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Exact potential-wave multiplier identity

Statement

Let $(M, g, N, o)$ , $U$ , $u$ , $V$ , $s$ , and $X$ satisfy the data and definitions of the wave-current definition, including $□_{g} u = V u + s$ . Then $\nabla_{a} J_{X}^{a} = K_{X} + s X (u) - \frac{1}{2} X (V) u^{2} .$ In particular, if $X$ satisfies the Killing equation $\nabla_{(a} X_{b)} = 0$ , $X (V) = 0$ , and $s = 0$ , then $\nabla_{a} J_{X}^{a} = 0$ . At a point where $n$ is a future unit timelike vector and $X$ is timelike with $- g (X, n) > 0$ , the density $T_{a b} X^{a} n^{b}$ is nonnegative whenever $V \geq 0$ .

Statement references

Wave stress and multiplier currents

Facts & Assumptions

Given: The Lorentzian data, smooth real fields $u, V, s$ , a smooth vector field $X$ , the equation $□_{g} u = V u + s$ , and the background countable-choice convention inherited from the cited geometry suppliers.

[F1]

Wave stress and multiplier currents: $T_{a b} = \nabla_{a} u \nabla_{b} u - \frac{1}{2} g_{a b} (∣ d u ∣_{g}^{2} + V u^{2})$ , $J_{X}^{a} = T^{a}_{b} X^{b}$ , and $K_{X} = T_{a b} \nabla^{(a} X^{b)}$ .

[F2]

Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model: the time orientation selects future timelike vectors at each point, and each tangent space has signature $(-, +, +, +)$ .

[F3]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): the background axiom is countable choice ${A C}_{ω}$ ; no additional choice principle is assumed.

[F4]

Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields: the metric-selected connection is metric compatible and torsion free, so $\nabla_{a} g_{b c} = 0$ and the Hessian of a scalar is symmetric.

[F5]

The observer rest-space decomposition and the spatial projection: for a unit future timelike $n$ , $T_{p} M = R n \oplus n^{⊥}$ , the restriction of $g$ to $n^{⊥}$ is positive definite, and $g (v, v) = - α^{2} + ∣ w ∣^{2}$ when $v = α n + w$ .

[F6]

Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors: on the positive-definite rest space $n^{⊥}$ , $∣ d u (w) ∣ \leq ∣ w ∣ ∣ d u ∣_{n^{⊥}}$ .

[F7]

Observers, rest spaces and the spatial projection at an event: an observer is a future unit timelike vector, and its rest space is $n^{⊥}$ .

[A1]

$□_{g} u = V u + s$ on $U$ , with $u, V, s$ smooth and real, as required in the Statement.

[A2]

For the positivity conclusion only, $V \geq 0$ at the point, $n$ is a future unit timelike vector, and $X$ is timelike with $- g (X, n) > 0$ .

Proof

1.1givenF1F3F4A1algebra

Metric compatibility and the product rule give $\nabla^{a} T_{a b} = (□_{g} u) \nabla_{b} u + \nabla^{a} u \nabla_{a} \nabla_{b} u - \frac{1}{2} \nabla_{b} (∣ d u ∣_{g}^{2}) - \frac{1}{2} \nabla_{b} (V u^{2}) .$ For a scalar, the Hessian is symmetric and $\nabla_{b} (∣ d u ∣_{g}^{2}) = 2 \nabla^{a} u \nabla_{b} \nabla_{a} u$ , so the middle two terms cancel. Substituting $□_{g} u = V u + s$ into the remaining terms yields $\nabla^{a} T_{a b} = s \nabla_{b} u - \frac{1}{2} u^{2} \nabla_{b} V$ .

2.1step 1.1F1algebra

By the product rule, $\nabla_{a} J_{X}^{a} = (\nabla^{a} T_{a b}) X^{b} + T_{a b} \nabla^{a} X^{b} .$ The tensor $T$ is symmetric, so its contraction with $\nabla^{a} X^{b}$ equals its contraction with $\nabla^{(a} X^{b)}$ , which is $K_{X}$ . Substituting step 1.1 gives the claimed divergence identity. If $\nabla_{(a} X_{b)} = 0$ , then $K_{X} = 0$ , and the stated divergence-free case follows.

3.1givenF2F5F6F7A2algebra∎

Decompose $X = α n + w$ with $w \in n^{⊥}$ by [F5]. By [A2], $α = - g (X, n) > 0$ , and timelikeness gives $∣ w ∣ < α$ . Put $A : = n (u)$ , and let $∣ d u ∣_{n^{⊥}}$ be the norm of the restriction of $d u$ to the positive-definite rest space, using its dual metric. By the definition of $T$ and the orthogonal decomposition, $T_{a b} n^{a} n^{b} = \frac{1}{2} (A^{2} + ∣ d u ∣_{n^{⊥}}^{2} + V u^{2}), T_{a b} w^{a} n^{b} = w (u) A .$ Therefore [F6] and $2 ∣ A ∣ ∣ d u ∣_{n^{⊥}} \leq A^{2} + ∣ d u ∣_{n^{⊥}}^{2}$ give $T_{a b} X^{a} n^{b} \geq \frac{1}{2} (α - ∣ w ∣) (A^{2} + ∣ d u ∣_{n^{⊥}}^{2}) + \frac{1}{2} α V u^{2} \geq 0.$ This is the required pointwise positivity.

Remarks

This is a conditional identity for the stated scalar equation. It does not assert that any particular coupled Einstein–Maxwell perturbation has this scalar form.

Exact potential-wave multiplier identity

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources