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Exact finite-frequency transverse Maxwell field

Example

Let M=R4 have length coordinates (x0,x1,x2,z), Minkowski metric η=diag⁡(−1,1,1,1), orientation dx0∧dx1∧dx2∧dz, and future direction ∂0, with x0=ct. Choose constants ν>0 with units m−1 and A∗≠0 with potential-component units T m. For 0<ϵ≤1, set

θ=ν(x0−z),a0=A∗ dx1,a1=0,AϵR:=Re⁡(ϵeiθ/ϵa0).

Then the real potential and field are

AϵR=ϵA∗cos⁡(θ/ϵ) dx1,(Fϵ)01=−νA∗sin⁡(θ/ϵ),(Fϵ)31=+νA∗sin⁡(θ/ϵ),

with all other independent field components zero. Let Ω=R3, I=R, x=(x1,x2,z) and t=x0/c. The corresponding smooth spatial fields E,B:Ω×I→R3 are

E(x,t)=cνA∗sin⁡ ⁣(ν(ct−z)ϵ)e1,B(x,t)=νA∗sin⁡ ⁣(ν(ct−z)ϵ)e2,

where e1=(1,0,0) and e2=(0,1,0), with units V m−1 and T, respectively. These are the SI fields of Smooth formal matter, EM tensors and observer quantities. This is an exact source-free Maxwell field for every fixed ϵ. For A∗≠0 it has nonzero electromagnetic stress on an open set, so it is not a fully coupled source-free Einstein–Maxwell solution on unchanged Minkowski space with Λ=0.

Facts & Assumptions

Given: The Minkowski chart, phase, constant transverse amplitude and parameter ranges in the Example.

[F1]

Smooth formal matter, EM tensors and observer quantities fixes F0i=−Ei/c, Fij=ϵijkBk and the smooth-field conventions.

[F2]

Finite WKB Maxwell residual, null rays and polarization quotient proves the exact finite-parameter potential and field expansions, residual identity, and conditional residual bounds under its four stated amplitude constraints.

[F3]

Smooth source/dust scaling and test-backreaction obstructions proves that a nonzero source-free Maxwell field with positive observer energy cannot solve the vacuum Einstein equation on the unchanged Minkowski metric; it also states the qualification against interpreting a test field as a coupled solution.

[F4]

Lorentzian exterior calculus and Hodge star gives the oriented Lorentzian Hodge-star component formula (⋆gF)ab=12(volg)abcdFcd; in a positively oriented orthonormal coframe its basis formula fixes the signs used in the direct calculation below.

[A1]

The real field is the real part of the complex potential specified in the Example. This is an algebraic construction, not an observation.

Verification

1.1givenA1F1F2algebra

The phase gradient is k=ν(dx0−dz) and g−1(k,k)=ν2(−1+1)=0. The constant a0 obeys g−1(k,a0)=0 and La0=0; a1=0 makes iLa1+Pa0=0 and ika(a1)a+∇a(a0)a=0. Here ∇ is the flat connection and the Ricci tensor vanishes. Thus all four hypotheses of [F2] hold.

1.2givenA1F1algebra

Differentiating AϵR in the displayed coordinates gives (Fϵ)01=∂0(AϵR)1=−νA∗sin⁡(θ/ϵ) and (Fϵ)31=∂z(AϵR)1=+νA∗sin⁡(θ/ϵ); no other independent component is produced. The SI component convention in [F1] therefore gives the displayed E1 and B2.

2.1F2F3F4step 1.1step 1.2algebra

In [F2], this example has a1=0, hence d1=0 and Pa1−∇d1=0. Its exact residual formula therefore gives ∇a(Fϵ)ab=0, and its exact closure identity gives dFϵ=0. Raising the free index in the first equality gives ∇aFϵab=0. To check the other source-free Maxwell equation directly, put f:=νA∗sin⁡(θ/ϵ). The displayed components give Fϵ=−f dx0∧dx1+f dz∧dx1. From the metric and orientation in the Example, [F4] yields ⋆Fϵ=f(dx0−dz)∧dx2. Since df is a scalar multiple of dθ=ν(dx0−dz), d(⋆Fϵ)=df∧(dx0−dz)∧dx2=0. Therefore dFϵ=0 and d(⋆Fϵ)=0 hold exactly for every fixed ϵ, as required by [F3].

3.1givenF3step 2.1algebra∎

Since νA∗≠0, there are points with sin⁡(θ/ϵ)≠0. At any such point the field is nonzero and [F3] gives positive electromagnetic observer energy. Minkowski has Gab=0, so the source-free Einstein equation with unchanged metric and Λ=0 fails there. This caveat concerns coupling only; the Maxwell calculation in step 2.1 remains exact.

Depends on

Dependency tree · 2 levels

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