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Transverse far field with a uniform remainder

Physical scope

Uniform far-field limits and retarded-time Poynting flux for smooth conserved volume sources supported in one fixed spatial ball at every time. The source may have an arbitrary smooth time history; the estimates use only stated derivative suprema on bounded retarded-time intervals. The result asserts a conditional prediction of the selected classical vacuum Maxwell model, not an empirical measurement, point-source field, self-force or radiation-reaction law.

Statement

Assume Countable Choice. Use the SI laboratory chart and source conventions: $ρ : R^{3} \times R \to R$ has units $C m^{- 3}$ and $J : R^{3} \times R \to R^{3}$ has units $A m^{- 2}$ . Let $a, c, μ_{0} > 0$ , $ε_{0} = (μ_{0} c^{2})^{- 1}$ , and let $ρ, J$ be jointly $C^{\infty}$ , vanish for $∣ y ∣ > a$ , and satisfy $\partial_{t} ρ + {div}_{y} J = 0$ . Let $K = [u_{-}, u_{+}] \subset R$ be a nondegenerate compact interval, $u_{-} < u_{+}$ , and put $K^{♯} = [u_{-} - 2 a / c, u_{+} + 2 a / c]$ . For $j = 0, 1, 2$ , define

$M_{ρ, j} : = \sup_{{\overline{B}}_{a} \times K^{♯}} ∣ \partial_{t}^{j} ρ ∣, M_{J, j} : = \sup_{{\overline{B}}_{a} \times K^{♯}} ∣ \partial_{t}^{j} J ∣ .$

For $n \in S^{2}$ and $u \in K$ , define the retarded current moment and its derivative by

$I (n, u) : = \int_{B_{a}} J (y, u + n \cdot y / c) d y, V (n, u) : = \partial_{u} I (n, u) = \int_{B_{a}} J_{t} (y, u + n \cdot y / c) d y .$

Let $E, B$ be the selected retarded fields of the preceding theorem, set $S = E \times B / μ_{0}$ , and define

$Γ_{E} : = \frac{μ_{0} λ_{3} (B_{a})}{4 π} (4 c^{2} M_{ρ, 0} + 13 c a M_{ρ, 1} + a^{2} M_{ρ, 2} + 3 a M_{J, 1} + \frac{a^{2}}{c} M_{J, 2}),$

$Γ_{B} : = \frac{μ_{0} λ_{3} (B_{a})}{4 π} (4 M_{J, 0} + \frac{13 a}{c} M_{J, 1} + \frac{a^{2}}{c^{2}} M_{J, 2}) .$

Then for every $n \in S^{2}$ , $u \in K$ and $r \geq 2 a$ , with $E_{l e a d} (n, u, r) : = \frac{μ_{0}}{4 π r} n \times (n \times V (n, u))$ ,

$∣ E (r n, u + r / c) - E_{l e a d} (n, u, r) ∣ \leq \frac{Γ_{E}}{r^{2}}, ∣ B (r n, u + r / c) - \frac{n \times E_{l e a d} (n, u, r)}{c} ∣ \leq \frac{Γ_{B}}{r^{2}} .$

Consequently the radiation data of the preceding definition exist uniformly on $S^{2} \times K$ and equal

$E_{r a d} (n, u) = \frac{μ_{0}}{4 π} n \times (n \times V (n, u)), B_{r a d} (n, u) = \frac{n \times E_{r a d} (n, u)}{c} .$

For each fixed retarded time $u$ , the outward retarded-time flux has the exact limit

$\lim_{r \to \infty} \int_{S^{2}} r^{2} n \cdot S (r n, u + r / c) d Ω (n) = \frac{μ_{0}}{16 π^{2} c} \int_{S^{2}} ∣ n \times V (n, u) ∣^{2} d Ω (n) .$

Statement references

Radiation data at fixed retarded time

Retarded Maxwell fields for spatially compact smooth histories

Vacuum electromagnetic energy density and energy flux (the Poynting vector)

Framework-local SI units, the vacuum constants and the exact/measured status of their values

Facts & Assumptions

Given: Countable Choice; the constants and source hypotheses in the Statement; the retarded potentials of the preceding theorem; and the fixed compact retarded-time interval $K$ with its enlargement $K^{♯}$ .

[A1]

Countable Choice is the measure convention used by the source integrals, divergence theorem and surface measure. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The source maps are smooth volume density and current density in the fixed laboratory frame; $C^{\infty}$ means all iterated coordinate derivatives exist and are continuous, and their spatial support is in ${\overline{B}}_{a}$ at every time. (Charge parameter, frame charge and current densities, volume charge and flux, and the four-current, Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields, $C^{k}$ Euclidean maps and diffeomorphisms)

[F2]

For these sources the retarded potentials are smooth, satisfy the wave equations and Lorenz gauge, and generate fields satisfying the smooth vacuum Maxwell system. (Retarded Maxwell fields for spatially compact smooth histories)

[F3]

The field and energy-flux definitions are $E = - \nabla φ - \partial_{t} A$ , $B = curl A$ , and $S = E \times B / μ_{0}$ ; $\nabla, div, curl$ are the spatial coordinate operators at fixed time; the constants obey $ε_{0} = (μ_{0} c^{2})^{- 1}$ . (The electric and magnetic fields as fixed-time spatial vector fields, Vacuum electromagnetic energy density and energy flux (the Poynting vector), The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Divergence and curl of a $C^{1}$ vector field, Framework-local SI units, the vacuum constants and the exact/measured status of their values)

[F4]

For a continuous integrand and its continuous time derivative on a neighborhood of a fixed compact set of finite measure, differentiation may be passed under the integral. (Differentiation under the integral sign over a compact measurable set and over a compact $C^{1}$ surface)

[F5]

The total-derivative chain rule and algebra of derivatives apply to $R$ , $1 / R$ and the retarded source compositions. The cross product is bilinear and alternating, $n \cdot (n \times w) = 0$ , $n \times (n \times w) = n (n \cdot w) - w$ , and $∣ n \times w ∣^{2} = ∣ n ∣^{2} ∣ w ∣^{2} - (n \cdot w)^{2}$ ; the Euclidean norm satisfies the triangle and Cauchy–Schwarz inequalities. (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The cross product in $R^{3}$ , The cross product is bilinear, alternating, and orthogonal to both factors, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, The squared cross-product norm is the Gram determinant of two vectors)

[F6]

The mean value theorem bounds a time-shift difference by a derivative supremum; the divergence theorem applies to a $C^{1}$ vector field on a bounded $C^{1}$ ball. (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ , Divergence on a bounded C1 Euclidean domain)

[F7]

On the finite-measure ball, continuous integrands are integrable; the Lebesgue integral is linear and satisfies the integral triangle inequality; the nonnegative integral is monotone. (Integrable real and complex functions, and their integrals, The Lebesgue integral is linear on $L^{1} (μ)$ , The modulus of an integral is bounded by the integral of the modulus, Monotonicity and nonnegative homogeneity of the nonnegative integral, Euclidean balls have positive finite Lebesgue measure)

[F8]

The polar surface measure on $S^{2}$ is the finite surface measure used in the surface-integral definition, and $\int_{S^{2}} 1 d Ω = 4 π$ . (Surface integration on compact C1 hypersurfaces, Zonal and quadratic sphere moments for the polar surface measure on $S^{2}$ (d))

[F9]

Continuous source derivatives on the compact product ${\overline{B}}_{a} \times K^{♯}$ are bounded and uniformly continuous. (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous)

[F10]

Radiation data, when they exist, are the uniform limits of $r E (r n, u + r / c)$ and $r B (r n, u + r / c)$ on compact retarded-time intervals. (Radiation data at fixed retarded time)

[F11]

A closed bounded subset of Euclidean $R^{3}$ is compact; the Borel sigma-algebra contains open sets and their complements, and under Countable Choice every Borel subset is Lebesgue measurable, in particular every closed set. For a measure $μ$ and measurable $A \subseteq B$ , one has $μ (A) \leq μ (B)$ . (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line (2), Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable, Measures are monotone)

Proof

technique · direct; differentiate the retarded potentials, use conservation to remove the longitudinal scalar-potential term, estimate each geometric and time-shift remainder uniformly, and pass the resulting uniform flux limit through the finite sphere integral

1.1givenF5algebraA1

Geometry of the observation point. Fix $n \in S^{2}$ , $u \in K$ , $r \geq 2 a$ , and $y \in B_{a}$ . Put $σ = n \cdot y$ , $R = ∣ r n - y ∣$ , $N = (r n - y) / R$ , $v = u + σ / c$ , and $δ = (r - σ - R) / c$ , so $u + r / c - R / c = v + δ$ . Since $r - σ \geq r - a \geq r / 2$ , the identity $R^{2} = (r - σ)^{2} + ∣ y ∣^{2} - σ^{2}$ gives $∣ R - (r - σ) ∣ = (∣ y ∣^{2} - σ^{2}) / (R + r - σ) \leq a^{2} / r$ ; hence $∣ δ ∣ \leq a^{2} / (r c) \leq a / (2 c)$ . Also $∣ r - R ∣ \leq ∣ σ ∣ + a^{2} / r \leq 3 a / 2$ , so $R \geq r / 2$ , $∣ N - n ∣ \leq ∣ (r - R) n - y ∣ / R \leq 5 a / r$ , $∣ R^{- 1} - r^{- 1} ∣ = ∣ r - R ∣ / (r R) \leq 3 a / r^{2}$ , and $∣ N / R - n / r ∣ \leq ∣ N - n ∣ / R + ∣ R^{- 1} - r^{- 1} ∣ \leq 13 a / r^{2}$ . Thus all retarded source arguments and the intervals between them lie in $K^{♯}$ .

1.2F2F3F4F5F7F11algebraA1

Retarded-field integrals. Put $C_{a} : = {\overline{B}}_{a} = {y : ∣ y ∣ \leq a}$ . If $∣ z ∣ > a$ , the ball of radius $(∣ z ∣ - a) / 2$ around $z$ is disjoint from $C_{a}$ : the triangle inequality gives $∣ y ∣ \geq ∣ z ∣ - ∣ y - z ∣ > a$ on that ball. Thus $C_{a}$ is closed. Also $C_{a} \subset B_{2 a}$ since $a > 0$ , so it is bounded and hence compact by [F11]. A closed set is Borel, because its complement is open and the Borel sigma-algebra is closed under complements; under [A1], [F11] therefore makes $C_{a}$ Lebesgue measurable. The open ball $B_{2 a}$ is measurable and has finite measure by [F7]. Applying measure monotonicity [F11] to $C_{a} \subset B_{2 a}$ gives $λ_{3} (C_{a}) \leq λ_{3} (B_{2 a}) < \infty$ . These are exactly the compactness, measurability and finite-measure hypotheses needed for the set form of [F4]. Since $ρ, J$ are jointly smooth and vanish for $∣ y ∣ > a$ , their values and their time derivatives vanish on $\partial B_{a}$ ; hence the potential integrals and the integrals obtained by one $x$ - or $t$ -derivative have the same values over $B_{a}$ and ${\overline{B}}_{a}$ . Fix an observation event with $∣ x ∣ > a$ and a product neighborhood with compact closure in the exterior region $∣ x ∣ > a$ . For each observation coordinate $x_{i}$ and $t$ , hold the other coordinates fixed and apply the set form of [F4] on ${\overline{B}}_{a}$ with that coordinate as parameter. The retarded integrands and their parameter derivatives are continuous on a neighborhood of ${\overline{B}}_{a}$ times the chosen parameter interval, since $R = ∣ x - y ∣ > 0$ there. Apply this argument to $ρ$ and separately to each component of $J$ , for each spatial-coordinate and time derivative needed in $\nabla φ$ , $\partial_{t} A$ and $curl A$ . The chain rule and algebra of derivatives give $\nabla_{x} R = N$ , $\nabla_{x} (t - R / c) = - N / c$ , and $\nabla_{x} (R^{- 1}) = - N / R^{2}$ . Substitution in $E = - \nabla φ - \partial_{t} A$ and $B = curl A$ therefore gives, at $x = r n$ , $t = u + r / c$ , $E = \frac{μ_{0}}{4 π} \int_{B_{a}} (\frac{c^{2} N ρ_{r e t}}{R^{2}} + \frac{c N ρ_{t, r e t}}{R} - \frac{J_{t, r e t}}{R}) d y, B = - \frac{μ_{0}}{4 π} \int_{B_{a}} N \times (\frac{J_{r e t}}{R^{2}} + \frac{J_{t, r e t}}{c R}) d y,$ where the subscript “ret” means evaluation at $v + δ$ . Here $R \geq r - a > 0$ for $r \geq 2 a$ .

1.3givenF1F6F7algebraA1

Continuity identity for the shifted source. Fix $n, u$ and choose $R_{0} > a$ . The $C^{1}$ vector field $y \mapsto J (y, u + n \cdot y / c)$ is zero near $\partial B_{R_{0}}$ . Its divergence is ${div}_{y} J (y, v) + (n / c) \cdot J_{t} (y, v) = - ρ_{t} (y, v) + (n / c) \cdot J_{t} (y, v)$ by the chain rule and continuity. The divergence theorem gives zero for its integral over $B_{R_{0}}$ , and the integrand is zero outside $B_{a}$ ; therefore $\int_{B_{a}} ρ_{t} (y, v) d y = (1 / c) n \cdot \int_{B_{a}} J_{t} (y, v) d y = (1 / c) n \cdot V (n, u)$ .

2.1givenF1F4F7step 1.2algebraA1

Differentiation of the retarded current moment. Choose a bounded open interval $U$ containing $K$ and a compact interval $K_{0}$ whose interior contains $U$ . Since $J$ and $J_{t}$ vanish on $\partial B_{a}$ by the smooth spatial-support hypothesis, the integral over $B_{a}$ agrees with that over the compact measurable finite-measure set ${\overline{B}}_{a}$ established in step 1.2. For fixed $n \in S^{2}$ , the map $(y, u) \mapsto J (y, u + n \cdot y / c)$ and its $u$ derivative $J_{t} (y, u + n \cdot y / c)$ are continuous on a neighborhood of ${\overline{B}}_{a} \times K_{0}$ . Apply the set form of [F4] separately to each of the three vector components with integration set ${\overline{B}}_{a}$ and parameter $u$ . Thus $\partial_{u} I (n, u) = \int_{B_{a}} J_{t} (y, u + n \cdot y / c) d y$ for every $u \in U$ , so the definition of $V$ in the Statement is well-defined on a neighborhood of $K$ .

2.2step 1.1step 1.2step 1.3F5F6F7algebraA1

Electric leading field and its remainder. Put $F_{l e a d} : = r^{- 1} (c n ρ_{t} (y, v) - J_{t} (y, v))$ inside the integrand of step 1.2. By step 1.3, its integral is $r^{- 1} (n (n \cdot V) - V) = r^{- 1} n \times (n \times V)$ . For the first term of the exact integrand, $c^{2} ∣ N ρ_{r e t} / R^{2} ∣ \leq 4 c^{2} M_{ρ, 0} / r^{2}$ . For its second term, split $\frac{c N ρ_{t, r e t}}{R} - \frac{c n ρ_{t} (y, v)}{r} = c (\frac{N}{R} - \frac{n}{r}) ρ_{t, r e t} + \frac{c n}{r} (ρ_{t, r e t} - ρ_{t} (y, v));$ the first summand is bounded by $13 c a M_{ρ, 1} / r^{2}$ , while the scalar mean value theorem and step 1.1 bound the second by $a^{2} M_{ρ, 2} / r^{2}$ . For the third term, set $Δ : = J_{t, r e t} - J_{t} (y, v)$ . If $Δ \neq 0$ , put $e = Δ / ∣ Δ ∣$ and apply the scalar mean value theorem to $s \mapsto e \cdot J_{t} (y, s)$ between $v$ and $v + δ$ ; Cauchy–Schwarz then gives $∣ Δ ∣ = ∣ e \cdot Δ ∣ \leq M_{J, 2} ∣ δ ∣$ . The same bound is trivial for $Δ = 0$ . Hence $∣ \frac{J_{t, r e t}}{R} - \frac{J_{t} (y, v)}{r} ∣ \leq ∣ \frac{1}{R} - \frac{1}{r} ∣ M_{J, 1} + \frac{1}{r} M_{J, 2} ∣ δ ∣ \leq \frac{3 a M_{J, 1} + a^{2} M_{J, 2} / c}{r^{2}} .$ Integrating these bounds with [F7] and multiplying by $μ_{0} / (4 π)$ proves the stated $Γ_{E} / r^{2}$ estimate.

