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Exact geometric electrovac normalization

Statement

Let $(M, g, N, o)$ , $F$ , $κ$ , $μ_{0}$ , $α$ and $F = α F$ be as in SI and geometrized Einstein–Maxwell parameters, with curvature tensors defined by The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors. Let $Λ \in R$ be a constant. Define the quadratic tensor

$Q_{a b} (g, F) : = F_{a c} F_{b}^{c} - \frac{1}{4} g_{a b} F_{c d} F^{c d},$

where indices are raised with $g^{- 1}$ . Then the following equations for the same metric and two-form are equivalent:

$G_{a b} (g) + Λ g_{a b} = \frac{κ}{μ_{0}} Q_{a b} (g, F),$

$G_{a b} (g) + Λ g_{a b} = 2 Q_{a b} (g, F),$

${Ric}_{a b} (g) - Λ g_{a b} = 2 Q_{a b} (g, F) .$

These are algebraic equivalences between tensor equations. No Maxwell equation, matter model, limiting procedure or physical solution is assumed.

Facts & Assumptions

Given: The smooth Lorentzian geometry and positive parameters of SI and geometrized Einstein–Maxwell parameters, a smooth real two-form $F$ , and the curvature convention of The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors. The inherited geometric setup carries ${A C}_{ω}$ as in [A1]; this finite tensor calculation uses no further choice.

[F1]

The geometric normalization has $ε_{0} = (μ_{0} c^{2})^{- 1}$ , $κ = 8 π G / c^{4}$ , $α^{2} = κ / (2 μ_{0})$ , and $F = α F$ . (SI and geometrized Einstein–Maxwell parameters)

[F2]

The curvature convention defines $G_{a b} = {Ric}_{a b} - \frac{1}{2} S g_{a b}$ and $S = g^{a b} {Ric}_{a b}$ on the four-dimensional Lorentzian manifold. (The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors)

[A1]

The only inherited choice assumption is Countable Choice ${A C}_{ω}$ , stated as choosing one element from each set in a countable family of nonempty sets. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

Proof

1.1givenF2A1algebra

Contracting the displayed definition of $Q$ gives $g^{a b} Q_{a b} = F_{a c} F^{a c} - \frac{1}{4} (g^{a b} g_{a b}) F_{c d} F^{c d} = F_{c d} F^{c d} - F_{c d} F^{c d} = 0$ , because the metric dimension is four. Thus $Q$ is trace-free.

1.2givenF1algebra

Each term of $Q (g, F)$ contains exactly two factors linear in $F$ , so for every scalar $s$ , $Q (g, s F) = s^{2} Q (g, F)$ . In particular, $Q (g, F) = α^{2} Q (g, F)$ .

2.1F1step 1.2algebra

Since $2 α^{2} = κ / μ_{0}$ by [F1], the right-hand side $(κ / μ_{0}) Q (g, F)$ equals $2 Q (g, F)$ ; replacing it in the first equation proves equivalence of the first two equations in both directions.

2.2F2step 1.1algebra

By [F2], tracing the second equation gives $- S + 4 Λ = 0$ , since $g^{a b} G_{a b} = S - \frac{1}{2} (4) S = - S$ and the right-hand side is trace-free by step 1.1. Hence $S = 4 Λ$ .

3.1F2step 1.1step 2.2algebra∎

Substituting $S = 4 Λ$ into $G_{a b} + Λ g_{a b}$ gives ${Ric}_{a b} - 2 Λ g_{a b} + Λ g_{a b} = {Ric}_{a b} - Λ g_{a b}$ , proving the second equation implies the third. Conversely, tracing the third equation and using step 1.1 gives $S - 4 Λ = 0$ ; substituting this into $G_{a b}$ recovers the second equation.

Exact geometric electrovac normalization

Statement

Facts & Assumptions

Proof

Depends on

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