Radiation asymptotics and controlled multipole expansion

Fix $(x,t)$. The source integrands vanish off ${\bar{B}}_a$. On this compact ball the retarded time $t-\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}c$ lies in the bounded interval $[t-(\mathrm{\mid }x\mathrm{\mid }+a)\mathrm{/}c,t]$, so smoothness bounds both source functions there by a finite constant $M$. The triangle inequality and [F5] give ${\int }_{B_a}\mathrm{\mid }h(y,t-\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}c)\mathrm{\mid }\mathrm{/}\mathrm{\mid }x-y\mathrm{\mid }dy\le M{\int }_{B(x,\mathrm{\mid }x\mathrm{\mid }+a)}\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}dy<\mathrm{\infty }$ for $h=\rho$ and each component of $J$; the possible point $y=x$ is harmless because $\mathrm{\mid }z{\mathrm{\mid }}^{-1}$ is locally integrable in three dimensions. Thus both retarded integrals are absolutely convergent.

Fix an event $(x_0,t_0)$ and choose a bounded open ball $O$ around it. Its closure is compact by [F3], as is the source ball ${\bar{B}}_a$. On $\bar{O}\times {\bar{B}}_a$, both observation times and all values $t-\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}c$ lie in one bounded interval because $\mathrm{\mid }x-y\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }+a$; take a closed finite interval containing these times. By [F3] choose a smooth compactly supported time cutoff $\chi$ equal to one on a neighborhood of this interval. Then $h_{\rho }(y,s):=\chi (s)\rho (y,s)$ and $h_J(y,s):=\chi (s)J(y,s)$ belong to $C_c^{\mathrm{\infty }}({\mathbb{R}}^3\times \mathbb{R})$ by [F1], since their supports are contained in ${\bar{B}}_a\times \mathrm{supp}\chi$. On $O$, every retarded source value and every observation-time value sees $\chi =1$, so the potentials in the Statement equal ${\mu }_0{c}^{2}K\ast h_{\rho }$ and ${\mu }_{0}K\ast h_J$, and the localized sources equal $\rho$ and $J$. By [F2] these potentials are smooth on $O$ and satisfy $W\phi ={\mu }_0{c}^2\rho =\rho \mathrm{/}{\epsilon }_0$ and $WA={\mu }_{0}J$ there. Since $(x_0,t_0)$ was arbitrary, the potentials are smooth globally and the two wave equations hold everywhere.

On a smaller neighborhood of the same event, change variables $z=x-y$ in the localized convolution: $K\ast h(x,t)=\int h(x-z,t-\mathrm{\mid }z\mathrm{\mid }\mathrm{/}c)\mathrm{/}(4\pi \mathrm{\mid }z\mathrm{\mid })dz$. For a spatial difference quotient, the mean value theorem bounds the quotient of $h(x-z,t-\mathrm{\mid }z\mathrm{\mid }\mathrm{/}c)$ by the supremum of $\mathrm{\mid }{\mathrm{\nabla }}_{y}h\mathrm{\mid }$ on a common compact neighborhood, times $1\mathrm{/}(4\pi \mathrm{\mid }z\mathrm{\mid })$; for a time difference quotient the same bound uses the supremum of $\mathrm{\mid }{\mathrm{\partial }}_{t}h\mathrm{\mid }$. These bounds are integrable on a fixed ball by [F5]. The pointwise limits are the corresponding derivatives of $h$, so [F4] permits dominated convergence and yields ${\mathrm{\partial }}_{x_i}(K\ast h)=K\ast ({\mathrm{\partial }}_{y_i}h)$ and ${\mathrm{\partial }}_t(K\ast h)=K\ast ({\mathrm{\partial }}_{t}h)$ on that neighborhood. For $h_J=\chi J$, ${\mathrm{div}}_y{h}_J=\chi {\mathrm{div}}_{y}J$; for $h_{\rho }=\chi \rho$, ${\mathrm{\partial }}_t{h}_{\rho }={\chi }^{\mathrm{\prime }}\rho +\chi {\mathrm{\partial }}_t\rho$. At every retarded argument sampled on the neighborhood, $\chi =1$ and ${\chi }^{\mathrm{\prime }}=0$ by the cutoff choice in step 2.1. Therefore ${\displaystyle \mathrm{div}A+c^{-2}{\mathrm{\partial }}_t\phi ={\mu }_{0}K\ast ({\mathrm{div}}_{y}J+{\mathrm{\partial }}_t\rho )=0}$ there by the assumed continuity equation. The event was arbitrary, so the Lorenz condition holds everywhere.

