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Charged dust balance equivalence and advected charge ratio

Statement

Let $Ω$ be an open subset of the oriented, time-oriented Lorentzian domain of Smooth formal matter, EM tensors and observer quantities. Let $ρ, σ \in C^{1} (Ω, R)$ with $ρ > 0$ , and let $U \in C^{2} (Ω, T Ω)$ be future timelike with $g (U, U) = - c^{2}$ . Set $T_{m}^{a b} = ρ U^{a} U^{b}$ , $j^{a} = σ U^{a}$ , and $a^{b} = U^{a} \nabla_{a} U^{b}$ . Assume the formal matter-exchange equation $\nabla_{a} T_{m}^{a b} = F^{b}_{c} j^{c}$ . Then this equation is equivalent on $Ω$ to

$\nabla_{a} (ρ U^{a}) = 0, ρ a^{b} = σ F^{b}_{c} U^{c} .$

On $Ω$ , define $α = σ / ρ$ . Under the continuity equation, charge-current conservation $\nabla_{a} j^{a} = 0$ is equivalent to $U^{a} \nabla_{a} α = 0$ . Thus the equation constrains the derivative of $α$ along the dust flow; it does not impose spatial constancy of $α$ . The claim is a pointwise conditional identity and supplies no coupled-field existence theorem or Maxwell equation.

Facts & Assumptions

Given: $Ω, ρ, σ, U, F$ with the regularity and sign conditions in the Statement, together with the formal exchange equation.

[F1]

Smooth formal matter, EM tensors and observer quantities: $T_{m}^{a b} = ρ U^{a} U^{b}$ , $j^{a} = σ U^{a}$ , $g (U, U) = - c^{2}$ , and the exchange equation is $\nabla_{a} T_{m}^{a b} = F^{b}_{c} j^{c}$ .

[F2]

Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields: $\nabla$ is the metric-compatible torsion-free connection; its induced tensor derivative commutes with contractions and annihilates $g$ and $g^{- 1}$ .

[F3]

Product connection on tensor and hom bundles: the induced tensor connection obeys the Leibniz rule on tensor products and on scalar multiples of tensor fields.

[F4]

Continuity and derivatives of positive-base real powers: $s \mapsto s^{- 1}$ is $C^{1}$ on $(0, \infty)$ with derivative $- s^{- 2}$ .

[F5]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ and $C^{k}$ Euclidean maps are closed under componentwise algebra and composition: compositions of $C^{1}$ coordinate maps are $C^{1}$ , and their total derivatives obey the chain rule.

[F6]

Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ : products of differentiable real functions obey the product rule.

[A1]

$ρ > 0$ on $Ω$ . This is the exact domain on which division by $ρ$ in the equivalence and the definition of $α$ are permitted.

Proof

1.1F1F2F3A1algebra

The product rule gives $\nabla_{a} T_{m}^{a b} = U^{b} \nabla_{a} (ρ U^{a}) + ρ a^{b}$ . Differentiating $g (U, U) = - c^{2}$ along $U$ and using metric compatibility gives $U_{b} a^{b} = \frac{1}{2} U^{a} \nabla_{a} (g (U, U)) = 0$ . Antisymmetry of $F$ gives $U_{b} F^{b}_{c} U^{c} = 0$ .

2.1F1step 1.1algebra

Contract the exchange equation with $U_{b}$ . By step 1.1, its left side is $- c^{2} \nabla_{a} (ρ U^{a})$ and its right side is $σ U_{b} F^{b}_{c} U^{c} = 0$ . Since $c > 0$ , continuity follows. Substituting it into the uncontracted equation from step 1.1 leaves $ρ a^{b} = σ F^{b}_{c} U^{c}$ .

2.2F1step 1.1algebra

Conversely, substituting continuity and the force equation into $U^{b} \nabla_{a} (ρ U^{a}) + ρ a^{b}$ gives $σ F^{b}_{c} U^{c} = F^{b}_{c} j^{c}$ , recovering the exchange equation. This proves the equivalence without dividing at zero density.

3.1F1F3F4F5F6A1algebra∎

Because $ρ > 0$ , the scalar function $s \mapsto s^{- 1}$ from [F4] composes with the coordinate representation of $ρ$ to a $C^{1}$ function by [F5]. Differentiating $ρ (1 / ρ) = 1$ with the product rule in [F6] gives $\partial_{a} (1 / ρ) = - (\partial_{a} ρ) / ρ^{2}$ , so $α = σ / ρ$ is a well-defined $C^{1}$ scalar field. Since $σ = ρ α$ , the product rule gives $\nabla_{a} j^{a} = ρ U^{a} \nabla_{a} α + α \nabla_{a} (ρ U^{a})$ . Under continuity this is $ρ U^{a} \nabla_{a} α$ ; by [A1] it vanishes exactly when $U^{a} \nabla_{a} α = 0$ . This condition is directional along $U$ and contains no transverse derivative.

Remarks

The hypotheses do not impose a constant charge-to-mass ratio. Example Advected nonconstant charge ratio without Maxwell closure ↗ gives an explicit nonconstant ratio satisfying this matter identity while failing sourced Maxwell closure.
On $ρ = 0$ , the proof makes no division and does not infer a dust velocity or force law from the exchange equation.

Charged dust balance equivalence and advected charge ratio

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources