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Wave multiplier divergence identity

Physical scope

A differential identity for a real test scalar on a fixed smooth Lorentzian background; it asserts no decay or coupled gravitational response.

Statement

Assume ${A C}_{ω}$ . Let $(M, g, N)$ be a time-oriented smooth Lorentzian four-manifold, let $U \subseteq M$ be open, let $ψ \in C^{\infty} (U, R)$ , and let $X$ be a smooth vector field on $U$ . For the tensor $Q_{a b} [ψ]$ and multiplier covector $J_{a}^{X} [ψ] = Q_{a b} [ψ] X^{b}$ defined from the metric-selected connection,

$\nabla^{a} J_{a}^{X} [ψ] = (□_{g} ψ) X^{b} \nabla_{b} ψ + \frac{1}{2} Q^{a b} [ψ] (L_{X} g)_{a b} .$

Consequently, when $□_{g} ψ = 0$ and $X$ is Killing, the multiplier current has zero divergence.

Statement references

Scalar test-wave stress and multiplier current

Killing fields of a Lorentzian metric

Facts & Assumptions

Given: The smooth Lorentzian background, open set, scalar field, vector field, and ${A C}_{ω}$ assumption in the Statement.

[F1]

Scalar test-wave stress and multiplier current: $Q_{a b} = \nabla_{a} ψ \nabla_{b} ψ - \frac{1}{2} g_{a b} \nabla^{c} ψ \nabla_{c} ψ$ , $J_{a}^{X} = Q_{a b} X^{b}$ , and $□_{g} ψ = g^{a b} \nabla_{a} \nabla_{b} ψ$ .

[F2]

Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields: $\nabla g = 0$ , the metric-selected connection is torsion-free, and its tensor derivative obeys the product rule and commutes with metric contractions.

[F3]

Killing fields of a Lorentzian metric: $X$ is Killing precisely when $L_{X} g = 0$ .

[F4]

Killing fields and conserved currents, and on-shell conservation of the Hilbert stress-energy tensor: for the metric-selected torsion-free connection, $(L_{X} g)_{a b} = \nabla_{a} X_{b} + \nabla_{b} X_{a}$ .

[A1]

${A C}_{ω}$ is the only choice principle assumed for the smooth tangent-bundle and connection conventions used by [F2] (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

1.1givenA1F1F2algebra

Differentiate the definition of $Q$ using $\nabla g = 0$ from [F2]: $\nabla^{a} Q_{a b} = (□_{g} ψ) \nabla_{b} ψ + \nabla^{a} ψ \nabla_{a} \nabla_{b} ψ - \nabla^{c} ψ \nabla_{b} \nabla_{c} ψ$ . Since $ψ$ is a scalar and the connection is torsion-free, $\nabla_{a} \nabla_{b} ψ = \nabla_{b} \nabla_{a} ψ$ ; after relabelling the contracted index, the last two terms cancel. Thus $\nabla^{a} Q_{a b} = (□_{g} ψ) \nabla_{b} ψ$ .

2.1step 1.1F1F2algebra

Apply the product rule to $J_{a}^{X} = Q_{a b} X^{b}$ and substitute step 1.1: $\nabla^{a} J_{a}^{X} = (□_{g} ψ) X^{b} \nabla_{b} ψ + Q_{a b} \nabla^{a} X^{b}$ .

3.1step 2.1F1F4algebra

The tensor $Q_{a b}$ is symmetric by [F1]. Therefore $Q_{a b} \nabla^{a} X^{b} = \frac{1}{2} Q^{a b} (\nabla_{a} X_{b} + \nabla_{b} X_{a}) = \frac{1}{2} Q^{a b} (L_{X} g)_{a b}$ by [F4], which proves the displayed identity.

4.1givenF3step 3.1algebra∎

If $□_{g} ψ = 0$ and $X$ is Killing, the first term in the identity vanishes by the equation and the second vanishes by [F3]. Hence $\nabla^{a} J_{a}^{X} = 0$ . This is a conditional model identity, not a decay estimate.

Wave multiplier divergence identity

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources