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Dominant energy inequality for the real scalar test tensor

Physical scope

A pointwise algebraic inequality for the mathematical test-scalar tensor on a Lorentzian tangent space; it does not assert that the tensor is coupled matter.

Statement

Assume ${A C}_{ω}$ . At a point $p$ of a time-oriented Lorentzian four-manifold, let $n$ be a future-directed unit timelike vector, let $X \in T_{p} M$ be future causal, and let $ψ$ be a smooth real scalar. Put $ℓ : = ♯_{g} (d ψ)$ , $a : = d ψ (n)$ , $b : = h_{n} (ℓ) \in n^{⊥}$ , and $∣ b ∣^{2} : = g (b, b)$ , where $h_{n}$ is the spatial projection relative to $n$ . Then

$Q (n, n) = \frac{1}{2} (a^{2} + ∣ b ∣^{2}), Q (n, X) \geq 0.$

Statement references

Scalar test-wave stress and multiplier current

Observers, rest spaces and the spatial projection at an event

The observer rest-space decomposition and the spatial projection

Facts & Assumptions

Given: A time-oriented Lorentzian tangent space of signature $(-, +, +, +)$ , a future unit timelike vector $n$ , a future causal vector $X$ , and $d ψ$ .

[F1]

Scalar test-wave stress and multiplier current: $Q (u, v) = d ψ (u) d ψ (v) - \frac{1}{2} g (u, v) g^{- 1} (d ψ, d ψ)$ .

[F2]

Observers, rest spaces and the spatial projection at an event: $g (n, n) = - 1$ , $n^{⊥}$ is the rest space, and $h_{n} (v) = v + g (n, v) n$ .

[F3]

The observer rest-space decomposition and the spatial projection: $n^{⊥}$ is positive definite and $h_{n}$ is the orthogonal projection onto it.

[F4]

Timelike, null, spacelike and causal vectors of a Lorentz vector space: causal vectors are nonzero and satisfy $g (X, X) \leq 0$ . By Classification of future and past causal vectors by a timelike field, future timelike vectors form the connected component $C_{p}^{+}$ containing $N_{p}$ , and future causal vectors are exactly $cl (C_{p}^{+}) ∖ {0}$ . In particular $n \in C_{p}^{+}$ and $X \in cl (C_{p}^{+})$ .

[F5]

Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors: in the positive-definite real inner product space $n^{⊥}$ , $∣ g (b, Y) ∣ \leq ∣ b ∣ ∣ Y ∣$ .

[A1]

${A C}_{ω}$ is the only choice principle assumed in the smooth tangent-bundle conventions used by [F2] and [F3] (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

1.1givenA1F1F2F3algebra

Set $ℓ = ♯_{g} (d ψ)$ and $a = g (ℓ, n)$ . Then $b = h_{n} (ℓ) = ℓ + a n$ , so $ℓ = b - a n$ and $g (b, n) = 0$ by [F2]–[F3]. It follows that $g^{- 1} (d ψ, d ψ) = g (ℓ, ℓ) = - a^{2} + ∣ b ∣^{2}$ . Substitution into [F1], using $g (n, n) = - 1$ , gives $Q (n, n) = a^{2} + \frac{1}{2} (- a^{2} + ∣ b ∣^{2}) = \frac{1}{2} (a^{2} + ∣ b ∣^{2})$ .

1.2givenF2F3F4algebra

Let $α : = - g (n, X)$ and $Y : = X - α n$ . The continuous linear function $v \mapsto g (n, v)$ never vanishes on the timelike component $C_{p}^{+}$ : a timelike vector in $n^{⊥}$ would contradict its positive definiteness in [F3]. Since $C_{p}^{+}$ is connected and contains $n$ , where this function is $- 1$ , it is negative throughout $C_{p}^{+}$ . Continuity makes it nonpositive on its closure, so [F4] gives $α \geq 0$ . By [F2], $Y = h_{n} (X) \in n^{⊥}$ . Since $g (X, X) = - α^{2} + ∣ Y ∣^{2} \leq 0$ , one has $∣ Y ∣ \leq α$ .

2.1F1step 1.1step 1.2algebra

By [F1] and step 1.1, $Q (n, X) = α a^{2} + a g (b, Y) + \frac{α}{2} (- a^{2} + ∣ b ∣^{2}) = \frac{α}{2} (a^{2} + ∣ b ∣^{2}) + a g (b, Y)$ .

3.1F5step 1.2step 2.1algebra∎

Apply [F5] and $∣ Y ∣ \leq α$ from step 1.2: $Q (n, X) \geq \frac{α}{2} (a^{2} + ∣ b ∣^{2}) - α ∣ a ∣ ∣ b ∣ = \frac{α}{2} (∣ a ∣ - ∣ b ∣)^{2} \geq 0$ . If $α = 0$ , step 1.2 gives $Y = 0$ and hence $X = 0$ , for which $Q (n, X) = 0$ as well. This proves the causal flux inequality, including the null case.

Dominant energy inequality for the real scalar test tensor

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources