Electrostatics in vacuum — Examples

By [F1] $\mathrm{\Omega }$ is open and $\rho$ is smooth and positive on it with ${\mathrm{\partial }}_i\rho =x_i\mathrm{/}\rho$ for $i=1,2$ and ${\mathrm{\partial }}_3\rho =0$; consequently $E$ is $C^{\mathrm{\infty }}$ on $\mathrm{\Omega }$ and $E\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathrm{\Omega })$ with $\mathrm{\mid }E\mathrm{\mid }=\frac{\mathrm{\mid }\lambda \mathrm{\mid }}{2\pi {\epsilon }_0\rho }$, since on every bounded set the transverse polar formula gives ${\int }_{\rho <A}{\rho }^{-1}d{\lambda }_2=2\pi A$ and Tonelli on ${\mathbb{R}}^2\times \mathbb{R}$ multiplies this nonnegative transverse integral (with value $0$ on the axis) by the finite longitudinal length of the containing cylinder to give a finite ${\lambda }_3$-integral ([F1], [F5], [F6]). Extending $E$ by $0$ on the axis does not change its $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ class or any regular distribution, because the axis is null ([F10]), and the functional ${\rho }_{\lambda }(\psi )=\lambda {\int }_{\mathbb{R}}\psi (0,0,z)dz$ is a distribution: on a compact test support contained in a cylinder $\rho \le A$, $\mathrm{\mid }z\mathrm{\mid }\le H$ its absolute pairing is at most $\mathrm{\mid }\lambda \mathrm{\mid }2H\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}$ ([F7], [F9], [F10]).

Clause (a): the chain rule and the derivative of the logarithm [F1], [F2] give $\mathrm{\nabla }\phi =-\frac{\lambda }{2\pi {\epsilon }_0}\frac{1}{\rho }\mathrm{\nabla }\rho =-\frac{\lambda }{2\pi {\epsilon }_0\rho }{\widehat{e}}_r=-E$ on $\mathrm{\Omega }$, since $\mathrm{\nabla }\rho =(x_1,x_2,0)\mathrm{/}\rho ={\widehat{e}}_r$.

Clause (b): writing $E=(E_1,E_2,0)$ with $E_i=\frac{\lambda x_i}{2\pi {\epsilon }_0{\rho }^2}$, the quotient rule and ${\mathrm{\partial }}_i\rho =x_i\mathrm{/}\rho$ give ${\mathrm{\partial }}_i{E}_i=\frac{\lambda }{2\pi {\epsilon }_0}\frac{{\rho }^2-2x_i^2}{{\rho }^4}$ for $i=1,2$, so $\mathrm{div}E=\frac{\lambda }{2\pi {\epsilon }_0}\frac{2{\rho }^2-2x_1^2-2x_2^2}{{\rho }^4}=0$; and $\mathrm{curl}E$ has components ${\mathrm{\partial }}_2{E}_3-{\mathrm{\partial }}_3{E}_2=0$, ${\mathrm{\partial }}_3{E}_1-{\mathrm{\partial }}_1{E}_3=0$ and ${\mathrm{\partial }}_1{E}_2-{\mathrm{\partial }}_2{E}_1=\frac{\lambda }{2\pi {\epsilon }_0}\frac{-2x_1{x}_2+2x_1{x}_2}{{\rho }^4}=0$, because $E_3=0$ and $E$ does not depend on $x_3$ ([F1], [F2]). Hence $(E,0)$ satisfies the reduced equations with zero volume source on $\mathrm{\Omega }$: the pair is time independent and the reduced smooth-branch equations are exactly $\mathrm{div}E=0$, $\mathrm{curl}E=0$ [F2].

Clause (c): let $u(\theta )=\mathrm{Re}{e}^{i\theta }$, $v(\theta )=\mathrm{Im}{e}^{i\theta }$, and let $\mathrm{\Phi }(\theta ,z)=(ru(\theta ),rv(\theta ),z)$ on $D=[0,2\pi ]\times [-L\mathrm{/}2,L\mathrm{/}2]$. This compact rectangle is the closure of its nonempty connected interior and is Jordan measurable, its boundary consisting of four line segments. By [F3], $\mathrm{\Phi }$ is $C^1$ on a neighbourhood of $D$ with ${\mathrm{\partial }}_{\theta }\mathrm{\Phi }=r(u^{\mathrm{\prime }},v^{\mathrm{\prime }},0)=r(-v,u,0)$ and ${\mathrm{\partial }}_z\mathrm{\Phi }=(0,0,1)$, so ${\mathrm{\partial }}_{\theta }\mathrm{\Phi }\times {\mathrm{\partial }}_z\mathrm{\Phi }=r(u,v,0)$ is nonzero on the interior because $r>0$ and $u^2+v^2=\mathrm{\mid }{e}^{i\theta }{\mathrm{\mid }}^2=1$; by $\ker (\exp )=2\pi i\mathbb{Z}$, and $\exp z=\exp w$ exactly when $z-w\in 2\pi i\mathbb{Z}$, equality of images forces equal $z$ and $\theta -{\theta }^{\mathrm{\prime }}\in 2\pi \mathbb{Z}$. If $0<\theta <2\pi$ and $0\le {\theta }^{\mathrm{\prime }}\le 2\pi$, their difference lies strictly between $-2\pi$ and $2\pi$, so it is zero. Thus no interior point shares its image with a distinct point of the whole rectangle; the seam $\theta =0\leftrightarrow 2\pi$ lies on the boundary. Hence $(D,\mathrm{\Phi })$ is an admissible regular patch for the lateral surface [F4]. Since $E(\mathrm{\Phi }(\theta ,z))=\frac{\lambda }{2\pi {\epsilon }_{0}r}(u,v,0)$, the flux integrand is $E\cdot ({\mathrm{\partial }}_{\theta }\mathrm{\Phi }\times {\mathrm{\partial }}_z\mathrm{\Phi })=\frac{\lambda }{2\pi {\epsilon }_{0}r}\cdot r=\frac{\lambda }{2\pi {\epsilon }_0}$, and the flux equals $\frac{\lambda }{2\pi {\epsilon }_0}{\lambda }_2(D)=\frac{\lambda }{2\pi {\epsilon }_0}\cdot 2\pi L=\frac{\lambda L}{{\epsilon }_0}$ by the box measure and the Riemann/Lebesgue agreement for the constant integrand ([F4], [F8]).

Clause (d), superposition representation: for a longitudinal coordinate $t\in \mathbb{R}$ (metres) put $z_t:=t{e}_3$ and $E_{1,t}:{\mathbb{R}}^3\to {\mathbb{R}}^3$ by $E_{1,t}(x):=\frac{x-z_t}{4\pi {\epsilon }_0\mathrm{\mid }x-z_t{\mathrm{\mid }}^3}$ for $x\mathrm{\ne }{z}_t$ and set $E_{1,t}(z_t):=0$ (the point-charge field per unit charge at $z_t$, obtained from [F7] by dividing the field of any nonzero charge $q$ by $q$, with units V C${}^{-1}$ m${}^{-1}$). For $x\in \mathrm{\Omega }$, direct computation with the primitive $t\mapsto \frac{t}{{\rho }^2\sqrt{{\rho }^2+t^2}}$ of $({\rho }^2+t^2{)}^{-3\mathrm{/}2}$ shows ${\int }_{-T}^T\frac{dt}{({\rho }^2+t^2{)}^{3\mathrm{/}2}}=\frac{2T}{{\rho }^2\sqrt{{\rho }^2+T^2}}\to \frac{2}{{\rho }^2}$ as $T\to \mathrm{\infty }$ and ${\int }_{x_3-T}^{x_3+T}\frac{x_3-t}{({\rho }^2+(x_3-t{)}^2{)}^{3\mathrm{/}2}}dt=0$ by oddness for every $T$ ([F8]); hence, for every $x\in \mathrm{\Omega }$, $\lambda {\int }_{\mathbb{R}}{E}_{1,t}(x)dt=\lambda (\frac{x_1}{2\pi {\epsilon }_0{\rho }^2},\frac{x_2}{2\pi {\epsilon }_0{\rho }^2},0)=E(x),$ the first two components because ${\int }_{\mathbb{R}}({\rho }^2+t^2{)}^{-3\mathrm{/}2}dt=2\mathrm{/}{\rho }^2$ and the third by oddness. Moreover $\mathrm{\mid }{E}_{1,t}(x)\mathrm{\mid }=\frac{1}{4\pi {\epsilon }_0({\rho }^2+(x_3-t{)}^2)}\le \frac{1}{4\pi {\epsilon }_0}(\frac{{\mathbf{1}}_{\mathrm{\mid }{x}_3-t\mathrm{\mid }\le \rho }}{{\rho }^2}+\frac{{\mathbf{1}}_{\mathrm{\mid }{x}_3-t\mathrm{\mid }>\rho }}{(x_3-t{)}^2})$, whose $t$-integral is at most $\frac{4}{4\pi {\epsilon }_0\rho }=\frac{1}{\pi {\epsilon }_0\rho }$, and $x\mapsto \mathrm{\mid }\mathrm{\nabla }\psi (x)\mathrm{\mid }\mathrm{/}\rho (x)$, extended by $0$ on the axis, is integrable for every compactly supported test: $\mathrm{\mid }\mathrm{\nabla }\psi \mathrm{\mid }$ is bounded on a containing cylinder and step 1.1 supplies the integral of $1\mathrm{/}\rho$ there ([F5], [F6], [F9]).

Clause (d), the distributional identities: let $\psi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$ and let $\mathrm{\Psi }\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3;{\mathbb{C}}^3)$. For $h=\mathrm{\nabla }\psi$ or $h=\mathrm{curl}\mathrm{\Psi }$, the function $g_h(x,t):=\mathrm{\mid }{E}_{1,t}(x)\mathrm{\mid }\mathrm{\mid }h(x)\mathrm{\mid }$ is nonnegative and Borel on ${\mathbb{R}}^3\times \mathbb{R}$: it is continuous off the closed set $\{(x,t):x=t{e}_3\}$ and is $0$ there. By [F6] it is product-measurable, and Tonelli, applied before any Fubini interchange, gives ${\int }_{{\mathbb{R}}^3\times \mathbb{R}}{g}_{h}d({\lambda }_3\times {\lambda }_1)={\int }_{{\mathbb{R}}^3}({\int }_{\mathbb{R}}{g}_h(x,t)dt)dx\le \frac{1}{\pi {\epsilon }_0}{\int }_{\mathrm{\Omega }}\frac{\mathrm{\mid }h(x)\mathrm{\mid }}{\rho (x)}dx<\mathrm{\infty }\mathrm{.}$ The bound holds off the null axis by step 5.1; the last integral is finite because $h$ is bounded and compactly supported and step 1.1 gives the transverse majorant. Thus $E_{1,t}\cdot h$ belongs to product $L^1$, and Fubini [F6] now licenses both orders of integration. Using step 5.1 almost everywhere, $⟨\mathrm{div}E,\psi ⟩=-{\int }_{{\mathbb{R}}^3}E\cdot \mathrm{\nabla }\psi d{\lambda }_3=-\lambda {\int }_{\mathbb{R}}({\int }_{{\mathbb{R}}^3}{E}_{1,t}\cdot \mathrm{\nabla }\psi d{\lambda }_3)dt=\lambda {\int }_{\mathbb{R}}⟨\mathrm{div}{E}_{1,t},\psi ⟩dt,$ and [F7] gives $\mathrm{div}{E}_{1,t}=\frac{1}{{\epsilon }_0}{\delta }_{z_t}$, so the last expression is $\frac{\lambda }{{\epsilon }_0}{\int }_{\mathbb{R}}\psi (0,0,t)dt=\frac{1}{{\epsilon }_0}{\rho }_{\lambda }(\psi )$ for every test function, which is the first identity. The same justified interchange for $h=\mathrm{curl}\mathrm{\Psi }$ gives $⟨\mathrm{curl}E,\mathrm{\Psi }⟩=\int E\cdot \mathrm{curl}\mathrm{\Psi }=\lambda {\int }_{\mathbb{R}}⟨\mathrm{curl}{E}_{1,t},\mathrm{\Psi }⟩dt=0$, since each translated unit-charge field has vanishing distributional curl by [F7]; hence $\mathrm{curl}E=0$ in ${\mathcal{D}}^{\mathrm{\prime }}({\mathbb{R}}^3)$.

Clause (e): with $\phi =-\frac{\lambda }{2\pi {\epsilon }_0}\log (\rho \mathrm{/}{r}_0)$, the limits $\rho \to \mathrm{\infty }$ and the $r_0$-shift are the elementary properties of the logarithm [F2]: replacing $r_0$ by $r_0^{\mathrm{\prime }}$ adds $-\frac{\lambda }{2\pi {\epsilon }_0}\log (r_0\mathrm{/}{r}_0^{\mathrm{\prime }})=\frac{\lambda }{2\pi {\epsilon }_0}\log (r_0^{\mathrm{\prime }}\mathrm{/}{r}_0)$ to $\phi$, which cancels in $\mathrm{\nabla }\phi$, and as $\rho \to \mathrm{\infty }$ the potential diverges to $-\mathrm{\infty }$ for $\lambda >0$ and to $+\mathrm{\infty }$ for $\lambda <0$, so for $\lambda \mathrm{\ne }0$ no normalisation at infinity is available ([F11]). If $\lambda =0$ then $E=0$, $\phi =0$, ${\rho }_{\lambda }=0$, and every flux and cutoff energy vanishes by the identities above. For non-star-shapedness, let $a\in \mathrm{\Omega }$; then $-a$ lies in $\mathrm{\Omega }$ and the segment $s\mapsto (1-2s)a$ at $s=1\mathrm{/}2$ is $0$, which is not in $\mathrm{\Omega }$, so no point of $\mathrm{\Omega }$ is a star centre for the whole set ([F1], [F11]).

Clause (f): $\mathrm{\mid }E{\mathrm{\mid }}^2=\frac{{\lambda }^2}{4{\pi }^2{\epsilon }_0^2{\rho }^2}$, and the two-dimensional polar formula of [F5], applied in the transverse variables and combined with Tonelli for the nonnegative integrand in the $x_3$-variable [F6], gives ${\int }_{\{\delta <\rho <R,\mathrm{\mid }{x}_3\mathrm{\mid }<L\mathrm{/}2\}}\mathrm{\mid }E{\mathrm{\mid }}^{2}d{\lambda }_3=L\cdot \frac{{\lambda }^2}{4{\pi }^2{\epsilon }_0^2}\cdot 2\pi \log \frac{R}{\delta }=\frac{{\lambda }^{2}L}{2\pi {\epsilon }_0^2}\log \frac{R}{\delta }$; multiplying by $\frac{{\epsilon }_0}{2}$ gives the displayed identity. For $\lambda \mathrm{\ne }0$ the factor $\log (R\mathrm{/}\delta )$ tends to $+\mathrm{\infty }$ as $\delta \downarrow 0$ and as $R\to \mathrm{\infty }$, so no finite energy per unit length exists ([F5], [F9]).

Clause (g) records the scope: [F11] declares only point and stationary-sheet sources, so the line distribution ${\rho }_{\lambda }$ is outside the framework's source types; the construction used the ideal point-charge identity of [F7] superimposed along the axis with the absolute majorant of step 5.1, the classical computations of steps 2.1-4.1 on the open set $\mathrm{\Omega }$, and the polar/Fubini computations of steps 6.1-8.1. No physical realisation, equilibrium, material model, self-force or empirical claim is made, and no compact-source corollary is asserted; this is exactly the scope stated.

Setup: the density is bounded and supported in ${\bar{B}}_a$, hence in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^3)$; its total charge is $\int \rho =\frac{3Q}{4\pi a^3}\frac{4\pi a^3}{3}=Q$ by [F1]; the sphere is null, so the value of $\rho$ there is immaterial ([F1]). Writing $r=\mathrm{\mid }x\mathrm{\mid }$ and $h(r):=\frac{Q}{4\pi {\epsilon }_0{a}^3}$ for $r<a$ and $h(r):=\frac{Q}{4\pi {\epsilon }_0{r}^3}$ for $r\ge a$, the field is $E(x)=h(r)x$ and the potential is the radial function $\phi (x)=f(r)$ with $f(r)=\frac{Q(3a^2-r^2)}{8\pi {\epsilon }_0{a}^3}$ for $r\le a$ and $f(r)=\frac{Q}{4\pi {\epsilon }_{0}r}$ for $r\ge a$. Both branches are smooth on their open regions, $h$ is continuous at $r=a$ with common value $\frac{Q}{4\pi {\epsilon }_0{a}^3}$, and $f$ is continuous with $f(a)=\frac{Q}{4\pi {\epsilon }_{0}a}$ from both branches; moreover $f^{\mathrm{\prime }}(r)=-rh(r)$ on each side, with common value $f^{\mathrm{\prime }}(a)=-\frac{Q}{4\pi {\epsilon }_0{a}^2}$, so the one-sided limits of $\mathrm{\nabla }\phi =f^{\mathrm{\prime }}(r)x\mathrm{/}r$ agree on $S_a$ ([F2]).

