Physics theoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Equality of total angular momentum integrals

Physical scope

Exact conditional identities for smooth classical electric fields and the stated vector potential in the fixed inertial SI laboratory chart. The all-space equality is restricted to source-free Schwartz vacuum packets in the specified Coulomb gauge. The finite-cube formula explicitly retains the divergence and boundary terms; no density-wise equality or universal gauge-independent split is claimed.

Statement

(a) For the real Schwartz vacuum fields and the fixed Coulomb-gauge potential defined in the preceding A-page item, let $J_{k i n}$ , $L_{c a n}$ , and $S_{c a n}$ be the three global functionals in that definition. Then

$J_{k i n} = L_{c a n} + S_{c a n} .$

(b) More generally, let $R > 0$ and let $E, A : \overline{Q_{R}} \to R^{3}$ be restrictions of $C^{1}$ maps on a neighborhood of the cube $Q_{R} = [- R, R]^{3}$ , let $B = curl A$ . Interpret every $R^{3}$ -valued volume integral below as the vector of its three ordinary real Riemann integrals, using the componentwise convention in the preceding angular-momentum Definition. The scalar integral over $\partial Q_{R}$ means the sum of the ordinary two-dimensional Riemann integrals over its six faces, with their outward unit normals $n$ . Define $J_{R} : = ε_{0} \int_{Q_{R}} x \times (E \times B) d^{3} x, L_{R} : = ε_{0} \sum_{i = 1}^{3} \int_{Q_{R}} E_{i} (x \times \nabla) A_{i} d^{3} x, S_{R} : = ε_{0} \int_{Q_{R}} E \times A d^{3} x .$ For each component $ℓ \in {1, 2, 3}$ ,

$(J_{R})_{ℓ} = (L_{R})_{ℓ} + (S_{R})_{ℓ} + ε_{0} \int_{Q_{R}} (div E) (x \times A)_{ℓ} d^{3} x - ε_{0} \int_{\partial Q_{R}} (n \cdot E) (x \times A)_{ℓ} d S .$

If the volume integrals converge absolutely in the componentwise convention of the preceding Fourier-field Definition and this boundary term tends to zero as $R \to \infty$ , the corresponding whole-space identity retains the term $ε_{0} \int_{R^{3}} (div E) (x \times A) d^{3} x$ . Thus the source-free Schwartz equality in (a) uses both $div E = 0$ and the vanishing surface limit; it is not an equality of local densities.

Statement references

Coulomb-gauge canonical and kinetic integrals

Classical transverse Fourier amplitudes

The cross product in $R^{3}$

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 $\int_{a}^{b} f = G (b) - G (a)$

Facts & Assumptions

Given: The fields $E, A, B$ and functionals of the statement, the fixed positive orientation and origin, and (for part (b)) a finite cube $Q_{R}$ .

[F1]

The global functionals are defined by $J_{k i n} = ε_{0} \int x \times (E \times B)$ , $L_{c a n} = ε_{0} \sum_{i} \int E_{i} (x \times \nabla) A_{i}$ , and $S_{c a n} = ε_{0} \int E \times A$ ; their packet fields are Schwartz. (Coulomb-gauge canonical and kinetic integrals)

[F2]

For these packets $B = curl A$ and $div E = 0$ ; $E, A, B$ are smooth Schwartz fields. (Classical transverse Fourier amplitudes)

[F3]

The component formula for the right-handed cross product is $(u \times v)_{i} = \sum_{j, k} ϵ_{i j k} u_{j} v_{k}$ , and its products are bilinear; every repeated coordinate index in the proof ranges from $1$ to $3$ and is summed. (The cross product in $R^{3}$ )

[F4]

Continuous functions on a cube and its faces have Riemann integrals; integrals on a product cube equal iterated integrals, and for a $C^{1}$ function on an interval the integral of its derivative is its endpoint increment. (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 $\int_{a}^{b} f = G (b) - G (a)$ )

Proof

1.1F3algebra

For $i \in {1, 2, 3}$ , expand the cross product by [F3]: $(E \times curl A)_{i} = \sum_{j, k, m, n} ϵ_{i j k} ϵ_{k m n} E_{j} \partial_{m} A_{n}$ $= \sum_{j} E_{j} \partial_{i} A_{j} - \sum_{j} E_{j} \partial_{j} A_{i}$ . Indeed, if either index pair repeats, both sides of $\sum_{k} ϵ_{i j k} ϵ_{k m n} = δ_{i m} δ_{j n} - δ_{i n} δ_{j m}$ are zero. For distinct $i, j$ , only the remaining coordinate $k$ contributes; the left side is $+ 1$ when $(m, n) = (i, j)$ , $- 1$ when $(m, n) = (j, i)$ , and zero otherwise, exactly as on the right. Multiplying by $ε_{0} \sum_{m, i} ϵ_{ℓ m i} x_{m}$ and summing over $i, m$ makes the first term exactly $(L_{R})_{ℓ}$ and leaves $(J_{R})_{ℓ} - (L_{R})_{ℓ} = - ε_{0} \sum_{m, i, j} \int_{Q_{R}} ϵ_{ℓ m i} x_{m} E_{j} \partial_{j} A_{i} d^{3} x$ .

2.1step 1.1F3F4algebra

For each fixed pair of the other coordinates, the one-dimensional FTC in [F4] applied to $x_{m} E_{j} A_{i}$ gives $\int_{Q_{R}} x_{m} E_{j} \partial_{j} A_{i} = \int_{\partial_{j} Q_{R}} n_{j} x_{m} E_{j} A_{i} d S - \int_{Q_{R}} (δ_{j m} E_{j} + x_{m} \partial_{j} E_{j}) A_{i} d^{3} x$ ; Riemann-Fubini in [F4] justifies integrating those section identities over the two remaining coordinates and summing faces. Substitution in step 1.1 gives the volume term $ε_{0} \int_{Q_{R}} (div E) (x \times A)_{ℓ}$ , the surface term $- ε_{0} \int_{\partial Q_{R}} (n \cdot E) (x \times A)_{ℓ}$ , and the remaining volume term $ε_{0} \int_{Q_{R}} (E \times A)_{ℓ} = (S_{R})_{ℓ}$ . This is exactly (b).

3.1step 2.1F1F2F4∎

For the Schwartz packets, every volume term in (b) converges absolutely by the estimates in the definition of $J_{k i n}, L_{c a n}, S_{c a n}$ . On each face of $Q_{R}$ , $∣ x ∣ \geq R$ ; choosing Schwartz decay order $N \geq 3$ for both $E$ and $A$ bounds the surface integrand by $C (1 + ∣ x ∣)^{- 2 N + 1}$ , and the total face area is $24 R^{2}$ . Hence the boundary term tends to zero. By [F2] the divergence term vanishes identically. Taking $R \to \infty$ in the finite-cube identity (b) proves (a).

Equality of total angular momentum integrals

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources