Finite-source assembly energy equals the electric field energy

Finiteness of the field energy: on the bounded ball ${\bar{B}}_{2R}$ the continuous function $\mathrm{\mid }E{\mathrm{\mid }}^2$ is bounded by [F4] and ${\lambda }_3({\bar{B}}_{2R})<\mathrm{\infty }$, so ${\int }_{\mathrm{\mid }x\mathrm{\mid }\le 2R}\mathrm{\mid }E{\mathrm{\mid }}^2<\mathrm{\infty }$; on the far field the bound of [F1] gives ${\int }_{\mathrm{\mid }x\mathrm{\mid }>2R}\mathrm{\mid }E{\mathrm{\mid }}^2\le \frac{\mathrm{\parallel }\rho {\mathrm{\parallel }}_1^2}{{\pi }^2{\epsilon }_0^2}{\int }_{\mathrm{\mid }x\mathrm{\mid }>2R}\mathrm{\mid }x{\mathrm{\mid }}^{-4}d{\lambda }_3$, and the polar formula computes ${\int }_{2R}^{\mathrm{\infty }}{r}^{-4}\cdot 4\pi r^{2}dr=4\pi \mathrm{/}(2R)<\mathrm{\infty }$ ([F1], [F4]).

Finiteness of the other two quantities: $\rho$ is bounded and supported in ${\bar{B}}_R$ and $\phi$ is continuous, so $\mathrm{\mid }\rho \phi \mathrm{\mid }$ is bounded with compact support and $\int \mathrm{\mid }\rho \phi \mathrm{\mid }<\mathrm{\infty }$. Put $M:=\sup \mathrm{\mid }\rho \mathrm{\mid }$. Splitting at the positive length $R$, translation invariance and the polar formula give, uniformly in $x$, ${\int }_{{\mathbb{R}}^3}\mathrm{\Phi }(x-y)\mathrm{\mid }\rho (y)\mathrm{\mid }d{\lambda }_3(y)\le M{\int }_{\mathrm{\mid }z\mathrm{\mid }<R}\mathrm{\Phi }(z)d{\lambda }_3(z)+\frac{\mathrm{\parallel }\rho {\mathrm{\parallel }}_1}{4\pi R}=\frac{M{R}^2}{2}+\frac{\mathrm{\parallel }\rho {\mathrm{\parallel }}_1}{4\pi R},$ since ${\int }_{\mathrm{\mid }z\mathrm{\mid }<R}\mathrm{\Phi }(z)d{\lambda }_3={\int }_0^{R}rdr=R^2\mathrm{/}2$. Tonelli therefore bounds the absolute double integral by $\mathrm{\parallel }\rho {\mathrm{\parallel }}_1(M{R}^2\mathrm{/}2+\mathrm{\parallel }\rho {\mathrm{\parallel }}_1\mathrm{/}(4\pi R))<\mathrm{\infty }$. This proves the remaining finiteness claims of (a), with both terms in the kernel estimate carrying charge per length.

Green's identity on the ball: fix $R^{\mathrm{\prime }}>R$. By [F2] the ball $B_{R^{\mathrm{\prime }}}$ is a bounded $C^1$ domain with outward normal $\nu (x)=x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ on $S_{R^{\mathrm{\prime }}}$, and $\phi \in C^2({\bar{B}}_{R^{\mathrm{\prime }}})$; Green's first identity with $u=v=\phi$ gives ${\int }_{B_{R^{\mathrm{\prime }}}}(\phi \mathrm{\Delta }\phi +\mathrm{\mid }D\phi {\mathrm{\mid }}^2)d{\lambda }_3={\int }_{S_{R^{\mathrm{\prime }}}}\phi {\mathrm{\partial }}_{\nu }\phi dS\mathrm{.}$ Substituting $\mathrm{\Delta }\phi =-\rho \mathrm{/}{\epsilon }_0$ and $\mathrm{\mid }D\phi {\mathrm{\mid }}^2=\mathrm{\mid }E{\mathrm{\mid }}^2$ from [F1], [F3] and rearranging gives ${\int }_{B_{R^{\mathrm{\prime }}}}\mathrm{\mid }E{\mathrm{\mid }}^{2}d{\lambda }_3=\frac{1}{{\epsilon }_0}{\int }_{B_{R^{\mathrm{\prime }}}}\rho \phi d{\lambda }_3+{\int }_{S_{R^{\mathrm{\prime }}}}\phi {\mathrm{\partial }}_{\nu }\phi dS\mathrm{.}$

Boundary estimate: take $R^{\mathrm{\prime }}>2R$. On $S_{R^{\mathrm{\prime }}}$ the decay bound of [F1] gives $\mathrm{\mid }\phi (x)\mathrm{\mid }\le C\mathrm{/}{R}^{\mathrm{\prime }}$ with $C:=(\mathrm{\mid }Q\mathrm{\mid }+2\mathrm{\parallel }\rho {\mathrm{\parallel }}_1)\mathrm{/}(4\pi {\epsilon }_0)$, because $R\mathrm{/}{R}^{\mathrm{\prime }}<1\mathrm{/}2$, the far-field bound gives $\mathrm{\mid }\mathrm{\nabla }\phi (x)\mathrm{\mid }\le \mathrm{\parallel }\rho {\mathrm{\parallel }}_1\mathrm{/}(\pi {\epsilon }_0{R}^{\mathrm{\prime }2})$, and ${\mathrm{\partial }}_{\nu }\phi =\mathrm{\nabla }\phi \cdot x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ has the same bound; with $\mathrm{\mid }{S}_{R^{\mathrm{\prime }}}\mathrm{\mid }=4\pi R^{\mathrm{\prime }2}$ from [F2] this yields $\mid {\int }_{S_{R^{\mathrm{\prime }}}}\phi {\mathrm{\partial }}_{\nu }\phi dS\mid \le \frac{C}{R^{\mathrm{\prime }}}\cdot \frac{\mathrm{\parallel }\rho {\mathrm{\parallel }}_1}{\pi {\epsilon }_0{R}^{\mathrm{\prime }2}}\cdot 4\pi R^{\mathrm{\prime }2}=\frac{4C\mathrm{\parallel }\rho {\mathrm{\parallel }}_1}{{\epsilon }_0{R}^{\mathrm{\prime }}},$ which tends to $0$ as $R^{\mathrm{\prime }}\to \mathrm{\infty }$ ([F1], [F2], [F4]).

