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Curved electromagnetic stress divergence

Statement

Let U be an oriented open subset of a smooth time-oriented Lorentzian four-manifold of signature (−,+,+,+), with metric-selected connection ∇. Let F∈Ω2(U) and j∈Γ(TU) be smooth, and let μ0>0. Assume

dF=0,∇aFab=−μ0jb,

with the sign convention of Sourced Maxwell sign and current conservation. Define the Maxwell Hilbert tensor by the compact metric variation of Compact potential and metric variation of the Maxwell action,

TabEM:=1μ0(FacFbc−14gabFcdFcd).

Then

∇aTEMab=−Fbcjc.

All fields and derivatives are smooth on U. In length-coordinate SI bookkeeping, assign [Fab]=Wb m−2, [μ0]=H m−1, [ja]=A m−2, and [TEM]=J m−3; these are formal units and not observed values. No quadratic stress is assigned to singular point self-fields or distributional sources.

Facts & Assumptions

Given: The Lorentzian metric, its metric-compatible torsion-free connection, a smooth real two-form F, a smooth vector current j, μ0>0, and the two Maxwell equations in the Statement.

[F1]

In the selected sourced convention, ∇aFab=−μ0jb and dF=0; equivalently the first equation is d⋆F=μ0ιjvolg. (Sourced Maxwell sign and current conservation)

[F2]

The Maxwell stress tensor obtained by compact variation at fixed lower F is TabEM=μ0−1(FacFbc−14gabFcdFcd). (Compact potential and metric variation of the Maxwell action)

[F3]

The induced connection on tensor fields satisfies the product rule and commutes with contractions; it is torsion-free and metric-compatible, so ∇agbc=0. (Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields)

[A1]

The geometric interface carries the ACω convention through its canonical smooth tangent-bundle supplier. This proof uses no choice after the smooth manifold, metric and fields are given. (Oriented Lorentzian fields, observers and coordinate conventions, The Axiom of Countable Choice (ACω))

Proof

technique · differentiate the variational Maxwell tensor, contract the homogeneous Maxwell identity with $F$, and cancel the derivative terms
1.1A1F1F3givenalgebra

Write F2:=FcdFcd. Because ∇ is torsion-free, dF=0 is equivalent to ∇aFdc+∇dFca+∇cFad=0. Metric compatibility and the product rule in [F3] give the raised-index derivatives used below.

2.1step 1.1F3algebra

Multiply the cyclic identity in step 1.1 by Fac. The first term is Xd:=Fac∇aFdc. In the third term, swapping the dummy indices a,c and using Fca=−Fac and Fad=−Fda gives another Xd. The middle term is Fac∇dFca=−12∇d(FacFac)=−12∇dF2. Thus 2Xd−12∇dF2=0, so Fac∇aFdc=14∇dF2.

3.1F1F2F3step 2.1algebra

Raise d in step 2.1 and differentiate the tensor in [F2]. The product rule yields μ0∇aTEMab=(∇aFac)Fbc+Fac∇aFbc−14∇bF2=−μ0jcFbc+14∇bF2−14∇bF2, where the sourced equation in [F1] gives the first term and step 2.1 gives the second.

4.1step 3.1algebra∎

The last two terms in step 3.1 cancel. Dividing by μ0>0 gives ∇aTEMab=−Fbcjc, as claimed. The calculation uses no regularity beyond the stated smoothness and no physical or empirical premise.

Remarks

Depends on

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