Clause (a): on each coordinate line the sphere has at most two intersection points. On every smooth subinterval the derivative of $\phi$ is the corresponding component of $-E$, with matching limits at the intersection points by step 1.1. Apply the FTC [F4] on those subintervals and pass their endpoints to an intersection; the resulting increment formula is the integral of the continuous component $-E_i$ along the whole short line segment. Dividing by the segment length proves the same partial derivative at the intersection. Thus all partial derivatives exist and are continuous, so the continuous-partial-derivatives characterization gives total differentiability at every point with $\mathrm{\nabla }\phi =-E$; $E$ is continuous because $h$ is, and both $E$ and $\phi$ are smooth on each open side ([F2]).

Clause (b): for $\mathrm{\mid }x\mathrm{\mid }<a$, $E(x)=\frac{Q}{4\pi {\epsilon }_0{a}^3}x$ is affine, so $\mathrm{div}E=\frac{3Q}{4\pi {\epsilon }_0{a}^3}=\rho (x)\mathrm{/}{\epsilon }_0$ and $\mathrm{curl}E=0$; for $\mathrm{\mid }x\mathrm{\mid }>a$, $E(x)=\frac{Qx}{4\pi {\epsilon }_0\mathrm{\mid }x{\mathrm{\mid }}^3}$ is radial of the Coulomb shape, and the quotient rule with ${\mathrm{\partial }}_i\mathrm{\mid }x\mathrm{\mid }=x_i\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ gives $\mathrm{div}E=\frac{Q}{4\pi {\epsilon }_0}{\sum }_i(\frac{1}{\mathrm{\mid }x{\mathrm{\mid }}^3}-\frac{3x_i^2}{\mathrm{\mid }x{\mathrm{\mid }}^5})=0=\rho (x)\mathrm{/}{\epsilon }_0$ and $\mathrm{curl}E=0$ componentwise, since for all indices $j,k$, ${\mathrm{\partial }}_j{E}_k=\frac{Q}{4\pi {\epsilon }_0}({\delta }_{jk}\mathrm{\mid }x{\mathrm{\mid }}^{-3}-3x_j{x}_k\mathrm{\mid }x{\mathrm{\mid }}^{-5})$ is symmetric in $j,k$ ([F2]).

Clause (c): the sphere $S_a$ is a compact embedded smooth two-sided hypersurface with outward normal $n(x)=x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$; the field $E$ is $C^1$ on $\{\mathrm{\mid }x\mathrm{\mid }<a\}$ and on $\{\mathrm{\mid }x\mathrm{\mid }>a\}$ with one-sided $C^1$ traces on $S_a$: inside, ${\mathrm{\partial }}_j{E}_k=\frac{Q}{4\pi {\epsilon }_0{a}^3}{\delta }_{jk}$; outside the derivative formula in step 3.1 extends continuously to $\mathrm{\mid }x\mathrm{\mid }=a>0$. By step 2.1 both traces are the continuous field $E$, so $[E]=E_{+}-E_{-}=0$ on $S_a$ ([F3]). The interface lemma [F3] therefore gives $\mathrm{div}E=\{\mathrm{div}E{\}}_{\mathrm{p}\mathrm{i}\mathrm{e}\mathrm{c}\mathrm{e}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}}+n\cdot [E]{\delta }_{S_a}$ and $\mathrm{curl}E=\{\mathrm{curl}E{\}}_{\mathrm{p}\mathrm{i}\mathrm{e}\mathrm{c}\mathrm{e}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}}+n\times [E]{\delta }_{S_a}$; the jump terms vanish, and the piecewise expressions equal $\rho \mathrm{/}{\epsilon }_0$ and $0$ respectively as locally integrable functions by step 3.1 and the nullity of the sphere ([F1], [F3]). Hence $\mathrm{div}E=\rho \mathrm{/}{\epsilon }_0$ and $\mathrm{curl}E=0$ in ${\mathcal{D}}^{\mathrm{\prime }}({\mathbb{R}}^3)$; by the static-to-distributional reading these are the weak equations for the source $(\rho ,0)$ with zero singular part ([F4]).