ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Charged configurations retain an additional term

Physical scope

A smooth sourced vacuum Maxwell solution in the SI laboratory chart, used only to demonstrate the divergence term in the componentwise angular-momentum identity. The displayed algebraic integrals extend the A-page formulas for this calculation; the example does not extend the A-page functionals' source-free Schwartz domain or make a gauge-invariant claim for general charged fields.

Example

Let $χ$ be the smooth cutoff equal to $1$ on the unit ball and $0$ outside the radius-two ball. Every integral over $R^{3}$ below means integration over any cube containing the compact support of its integrand; the value is independent of the chosen cube. For vector-valued integrands, take the three ordinary real Riemann integrals componentwise, as in the preceding angular- momentum Definition. Fix $L > 0$ metres, $E_{0} > 0$ volts per metre, and $A_{0} > 0$ tesla metres, and put $f (x) : = χ (x / L)$ and $g (x) : = χ (x / (2 L))$ . For all $(x, t) \in R^{3} \times R$ , define

$E (x, t) : = E_{0} f (x) e_{1}, A (x, t) : = A_{0} g (x) e_{2} - t E_{0} f (x) e_{1}, φ : = 0, B : = curl A, ρ : = ε_{0} E_{0} \partial_{1} f, J : = μ_{0}^{- 1} curl B .$

These maps are smooth, compactly supported in space, and satisfy all four smooth-branch Maxwell equations. At $t = 0$ the boundary term in the finite-cube angular-momentum identity vanishes, but the charge correction is

$\int_{R^{3}} ρ x \times A (x, 0) d^{3} x = - ε_{0} E_{0} A_{0} L^{3} (\int_{R^{3}} χ (y) d^{3} y) e_{3} \neq 0.$

Thus the source-free total-integral equality cannot be extended to these sourced fields by deleting the divergence term. This is a model calculation, not a reported measurement.

Example references

Equality of total angular momentum integrals

Coulomb-gauge canonical and kinetic integrals

The microscopic vacuum Maxwell equations: smooth and weak branches

Facts & Assumptions

Given: The positive length and SI amplitude scales, the cutoff $χ$ , the fields and sources defined above, and the adopted smooth Maxwell branch.

[F1]

The source-free packet domain and its functionals are defined in Coulomb-gauge canonical and kinetic integrals. On each cube, the exact component identity is $J_{R} = L_{R} + S_{R} + ε_{0} \int_{Q_{R}} (div E) (x \times A) - ε_{0} \int_{\partial Q_{R}} (n \cdot E) (x \times A)$ . (Equality of total angular momentum integrals)

[A1]

Adopt the smooth branch of the vacuum Maxwell postulate: $div E = ρ / ε_{0}$ , $div B = 0$ , $curl E = - \partial_{t} B$ , and $curl B = μ_{0} J + c^{- 2} \partial_{t} E$ for smooth fields and continuous sources. (The microscopic vacuum Maxwell equations: smooth and weak branches)

[F2]

The fields are maps $E, B : R^{3} \times R \to R^{3}$ ; the laboratory chart is positively oriented, and the source fields $ρ, J$ are continuous maps to $R$ and $R^{3}$ with units $C m^{- 3}$ and $A m^{- 2}$ . The divergence and curl use their coordinate partial-derivative formulas; for a $C^{2}$ vector field, divergence of curl vanishes, and continuous second partial derivatives commute. (The electric and magnetic fields as fixed-time spatial vector fields, Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields, Charge parameter, frame charge and current densities, volume charge and flux, and the four-current, Divergence and curl of a $C^{1}$ vector field, The divergence of the curl of a $C^{2}$ field vanishes, Clairaut--Schwarz theorem for continuous second partial derivatives)

[F3]

The SI constants satisfy $ε_{0} = (μ_{0} c^{2})^{- 1}$ , and $[E_{0}] = V m^{- 1}$ , $[A_{0}] = T m$ , $[c] = m s^{- 1}$ . (Framework-local SI units, the vacuum constants and the exact/measured status of their values)

[F4]