3.1step 1.1step 1.2step 2.2F5F6F7algebraA1

Magnetic leading field and its remainder. By $n \times (n \times V) = n (n \cdot V) - V$ , one has $n \times E_{l e a d} / c = - μ_{0} (4 π c r)^{- 1} n \times V$ , which is the leading part of the magnetic integral in step 1.2. The $J_{r e t} / R^{2}$ term has norm at most $4 M_{J, 0} / r^{2}$ . For the $J_{t}$ term, split its coefficient as $N / (c R) - n / (c r)$ plus the time shift $n / (c r)$ . The geometric coefficient is bounded by $13 a / (c r^{2})$ . For the time shift, the vector mean value argument of step 2.2 gives $∣ J_{t, r e t} - J_{t} (y, v) ∣ \leq M_{J, 2} ∣ δ ∣$ , so its contribution is at most $a^{2} M_{J, 2} / (c^{2} r^{2})$ . The cross product with a unit vector does not increase norm by [F5], so [F7] gives the stated $Γ_{B} / r^{2}$ estimate.

4.1step 2.2step 3.1F5F10algebraA1

Radiation-data limits. Multiplying the electric estimate of step 2.2 by $r$ gives a uniform error at most $Γ_{E} / r$ on $S^{2} \times K$ , which tends to zero; thus the uniform limit in [F10] is $E_{r a d} = μ_{0} n \times (n \times V) / (4 π)$ . Multiplying the magnetic estimate of step 3.1 by $r$ likewise gives the uniform limit $B_{r a d} = n \times E_{r a d} / c$ . Since $n \cdot (n \times (n \times V)) = 0$ , the leading electric field is transverse.

5.1step 2.2step 3.1step 4.1F5F7F8F9algebraA1∎

Retarded-time flux. The map $(n, u) \mapsto V (n, u)$ is continuous on $S^{2} \times K$ : its integrand is continuous on the compact product ${\overline{B}}_{a} \times S^{2} \times K^{♯}$ , hence uniformly continuous by [F9], and the integral difference is bounded by $λ_{3} (B_{a})$ times the uniform integrand difference, using linearity and the integral triangle inequality [F7]. It is also bounded there by $λ_{3} (B_{a}) M_{J, 1}$ . The field estimates of steps 2.2 and 3.1 yield $E = E_{r a d} / r + e$ and $B = B_{r a d} / r + b$ , where $∣ e ∣ \leq Γ_{E} / r^{2}$ and $∣ b ∣ \leq Γ_{B} / r^{2}$ uniformly on $S^{2} \times K$ . Put $C_{E} : = μ_{0} λ_{3} (B_{a}) M_{J, 1} / (4 π)$ and $C_{B} : = C_{E} / c$ ; then $∣ E_{r a d} ∣ \leq C_{E}$ and $∣ B_{r a d} ∣ \leq C_{B}$ . Bilinearity and the cross-product norm bound [F5] give $∣ r^{2} n \cdot S (r n, u + r / c) - \frac{1}{μ_{0}} n \cdot (E_{r a d} \times B_{r a d}) ∣ \leq \frac{C_{E} Γ_{B} + C_{B} Γ_{E}}{μ_{0} r} + \frac{Γ_{E} Γ_{B}}{μ_{0} r^{2}},$ which tends to zero uniformly for $r \geq 2 a$ . Bilinearity of the cross product gives $r^{2} n \cdot S (r n, u + r / c) = \frac{1}{μ_{0}} n \cdot (E_{r a d} \times B_{r a d}) + R (n, u, r) = \frac{∣ E_{r a d} ∣^{2}}{μ_{0} c} + R (n, u, r),$ uniformly, where $∣ R (n, u, r) ∣$ is bounded by the preceding display. Since $B_{r a d} = n \times E_{r a d} / c$ and $n \cdot E_{r a d} = 0$ , the determinant formula for the cross product gives $n \cdot (E_{r a d} \times (n \times E_{r a d})) = \det [E_{r a d}, n \times E_{r a d}, n] = \det [n, E_{r a d}, n \times E_{r a d}] = ∣ n \times E_{r a d} ∣^{2} = ∣ E_{r a d} ∣^{2},$ where the middle equality is a cyclic permutation and the last follows from [F5]. The sphere has finite area $4 π$ by [F8], so integrating the explicit uniform remainder makes it tend to zero. Finally $∣ E_{r a d} ∣^{2} = μ_{0}^{2} ∣ n \times V ∣^{2} / (16 π^{2})$ by [F5], which gives the asserted flux limit at fixed retarded time.

Transverse far field with a uniform remainder

Statement

Statement references

Facts & Assumptions

Proof

Depends on

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