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Stationary first law from the covariant boundary identity

Physical scope

Conditional first law for a classical stationary nondegenerate bifurcate horizon with the specified covariant potential, globally compatible Noether charge, vanishing linearized constraint variation, and integrable asymptotic Hamiltonian charge. It is not a dynamical law, a quantum result, or empirical confirmation of a gravity model.

Statement

For the stationary Noether data of the preceding Definition, choose Qξ in the canonical local representative class of the preceding Noether-charge decomposition theorem. Let δϕ be an admitted linearized solution with fixed ξ and δCξ=0. Let CR be finite oriented smooth singular hypersurface chains with ∂CR=SR−B, where the boundary orientations are outward at SR and inward at B. Then d(δQξ−ιξθ)=0 on each CR, and δHξ=lim⁡R→∞∫SR(δQξ−ιξθ)=∫BδQξ=ℏκ2πkB δSW. The last equality uses the derivative-free curvature hypothesis and the Noether entropy functional of the preceding theorem. The identity is conditional on the exact boundary-charge and vanishing-constraint assumptions; κ is the unperturbed surface gravity. For the Einstein-Hilbert Lagrangian LEH=Rϵg/(16πG) with Newton constant G>0 in the same c=1 action normalization, it gives SW=kBAB/(4Gℏ), where AB:=∫BdAB.

Facts & Assumptions

Given: The stationary Noether data, the compact boundaryless bifurcation surface and the admitted smooth linearized perturbation in the Statement.

[A1]

Stationary covariant Noether boundary data: Jξ=θ(ϕ;Lξϕ)−ιξL, Cξ=Jξ−dQξ, Lξϕ=0, Eϕ=0, and δCξ=0. The boundary one-form is exactly lim⁡R∫SR(δQξ−ιξθ) and satisfies dSHξ=αξ.

[F1]

Cartan's magic formula: Lξα=d(ιξα)+ιξdα for every differential form α.

[F2]

Stokes theorem for smooth singular chains: ∫CRdα=∫∂CRα for each finite smooth singular chain CR and smooth form α.

[F3]

Noether charge decomposition for algebraic-curvature Lagrangians: the selected charge is in the local representative class with Xa3…ancd=−Pabcd(ϵg)aba3…an.

[L1]

Stationary horizon Noether-charge entropy variation: for that representative, δ∫BQξ=(ℏκ/(2πkB))δSW.

Proof

1.1A1F1algebra

Differentiate Jξ at fixed ξ. Since Lξϕ=0 on the stationary background and θ is linear in its variation argument, δJξ=θ(ϕ;δ(Lξϕ))−ιξδL=θ(ϕ;Lξδϕ)−ιξ(Eϕ[δϕ]+dθ(ϕ;δϕ)). The first equality uses that the coefficient variation of θ(ϕ;0) is zero; the second uses δ(Lξϕ)=Lξδϕ because ξ is fixed. Covariance of θ identifies θ(ϕ;Lξδϕ)=Lξθ(ϕ;δϕ). Since Eϕ=0, Cartan's identity [F1] yields δJξ=d(ιξθ).

2.1A1step 1.1algebra

By [A1], Cξ=Jξ−dQξ and δCξ=0. Differentiating this identity gives δJξ=d(δQξ) because d is independent of the field variation and the spacetime coordinates. Combining with step 1.1 gives d(δQξ−ιξθ)=0. This conclusion uses the explicit constraint-variation premise; it is not inferred from the closure of a presymplectic current.

3.1A1F2step 2.1algebra

Apply [F2] to α=δQξ−ιξθ on CR. By step 2.1 and ∂CR=SR−B, 0=∫CRdα=∫SR(δQξ−ιξθ)−∫B(δQξ−ιξθ). On B, ξ=0, so ιξθ=0. Taking the limit in the first term is permitted by the boundary convergence assumption in [A1], and its value is δHξ by the specified definition of the Hamiltonian variation. Hence δHξ=∫BδQξ.

4.1F3step 3.1L1

By [F3], the charge used here is in the representative class required by [L1]. Apply [L1] to the right-hand side of step 3.1. This gives δHξ=(ℏκ/(2πkB))δSW, with all field variations taken at fixed generator and fixed background κ.

5.1L1algebra∎

For ℓEH=R/(16πG), the curvature convention in the preceding entropy-variation theorem gives R=gacgbdRabcd. Holding g fixed and using δRabcd=−δRabdc, the Riemann-symmetric derivative is Pabcd=132πG(gacgbd−gadgbc). Its binormal contraction is Pabcdϵabϵcd=(−2−2)/(32πG)=−1/(8πG), because ϵabϵab=−2 and ϵabϵba=2. Substitution in the definition of SW yields SW=kB∫BdAB/(4Gℏ)=kBAB/(4Gℏ).

Remarks

The stationary relation is a model deduction. Calling SW the physical thermodynamic state entropy requires an additional physical identification; the restricted Kerr–Newman assumption on the earlier thermodynamic page is not generalized here.

Depends on

Dependency tree · 2 levels

Sources