The separated TE and TM candidate modes of the lossless rectangular PEC guide: exact verification, index restrictions and cutoff

[Equivalence of the real and the amplitude equations] Let $E,H$ be separated fields with amplitudes $\tilde{E},\tilde{H}$ on the guide region, and use the combinations of [F3]. Then the real equations $\mathrm{div}D=0$, $\mathrm{div}B=0$, $\mathrm{curl}E=-{\mathrm{\partial }}_{t}B$, $\mathrm{curl}H={\mathrm{\partial }}_{t}D$ of [F1] hold if and only if the four amplitude equations $\stackrel{}{\mathrm{div}}(\tilde{E})=0$, $\stackrel{}{\mathrm{div}}(\tilde{H})=0$, $\stackrel{}{\mathrm{curl}}(\tilde{E})=i\omega \mu \tilde{H}$, $\stackrel{}{\mathrm{curl}}(\tilde{H})=-i\omega \epsilon \tilde{E}$ hold on $(0,a)\times (0,b)$: indeed, since $\epsilon ,\mu$ are constants, F3,(c) give $\mathrm{div}D=\epsilon \mathrm{Re}(\stackrel{}{\mathrm{div}}(\tilde{E})e^{i\theta })$, $\mathrm{div}B=\mu \mathrm{Re}(\stackrel{}{\mathrm{div}}(\tilde{H})e^{i\theta })$, $\mathrm{curl}E=\mathrm{Re}(\stackrel{}{\mathrm{curl}}(\tilde{E})e^{i\theta })$, $\mathrm{curl}H=\mathrm{Re}(\stackrel{}{\mathrm{curl}}(\tilde{H})e^{i\theta })$ and F3 gives ${\mathrm{\partial }}_{t}B=\mu \mathrm{Re}(-i\omega \tilde{H}{e}^{i\theta })$, ${\mathrm{\partial }}_{t}D=\epsilon \mathrm{Re}(-i\omega \tilde{E}{e}^{i\theta })$, so substituting the amplitude equations and taking real parts reproduces the real equations; conversely, the four actual real residuals are $\mathrm{Re}(\epsilon \stackrel{}{\mathrm{div}}(\tilde{E})e^{i\theta })$, $\mathrm{Re}(\mu \stackrel{}{\mathrm{div}}(\tilde{H})e^{i\theta })$, $\mathrm{Re}((\stackrel{}{\mathrm{curl}}(\tilde{E})-i\omega \mu \tilde{H})e^{i\theta })$ and $\mathrm{Re}((\stackrel{}{\mathrm{curl}}(\tilde{H})+i\omega \epsilon \tilde{E})e^{i\theta })$. If the real equations hold these residuals vanish for all times in a nonempty open interval. At each fixed spatial point, varying time supplies two real phases whose difference is not in $\pi \mathbb{Z}$ because $\omega >0$; the nonzero spatial factor $e^{-\mathrm{Im}\beta x_3}$ may be cancelled. By F3 each amplitude residual is therefore zero, proving the converse. All steps are complex-linear algebra in the amplitudes.

[Smoothness of the constructed fields] In either family the amplitudes $\tilde{E},\tilde{H}$ are finite sums of constant multiples of the components of ${\mathrm{\nabla }}_{t}h$ or ${\mathrm{\nabla }}_{t}e$ and of $h$ or $e$; the profiles $h,e$ are finite products of functions $\cos \frac{m\pi x_1}{a}$, $\cos \frac{n\pi x_2}{b}$, $\sin \frac{m\pi x_1}{a}$, $\sin \frac{n\pi x_2}{b}$ composed with affine functions, hence are $C^{\mathrm{\infty }}$ by [F6] (chain rule for the affine substitutions composed with the sine and cosine derivatives, and the product rule); and $e^{i\theta }$ is $C^{\mathrm{\infty }}$ by [F6] (the cartesian form exhibits it as $e^{-q{x}_3}(\cos \phi +i\sin \phi )$ with affine $\phi$, a finite combination of products of smooth functions). Therefore $E=\mathrm{Re}(\tilde{E}{e}^{i\theta })$ and $B=\mu \mathrm{Re}(\tilde{H}{e}^{i\theta })$, as finite sums of products of $C^{\mathrm{\infty }}$ complex functions and of their real parts, have all iterated partial derivatives existing and continuous: differentiating a product once in a coordinate lowers no regularity by [F6], so induction on the order gives $C^{\mathrm{\infty }}$, and in particular every derivative used below exists and is continuous. The factor $(\beta \mathrm{/}{k}^2){\mathrm{\nabla }}_t$ is dimensionless, $(\omega \mu \mathrm{/}{k}^2){\mathrm{\nabla }}_t$ carries impedance $\mathrm{V}{\mathrm{A}}^{-1}$, and $(\omega \epsilon \mathrm{/}{k}^2){\mathrm{\nabla }}_t$ carries admittance $\mathrm{A}{\mathrm{V}}^{-1}$; multiplication by $H_{\ast }$ or $E_{\ast }$ therefore gives the stated electric and magnetic amplitude units.