The pair is jointly smooth by step 2.1, satisfies the two wave equations there, and is in Lorenz gauge by step 3.1. Applying [F6] on ${\mathbb{R}}^3\times \mathbb{R}$ gives the two sourced Maxwell equations; the other two equations are the automatic identities of the potential representation included in [F6]. The selection postulate [F7] identifies this retarded solution with the model's physical branch and excludes an independent homogeneous addition.

Geometry of the observation point. Fix $n\in S^2$, $u\in K$, $r\ge 2a$, and $y\in B_a$. Put $\sigma =n\cdot y$, $R=\mathrm{\mid }rn-y\mathrm{\mid }$, $N=(rn-y)\mathrm{/}R$, $v=u+\sigma \mathrm{/}c$, and $\delta =(r-\sigma -R)\mathrm{/}c$, so $u+r\mathrm{/}c-R\mathrm{/}c=v+\delta$. Since $r-\sigma \ge r-a\ge r\mathrm{/}2$, the identity $R^2=(r-\sigma {)}^2+\mathrm{\mid }y{\mathrm{\mid }}^2-{\sigma }^2$ gives $\mathrm{\mid }R-(r-\sigma )\mathrm{\mid }=(\mathrm{\mid }y{\mathrm{\mid }}^2-{\sigma }^2)\mathrm{/}(R+r-\sigma )\le a^2\mathrm{/}r$; hence $\mathrm{\mid }\delta \mathrm{\mid }\le a^2\mathrm{/}(rc)\le a\mathrm{/}(2c)$. Also $\mathrm{\mid }r-R\mathrm{\mid }\le \mathrm{\mid }\sigma \mathrm{\mid }+a^2\mathrm{/}r\le 3a\mathrm{/}2$, so $R\ge r\mathrm{/}2$, $\mathrm{\mid }N-n\mathrm{\mid }\le \mathrm{\mid }(r-R)n-y\mathrm{\mid }\mathrm{/}R\le 5a\mathrm{/}r$, $\mathrm{\mid }{R}^{-1}-r^{-1}\mathrm{\mid }=\mathrm{\mid }r-R\mathrm{\mid }\mathrm{/}(rR)\le 3a\mathrm{/}{r}^2$, and $\mathrm{\mid }N\mathrm{/}R-n\mathrm{/}r\mathrm{\mid }\le \mathrm{\mid }N-n\mathrm{\mid }\mathrm{/}R+\mathrm{\mid }{R}^{-1}-r^{-1}\mathrm{\mid }\le 13a\mathrm{/}{r}^2$. Thus all retarded source arguments and the intervals between them lie in $K^{\mathrm{♯}}$.

Retarded-field integrals. Put $C_a:={\bar{B}}_a=\{y:\mathrm{\mid }y\mathrm{\mid }\le a\}$. If $\mathrm{\mid }z\mathrm{\mid }>a$, the ball of radius $(\mathrm{\mid }z\mathrm{\mid }-a)\mathrm{/}2$ around $z$ is disjoint from $C_a$: the triangle inequality gives $\mathrm{\mid }y\mathrm{\mid }\ge \mathrm{\mid }z\mathrm{\mid }-\mathrm{\mid }y-z\mathrm{\mid }>a$ on that ball. Thus $C_a$ is closed. Also $C_a\subset B_{2a}$ 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 $B_{2a}$ is measurable and has finite measure by [F7]. Applying measure monotonicity [F11] to $C_a\subset B_{2a}$ gives ${\lambda }_3(C_a)\le {\lambda }_3(B_{2a})<\mathrm{\infty }$. These are exactly the compactness, measurability and finite-measure hypotheses needed for the set form of [F4]. Since $\rho ,J$ are jointly smooth and vanish for $\mathrm{\mid }y\mathrm{\mid }>a$, their values and their time derivatives vanish on $\mathrm{\partial }{B}_a$; hence the potential integrals and the integrals obtained by one $x$- or $t$-derivative have the same values over $B_a$ and ${\bar{B}}_a$. Fix an observation event with $\mathrm{\mid }x\mathrm{\mid }>a$ and a product neighborhood with compact closure in the exterior region $\mathrm{\mid }x\mathrm{\mid }>a$. For each observation coordinate $x_i$ and $t$, hold the other coordinates fixed and apply the set form of [F4] on ${\bar{B}}_a$ with that coordinate as parameter. The retarded integrands and their parameter derivatives are continuous on a neighborhood of ${\bar{B}}_a$ times the chosen parameter interval, since $R=\mathrm{\mid }x-y\mathrm{\mid }>0$ there. Apply this argument to $\rho$ and separately to each component of $J$, for each spatial-coordinate and time derivative needed in $\mathrm{\nabla }\phi$, ${\mathrm{\partial }}_{t}A$ and $\mathrm{curl}A$. The chain rule and algebra of derivatives give ${\mathrm{\nabla }}_{x}R=N$, ${\mathrm{\nabla }}_x(t-R\mathrm{/}c)=-N\mathrm{/}c$, and ${\mathrm{\nabla }}_x(R^{-1})=-N\mathrm{/}{R}^2$. Substitution in $E=-\mathrm{\nabla }\phi -{\mathrm{\partial }}_{t}A$ and $B=\mathrm{curl}A$ therefore gives, at $x=rn$, $t=u+r\mathrm{/}c$, ${\displaystyle E=\frac{{\mu }_0}{4\pi }{\int }_{B_a}(\frac{c^{2}N{\rho }_{\mathrm{r}\mathrm{e}\mathrm{t}}}{R^2}+\frac{cN{\rho }_{t,\mathrm{r}\mathrm{e}\mathrm{t}}}{R}-\frac{J_{t,\mathrm{r}\mathrm{e}\mathrm{t}}}{R})dy,B=-\frac{{\mu }_0}{4\pi }{\int }_{B_a}N\times (\frac{J_{\mathrm{r}\mathrm{e}\mathrm{t}}}{R^2}+\frac{J_{t,\mathrm{r}\mathrm{e}\mathrm{t}}}{cR})dy,}$ where the subscript “ret” means evaluation at $v+\delta$. Here $R\ge r-a>0$ for $r\ge 2a$.

Continuity identity for the shifted source. Fix $n,u$ and choose $R_0>a$. The $C^1$ vector field $y\mapsto J(y,u+n\cdot y\mathrm{/}c)$ is zero near $\mathrm{\partial }{B}_{R_0}$. Its divergence is ${\mathrm{div}}_{y}J(y,v)+(n\mathrm{/}c)\cdot J_t(y,v)=-{\rho }_t(y,v)+(n\mathrm{/}c)\cdot J_t(y,v)$ 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 ${\int }_{B_a}{\rho }_t(y,v)dy=(1\mathrm{/}c)n\cdot {\int }_{B_a}{J}_t(y,v)dy=(1\mathrm{/}c)n\cdot V(n,u)$.