First equality: the sets $B_{R^{\mathrm{\prime }}}$ increase to ${\mathbb{R}}^3$ along the increasing sequence $R_n^{\mathrm{\prime }}=(n+2)R$, $n\ge 1$, and $\mathrm{\mid }E{\mathrm{\mid }}^2\ge 0$, so monotone convergence gives ${\int }_{B_{R^{\mathrm{\prime }}}}\mathrm{\mid }E{\mathrm{\mid }}^2\to {\int }_{{\mathbb{R}}^3}\mathrm{\mid }E{\mathrm{\mid }}^2$; since $\rho$ is supported in ${\bar{B}}_R\subset B_{R^{\mathrm{\prime }}}$, the integrals ${\int }_{B_{R^{\mathrm{\prime }}}}\rho \phi$ are eventually constant equal to ${\int }_{{\mathbb{R}}^3}\rho \phi$, which is finite by step 2.1; the boundary term vanishes by step 4.1. Passing to the limit in the identity of step 3.1 therefore gives $\int \mathrm{\mid }E{\mathrm{\mid }}^2=\frac{1}{{\epsilon }_0}\int \rho \phi$, that is $\frac{{\epsilon }_0}{2}\int \mathrm{\mid }E{\mathrm{\mid }}^2=\frac{1}{2}\int \rho \phi$ ([F4]).

Second equality: by [F1] the potential is $\phi ={\epsilon }_0^{-1}N\rho$, and the product $\rho (x)\mathrm{\Phi }(x-y)\rho (y)$ is absolutely integrable over ${\mathbb{R}}^3\times {\mathbb{R}}^3$ by step 2.1, so Fubini applies: ${\int }_{{\mathbb{R}}^3}\rho \phi d{\lambda }_3=\frac{1}{{\epsilon }_0}{\int }_{{\mathbb{R}}^3}\rho (x)({\int }_{{\mathbb{R}}^3}\mathrm{\Phi }(x-y)\rho (y)d{\lambda }_3(y))d{\lambda }_3(x)=\frac{1}{{\epsilon }_0}\iint \mathrm{\Phi }(x-y)\rho (x)\rho (y)d({\lambda }_3\times {\lambda }_3)\mathrm{.}$ Since $\mathrm{\Phi }(x-y)=\mathrm{\Phi }(y-x)$ and $\rho (x)\rho (y)=\rho (y)\rho (x)$, the double integral is unchanged by exchanging $x$ and $y$, which is the ordered-pair reading of the factor; multiplying the identity by $\frac{1}{2}$ gives $\frac{1}{2}\int \rho \phi =\frac{1}{2{\epsilon }_0}\iint \mathrm{\Phi }(x-y)\rho (x)\rho (y)d({\lambda }_3\times {\lambda }_3)$, the second equality of (b) ([F1], [F4]).

Assembly reading: for $0\le s\le 1$ put ${\rho }_s:=s\rho$ and ${\phi }_s:={\epsilon }_0^{-1}N{\rho }_s=s\phi$, so that $E_s=sE$ and all the above estimates hold with ${\rho }_s$ in place of $\rho$ ([F1], [F5]). An increment of charge $d{\rho }_s=\rho ds$ added at $x$ to a configuration whose potential is ${\phi }_s$ has electric-force work $-d{\rho }_s{\phi }_s(x)$ when the charge is brought from infinity, by the work formula of [F5] together with the decay ${\phi }_s(x)\to 0$ at infinity; the balancing external work is $+d{\rho }_s{\phi }_s(x)$, so the total external assembly work is $W_{\mathrm{e}\mathrm{x}\mathrm{t}}={\int }_0^1({\int }_{{\mathbb{R}}^3}\rho (x){\phi }_s(x)d{\lambda }_3(x))ds={\int }_0^{1}sds{\int }_{{\mathbb{R}}^3}\rho \phi d{\lambda }_3=\frac{1}{2}{\int }_{{\mathbb{R}}^3}\rho \phi d{\lambda }_3,$ the interchange being justified by the absolute finiteness of step 2.1 applied to ${\rho }_s$; both signs of $\rho$ are allowed because all integrals involved are absolutely convergent ([F4], [F5]).

Clauses (a), (b) and (c): finiteness is steps 1.1-2.1, the two equalities are steps 5.1 and 6.1, the assembly reading is step 7.1, and the scope clause records that the construction is an ideal quasi-static bookkeeping with external support, uses the pure electrostatic regime, and neither models radiation or material dynamics nor identifies the field energy with a particle rest mass; the divergent point-source limit is treated separately below and is not asserted here.