TEM modes of a coaxial guide and the single-conductor obstruction for harmonic transverse potentials

[Potential form and the PEC boundary equivalence] Let $\phi \in C^2(\bar{A};\mathbb{C})$ with $\mathrm{\Delta }\phi =0$, fix $\zeta \in \{+1,-1\}$, and put $\tilde{E}:=-\mathrm{\nabla }\phi$, $\beta :=\zeta \omega \mathrm{/}v$, $\tilde{H}:=\zeta e_z\times \tilde{E}\mathrm{/}Z$, with real fields $E=\mathrm{Re}(\tilde{E}{e}^{i(\beta x_3-\omega t)})$, $B=\mu \mathrm{Re}(\tilde{H}{e}^{i(\beta x_3-\omega t)})$. Then ${\tilde{E}}_3=0$ and ${\tilde{H}}_3=0$ because $e_z\times \tilde{E}$ and $\mathrm{\nabla }\phi$ are transverse [F2], so $E_z=B_z=0$ (TEM); and the four amplitude equations of [F1] hold: $\stackrel{}{\mathrm{div}}(\tilde{E})=-\mathrm{\Delta }\phi =0$; $\stackrel{}{\mathrm{div}}(\tilde{H})=\frac{\zeta }{Z}({\mathrm{\partial }}_1(-{\tilde{E}}_2)+{\mathrm{\partial }}_2({\tilde{E}}_1))=-\frac{\zeta }{Z}{\mathrm{curl}}_2\tilde{E}=0$ because $\tilde{E}=-\mathrm{\nabla }\phi$ has equal mixed partials [F3]; $\stackrel{}{\mathrm{curl}}(\tilde{E})=(i\beta {\mathrm{\partial }}_2\phi ,-i\beta {\mathrm{\partial }}_1\phi ,{\mathrm{curl}}_2\tilde{E})=i\beta e_z\times \tilde{E}=i\omega \mu \tilde{H}$, using $\tilde{H}=\frac{\beta }{\omega \mu }{e}_z\times \tilde{E}$, which holds because $\frac{\beta }{\omega \mu }=\frac{\zeta \omega \mathrm{/}v}{\omega \mu }=\frac{\zeta }{v\mu }=\frac{\zeta }{Z}$ since $v\mu =\sqrt{\mu \mathrm{/}\epsilon }=Z$; and ${\mathrm{curl}}_2\tilde{H}=(\zeta \mathrm{/}Z){\mathrm{div}}_t\tilde{E}=0$, hence $\stackrel{}{\mathrm{curl}}(\tilde{H})=i\beta e_z\times \tilde{H}=i\beta \frac{\zeta }{Z}{e}_z\times (e_z\times \tilde{E})=-\frac{i\zeta \beta }{Z}\tilde{E}=-i\omega \epsilon \tilde{E}$ by [F2] and $\frac{\zeta \beta }{Z}=\frac{\omega \mathrm{/}v}{Z}=\frac{\omega }{vZ}=\omega \epsilon$ because $vZ=1\mathrm{/}\epsilon$. Hence the real fields satisfy the source-free lossless macroscopic equations by [F1]. If the potential has a differentiable boundary trace on a smooth wall of $A$ with outward unit normal $n$ (an in-plane unit vector) and tangent $\tau :=e_z\times n$, then $n\times \tilde{E}=(0,0,n_1{\tilde{E}}_2-n_2{\tilde{E}}_1)$ and, since $\tilde{E}=-\mathrm{\nabla }\phi$, $n_1{\tilde{E}}_2-n_2{\tilde{E}}_1=n_2{\mathrm{\partial }}_1\phi -n_1{\mathrm{\partial }}_2\phi =-\mathrm{\nabla }\phi \cdot \tau =-{\mathrm{\partial }}_{\tau }\phi$, so $n\times \tilde{E}=-e_z{\mathrm{\partial }}_{\tau }\phi$ (only the vanishing of this normal-cross field is used below); likewise $n\cdot \tilde{H}=\frac{\zeta }{Z}n\cdot (e_z\times \tilde{E})=\frac{\zeta }{Z}\tilde{E}\cdot (n\times e_z)=-\frac{\zeta }{Z}\tilde{E}\cdot \tau =\frac{\zeta }{Z}{\mathrm{\partial }}_{\tau }\phi$ by the scalar triple product and $n\times e_z=-\tau$ [F2]. Since a separated real field vanishes identically exactly when its amplitude does [F1], the PEC conditions $n\times E=0$ and $n\cdot B=0$ hold exactly when ${\mathrm{\partial }}_{\tau }\phi =0$ along the wall. In a $C^1$ boundary parametrization the derivative of the potential trace is the parametrization speed times ${\mathrm{\partial }}_{\tau }\phi$ (up to orientation); the chain rule and FTC [F6] therefore make it locally constant when PEC is imposed. A locally constant function on a connected boundary component is constant there, since any one of its value fibres and its complement are both relatively open.