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 $\mathrm{\partial }{B}_a$ by the smooth spatial-support hypothesis, the integral over $B_a$ agrees with that over the compact measurable finite-measure set ${\bar{B}}_a$ established in step 1.2. For fixed $n\in S^2$, the map $(y,u)\mapsto J(y,u+n\cdot y\mathrm{/}c)$ and its $u$ derivative $J_t(y,u+n\cdot y\mathrm{/}c)$ are continuous on a neighborhood of ${\bar{B}}_a\times K_0$. Apply the set form of [F4] separately to each of the three vector components with integration set ${\bar{B}}_a$ and parameter $u$. Thus ${\mathrm{\partial }}_{u}I(n,u)={\int }_{B_a}{J}_t(y,u+n\cdot y\mathrm{/}c)dy$ for every $u\in U$, so the definition of $V$ in the Statement is well-defined on a neighborhood of $K$.

Electric leading field and its remainder. Put $F_{\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{d}}:=r^{-1}(cn{\rho }_t(y,v)-J_t(y,v))$ 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, $c^2\mathrm{\mid }N{\rho }_{\mathrm{r}\mathrm{e}\mathrm{t}}\mathrm{/}{R}^2\mathrm{\mid }\le 4c^2{M}_{\rho ,0}\mathrm{/}{r}^2$. For its second term, split ${\displaystyle \frac{cN{\rho }_{t,\mathrm{r}\mathrm{e}\mathrm{t}}}{R}-\frac{cn{\rho }_t(y,v)}{r}=c(\frac{N}{R}-\frac{n}{r}){\rho }_{t,\mathrm{r}\mathrm{e}\mathrm{t}}+\frac{cn}{r}({\rho }_{t,\mathrm{r}\mathrm{e}\mathrm{t}}-{\rho }_t(y,v));}$ the first summand is bounded by $13ca{M}_{\rho ,1}\mathrm{/}{r}^2$, while the scalar mean value theorem and step 1.1 bound the second by $a^2{M}_{\rho ,2}\mathrm{/}{r}^2$. For the third term, set $\mathrm{\Delta }:=J_{t,\mathrm{r}\mathrm{e}\mathrm{t}}-J_t(y,v)$. If $\mathrm{\Delta }\mathrm{\ne }0$, put $e=\mathrm{\Delta }\mathrm{/}\mathrm{\mid }\mathrm{\Delta }\mathrm{\mid }$ and apply the scalar mean value theorem to $s\mapsto e\cdot J_t(y,s)$ between $v$ and $v+\delta$; Cauchy–Schwarz then gives $\mathrm{\mid }\mathrm{\Delta }\mathrm{\mid }=\mathrm{\mid }e\cdot \mathrm{\Delta }\mathrm{\mid }\le M_{J,2}\mathrm{\mid }\delta \mathrm{\mid }$. The same bound is trivial for $\mathrm{\Delta }=0$. Hence ${\displaystyle \mid \frac{J_{t,\mathrm{r}\mathrm{e}\mathrm{t}}}{R}-\frac{J_t(y,v)}{r}\mid \le \mid \frac{1}{R}-\frac{1}{r}\mid M_{J,1}+\frac{1}{r}{M}_{J,2}\mathrm{\mid }\delta \mathrm{\mid }\le \frac{3a{M}_{J,1}+a^2{M}_{J,2}\mathrm{/}c}{r^2}\mathrm{.}}$ Integrating these bounds with [F7] and multiplying by ${\mu }_0\mathrm{/}(4\pi )$ proves the stated ${\mathrm{\Gamma }}_E\mathrm{/}{r}^2$ estimate.

