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Observer energy conditions for selected matter plus EM

Physical scope

Pointwise observer interpretation of the selected smooth electromagnetic, charged-dust and perfect-fluid stress tensors. Electromagnetic and nonnegative-dust energy conditions are conditional tensor deductions; perfect-fluid conclusions require the exact stated inequalities. No general matter positivity or empirical claim is made.

Statement

Use the smooth electromagnetic stress, observer contractions and selected matter tensors of Smooth matter quantities, current and observer observables, with μ0>0 and a future unit observer n. The electromagnetic tensor is symmetric and trace-free, and it satisfies WEC, NEC, SEC and DEC. In a positively oriented orthonormal rest tetrad, with the cross product taken in the oriented Euclidean rest space, its energy density and spatial energy-current components are

u=ϵ0∣En∣2+∣Bn∣2/μ02,Sn=En×Bnμ0c.

For charged dust with ρ≥0, the matter tensor and its sum with the electromagnetic tensor satisfy all four conditions. For a selected perfect-fluid tensor with rest energy density ϵ and pressure p, the exact pointwise conditions on the matter tensor are

WEC  ⟺  ϵ≥0 and ϵ+p≥0,NEC  ⟺  ϵ+p≥0,

SEC  ⟺  ϵ+p≥0 and ϵ+3p≥0,DEC  ⟺  ϵ≥∣p∣.

If the perfect-fluid tensor satisfies any one of these conditions, adding the electromagnetic tensor preserves that condition. The inequalities are not assumed universally, and adding the electromagnetic tensor to arbitrary matter need not repair a violated condition.

Facts & Assumptions

Given: The smooth Lorentzian metric, positive μ0, observer, electromagnetic field and selected dust or perfect-fluid tensors in the Statement.

[F1]

Smooth matter quantities, current and observer observables: TEMab=μ0−1(FacFbc−14gabFcdFcd), F0i=−Ei/c, Fij=ϵijkBk, and the dust and perfect-fluid matter tensor definitions.

[F2]

EM, dust and perfect-fluid energy inequalities: the electromagnetic tensor satisfies WEC, NEC, SEC and DEC with rest-frame density u=(∣E∣2/c2+∣B∣2)/(2μ0) and energy flux (E×B)/(μ0c); dust with ρ≥0 satisfies all four; the displayed perfect-fluid inequalities are exact; sums preserve each satisfied condition. This draft mathematical proof is used only at its stated algebraic interface.

[A1]

The dust branch, when used, assumes ρ≥0. No sign restriction is added to ϵ or p except when a particular perfect-fluid condition is asserted.

Proof

1.1F1F2A1given

The definitions in [F1] fix the electromagnetic stress and observer convention. The mathematical component and cone argument [F2] establishes its symmetry, trace, observer density and flux, and all four energy conditions. The same supplier gives the exact inequalities for the selected dust and perfect-fluid tensors under [A1].

2.1F2step 1.1algebra∎

At each point, [F2] proves the future-causal cone and each quadratic condition are preserved under addition of tensors that satisfy the same condition. Therefore nonnegative dust plus the electromagnetic tensor satisfies all four; a perfect-fluid sum satisfies each condition asserted only when the corresponding listed fluid inequalities hold. No conclusion is made for an arbitrary matter tensor.

Depends on

Dependency tree · 2 levels

Sources