[Converse classification and the star-shaped potential] Let $(\tilde{E},\tilde{H})$ be $C^1$ transverse amplitudes of a separated TEM field satisfying the four amplitude equations of [F1]. The third components of the two curl equations read ${\mathrm{curl}}_2\tilde{E}=i\omega \mu {\tilde{H}}_3=0$ and ${\mathrm{curl}}_2\tilde{H}=-i\omega \epsilon {\tilde{E}}_3=0$, while the two divergence equations read ${\mathrm{div}}_t\tilde{E}=0$ and ${\mathrm{div}}_t\tilde{H}=0$ (the $i\beta$-terms vanish with the third components). The transverse part of Faraday reads $(0,0,0)$ in its third component and $(-i\beta {\tilde{E}}_2,i\beta {\tilde{E}}_1,0)=i\beta e_z\times \tilde{E}$ in its transverse components, so $i\beta e_z\times \tilde{E}=i\omega \mu \tilde{H}$ and, since $\omega \mu >0$, $\tilde{H}=\frac{\beta }{\omega \mu }{e}_z\times \tilde{E}$ with no division by $\beta$. Substituting into the transverse part of Ampere and using $e_z\times (e_z\times \tilde{E})=-\tilde{E}$ [F2] gives $-i\frac{{\beta }^2}{\omega \mu }\tilde{E}=-i\omega \epsilon \tilde{E}$, that is $({\beta }^2-\epsilon \mu {\omega }^2)\tilde{E}=0$. If $\tilde{E}\mathrm{\ne }0$ then ${\beta }^2=\epsilon \mu {\omega }^2$, so $\beta =\pm \omega \mathrm{/}v$ by [F6]; if $\tilde{E}=0$ then Faraday gives $\tilde{H}=0$, and the zero field solves the equations for every longitudinal label $\beta$ with no dispersion conclusion. Finally, if $A$ is star-shaped and $\tilde{E}$ is a $C^1$ transverse closed field (${\mathrm{curl}}_2\tilde{E}=0$), then its real and imaginary parts are closed $C^1$ fields on the star-shaped domain, so [F5] supplies $C^2$ potentials for each, hence a complex $\phi \in C^2(A;\mathbb{C})$ with $\tilde{E}=-\mathrm{\nabla }\phi$ after a sign change, and ${\mathrm{div}}_t\tilde{E}=0$ makes $\mathrm{\Delta }\phi =0$; no real-only or boundary-extension conclusion is inferred from interior exactness.

[Single-conductor obstruction] Suppose $\phi \in C^2(\bar{A};\mathbb{C})$ is harmonic on the bounded connected domain $A$ and has constant boundary trace on each connected component of $\mathrm{\partial }A$, with the PEC potential description of step 1.1 available. If $\mathrm{\partial }A$ is connected, the trace is a single constant $c$, and applying [F4] to the real and imaginary parts of $\phi$ and of the constant function $c$ gives $\phi =c$ on $\bar{A}$; then $\tilde{E}=-\mathrm{\nabla }\phi =0$ and $\tilde{H}\propto e_z\times \tilde{E}=0$. For a rectangular cross-section with a potential in $C^2(A)\cap C(\bar{A})$ having constant boundary trace the same argument applies, because [F4] requires only a bounded nonempty open set and no boundary regularity. For rectangular separated TEM amplitudes that are $C^1$ on the interior and continuous on the closure with the classical PEC traces on the open edges, fix a star centre $x_{\ast }$ and use the radial primitive $\psi (x)={\int }_0^1\tilde{E}(x_{\ast }+t(x-x_{\ast }))\cdot (x-x_{\ast })dt$ furnished by [F5]. It is $C^2$ on the interior with $\mathrm{\nabla }\psi =\tilde{E}$; the displayed formula extends continuously to the closure by uniform continuity of the electric amplitude there. For two points on one open edge, apply the interior FTC line-integral identity on parallel interior segments and pass to the edge using this continuity. Thus the boundary potential increment is the integral of the continuous tangential electric trace, whose difference quotients give the edge derivative; PEC makes that derivative zero; by the one-variable constancy of [F6] the potential is constant on each open edge, and continuity at the corners joins those constants, so the trace is constant on $\mathrm{\partial }A$ and [F4] gives $\psi$ constant, hence $\tilde{E}=0$ and $\tilde{H}=0$. Thus a single-conductor cross-section with connected boundary admits no nonzero TEM field in the stated potential form, and the same holds for the stated rectangular class.

[Coaxial annulus: potential and fields] In the annulus $A=\{a<\mathrm{\mid }x\mathrm{\mid }<b\}$ put $\phi (x):=V\ln (b\mathrm{/}\mathrm{\mid }x\mathrm{\mid })\mathrm{/}\ln (b\mathrm{/}a)$ and choose the forward label $k:=\omega \mathrm{/}v$ (so $\zeta =+1$). Since ${\mathrm{\partial }}_i\ln \mathrm{\mid }x\mathrm{\mid }=x_i\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2$ by [F6] and the chain rule, ${\mathrm{\partial }}_i{\mathrm{\partial }}_i\ln \mathrm{\mid }x\mathrm{\mid }=1\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2-2x_i^2\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^4$ and $\mathrm{\Delta }\ln \mathrm{\mid }x\mathrm{\mid }=2\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2-2\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2=0$, so $\mathrm{\Delta }\phi =0$ on $A$; also $\phi =V$ on $\mathrm{\mid }x\mathrm{\mid }=a$ and $\phi =0$ on $\mathrm{\mid }x\mathrm{\mid }=b$, so the trace is constant on each boundary circle. The constructed fields of step 1.1 are then $\tilde{E}=-\mathrm{\nabla }\phi =V{e}_r\mathrm{/}(\mathrm{\mid }x\mathrm{\mid }\ln (b\mathrm{/}a))$ with $e_r=x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$, and $\tilde{H}=e_z\times \tilde{E}\mathrm{/}Z=(V\mathrm{/}(Z\mathrm{\mid }x\mathrm{\mid }\ln (b\mathrm{/}a)))e_{\phi }$ with $e_{\phi }:=e_z\times e_r$; by step 1.1 they satisfy the source-free lossless macroscopic equations and the PEC conditions, and the longitudinal label satisfies $k=\omega \mathrm{/}v$, nondispersive with no cutoff, so $k$ is real and nonzero for every $\omega >0$.