[Dispersion relation and cutoff] Let $k_c^2={\omega }^2\epsilon \mu -{\beta }^2>0$ and put ${\omega }_c:=k_c\mathrm{/}\sqrt{\epsilon \mu }>0$; the number ${\omega }_c$ is well defined because $\epsilon \mu >0$ and $k_c^2>0$ [F7]. Real $\beta$ with ${\beta }^2={\omega }^2\epsilon \mu -k_c^2$ exist exactly when ${\omega }^2\epsilon \mu -k_c^2\ge 0$, i.e. ${\omega }^2\ge k_c^2\mathrm{/}(\epsilon \mu )={\omega }_c^2$, i.e. $\omega \ge {\omega }_c$ (both $\omega$ and ${\omega }_c$ are positive); then the real solutions are $\beta =\pm \sqrt{{\omega }^2\epsilon \mu -k_c^2}$ [F7], and $\beta \mathrm{\ne }0$ exactly when $\omega >{\omega }_c$, while $\beta =0$ occurs exactly at $\omega ={\omega }_c$. If $\omega <{\omega }_c$ the right side is negative and the two solutions of ${\beta }^2=-(\text{\hspace{0.17em}⁣}{k}_c^2-{\omega }^2\epsilon \mu )$ are $\beta =\pm i\sqrt{k_c^2-{\omega }^2\epsilon \mu }$ because $(i\gamma {)}^2=-{\gamma }^2$ for real $\gamma$ [F7]. For any such root the phase factor obeys $e^{i\theta }=e^{i\beta x_3}{e}^{-i\omega t}$, and $e^{i\beta x_3}=e^{-\mathrm{Im}\beta x_3}(\cos (\mathrm{Re}\beta x_3)+i\sin (\mathrm{Re}\beta x_3))$ by [F6], so the field carries the real modulation factor $e^{-\mathrm{Im}\beta x_3}$ in the declared phase convention; on a declared half-guide $x_3\ge 0$ with $\mathrm{Im}\beta >0$ this factor decays, and no propagating-power claim is made. The verification of clauses (a) and (b) below uses only ${\beta }^2={\omega }^2\epsilon \mu -k^2$ (equivalently $k^2+{\beta }^2={\omega }^2\epsilon \mu$), so it covers imaginary and complex $\beta$ as well.

[The TE family] Fix $m,n\ge 0$ with $(m,n)\mathrm{\ne }(0,0)$, put $h:=H_{\ast }{h}_{mn}$, $k^2:=k_{mn}^2$ (so $k^2>0$ by [F5] and $k^2={\omega }^2\epsilon \mu -{\beta }^2=k_c^2$ by the required dispersion relation), and let $\tilde{E}:=-\frac{i\omega \mu }{k^2}{e}_z\times {\mathrm{\nabla }}_{t}h$, $\tilde{H}:=\frac{i\beta }{k^2}{\mathrm{\nabla }}_{t}h+e_{z}h$, $E=\mathrm{Re}(\tilde{E}{e}^{i\theta })$, $B=\mu \mathrm{Re}(\tilde{H}{e}^{i\theta })$. First, $e_z\times {\mathrm{\nabla }}_{t}h$ is transverse by [F4], so ${\tilde{E}}_3=0$ and $E_z=\mathrm{Re}(0)=0$: the field is TE. Second, the amplitude equations of step 1.1 hold: $\stackrel{}{\mathrm{div}}(\tilde{E})=-\frac{i\omega \mu }{k^2}({\mathrm{\partial }}_1(-{\mathrm{\partial }}_{2}h)+{\mathrm{\partial }}_2({\mathrm{\partial }}_{1}h)+i\beta \cdot 0)=-\frac{i\omega \mu }{k^2}\mathrm{div}(e_z\times {\mathrm{\nabla }}_{t}h)=0$ by [F4]; $\stackrel{}{\mathrm{div}}(\tilde{H})=\frac{i\beta }{k^2}({\mathrm{\partial }}_1^{2}h+{\mathrm{\partial }}_2^{2}h)+i\beta h=\frac{i\beta }{k^2}{\mathrm{\Delta }}_{t}h+i\beta h=\frac{i\beta }{k^2}(-k^{2}h)+i\beta h=0$ by [F5]; $\stackrel{}{\mathrm{curl}}(\tilde{E})=(-i\beta {\tilde{E}}_2,i\beta {\tilde{E}}_1,{\mathrm{\partial }}_1{\tilde{E}}_2-{\mathrm{\partial }}_2{\tilde{E}}_1)=(-\frac{\beta \omega \mu }{k^2}{\mathrm{\partial }}_{1}h,-\frac{\beta \omega \mu }{k^2}{\mathrm{\partial }}_{2}h,-\frac{i\omega \mu }{k^2}({\mathrm{\partial }}_1^{2}h+{\mathrm{\partial }}_2^{2}h))=(\frac{i\beta }{k^2}i\omega \mu {\mathrm{\partial }}_{1}h,\frac{i\beta }{k^2}i\omega \mu {\mathrm{\partial }}_{2}h,i\omega \mu h)=i\omega \mu \tilde{H}$, using $i\beta {\tilde{E}}_1=i\beta \frac{i\omega \mu }{k^2}{\mathrm{\partial }}_{2}h=-\frac{\beta \omega \mu }{k^2}{\mathrm{\partial }}_{2}h$ and $-\frac{i\omega \mu }{k^2}(-k^{2}h)=i\omega \mu h$; and $\stackrel{}{\mathrm{curl}}(\tilde{H})=({\mathrm{\partial }}_{2}h-i\beta \frac{i\beta }{k^2}{\mathrm{\partial }}_{2}h,i\beta \frac{i\beta }{k^2}{\mathrm{\partial }}_{1}h-{\mathrm{\partial }}_{1}h,\frac{i\beta }{k^2}({\mathrm{\partial }}_1{\mathrm{\partial }}_{2}h-{\mathrm{\partial }}_2{\mathrm{\partial }}_{1}h))=(1+\frac{{\beta }^2}{k^2})({\mathrm{\partial }}_{2}h,-{\mathrm{\partial }}_{1}h,0)=\frac{k^2+{\beta }^2}{k^2}({\mathrm{\partial }}_{2}h,-{\mathrm{\partial }}_{1}h,0)=\frac{{\omega }^2\epsilon \mu }{k^2}({\mathrm{\partial }}_{2}h,-{\mathrm{\partial }}_{1}h,0)=-i\omega \epsilon \tilde{E}$, where the last mixed-partial difference vanishes by the identity $\mathrm{div}(e_z\times {\mathrm{\nabla }}_{t}h)=0$ of [F4] and the final equality uses $\tilde{E}=-\frac{i\omega \mu }{k^2}(-{\mathrm{\partial }}_{2}h,{\mathrm{\partial }}_{1}h,0)$. Third, the open-wall conditions: on $\{x_1=0\}$ and $\{x_1=a\}$ (normals $\mp e_1$) the tangential components of $E$ are $E_2=\mathrm{Re}(-\frac{i\omega \mu }{k^2}{\mathrm{\partial }}_{1}h{e}^{i\theta })=-\frac{\omega \mu }{k^2}\mathrm{Re}(i{\mathrm{\partial }}_{1}h{e}^{i\theta })=0$ and $E_3=0$, because ${\mathrm{\partial }}_{1}h=0$ there by [F5], and $n\cdot B=\mp B_1=\mp \mu \mathrm{Re}\left(\frac{i\beta }{k^2}{\mathrm{\partial }}_{1}h{e}^{i\theta }\right)=0$; on $\{x_2=0\}$ and $\{x_2=b\}$ (normals $\mp e_2$) the tangential components are $E_1=\mathrm{Re}\left(\frac{i\omega \mu }{k^2}{\mathrm{\partial }}_{2}h{e}^{i\theta }\right)=0$, since ${\mathrm{\partial }}_{2}h=0$ there by [F5], and $E_3=0$, while $n\cdot B=\mp B_2=\mp \mu \mathrm{Re}\left(\frac{i\beta }{k^2}{\mathrm{\partial }}_{2}h{e}^{i\theta }\right)=0$. Hence $n\times E=0$ and $n\cdot B=0$ on every open wall, and clause (a) holds.

[The TM family] Fix $m,n\ge 1$, put $e:=E_{\ast }{e}_{mn}$ and $k^2:=k_{mn}^2=k_c^2>0$, and let $\tilde{E}:=\frac{i\beta }{k^2}{\mathrm{\nabla }}_{t}e+e_{z}e$, $\tilde{H}:=\frac{i\omega \epsilon }{k^2}{e}_z\times {\mathrm{\nabla }}_{t}e$, $E=\mathrm{Re}(\tilde{E}{e}^{i\theta })$, $B=\mu \mathrm{Re}(\tilde{H}{e}^{i\theta })$. Here $e_z\times {\mathrm{\nabla }}_{t}e$ is transverse by [F4], so ${\tilde{H}}_3=0$ and $B_z=\mu \mathrm{Re}(0)=0$: the field is TM. The amplitude equations hold: $\stackrel{}{\mathrm{div}}(\tilde{E})=\frac{i\beta }{k^2}({\mathrm{\partial }}_1^{2}e+{\mathrm{\partial }}_2^{2}e)+i\beta e=\frac{i\beta }{k^2}(-k^{2}e)+i\beta e=0$ by [F5]; $\stackrel{}{\mathrm{div}}(\tilde{H})=\frac{i\omega \epsilon }{k^2}({\mathrm{\partial }}_1(-{\mathrm{\partial }}_{2}e)+{\mathrm{\partial }}_2({\mathrm{\partial }}_{1}e))+i\beta \cdot 0=0$ by the identity $\mathrm{div}(e_z\times {\mathrm{\nabla }}_{t}e)=0$ of [F4]; $\stackrel{}{\mathrm{curl}}(\tilde{E})=({\mathrm{\partial }}_{2}e-i\beta \frac{i\beta }{k^2}{\mathrm{\partial }}_{2}e,i\beta \frac{i\beta }{k^2}{\mathrm{\partial }}_{1}e-{\mathrm{\partial }}_{1}e,\frac{i\beta }{k^2}({\mathrm{\partial }}_1{\mathrm{\partial }}_{2}e-{\mathrm{\partial }}_2{\mathrm{\partial }}_{1}e))=\frac{k^2+{\beta }^2}{k^2}({\mathrm{\partial }}_{2}e,-{\mathrm{\partial }}_{1}e,0)=\frac{{\omega }^2\epsilon \mu }{k^2}({\mathrm{\partial }}_{2}e,-{\mathrm{\partial }}_{1}e,0)=i\omega \mu \tilde{H}$ (the mixed difference vanishes by [F4] and the final equality uses $\tilde{H}=\frac{i\omega \epsilon }{k^2}(-{\mathrm{\partial }}_{2}e,{\mathrm{\partial }}_{1}e,0)$); and $\stackrel{}{\mathrm{curl}}(\tilde{H})=(-i\beta {\tilde{H}}_2,i\beta {\tilde{H}}_1,{\mathrm{\partial }}_1{\tilde{H}}_2-{\mathrm{\partial }}_2{\tilde{H}}_1)=(\frac{\beta \omega \epsilon }{k^2}{\mathrm{\partial }}_{1}e,\frac{\beta \omega \epsilon }{k^2}{\mathrm{\partial }}_{2}e,\frac{i\omega \epsilon }{k^2}({\mathrm{\partial }}_1^{2}e+{\mathrm{\partial }}_2^{2}e))=\frac{\omega \epsilon \beta }{k^2}{\mathrm{\nabla }}_{t}e-i\omega \epsilon e_{z}e=-i\omega \epsilon \tilde{E}$, using $\frac{i\omega \epsilon }{k^2}(-k^{2}e)=-i\omega \epsilon e$, in agreement with $\tilde{E}=\frac{i\beta }{k^2}{\mathrm{\nabla }}_{t}e+e_{z}e$. The open-wall conditions: $e=0$ on $\mathrm{\partial }R$ by [F5], so $E_3=\mathrm{Re}(e{e}^{i\theta })=0$ on the walls; on $\{x_1=0\}$ and $\{x_1=a\}$ (normals $\mp e_1$) the tangential components are $E_2=\mathrm{Re}\left(\frac{i\beta }{k^2}{\mathrm{\partial }}_{2}e{e}^{i\theta }\right)=0$, since ${\mathrm{\partial }}_{2}e$ carries the factor $\sin \frac{m\pi x_1}{a}$ vanishing there, and $E_3=0$, while $n\cdot B=\mp B_1=\mp \mu \mathrm{Re}(\frac{i\omega \epsilon }{k^2}(-{\mathrm{\partial }}_{2}e)e^{i\theta })=0$; on $\{x_2=0\}$ and $\{x_2=b\}$ (normals $\mp e_2$) the tangential components are $E_1=\mathrm{Re}\left(\frac{i\beta }{k^2}{\mathrm{\partial }}_{1}e{e}^{i\theta }\right)=0$, since ${\mathrm{\partial }}_{1}e$ carries the factor $\sin \frac{n\pi x_2}{b}$ vanishing there, and $E_3=0$, while $n\cdot B=\mp B_2=\mp \mu \mathrm{Re}\left(\frac{i\omega \epsilon }{k^2}{\mathrm{\partial }}_{1}e{e}^{i\theta }\right)=0$. Hence the wall conditions hold and clause (b) follows.

