Explicit checks of the causal wave kernel, the gauge conventions and the static retarded limit

Static specialization (i). With $z\equiv 0$ the root equation reads $t-s=\mathrm{\mid }x\mathrm{\mid }\mathrm{/}c$, whose unique solution is $s=t-\mathrm{\mid }x\mathrm{\mid }\mathrm{/}c$; then $R=x$, $r=\mathrm{\mid }x\mathrm{\mid }$, $n=x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$, $\beta =0$, $N=0$, $\kappa =1$, $D=\mathrm{\mid }x\mathrm{\mid }$ and $\dot{\beta }=0$. By [F2] the potentials are $\phi =C\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ and $A=0$ (since $\beta =0$), and the field formulas give $E=C(1-0)(n-0)\mathrm{/}(1\cdot \mathrm{\mid }x{\mathrm{\mid }}^2)=Cx\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^3$ and $B=n\times E\mathrm{/}c=(C\mathrm{/}c)(n\times n)\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2=0$, using $n\times n=0$.

Static Poisson kernel (i). Let $h\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$, fix a compact observation region $Q_0\subseteq {\mathbb{R}}^3\times \mathbb{R}$ and note that for $(x,t)\in Q_0$ and $y$ in the support of $h$ the retarded time $s=t-\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}c$ stays in a bounded interval; choose a smooth nonnegative time cutoff $\sigma$ equal to one on that interval with compact support, and put $h_{\mathrm{\#}}(x,t):=\sigma (t)h(x)\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3\times \mathbb{R})$, which agrees with the time-independent extension of $h$ on the contributing region. By [F3], $(K\ast h_{\mathrm{\#}})(x,t)=\int h_{\mathrm{\#}}(y,t-\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}c)(4\pi \mathrm{\mid }x-y\mathrm{\mid }{)}^{-1}dy=\frac{1}{4\pi }\int h(y)\mathrm{/}\mathrm{\mid }x-y\mathrm{\mid }dy$ on $Q_0$, since $\sigma$ equals one at the retarded times and the static source does not depend on time; the remaining integral is finite: $h$ is bounded with compact support, the polar formula bounds its absolute integral on $B_1(x)$ by $\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{B_1(x)}\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}dy<\mathrm{\infty }$, and on the remaining compact support one has $\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}\le 1$, giving an upper bound $\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}$ times a finite ball volume. This is the stated extension of the compact spacetime-source formula by source-time cutoffs; the uncut static source is not compactly supported in spacetime.

The gauge transformation (ii). For $\phi \equiv 0$ and $A=(x_1,0,0)$: $\mathrm{div}A={\mathrm{\partial }}_1{x}_1=1$, so $G=1$; $\mathrm{curl}A=0$; $E=-{\mathrm{\partial }}_{t}A=0$. For $\chi =-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}6$: ${\mathrm{\partial }}_t^2\chi =0$, and $\mathrm{\Delta }\chi =-\frac{1}{6}\mathrm{\Delta }\mathrm{\mid }x{\mathrm{\mid }}^2=-1$ with $\mathrm{\Delta }\mathrm{\mid }x{\mathrm{\mid }}^2=6$, so $W\chi =c^{-2}\cdot 0-(-1)=1=G$. Since $\mathrm{\nabla }\chi =(-x_1\mathrm{/}3,-x_2\mathrm{/}3,-x_3\mathrm{/}3)$, the gauge transform is $T_{\chi }(\phi ,A)=(0-{\mathrm{\partial }}_t\chi ,A+\mathrm{\nabla }\chi )=(0,((2\mathrm{/}3)x_1,-x_2\mathrm{/}3,-x_3\mathrm{/}3))$, whose divergence is $\frac{2}{3}-\frac{1}{3}-\frac{1}{3}=0$ and whose curl vanishes because it is the gradient of $\frac{1}{3}{x}_1^2-\frac{1}{6}{x}_2^2-\frac{1}{6}{x}_3^2$ by the curl-of-gradient theorem; hence $E(T_{\chi }(\phi ,A))=-\mathrm{\nabla }0-{\mathrm{\partial }}_t(0,\dots )=0=E(\phi ,A)$ and $B(T_{\chi }(\phi ,A))=0=B(\phi ,A)$, and the transformed pair is in Lorenz gauge. This instantiates clauses (b) and (c) of [F5] with a nonzero constant $G$ and a gauge function that does not have the zero Cauchy data of the existence construction; by clause (d) of [F5], $W(\chi -{\chi }_0)=1-1=0$ for the zero-data solution ${\chi }_0$ of $W{\chi }_0=1$, so the two are compatible.

Sharpness at the apex (iii). Let $0\le \psi \in \mathcal{D}({\mathbb{R}}^3\times \mathbb{R})$ with $\psi (0,0)>0$. By continuity and $\psi (0,0)>0$ there are $a>0$ and $\delta >0$ with $\psi (y,\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c)\ge a$ whenever $\mathrm{\mid }y\mathrm{\mid }<\delta$. On that ball $y\mapsto \psi (y,\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c)\mathrm{/}\mathrm{\mid }y\mathrm{\mid }$ is nonnegative and bounded below by $a\mathrm{/}\mathrm{\mid }y\mathrm{\mid }$, so by the polar formula and the positivity of the integral of a nonnegative measurable function that is positive on a set of positive measure, $⟨K,\psi ⟩=\frac{1}{4\pi }\int \psi (y,\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c)\mathrm{\mid }y{\mathrm{\mid }}^{-1}dy\ge \frac{a}{4\pi }{\int }_{B_{\delta }}\mathrm{\mid }y{\mathrm{\mid }}^{-1}dy=\frac{a}{4\pi }\cdot \frac{\sigma (S^2){\delta }^2}{2}>0$; [F8] supplies such nonnegative bumps supported in every prescribed neighbourhood of the origin, so this nonvanishing proves $(0,0)\in \mathrm{supp}K$. For $y_0\mathrm{\ne }0$ and a nonnegative bump $\psi$ positive at $(y_0,\mathrm{\mid }{y}_0\mathrm{\mid }\mathrm{/}c)$ with support in a small ball, continuity gives $\psi (y,\mathrm{\mid }y\mathrm{\mid }\mathrm{/}c)\ge a>0$ on a neighbourhood of $y_0$, while $\mathrm{\mid }y{\mathrm{\mid }}^{-1}\ge 1\mathrm{/}(2\mathrm{\mid }{y}_0\mathrm{\mid })$ there, and the same computation gives $⟨K,\psi ⟩>0$, so every such cone point lies in the support in accordance with the support computation of [F3]. Both computations use only nonnegative tests, and no factorization of an arbitrary test into radial and time factors is assumed.