[Coaxial energies in polar coordinates] Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. With $W_e:=\frac{1}{2}\epsilon {\int }_A\mathrm{\mid }\tilde{E}{\mathrm{\mid }}^{2}dA$ and $W_m:=\frac{1}{2}\mu {\int }_A\mathrm{\mid }\tilde{H}{\mathrm{\mid }}^{2}dA$, the polar formula of [F6] (Jacobian $r$; the omitted radial segment is null) and $\mathrm{\mid }\tilde{E}{\mathrm{\mid }}^2=Z^2\mathrm{\mid }\tilde{H}{\mathrm{\mid }}^2=V^2\mathrm{/}(\mathrm{\mid }x{\mathrm{\mid }}^2{\ln }^2(b\mathrm{/}a))$ give ${\int }_A\mathrm{\mid }\tilde{E}{\mathrm{\mid }}^{2}dA={\int }_0^{2\pi }{\int }_a^b\frac{V^2}{r^2{\ln }^2(b\mathrm{/}a)}rdrd\theta =\frac{2\pi V^2}{\ln (b\mathrm{/}a)}$, using ${\int }_a^{b}dr\mathrm{/}r=\ln (b\mathrm{/}a)$ by [F6]. Hence $W_e=\frac{1}{2}\epsilon V^2\frac{2\pi }{\ln (b\mathrm{/}a)}=\frac{1}{2}C{V}^2$ with $C:=\frac{2\pi \epsilon }{\ln (b\mathrm{/}a)}$, and $W_m=\frac{1}{2}\mu \frac{V^2}{Z^2}\frac{2\pi }{\ln (b\mathrm{/}a)}=\frac{1}{2}C{V}^2$ because $\mu \mathrm{/}{Z}^2=\epsilon$. For the sheet-current reading additionally assume zero conductor-side magnetic intensity trace. On the inner wall the filling normal is $n=-e_r$, so Rectangular PEC guide and cavity geometry, the zero-normal-B reflection class, and axis-relative TE/TM/TEM(b) gives ${\tilde{K}}_f=n\times (0-\tilde{H})={\tilde{H}}_{\phi }(a)e_z$. Without that additional trace assumption the following $I$ is only the filling-side wall circulation. Integrating that axial circulation around the circle gives $I=2\pi a{\tilde{H}}_{\phi }(a)=2\pi V\mathrm{/}(Z\ln (b\mathrm{/}a))$. With $L:=\frac{\mu \ln (b\mathrm{/}a)}{2\pi }$ one gets $\frac{1}{2}L{I}^2=\frac{1}{2}\frac{\mu \ln (b\mathrm{/}a)}{2\pi }\frac{4{\pi }^2{V}^2}{Z^2{\ln }^2(b\mathrm{/}a)}=\frac{1}{2}\frac{2\pi \mu V^2}{Z^2\ln (b\mathrm{/}a)}=\frac{1}{2}C{V}^2=W_m$, and $LC=\frac{\mu \ln (b\mathrm{/}a)}{2\pi }\frac{2\pi \epsilon }{\ln (b\mathrm{/}a)}=\epsilon \mu$. The instantaneous per-length energies are $\frac{1}{2}C{V}^2{\cos }^2\theta$ and $\frac{1}{2}L{I}^2{\cos }^2\theta$ with $\theta =k{x}_3-\omega t$, by the definitions of $E$ and $B$ and the identity $\mathrm{Re}(w{e}^{i\theta }{)}^2+\mathrm{\mid }\mathrm{Im}(w{e}^{i\theta }){\mathrm{\mid }}^2=\mathrm{\mid }w{\mathrm{\mid }}^2$; for $V=0$ all fields and energies vanish. For $V\mathrm{\ne }0$ the two boundary components carry distinct constants $V$ and $0$, whereas $V=0$ gives the zero field with equal boundary constants; this respects the obstruction of step 2.1.

[Scope] Clause (a) is steps 1.1 and 1.2: the potential form gives TEM fields satisfying the equations and turns PEC on a smooth wall into the vanishing tangential potential derivative, the converse classification gives the two nonzero signs $\beta =\pm \omega \mathrm{/}v$ and the unrestricted label for the zero field, and the star-shaped interior exactness is used only with the stated potential-trace hypotheses for boundary conclusions. Clause (b) is step 2.1, proved for the continuous-boundary harmonic-potential form and the stated rectangular class; the general reduction of an arbitrary TEM field to the potential form on a merely simply connected non-star-shaped cross-section is not asserted. Clause (c) is steps 2.2 and 3.1: the coaxial computation is conditional, in the declared lossless model, with the polar-measure branch the only place Countable Choice is used, and no lossy, radiation, empirical or signal-speed claim is made.