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Smooth matter quantities, current and observer observables — Examples

1 · Prerequisites

2 · Summary

Scope

This examples page works out the selected electromagnetic stress in two observer frames. A pure magnetic field has unequal principal stresses, while crossed equal-normalized electric and magnetic fields give a null energy current and saturate the dominant-energy causal-flux bound. These are exact pointwise calculations under the A-page definitions. They are ideal test-background statements, not measurements or full coupled Einstein–Maxwell solutions.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

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5 · Examples, experiments and counterexamples

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Pure magnetic anisotropic stress and null-flux equality

Physical scope

Two exact pointwise electromagnetic-stress calculations in an orthonormal observer frame: a pure magnetic field has anisotropic principal stresses, and crossed equal-normalized electric and magnetic fields saturate the DEC flux bound. These are ideal local/test-background calculations, not measurements or full coupled Einstein–Maxwell solutions.

Example

At a point in a Minkowski orthonormal observer frame, use the SI electromagnetic stress and observer conventions of Smooth matter quantities, current and observer observables.

For a pure magnetic field $E = 0$ , $B = B_{0} e_{3}$ with $B_{0} \neq 0$ , the energy density is $u = B_{0}^{2} / (2 μ_{0})$ , the energy current is $(u, 0, 0, 0)$ , and the spatial principal stresses are $(u, u, - u)$ . The tensor is trace-free, obeys WEC, NEC, SEC and DEC, and is not an isotropic perfect-fluid stress because $u \neq - u$ .

For crossed fields $E = E_{0} e_{1}$ , $B = (E_{0} / c) e_{2}$ with $E_{0} \neq 0$ , the energy density is $u = E_{0}^{2} / (μ_{0} c^{2})$ and the spatial energy current is $u e_{3}$ . The observer energy-current four-vector is therefore null, so the DEC causal-flux inequality is saturated. Both calculations are exact ideal tensor evaluations at an observer event; they are not experimental reports or claims that the fields solve the full Einstein equation on an unchanged Minkowski metric.

Facts & Assumptions

Given: A point with Minkowski metric, a positively oriented orthonormal observer tetrad, $μ_{0} > 0$ , and one of the two stated nonzero field configurations.

[F1]

Smooth matter quantities, current and observer observables: the electromagnetic tensor is $μ_{0}^{- 1} (F^{a c} F^{b}_{c} - \frac{1}{4} g^{a b} F_{c d} F^{c d})$ , and the frame convention is $F_{0 i} = - E_{i} / c$ , $F_{i j} = ϵ_{i j k} B_{k}$ .

[F2]

Observer energy conditions for selected matter plus EM: the electromagnetic stress has observer density $u = (ϵ_{0} ∣ E ∣^{2} + ∣ B ∣^{2} / μ_{0}) / 2$ , energy current $(E \times B) / (μ_{0} c)$ , and satisfies the four stated energy conditions with causal DEC current.

[A1]

The first configuration has $E = 0$ , $B = B_{0} e_{3}$ , $B_{0} \neq 0$ ; the second has $E = E_{0} e_{1}$ , $B = (E_{0} / c) e_{2}$ , $E_{0} \neq 0$ . These are pointwise ideal field values, not reported measurements.

Verification

1.1F1F2A1algebra

In the first configuration the tensor formula [F1] gives $T_{E M}^{00} = B_{0}^{2} / (2 μ_{0}) = u$ , $T_{E M}^{11} = T_{E M}^{22} = u$ , $T_{E M}^{33} = - u$ , and zero mixed components. Thus its trace is $- u + u + u - u = 0$ and its spatial principal stresses are unequal. To exclude every perfect-fluid representation, suppose $T^{a b} = (ϵ + p) U^{a} U^{b} / c^{2} + p η^{a b}$ with $η (U, U) = - c^{2}$ . Then $U^{0} \neq 0$ and $0 = T^{0 i} = (ϵ + p) U^{0} U^{i} / c^{2}$ . If $ϵ + p \neq 0$ , all $U^{i}$ vanish and $T^{i j} = p δ^{i j}$ , contradicting $(u, u, - u)$ with $u > 0$ . If $ϵ + p = 0$ , the same isotropic spatial matrix follows directly from $T = p η$ and gives the same contradiction. Hence no such representation exists; [F2] supplies WEC, NEC, SEC and DEC for this electromagnetic tensor.

2.1F1F2A1step 1.1algebra∎

In the crossed-field configuration, $∣ E ∣^{2} = E_{0}^{2}$ and $∣ B ∣^{2} = E_{0}^{2} / c^{2}$ , so [F2] gives $u = E_{0}^{2} / (μ_{0} c^{2})$ . Also $E \times B = (E_{0}^{2} / c) e_{3}$ , hence $S_{n} = (E \times B) / (μ_{0} c) = u e_{3}$ . The observer energy-current four-vector $(u, S_{n})$ is nonzero null, so its DEC norm bound is attained.