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Covariant stationary Noether charge and Wald entropy — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

f(R) stationary entropy coefficient

Physical scope

Classical f(R) metric Lagrangians on smooth stationary nondegenerate bifurcate horizons under the algebraic-curvature assumptions of the stationary Noether first law. This is not a dynamical entropy law.

Example

In units with c=1, let L=f(R)ϵg/(16πG), where G>0 is the Newton coupling in the given n-dimensional action normalization, with units Ln−2/action. Choose an open interval J⊆R and f∈C∞(J), and admit only smooth background fields and finite-dimensional smooth field families for which Rϕp(x)∈J for every parameter p and spacetime point x∈M. Thus this composed Lagrangian is smooth throughout the spacetime and variation domain required by the preceding first-law theorem; regularity only at curvature values on B is not assumed. The scalar curvature R has units L−2, so f has those units and f′(R) is dimensionless. Under the same smooth stationary horizon assumptions as the preceding theorem, SW=kB4Gℏ∫Bf′(R) dAB. The example computes the Noether-charge functional only; it asserts no monotonicity along a dynamical horizon.

Facts & Assumptions

Given: The algebraic-curvature stationary data and C1 function f in the Example.

[F1]

Stationary horizon Noether-charge entropy variation: for an algebraic-curvature Lagrangian, SW=−(2πkB/ℏ)∫BPabcdϵabϵcd dAB.

[F2]

The curvature convention and scalar R=gacgbdRabcd are those explicitly fixed in the Statement of Stationary horizon Noether-charge entropy variation.

Verification

Given: The Lagrangian and curvature convention of the Example and [F1]–[F2].

1.1F1F2algebra

Holding g fixed and varying only the algebraic curvature slot gives δℓ=f′(R)16πGδR=f′(R)32πG(gacgbd−gadgbc)δRabcd, because δRabcd=−δRabdc and exchanging c,d in the second summand changes its sign. Thus Pabcd=f′(R)32πG(gacgbd−gadgbc).

2.1F1step 1.1algebra∎

Contracting with the normalized binormal gives Pabcdϵabϵcd=f′(R)32πG(ϵcdϵcd−ϵdcϵcd)=−f′(R)8πG, since ϵcdϵcd=−2 and ϵdcϵcd=2. Substitute this pointwise contraction into [F1] and integrate over B to obtain the displayed expression. The integral allows R and f′(R) to vary over the horizon.

Sources