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Transverse fields, helicity and angular momentum — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Nonzero plane waves lie outside the finite-energy angular-momentum domain

Physical scope

Ideal nonzero monochromatic plane waves in source-free classical vacuum electromagnetism on all of R^3, in the SI laboratory chart. The claim concerns only the domain of finite-energy total angular-momentum functionals; it does not reinterpret a plane wave by box normalization or claim a finite-beam approximation.

Statement refuted

Every nonzero monochromatic plane wave in vacuum belongs to the Schwartz Fourier-packet domain on which the all-space angular-momentum functionals are defined.

Facts & Assumptions

Given: The positive dimensional scales κ and E0, the coordinate unit vectors, k=κe3, ω=cκ, e=E0e1, b=(E0/c)e2, and the corresponding real plane wave.

[F1]

The plane-wave encoding is E=Re⁡(eexp⁡(i(k⋅x−ωt))) and B=Re⁡(bexp⁡(i(k⋅x−ωt))); its fields are identically zero exactly when both amplitudes vanish. (Real plane-wave fields encoded by complex amplitudes)

[F2]

For these transverse vacuum modes, the exact period average is ⟨u⟩=ε0∣e∣2/2, and every nonzero amplitude has infinite all-space box-exhaustion energy. (Period-averaged energy density, Poynting vector and momentum flux of a plane wave)

[F3]

With ω=c∣k∣, k⋅e=0, and b=k×e/ω, these plane waves satisfy the smooth source-free Maxwell equations; the plane-wave amplitudes determine the fields, and the pair is nonzero when e≠0. (Dispersion, transversality and the two polarization conditions for plane-wave solutions) The SI conventions give c>0 and ε0>0. (Framework-local SI units, the vacuum constants and the exact/measured status of their values)

[F4]

The angular-momentum functionals in question are defined on the real Schwartz vacuum packet fields of the Coulomb-gauge construction; their integrands are globally integrable on that domain, and every field in the Fourier-packet class has finite total energy. (Coulomb-gauge canonical and kinetic integrals, Classical transverse Fourier amplitudes)

Counterexample

Choose κ>0 in m−1 and E0>0 in V m−1. Set k=κe3, ω=cκ, e=E0e1, and b=(E0/c)e2, where e1=(1,0,0), e2=(0,1,0) and e3=(0,0,1) are the positively oriented coordinate unit vectors. Let E,B be the real monochromatic plane-wave fields with phase k⋅x−ωt.

1.1givenF1F2F3algebra

Since E0>0, e≠0, and direct calculation gives k⋅e=0, k×e/ω=(E0/c)e2=b, and ω=c∣k∣. Therefore [F3] gives a smooth source-free Maxwell solution and [F1] makes it nonzero. By [F2], its period-averaged energy density is ε0E02/2>0 and its total energy over R3 is infinite.

2.1step 1.1F4algebra∎

Every member of the Fourier-packet domain in [F4] has finite energy. Thus the nonzero wave of step 1.1 does not belong to that domain, refuting the universal statement. Componentwise formal cancellations do not make these all-space functionals defined on this field.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

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 R3 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, E0>0 volts per metre, and A0>0 tesla metres, and put f(x):=χ(x/L) and g(x):=χ(x/(2L)). For all (x,t)∈R3×R, define

E(x,t):=E0f(x)e1,A(x,t):=A0g(x)e2−tE0f(x)e1,φ:=0,B:=curl⁡A,ρ:=ε0E0∂1f,J:=μ0−1curl⁡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

∫R3ρ x×A(x,0) d3x=−ε0E0A0L3(∫R3χ(y) d3y)e3≠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.

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 JR=LR+SR+ε0∫QR(div⁡E)(x×A)−ε0∫∂QR(n⋅E)(x×A). (Equality of total angular momentum integrals)

[A1]

Adopt the smooth branch of the vacuum Maxwell postulate: div⁡E=ρ/ε0, div⁡B=0, curl⁡E=−∂tB, and curl⁡B=μ0J+c−2∂tE for smooth fields and continuous sources. (The microscopic vacuum Maxwell equations: smooth and weak branches)

[F2]

The fields are maps E,B:R3×R→R3; the laboratory chart is positively oriented, and the source fields ρ,J are continuous maps to R and R3 with units C m−3 and A m−2. The divergence and curl use their coordinate partial-derivative formulas; for a C2 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 C1 vector field, The divergence of the curl of a C2 field vanishes, Clairaut--Schwarz theorem for continuous second partial derivatives)

[F3]

The SI constants satisfy ε0=(μ0c2)−1, and [E0]=V m−1, [A0]=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=−∇φ−∂tA and B=curl⁡A. (Potential representations, gauge transformations and the Lorenz and Coulomb conditions)

[F5]

There exists a smooth cutoff χ:R3→[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×(0,g,0)=(−x3g,0,x1g) and is bilinear. (The cross product in R3)

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/3,L/3]3 lies where f=1. Its positive volume and monotonicity of the Riemann integral [F7] imply ∫f d3x>0. The maps E,A,B,ρ,J in the Example are smooth, and [F4] gives E=−∂tA and B=curl⁡A.

2.1step 1.1A1F2F3F4algebra

Direct differentiation gives div⁡E=E0∂1f=ρ/ε0 and div⁡B=0, since mixed partial derivatives cancel in the divergence of a curl. Also ∂tB=−curl⁡E, because ∂tA=−E, and ∂tE=0. The definition J=μ0−1curl⁡B gives curl⁡B=μ0J+c−2∂tE. Thus all four equations in [A1] hold, with continuous sources. Moreover ∂tρ=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=ε0div⁡E, so [F1] leaves exactly the correction ∫R3ρ x×A(x,0) d3x.

4.1step 1.1step 3.1F2F3F6F7algebra

At t=0, A=A0ge2 and x×A=A0g(−x3,0,x1) by [F6]. On the support of f and ∂1f, g=1 and ∂1g=0. The FTC and iterated integration in [F7] give ∫x3g∂1f d3x=0 and ∫x1g∂1f d3x=−∫f d3x. Rescaling y=x/L gives ∫f(x) d3x=L3∫χ(y) d3y>0. Hence ∫ρ x×A(x,0) d3x=−ε0E0A0L3(∫χ(y) d3y)e3≠0.

5.1step 1.1step 4.1F1F2F7∎

The field is sourced: if ∂1f vanished identically, the FTC in [F7] would make f constant along every x1-line. Compact support would then make that constant zero, contradicting f=1 near the origin from step 1.1. Hence ρ=ε0E0∂1f 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.

Sources