormalization is part of a vector-defined quantity. A rescaling analogy does not calculate kappa or change a black-hole geometry.

Check your understanding

Why is a normalization of the stationary time translation at infinity needed when quoting kappa?

Try it yourself

For a Schwarzschild hole with M=1L, a static observer is held at r=4L. Compute f, the proper acceleration, and its redshifted value sqrt(f)astat. Compare the latter with kappa=1/(4L). Use sqrt(1/2)≈0.7071.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The redshifted acceleration equals M/r² at finite radius and reaches kappa only as r approaches 2M.

Compare with the solution

Here f=1−2/4=1/2f=1-2/4=1/2. The proper acceleration is astat=1/[161/2]≃0.08839/La_{\mathrm{stat}}=1/[16\sqrt{1/2}]\simeq0.08839/L. Multiplication by sqrt(f) gives 1/(16L)=0.0625/L1/(16L)=0.0625/L, whereas κ=0.25/L\kappa=0.25/L. The finite-radius redshifted acceleration is therefore one quarter of kappa. As r approaches 2L, the proper acceleration diverges while its redshifted value approaches 0.25/L. None of these hovering accelerations is the proper acceleration of a geodesic freely falling through the horizon.

Take it with you

  • Surface gravity is tied to a normalized stationary horizon generator.
  • Zeroth-law constancy and first-law variations have specific equilibrium or perturbative hypotheses.
  • The redshifted hovering-acceleration limit is finite even when local hovering acceleration diverges.

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