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Observer electric/magnetic reconstruction and invariants

Statement

Let $U \subseteq M$ be an open set with smooth observer congruence $n$ , let $F$ be a smooth real two-form on $U$ , and let $c > 0$ be the coordinate conversion parameter of Oriented Lorentzian fields, observers and coordinate conventions. Define covectors

$E_{a} : = c F_{a b} n^{b}, B_{a} : = - (⋆_{g} F)_{a b} n^{b} .$

Here $c$ is the positive coordinate-conversion parameter of Oriented Lorentzian fields, observers and coordinate conventions, with SI unit $m s^{- 1}$ in its length-coordinate convention. The names electric and magnetic distinguish these two covectors; for arbitrary mathematical $F$ no physical field unit or measurement is assigned.

For each $p \in U$ , define the rest covector space by

$R_{n, p}^{*} : = {α \in T_{p}^{*} M : α (n_{p}) = 0} .$

The musical isomorphism $♭_{p}$ maps the vector rest space $n_{p}^{⊥}$ onto $R_{n, p}^{*}$ : $α \in R_{n, p}^{*}$ iff $g_{p} (♯_{p} α, n_{p}) = α (n_{p}) = 0$ . Moreover, $g_{p}^{- 1} (v^{♭}, w^{♭}) = g_{p} (v, w)$ , so the restriction of $g_{p}^{- 1}$ to $R_{n, p}^{*}$ is positive definite by The observer rest-space decomposition and the spatial projection. The covectors $E_{p}$ and $B_{p}$ annihilate $n_{p}$ ; define $⟨ E, B ⟩_{n} (p) : = g_{p}^{- 1} (E_{p}, B_{p})$ . Then

$F = n^{♭} \land E / c + ⋆_{g} (n^{♭} \land B),$

$F_{a b} F^{a b} = 2 (⟨ B, B ⟩_{n} - ⟨ E, E ⟩_{n} / c^{2}), F \land F = - \frac{2}{c} ⟨ E, B ⟩_{n} {v o l}_{g} .$

Equivalently, in any positively oriented future tetrad with $E_{0} = n$ , and with $ϵ_{123} = + 1$ in its oriented spatial triad, the components satisfy $F_{0 i} = - E_{i} / c$ and $F_{i j} = ϵ_{i j k} B_{k}$ . These are pointwise mathematical decompositions of an arbitrary two-form; no Maxwell equation or physical observation is assumed.

Facts & Assumptions

Given: An oriented, time-oriented Lorentzian four-manifold, a smooth observer congruence $n$ , a smooth two-form $F$ , and a positive scalar $c$ .

[F1]

Every point in the domain of $n$ has a neighborhood with a positively oriented $n$ -adapted orthonormal tetrad. The local assertion does not give a global tetrad or an orthogonal foliation. (Oriented Lorentzian fields, observers and coordinate conventions, Observers, rest spaces and the spatial projection at an event, The observer rest-space decomposition and the spatial projection)

[F2]

In a positively oriented orthonormal coframe $(e^{0}, e^{1}, e^{2}, e^{3})$ , put $ϵ_{0} = - 1$ and $ϵ_{1} = ϵ_{2} = ϵ_{3} = 1$ . For an increasing index set $I \subseteq {0, 1, 2, 3}$ , let $I^{c}$ be its increasing complement, let $s (I, I^{c})$ be the sign of the permutation $(I, I^{c})$ to $(0, 1, 2, 3)$ , and set $ϵ_{I} = \prod_{i \in I} ϵ_{i}$ . Then $⋆_{g} e^{I} = ϵ_{I} s (I, I^{c}) e^{I^{c}}$ ; in particular $⋆_{g} e^{01} = - e^{23}$ and $⋆_{g} e^{23} = e^{01}$ . For two-forms $(⋆_{g} F)_{a b} = \frac{1}{2} ({v o l}_{g})_{a b c d} F^{c d}$ and $⋆_{g}^{2} = - id$ . (Lorentzian exterior calculus and Hodge star)

[F3]

The musical isomorphism is $v^{♭} = g (v, \cdot)$ with inverse $♯$ , and $g^{- 1} (α, β) = g (♯ α, ♯ β)$ ; hence $g^{- 1} (v^{♭}, w^{♭}) = g (v, w)$ , and $g^{- 1}$ raises the indices in $F^{a b}$ . (The inverse metric, musical isomorphisms and index raising)

[F4]

Wedges of two-forms commute, and the wedge product is associative and graded-commutative. (Differential forms form a graded commutative algebra)

[A1]

The smooth geometric background inherits ${A C}_{ω}$ from The Axiom of Countable Choice ( ${A C}_{ω}$ ). The finite component argument uses no full Axiom of Choice.

Proof

1.1givenF2algebra

Antisymmetry of $F$ gives $E_{a} n^{a} = c F_{a b} n^{a} n^{b} = 0$ . The Hodge dual is also a two-form, so $B_{a} n^{a} = - (⋆_{g} F)_{a b} n^{a} n^{b} = 0$ . Thus both covectors restrict to the positive rest covector space.

2.1step 1.1F1F2F3A1algebra

Fix a point and use [F1], on the inherited background [A1], to choose a positively oriented orthonormal coframe $(e^{0}, e^{1}, e^{2}, e^{3})$ with $n^{♭} = - e^{0}$ . The definition gives $F_{0 i} = - E_{i} / c$ . For $i = 1, 2, 3$ , the component formula in [F2] gives $(⋆_{g} F)_{10} = - F_{23}$ , $(⋆_{g} F)_{20} = - F_{31}$ , and $(⋆_{g} F)_{30} = - F_{12}$ ; therefore $B_{1} = F_{23}$ , $B_{2} = F_{31}$ , and $B_{3} = F_{12}$ , equivalently $F_{i j} = ϵ_{i j k} B_{k}$ . The basis formula in [F2] gives $⋆_{g} (- e^{0} \land B) = B_{1} e^{2} \land e^{3} + B_{2} e^{3} \land e^{1} + B_{3} e^{1} \land e^{2}$ . Together with $n^{♭} \land E / c = - \sum_{i} (E_{i} / c) e^{0} \land e^{i}$ , this matches all six independent components of $F$ and proves $F = n^{♭} \land E / c + ⋆_{g} (n^{♭} \land B)$ .

3.1step 2.1F1F3algebra

In this coframe, $F^{0 i} = - F_{0 i} = E_{i} / c$ and $F^{i j} = F_{i j}$ . Antisymmetry counts each independent pair twice, so $F_{a b} F^{a b} = 2 \sum_{i < j} F_{i j}^{2} + 2 \sum_{i} F_{0 i} F^{0 i} = 2 \sum_{i} B_{i}^{2} - \frac{2}{c^{2}} \sum_{i} E_{i}^{2}$ . The spatial coframe induces the Euclidean inner product $g^{- 1}$ on the rest covectors, so these sums are $⟨ B, B ⟩_{n}$ and $⟨ E, E ⟩_{n}$ . This proves the first invariant.

4.1step 2.1F1F4algebra∎

Write $F = - \sum_{i} (E_{i} / c) e^{0} \land e^{i} + B_{1} e^{23} + B_{2} e^{31} + B_{3} e^{12}$ . The pure electric and pure magnetic terms wedge to zero. By [F4], each matching electric/magnetic pair occurs twice and contributes $- 2 E_{i} B_{i} {v o l}_{g} / c$ . Therefore $F \land F = - (2 / c) ⟨ E, B ⟩_{n} {v o l}_{g}$ . Both invariant identities are tensorial, so the local calculation holds at every point of $U$ .

Observer electric/magnetic reconstruction and invariants

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources