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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.

Pure electric and magnetic observer stress

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources