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Divergence of the measured energy-momentum current and the finite-chain balance identity

Statement

Let (M,g,N) be a time-oriented smooth Lorentzian four-manifold with metric-selected connection ∇, let U⊆M be a nonempty open subset, let T be a smooth symmetric covariant 2-tensor field on U with ∇aTab=0 on U, and let n be a smooth observer congruence on U. Let μ, J, Φ, S be the fields of The observer split of a symmetric two-tensor: energy density, energy-momentum current, flux and stress and let a, θ, D be the kinematic fields of Acceleration, expansion and the along-congruence derivative of an observer congruence.

(a) Divergence identity. On U,

∇aJa=−Tab∇anb.

(b) Balance identity. On U,

Dμ+μθ+∇aΦa=−Tab∇anb.

(c) Finite-chain balance. For this clause choose an orientation on U and use its metric volume form volg∣U, written volg. Let C be a finite smooth singular 4-chain of M contained in U with induced boundary ∂C=Σ2−Σ1+B, where Σ1,Σ2,B are smooth singular 3-chains (The metric volume form, current flux through 3-faces and finite-chain charges clause (b)). Then

ΦJ(Σ2)−ΦJ(Σ1)=−ΦJ(B)+∫C(−Tab∇anb)volg.

In particular, if Tab∇anb=0 on an open subset containing the image of C, then ΦJ(Σ2)−ΦJ(Σ1)=−ΦJ(B).

(d) Killing contraction. If ξ is a smooth vector field on U with ∇aξb+∇bξa=0 (a Killing field, Killing fields of a Lorentzian metric), then Tab∇aξb=0 on U.

Facts & Assumptions

Given: The time-oriented smooth Lorentzian four-manifold (M,g,N) with metric-selected connection ∇, the open subset U, the smooth symmetric covariant 2-tensor field T with ∇aTab=0 on U, the smooth observer congruence n on U, the fields μ, J, Φ, S of The observer split of a symmetric two-tensor: energy density, energy-momentum current, flux and stress, and the fields a, θ, D of Acceleration, expansion and the along-congruence derivative of an observer congruence. In (c) additionally a chosen orientation on U, its metric volume form, and the finite smooth singular 4-chain C⊆U with boundary ∂C=Σ2−Σ1+B; in (d) additionally the vector field ξ with ∇aξb+∇bξa=0.

[L1]

The extension of ∇ to tensor fields satisfies the product rule, and ∇g=0, ∇g−1=0; consequently raising and lowering indices commute with ∇, and for a smooth covariant 2-tensor field T with ∇aTab=0 one has ∇aTab=0 (Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields clause (c), The inverse metric, musical isomorphisms and index raising, Tensor fields of type (r,s) on a Lorentzian four-manifold and their components, Metric compatibility, torsion freedom and the metric-selected connection class).

[L3]

On the oriented open submanifold U with restricted metric and connection, for a smooth vector field K on U and a finite smooth singular 4-chain C of U, ΦK(∂C)=∫C(∇aKa)volg; consequently, if ∂C=Σ2−Σ1+B, then ΦK(Σ2)−ΦK(Σ1)=−ΦK(B)+∫C(∇aKa)volg (Divergence, Stokes flux balance and the non-null face formula for a Lorentzian current clause (b), The metric volume form, current flux through 3-faces and finite-chain charges clause (b)).

Proof

technique · direct; differentiate the current, substitute the orthogonal decomposition of $T$, and apply the finite-chain flux balance to the current
1.1L1givenalgebra

(Divergence identity.) By the definition Ja=−Tabnb and the product rule [L1], ∇aJa=−∇a(Tabnb)=−(∇aTab)nb−Tab∇anb; the hypothesis ∇aTab=0 gives ∇aTab=0 by [L1], so the first term vanishes and ∇aJa=−Tab∇anb. This proves (a).

2.1step 1.1L2L4algebra

(Balance identity.) By [L2], J=μn+Φ; hence ∇aJa=∇a(μna)+∇aΦa=na∇aμ+μ∇ana+∇aΦa, so by [L4] ∇aJa=Dμ+μθ+∇aΦa. Substituting step 1.1 gives (b).

2.2step 1.1L3algebra

(Finite-chain balance.) A simplex of M with image in U is a simplex of U: restrict its smooth extension to the open inverse image of U, which contains its simplex domain. Apply [L3] on the oriented manifold U, where J is defined, to this chain C. Then ΦJ(∂C)=∫C(∇aJa)volg, and the additivity of the flux over chains gives ΦJ(∂C)=ΦJ(Σ2)−ΦJ(Σ1)+ΦJ(B). Substituting step 1.1 yields ΦJ(Σ2)−ΦJ(Σ1)=−ΦJ(B)+∫C(−Tab∇anb)volg; if the integrand vanishes on an open set containing the image of C, the integral vanishes and the second assertion follows. This proves (c).

3.1givenalgebra∎

(Killing contraction.) Write Tab∇aξb=12Tab(∇aξb+∇bξa)+12Tab(∇aξb−∇bξa). The first summand vanishes by the Killing equation. In the second, the symmetry Tab=Tba and a relabelling a↔b give Tab(∇aξb−∇bξa)=Tab∇aξb−Tba∇aξb=0. Hence Tab∇aξb=0. This proves (d).

Remarks

  • Source of the balance. The right-hand side −Tab∇anb is a source term produced by the congruence: it vanishes identically for a parallel congruence (∇n=0), and for a normalized timelike Killing congruence it reduces to −Φaaa by Kinematics of a timelike Killing congruence and conservation of the redshift-weighted energy current clause (c). The measured energy current alone therefore need not be conserved; the Killing-weighted current of that lemma is guaranteed to be conserved under its stated hypotheses.

  • No positivity or dynamics. The lemma uses only ∇aTab=0 and the congruence property; no energy condition, no equation of state and no existence of a solution of the field equation is used or asserted.

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