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Nonzero flat test field is not a coupled flat solution

Statement refuted

For Λ=0, every smooth source-free Maxwell field on a prescribed Minkowski metric is an electrovacuum solution with that same flat metric.

Facts & Assumptions

Given: The tensor system and constraint conventions of Electrovacuum tensor Cauchy system and Constrained electrovacuum initial data, with κ>0, μ0>0, Λ=0, and B0≠0.

[F1]

Electrovacuum tensor Cauchy system: a coupled solution must satisfy both Maxwell equations and G+Λg=κTEM.

[F2]

Maxwell stress algebra and covariant conservation: for e=0 and b=(0,0,B0), TEM(n,n)=B02/(2μ0).

[F3]

Gauss-Codazzi and Maxwell initial constraints: any smooth coupled solution induces data satisfying the Hamiltonian constraint.

Counterexample

Given: The system and constraint conventions above; κ,μ0>0, Λ=0, and a fixed B0≠0 (tesla in the SI interpretation).

1.1F1algebra

Take M=R4 with length coordinates (x0,x1,x2,x3), metric g=−(dx0)2+(dx1)2+(dx2)2+(dx3)2, orientation dx0∧dx1∧dx2∧dx3, and future field N=∂x0; in the SI interpretation x0=ct. Set F=B0 dx1∧dx2. The metric coefficients and those of F and ⋆gF are constant, so the curvature vanishes and dF=d(⋆gF)=0. Hence this is a smooth source-free Maxwell field on the prescribed Minkowski background.

2.1F1F2F3algebra∎

On the slice x0=0, the data are h=δ, K=0, e=0, and b=(0,0,B0). The Hamiltonian constraint from [F3] would require 0=(κ/μ0)B02, because R(h)=0 and Λ=0. This is impossible since κ,μ0>0 and B0≠0. Hence the given flat Maxwell test field is not a coupled electrovacuum solution with that metric.

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