A potential pair satisfies $E = - \nabla φ - \partial_{t} A$ and $B = curl A$ . (Potential representations, gauge transformations and the Lorenz and Coulomb conditions)

[F5]

There exists a smooth cutoff $χ : R^{3} \to [0, 1]$ equal to $1$ on the unit ball and $0$ outside the radius-two ball. (Explicit compactly supported smooth cutoffs)

[F6]

The coordinate cross product satisfies $x \times (0, g, 0) = (- x_{3} g, 0, x_{1} g)$ and is bilinear. (The cross product in $R^{3}$ )

[F7]

Continuous functions on cubes and their faces are Riemann integrable, iterated integrals on product rectangles agree, the integral of a derivative is the endpoint increment, and the integral is monotone under pointwise inequalities. (Axis-parallel rectangles in $R^{m}$ and their volume, The lower and upper Darboux integrals over a nondegenerate rectangle in $R^{m}$ , Every continuous function on a closed nondegenerate rectangle in $R^{m}$ is Riemann integrable, Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ )

Verification

1.1givenF4F5F7construct

By [F5], $f$ and $g$ are smooth and compactly supported; $g = 1$ on the support of $f$ . Scaling shows that $f = 1$ on the ball of radius $L$ , so the cube $[- L / \sqrt{3}, L / \sqrt{3}]^{3}$ lies where $f = 1$ . Its positive volume and monotonicity of the Riemann integral [F7] imply $\int f d^{3} x > 0$ . The maps $E, A, B, ρ, J$ in the Example are smooth, and [F4] gives $E = - \partial_{t} A$ and $B = curl A$ .

2.1step 1.1A1F2F3F4algebra

Direct differentiation gives $div E = E_{0} \partial_{1} f = ρ / ε_{0}$ and $div B = 0$ , since mixed partial derivatives cancel in the divergence of a curl. Also $\partial_{t} B = - curl E$ , because $\partial_{t} A = - E$ , and $\partial_{t} E = 0$ . The definition $J = μ_{0}^{- 1} curl B$ gives $curl B = μ_{0} J + c^{- 2} \partial_{t} E$ . Thus all four equations in [A1] hold, with continuous sources. Moreover $\partial_{t} ρ = 0$ and $div J = 0$ , so these sources satisfy continuity as well.

3.1step 2.1A1F1

At $t = 0$ , all fields have compact support, so the boundary term in [F1] vanishes on sufficiently large cubes. Gauss's equation in [A1] identifies $ρ = ε_{0} div E$ , so [F1] leaves exactly the correction $\int_{R^{3}} ρ x \times A (x, 0) d^{3} x$ .

4.1step 1.1step 3.1F2F3F6F7algebra

At $t = 0$ , $A = A_{0} g e_{2}$ and $x \times A = A_{0} g (- x_{3}, 0, x_{1})$ by [F6]. On the support of $f$ and $\partial_{1} f$ , $g = 1$ and $\partial_{1} g = 0$ . The FTC and iterated integration in [F7] give $\int x_{3} g \partial_{1} f d^{3} x = 0$ and $\int x_{1} g \partial_{1} f d^{3} x = - \int f d^{3} x$ . Rescaling $y = x / L$ gives $\int f (x) d^{3} x = L^{3} \int χ (y) d^{3} y > 0$ . Hence $\int ρ x \times A (x, 0) d^{3} x = - ε_{0} E_{0} A_{0} L^{3} (\int χ (y) d^{3} y) e_{3} \neq 0$ .

5.1step 1.1step 4.1F1F2F7∎

The field is sourced: if $\partial_{1} f$ vanished identically, the FTC in [F7] would make $f$ constant along every $x^{1}$ -line. Compact support would then make that constant zero, contradicting $f = 1$ near the origin from step 1.1. Hence $ρ = ε_{0} E_{0} \partial_{1} f$ is not identically zero, and this example lies outside the source-free packet domain in [F1]. The explicit nonzero correction is the conditional sourced identity in part (b), not a gauge-independent split for charged fields.

Charged configurations retain an additional term

Example

Example references

Facts & Assumptions

Verification

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