Magnetic leading field and its remainder. By $n\times (n\times V)=n(n\cdot V)-V$, one has $n\times E_{\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{d}}\mathrm{/}c=-{\mu }_0(4\pi cr{)}^{-1}n\times V$, which is the leading part of the magnetic integral in step 1.2. The $J_{\mathrm{r}\mathrm{e}\mathrm{t}}\mathrm{/}{R}^2$ term has norm at most $4M_{J,0}\mathrm{/}{r}^2$. For the $J_t$ term, split its coefficient as $N\mathrm{/}(cR)-n\mathrm{/}(cr)$ plus the time shift $n\mathrm{/}(cr)$. The geometric coefficient is bounded by $13a\mathrm{/}(c{r}^2)$. For the time shift, the vector mean value argument of step 2.2 gives $\mathrm{\mid }{J}_{t,\mathrm{r}\mathrm{e}\mathrm{t}}-J_t(y,v)\mathrm{\mid }\le M_{J,2}\mathrm{\mid }\delta \mathrm{\mid }$, so its contribution is at most $a^2{M}_{J,2}\mathrm{/}(c^2{r}^2)$. The cross product with a unit vector does not increase norm by [F5], so [F7] gives the stated ${\mathrm{\Gamma }}_B\mathrm{/}{r}^2$ estimate.

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

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 ${\bar{B}}_a\times S^2\times K^{\mathrm{♯}}$, hence uniformly continuous by [F9], and the integral difference is bounded by ${\lambda }_3(B_a)$ times the uniform integrand difference, using linearity and the integral triangle inequality [F7]. It is also bounded there by ${\lambda }_3(B_a)M_{J,1}$. The field estimates of steps 2.2 and 3.1 yield $E=E_{\mathrm{r}\mathrm{a}\mathrm{d}}\mathrm{/}r+e$ and $B=B_{\mathrm{r}\mathrm{a}\mathrm{d}}\mathrm{/}r+b$, where $\mathrm{\mid }e\mathrm{\mid }\le {\mathrm{\Gamma }}_E\mathrm{/}{r}^2$ and $\mathrm{\mid }b\mathrm{\mid }\le {\mathrm{\Gamma }}_B\mathrm{/}{r}^2$ uniformly on $S^2\times K$. Put $C_E:={\mu }_0{\lambda }_3(B_a)M_{J,1}\mathrm{/}(4\pi )$ and $C_B:=C_E\mathrm{/}c$; then $\mathrm{\mid }{E}_{\mathrm{r}\mathrm{a}\mathrm{d}}\mathrm{\mid }\le C_E$ and $\mathrm{\mid }{B}_{\mathrm{r}\mathrm{a}\mathrm{d}}\mathrm{\mid }\le C_B$. Bilinearity and the cross-product norm bound [F5] give ${\displaystyle \mid r^{2}n\cdot S(rn,u+r\mathrm{/}c)-\frac{1}{{\mu }_0}n\cdot (E_{\mathrm{r}\mathrm{a}\mathrm{d}}\times B_{\mathrm{r}\mathrm{a}\mathrm{d}})\mid \le \frac{C_E{\mathrm{\Gamma }}_B+C_B{\mathrm{\Gamma }}_E}{{\mu }_{0}r}+\frac{{\mathrm{\Gamma }}_E{\mathrm{\Gamma }}_B}{{\mu }_0{r}^2},}$ which tends to zero uniformly for $r\ge 2a$. Bilinearity of the cross product gives ${\displaystyle r^{2}n\cdot S(rn,u+r\mathrm{/}c)=\frac{1}{{\mu }_0}n\cdot (E_{\mathrm{r}\mathrm{a}\mathrm{d}}\times B_{\mathrm{r}\mathrm{a}\mathrm{d}})+R(n,u,r)=\frac{\mathrm{\mid }{E}_{\mathrm{r}\mathrm{a}\mathrm{d}}{\mathrm{\mid }}^2}{{\mu }_{0}c}+R(n,u,r),}$ uniformly, where $\mathrm{\mid }R(n,u,r)\mathrm{\mid }$ is bounded by the preceding display. Since $B_{\mathrm{r}\mathrm{a}\mathrm{d}}=n\times E_{\mathrm{r}\mathrm{a}\mathrm{d}}\mathrm{/}c$ and $n\cdot E_{\mathrm{r}\mathrm{a}\mathrm{d}}=0$, the determinant formula for the cross product gives ${\displaystyle n\cdot (E_{\mathrm{r}\mathrm{a}\mathrm{d}}\times (n\times E_{\mathrm{r}\mathrm{a}\mathrm{d}}))=\det [E_{\mathrm{r}\mathrm{a}\mathrm{d}},n\times E_{\mathrm{r}\mathrm{a}\mathrm{d}},n]=\det [n,E_{\mathrm{r}\mathrm{a}\mathrm{d}},n\times E_{\mathrm{r}\mathrm{a}\mathrm{d}}]=\mathrm{\mid }n\times E_{\mathrm{r}\mathrm{a}\mathrm{d}}{\mathrm{\mid }}^2=\mathrm{\mid }{E}_{\mathrm{r}\mathrm{a}\mathrm{d}}{\mathrm{\mid }}^2,}$ where the middle equality is a cyclic permutation and the last follows from [F5]. The sphere has finite area $4\pi$ by [F8], so integrating the explicit uniform remainder makes it tend to zero. Finally $\mathrm{\mid }{E}_{\mathrm{r}\mathrm{a}\mathrm{d}}{\mathrm{\mid }}^2={\mu }_0^2\mathrm{\mid }n\times V{\mathrm{\mid }}^2\mathrm{/}(16{\pi }^2)$ by [F5], which gives the asserted flux limit at fixed retarded time.

