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Observer Maxwell equations and conserved source charge — Examples

1 · Prerequisites

2 · Summary

This worked pointwise calculation evaluates the electromagnetic stress tensor for pure electric and pure magnetic observer fields. The component formulas are conditional on the current draft SI stress-component lemma, whose proof closure remains unresolved; the example reports no measurement and asserts no global field solution.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

Pure electric and magnetic observer stress

Physical scope

Pointwise algebraic computation of the pure-electric and pure-magnetic observer components of the adopted SI electromagnetic stress tensor, conditional on the planned stress-component lemma. It is not a global field solution or measured result.

Example

Let $p$ be an event in an open neighborhood $U$ on which a smooth future-directed unit timelike observer field $n$ is defined. At $p$ choose a positively oriented orthonormal tetrad with time vector $n (p)$ and signature $(-, +, +, +)$ . Let $E_{0} \in R$ have units $V m^{- 1}$ and $B_{0} \in R$ have units tesla. For the pure-electric observer field $E = (E_{0}, 0, 0)$ , $B = 0$ , the SI stress components of [F1], with the stated Countable Choice background premise [A2], are

$T^{00} = \frac{ϵ_{0} E_{0}^{2}}{2}, T^{0 i} = 0, (T^{i j}) = diag ⁣ (- \frac{ϵ_{0} E_{0}^{2}}{2}, \frac{ϵ_{0} E_{0}^{2}}{2}, \frac{ϵ_{0} E_{0}^{2}}{2}) .$

For the pure-magnetic observer field $E = 0$ , $B = (0, 0, B_{0})$ , they are

$T^{00} = \frac{B_{0}^{2}}{2 μ_{0}}, T^{0 i} = 0, (T^{i j}) = diag ⁣ (\frac{B_{0}^{2}}{2 μ_{0}}, \frac{B_{0}^{2}}{2 μ_{0}}, - \frac{B_{0}^{2}}{2 μ_{0}}) .$

In each case $T^{a}_{a} = 0$ . The spatial stress is the matrix $(T^{i j})$ in the chosen observer frame; it is called isotropic when $T^{i j} = q δ_{i j}$ for some real $q$ , and anisotropic otherwise. The electric case is anisotropic exactly when $E_{0} \neq 0$ , and the magnetic case exactly when $B_{0} \neq 0$ . At zero amplitude the corresponding tensor vanishes and its spatial stress is isotropic; no perfect-fluid identification is made.

Facts & Assumptions

Given: The event and smooth observer field stated in the Example, its oriented orthonormal tetrad at $p$ , real numbers $E_{0}, B_{0}$ , the two pointwise field-value choices in the Example, and Countable Choice ${A C}_{ω}$ . The fields and tensor are smooth on the surrounding model domain, although only their values at the chosen event enter this computation.

[F1]

Draft mathematical supplier Maxwell stress tensor in an observer frame: its Statement gives, for real observer fields $E_{i}$ and $B_{i}$ , the adopted SI tensor satisfies $T^{00} = (ϵ_{0} ∣ E ∣^{2} + μ_{0}^{- 1} ∣ B ∣^{2}) / 2$ , $T^{0 i} = (E \times B)_{i} / (μ_{0} c)$ , and $T^{i j} = - ϵ_{0} (E_{i} E_{j} - δ_{i j} ∣ E ∣^{2} / 2) - μ_{0}^{- 1} (B_{i} B_{j} - δ_{i j} ∣ B ∣^{2} / 2)$ . Here $T^{a b}$ is a contravariant tensor with units $J m^{- 3}$ , $c, μ_{0} > 0$ , $ϵ_{0} = (μ_{0} c^{2})^{- 1} > 0$ , and the observer components use $F_{0 i} = - E_{i} / c$ , $F_{i j} = ε_{i j k} B_{k}$ . The required local-potential and circle-gluing carrier statements are present and their interfaces match; those items remain drafts, and the circle-gluing proof now has its typed prerequisites authored on emg-mg02. This formula remains a draft supplier use; independent proof acceptance is not asserted. Its dependency closure includes ${A C}_{ω}$ .

[A2]

The Axiom of Countable Choice The Axiom of Countable Choice ( ${A C}_{ω}$ ) ( ${A C}_{ω}$ ): every sequence of nonempty sets has a choice function. It is carried by the planned mathematical supplier closure; the exact inherited use is selecting a countable subatlas in the tangent/tensor-bundle construction documented in the geometry supplier record. The component arithmetic below uses no stronger choice principle.

Verification

1.1givenA2F1algebra

For $E = (E_{0}, 0, 0)$ and $B = 0$ , [F1] gives $T^{00} = ϵ_{0} E_{0}^{2} / 2$ and $T^{0 i} = 0$ . The electric quadratic matrix $E_{i} E_{j} - δ_{i j} ∣ E ∣^{2} / 2$ is $diag (E_{0}^{2} / 2, - E_{0}^{2} / 2, - E_{0}^{2} / 2)$ , so the minus sign in the spatial formula of [F1] gives the displayed spatial diagonal. Contracting with $g_{a b} = diag (- 1, 1, 1, 1)$ yields $T^{a}_{a} = - ϵ_{0} E_{0}^{2} / 2 - ϵ_{0} E_{0}^{2} / 2 + ϵ_{0} E_{0}^{2} / 2 + ϵ_{0} E_{0}^{2} / 2 = 0$ . The spatial diagonal entries are all equal exactly when $ϵ_{0} E_{0}^{2} / 2 = 0$ , equivalently $E_{0} = 0$ since $ϵ_{0} > 0$ . Then all tensor components vanish; otherwise the spatial stress is anisotropic by the definition in the Example.

2.1givenA2F1algebra∎

For $E = 0$ and $B = (0, 0, B_{0})$ , [F1] gives $T^{00} = B_{0}^{2} / (2 μ_{0})$ and $T^{0 i} = 0$ . The magnetic quadratic matrix $B_{i} B_{j} - δ_{i j} ∣ B ∣^{2} / 2$ is $diag (- B_{0}^{2} / 2, - B_{0}^{2} / 2, B_{0}^{2} / 2)$ , so the spatial minus sign in [F1] gives the displayed magnetic diagonal. Its metric trace is $- B_{0}^{2} / (2 μ_{0}) + B_{0}^{2} / (2 μ_{0}) + B_{0}^{2} / (2 μ_{0}) - B_{0}^{2} / (2 μ_{0}) = 0$ . The spatial diagonal entries are all equal exactly when $B_{0}^{2} / (2 μ_{0}) = 0$ , equivalently $B_{0} = 0$ since $μ_{0} > 0$ . Then all tensor components vanish; otherwise the spatial stress is anisotropic by the definition in the Example. The calculation is pointwise and supplies no global or empirical claim.