[Zero indices and nonvanishing] For $m=n=0$ one has $h_{00}\equiv 1$, so ${\mathrm{\nabla }}_t{h}_{00}=0$ and $k_{00}^2=0$: the divided reconstruction formulas of the TE family are undefined, and this exclusion is by the reconstruction denominator. A purely longitudinal constant candidate has $\tilde{E}=0$ and $\tilde{H}=c{e}_z$; the Faraday amplitude equation of step 1.1 then reads $\stackrel{}{\mathrm{curl}}(0)=(0,0,0)=i\omega \mu c{e}_z$, so $c=0$ because $\omega \mu >0$: it is excluded by the full field equations, not by divergence. Consistently $\stackrel{}{\mathrm{div}}(c{e}_z)=i\beta c$, which vanishes at $\beta =0$, so the divergence equation alone would not exclude a constant longitudinal amplitude at $\beta =0$; the exclusion uses Faraday as displayed. For $m=0$ or $n=0$ the TM profile vanishes identically, $e_{mn}\equiv 0$, so $\tilde{E}=e_{z}e=0$ and $\tilde{H}=0$; hence the TM indices are restricted to $m,n\ge 1$. Conversely, for $(m,n)\mathrm{\ne }(0,0)$ one has ${\mathrm{\nabla }}_t{h}_{mn}\mathrm{\ne }0$: if $m\mathrm{\ne }0$ then ${\mathrm{\partial }}_1{h}_{mn}=-\frac{m\pi }{a}\sin \frac{m\pi x_1}{a}\cos \frac{n\pi x_2}{b}$ is nonzero at $x_1=\frac{a}{2m}$ and an interior $x_2\in (0,b)$ with $\cos (n\pi x_2\mathrm{/}b)\mathrm{\ne }0$ (take $x_2=b\mathrm{/}(4n)$ when $n>0$, and $x_2=b\mathrm{/}2$ when $n=0$), and if $m=0$ then $n\mathrm{\ne }0$ and the analogous ${\mathrm{\partial }}_2{h}_{mn}$ is nonzero; and for $m,n\ge 1$ one has $e_{mn}\mathrm{\ne }0$, for instance at $x_1=\frac{a}{2m},x_2=\frac{b}{2n}$. Since $H_{\ast },E_{\ast }>0$, the TE electric amplitude is a nonzero multiple of $e_z\times {\mathrm{\nabla }}_t{h}_{mn}$ and its longitudinal magnetic amplitude is $H_{\ast }{h}_{mn}\mathrm{\ne }0$; the TM longitudinal electric amplitude is $E_{\ast }{e}_{mn}\mathrm{\ne }0$ and its magnetic amplitude is a nonzero multiple of $e_z\times {\mathrm{\nabla }}_t{e}_{mn}$. Thus both displayed fields are nonzero for all admissible indices, even at $\beta =0$ where the other transverse amplitude vanishes, and by the consequence clause of F3 the real fields $E=\mathrm{Re}(\tilde{E}{e}^{i\theta })$, $B=\mu \mathrm{Re}(\tilde{H}{e}^{i\theta })$ are not identically zero.

[Conclusion and scope] Clauses (a) and (b) are steps 2.1 and 2.2: for every admissible index pair the displayed fields are $C^{\mathrm{\infty }}$ by step 1.2, satisfy the four source-free equations by steps 1.1 and the amplitude verifications, have their stated TE or TM property, and satisfy the PEC wall conditions on every open wall by the trace identities of [F5]; clause (c) is step 2.3; clause (d) is step 1.3, whose algebra covers real, zero and imaginary roots alike with the factor $e^{-\mathrm{Im}\beta x_3}$ and the declared half-guide reading; and the argument used only the declared lossless isotropic scalar branch, the PEC model of the definition, the four macroscopic equations with ${\rho }_f=J_f=0$ and the separated phasor calculus, with no completeness, uniqueness, lossy, dispersive, open-end or empirical assertion and no propagating-power claim below cutoff. This is clause (e).