Dipole moment identities. The compact-ball verification in the preceding transverse-radiation theorem [F1] applies to the same $a>0$ and Countable Choice hypothesis: ${\bar{B}}_a$ is closed and bounded, hence compact; it is Borel and Lebesgue measurable; and ${\lambda }_3({\bar{B}}_a)\le {\lambda }_3(B_{2a})<\mathrm{\infty }$. Thus it is an admissible fixed integration set for [F3]. For each $i\in \{1,2,3\}$ put $p_i(u)={\int }_{B_a}{y}_i\rho (y,u)dy$. Smoothness and vanishing of $\rho ,J$ for $\mathrm{\mid }y\mathrm{\mid }>a$ imply that $\rho$, $J_i$ and their time derivatives vanish on $\mathrm{\partial }{B}_a$. Thus the integrals of $y_i\rho$ and $y_i{\rho }_t$ over $B_a$ agree with those over that compact measurable finite-measure set ${\bar{B}}_a$. Apply the set form of [F3] with integration set ${\bar{B}}_a$ and parameter $u$ to $f_i(y,u)=y_i\rho (y,u)$; its parameter derivative is $y_i{\rho }_t(y,u)$, and both are continuous on a neighborhood of ${\bar{B}}_a$ times any compact time interval. This gives $p_i^{\mathrm{\prime }}(u)={\int }_{B_a}{y}_i{\rho }_t(y,u)dy$. Fix $R_0>a$ and apply the divergence theorem to $y\mapsto y_{i}J(y,u)$ on $B_{R_0}$. The boundary flux is zero because $J=0$ near $\mathrm{\partial }{B}_{R_0}$, while $\mathrm{div}(y_{i}J)=J_i+y_i\mathrm{div}J$. Using ${\rho }_t=-\mathrm{div}J$ and the vanishing of the source outside $B_a$ gives $p_i^{\mathrm{\prime }}(u)={\int }_{B_a}{J}_i(y,u)dy$. For each $i$, the integrals of $J_i$ and $J_{i,t}$ over $B_a$ agree with those over ${\bar{B}}_a$; the closed ball has the compactness, measurability and finite measure just verified using [F1]. Both integrands are continuous on a neighborhood of ${\bar{B}}_a$ times any compact time interval. Apply the set form of [F3] on ${\bar{B}}_a$ separately to each component $J_i$ with parameter $u$; differentiating the preceding identity gives $p_i^{\mathrm{\prime }\mathrm{\prime }}(u)={\int }_{B_a}{J}_{i,t}(y,u)dy$. Collecting the three component identities yields $p^{\mathrm{\prime }\mathrm{\prime }}(u)={\int }_{B_a}{J}_t(y,u)dy$.