Physics theoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Convective/Ohmic comoving mechanical work prediction

Physical scope

A conditional local energy and pressure-force projection for the selected smooth charged perfect fluid. Comoving electromagnetic work is zero for the convective current and is nonnegative for the separately adopted smooth Ohmic law. This is a mechanical balance, not an entropy theorem or global fluid solution.

Statement

Let $Ω$ be a smooth open state domain of Selected charged perfect-fluid exchange/Ohmic branch. Assume $ϵ, p, σ \in C^{1} (Ω, R)$ , $U \in C^{2} (Ω, T Ω)$ is future timelike with $g (U, U) = - c^{2}$ , $F$ is a smooth electromagnetic two-form, and the matter stress and current satisfy the selected fluid closure and exchange equation. Define

$D : = U^{a} \nabla_{a}, θ : = \nabla_{a} U^{a}, a^{b} : = D U^{b}, h^{b}_{c} : = δ^{b}_{c} + \frac{U^{b} U_{c}}{c^{2}}, f^{b} : = F^{b}_{c} j^{c}, E_{U}^{a} : = F^{a}_{b} U^{b} .$

Then the longitudinal work and rest-space force equations are

$D ϵ + (ϵ + p) θ = - U_{b} f^{b}, \frac{ϵ + p}{c^{2}} a^{b} + h^{b a} \nabla_{a} p = h^{b}_{c} f^{c} .$

For the convective branch $j^{a} = σ U^{a}$ , the comoving electromagnetic work $- U_{b} f^{b}$ is zero. For the optional smooth Ohmic branch $j^{a} = σ U^{a} + q^{a}$ , $q^{a} = χ E_{U}^{a}$ , with $χ \geq 0$ , it is

$- U_{b} f^{b} = χ g (E_{U}, E_{U}) \geq 0.$

Here $E_{U}$ is the rest-frame electric field and the induced metric on $U^{⊥}$ is positive definite. This is a conditional mechanical energy identity; it does not assert an entropy-production law, equation-of-state evolution, causal constitutive theory, or fluid existence.

Facts & Assumptions

Given: The regularity, timelike normalization, perfect-fluid stress, selected current, smooth electromagnetic field and matter-exchange equation in the Statement.

[A1]

Selected charged perfect-fluid exchange/Ohmic branch: the selected smooth model adopts $T_{m}^{a b} = (ϵ + p) U^{a} U^{b} / c^{2} + p g^{a b}$ and $\nabla_{a} T_{m}^{a b} = F^{b}_{c} j^{c}$ , with either $j^{a} = σ U^{a}$ or the separately specified smooth conduction current $j^{a} = σ U^{a} + q^{a}$ and $q^{a} U_{a} = 0$ .

[F1]

Perfect-fluid exchange projections and conduction heating: for these $C^{1}$ fluid scalars, $C^{2}$ unit-normalized $U$ , and the exchange equation, the two displayed projections are equivalent to the full tensor balance; if $q^{a} = χ E_{U}^{a}$ with $χ \geq 0$ , then $- U_{b} f^{b} = χ g (E_{U}, E_{U}) \geq 0$ , while $q = 0$ gives zero work. This is a draft mathematical supplier and is used only at its exact current statement.

Proof

1.1A1F1given

The physical stress, current, smoothness and normalization in [A1] match the hypotheses of [F1]. Applying its projection identity to $\nabla_{a} T_{m}^{a b} = f^{b}$ gives the stated longitudinal energy equation and orthogonal pressure/acceleration equation, with no equation-of-state or entropy premise.

2.1A1F1step 1.1algebra∎

For $j^{a} = σ U^{a}$ , antisymmetry of $F_{a b}$ gives $U_{b} F^{b}_{c} U^{c} = 0$ . For $j^{a} = σ U^{a} + q^{a}$ , the same antisymmetry gives $- U_{b} F^{b}_{c} q^{c} = E_{U, c} q^{c}$ . Under the optional law $q^{a} = χ E_{U}^{a}$ , this equals $χ g (E_{U}, E_{U}) \geq 0$ because $χ \geq 0$ and $U^{⊥}$ is positive definite; hence the two stated work branches follow.

Convective/Ohmic comoving mechanical work prediction

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources