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A rectangular-cavity standing mode at resonance: explicit fields and the PEC end-cap check

Physical scope

An explicit conditional standing-wave cavity mode of the lossless isotropic scalar branch in the rectangular perfectly conducting cavity, with its exact resonance condition and end-cap check. No measured resonance or empirical claim.

Example

All displayed fields and parameters below are numerical values in fixed coherent SI units: $a, b, d, x_{j}$ in m, $t$ in s, $ω$ in $s^{- 1}$ , $ε$ in $F m^{- 1}$ , $μ$ in $H m^{- 1}$ , $E, E_{s p}$ in $V m^{- 1}$ and $B$ in T. The TM longitudinal normalization is $E_{*} = 1 V m^{- 1}$ in The separated TE and TM candidate modes of the lossless rectangular PEC guide: exact verification, index restrictions and cutoff(b), so the dimensionless profile below specifies the numerical electric-field components.

Let $a, b, d > 0$ and $ε, μ > 0$ be constants, put $E_{s p} (x) : = e_{1} \sin \frac{π x_{2}}{b} \sin \frac{π x_{3}}{d}$ , and for $ω > 0$ define $E (x, t) : = E_{s p} (x) \cos (ω t)$ and $B (x, t) : = - \frac{1}{ω} (curl E_{s p}) (x) \sin (ω t)$ , where the curl is the fixed-time spatial curl of Divergence and curl of a $C^{1}$ vector field. Then $E, B$ are smooth, $div E_{s p} = 0$ , $Δ E_{s p} = - [(π / b)^{2} + (π / d)^{2}] E_{s p}$ , and $E, B$ satisfy the four source-free macroscopic equations on the box $Ω_{c} = (0, a) \times (0, b) \times (0, d)$ in the lossless isotropic scalar branch exactly when $ω^{2} = \frac{1}{ε μ} [(\frac{π}{b})^{2} + (\frac{π}{d})^{2}];$ with that resonance, $n \times E = 0$ and $n \cdot B = 0$ hold on all six walls, so $(E, B)$ is a PEC cavity mode. It is the cutoff instance $β = 0$ of the TM family of The separated TE and TM candidate modes of the lossless rectangular PEC guide: exact verification, index restrictions and cutoff(b) taken with axis $e_{1}$ , cross-section $(0, b) \times (0, d)$ and indices $(m, n) = (1, 1)$ : the field is independent of $x_{1}$ , which is why it also satisfies the $x_{1} = 0$ and $x_{1} = a$ end-cap conditions for every box length $a > 0$ . The example asserts a conditional model solution, not an observed resonance.

Facts & Assumptions

Given: The SI numerical-value convention and $E_{*} = 1 V m^{- 1}$ normalization of the Example; constants $a, b, d > 0$ and $ε, μ > 0$ ; $E_{s p} (x) : = e_{1} \sin \frac{π x_{2}}{b} \sin \frac{π x_{3}}{d}$ ; for $ω > 0$ the fields $E (x, t) : = E_{s p} (x) \cos (ω t)$ and $B (x, t) : = - \frac{1}{ω} (curl E_{s p}) (x) \sin (ω t)$ on the box $Ω_{c} = (0, a) \times (0, b) \times (0, d)$ ; the fixed-time spatial curl and divergence of Divergence and curl of a $C^{1}$ vector field; the box geometry, its six open walls and the PEC conditions of Rectangular PEC guide and cavity geometry, the zero-normal-B reflection class, and axis-relative TE/TM/TEM.

[F1]

Profile identities: with $e_{11} (y_{1}, y_{2}) : = \sin \frac{π y_{1}}{b} \sin \frac{π y_{2}}{d}$ on the rectangle $(0, b) \times (0, d)$ one has $Δ e_{11} = - k_{11}^{2} e_{11}$ with $k_{11}^{2} = (π / b)^{2} + (π / d)^{2}$ , $e_{11} = 0$ on the rectangle boundary; in particular $\sin 0 = \sin π = 0$ and the second derivatives of the sine factors are $- \frac{π^{2}}{b^{2}} \sin \frac{π y_{1}}{b}$ and $- \frac{π^{2}}{d^{2}} \sin \frac{π y_{2}}{d}$ (The separated sine and cosine rectangle profiles: eigen-identities, boundary traces, orthogonality and Dirichlet integrals (b)).

[F2]

Use the positively oriented cyclic coordinates $(y_{1}, y_{2}, y_{3}) = (x_{2}, x_{3}, x_{1})$ with basis $(e_{2}, e_{3}, e_{1})$ : cyclic permutation leaves the coordinate cross-product, divergence and curl formulas unchanged. Thus the transverse lemma and candidate theorem apply with longitudinal axis $e_{1}$ and transverse coordinates $(x_{2}, x_{3})$ . For an $x_{1}$ -independent profile $h$ regarded on the splitting $R^{3} = R^{2} \oplus R e_{1}$ with coordinates $(x_{2}, x_{3})$ , $curl (e_{1} h) = - e_{1} \times \nabla^{(2, 3)} h$ , $div (e_{1} h) = \partial_{1} h = 0$ , and $e_{1} \times (e_{1} \times v) = - v$ for transverse $v$ (Transverse vector-calculus identities for the splitting $R^{3} = R^{2} \oplus R e_{z}$ (a),(b) applied with $e_{z} : = e_{1}$ ).

[F3]

With the stated SI numerical-value convention and $E_{*} = 1 V m^{- 1}$ , the TM family of The separated TE and TM candidate modes of the lossless rectangular PEC guide: exact verification, index restrictions and cutoff (b) with axis $e_{1}$ , cross-section $(0, b) \times (0, d)$ , indices $(1, 1)$ and $β = 0$ : $k_{c}^{2} = k_{11}^{2}$ , and the amplitudes $\tilde{E} = \frac{i β}{k_{c}^{2}} \nabla_{t} e + e_{z} e$ , $\tilde{H} = \frac{i ω ε}{k_{c}^{2}} e_{z} \times \nabla_{t} e$ at $β = 0$ become $\tilde{E} = e_{1} e_{11}$ and $\tilde{H} = \frac{i ω ε}{k_{11}^{2}} e_{1} \times \nabla e_{11}$ ; the theorem gives smoothness, the four source-free equations at the cutoff $ω^{2} = k_{c}^{2} / (ε μ)$ , and the PEC conditions on the four faces with normals $\pm e_{2}, \pm e_{3}$ (The separated TE and TM candidate modes of the lossless rectangular PEC guide: exact verification, index restrictions and cutoff (b),(d)).

Verification

technique · direct: compute the divergence, Laplacian and curl of the explicit spatial profile, verify the four equations and the six wall conditions, and identify the field with the cutoff TM instance

1.1F1F2algebra

[The spatial profile, the equations and the resonance] The profile has only its first component, independent of $x_{1}$ , so $div E_{s p} = \partial_{1} (\sin \frac{π x_{2}}{b} \sin \frac{π x_{3}}{d}) = 0$ and $Δ E_{s p} = e_{1} (\partial_{2}^{2} + \partial_{3}^{2}) e_{11} = - [(π / b)^{2} + (π / d)^{2}] E_{s p} = - k_{11}^{2} E_{s p}$ by [F1]. Hence $div E = div E_{s p} \cos ω t = 0$ and $div B = - \frac{1}{ω} div (curl E_{s p}) \sin ω t = 0$ (a divergence of a curl vanishes componentwise by the equality of mixed partials of the smooth profile). Further $curl E = (curl E_{s p}) \cos ω t$ and $- \partial_{t} B = (curl E_{s p}) \cos ω t$ , so Faraday's law holds identically. For Amp`ere's law compute $curl curl E_{s p}$ : by [F2] $curl E_{s p} = curl (e_{1} e_{11}) = - e_{1} \times \nabla e_{11} = (0, \partial_{3} e_{11}, - \partial_{2} e_{11})$ , whose curl is $(\partial_{2} (- \partial_{2} e_{11}) - \partial_{3} (\partial_{3} e_{11}), 0, 0) = - (Δ e_{11}) e_{1} = k_{11}^{2} e_{1} e_{11} = k_{11}^{2} E_{s p}$ by [F1]; therefore $curl H = \frac{1}{μ} curl B = - \frac{1}{ω μ} k_{11}^{2} E_{s p} \sin ω t$ and $\partial_{t} D = ε \partial_{t} E = - ε ω E_{s p} \sin ω t$ , which agree exactly when $k_{11}^{2} = ω^{2} ε μ$ , i.e. when $ω^{2} = \frac{1}{ε μ} [(π / b)^{2} + (π / d)^{2}]$ . The fields are smooth because $E_{s p}$ is a product of sines and $curl E_{s p}$ a finite combination of such products, and both are independent of $x_{1}$ .

2.1step 1.1F1F2algebra

[The six wall conditions] On the walls $x_{1} = 0$ and $x_{1} = a$ with normals $\mp e_{1}$ : $n \times E = \mp e_{1} \times E = \mp \cos ω t e_{1} \times E_{s p} = 0$ since $E_{s p}$ is parallel to $e_{1}$ , and $n \cdot B = \mp B_{1} = \pm \frac{1}{ω} (curl E_{s p})_{1} \sin ω t = 0$ because $(curl E_{s p})_{1} = \partial_{2} (E_{s p})_{3} - \partial_{3} (E_{s p})_{2} = 0$ . On the walls $x_{2} = 0$ and $x_{2} = b$ with normals $\mp e_{2}$ : $e_{2} \times E = E_{1} (e_{2} \times e_{1}) = - E_{1} e_{3}$ and $E_{1} = \sin \frac{π x_{2}}{b} \sin \frac{π x_{3}}{d} \cos ω t$ vanishes because $\sin \frac{π x_{2}}{b} = 0$ there, so $n \times E = 0$ ; and $n \cdot B = \mp B_{2} = \pm \frac{1}{ω} (curl E_{s p})_{2} \sin ω t$ with $(curl E_{s p})_{2} = \partial_{3} (E_{s p})_{1} - \partial_{1} (E_{s p})_{3} = \frac{π}{d} \sin \frac{π x_{2}}{b} \cos \frac{π x_{3}}{d}$ , which vanishes at $x_{2} = 0, b$ . On the walls $x_{3} = 0$ and $x_{3} = d$ with normals $\mp e_{3}$ : $e_{3} \times E = E_{1} e_{2}$ , whose coefficient $E_{1}$ vanishes because $\sin \frac{π x_{3}}{d} = 0$ there, so $n \times E = 0$ ; and $n \cdot B = \mp B_{3} = \pm \frac{1}{ω} (curl E_{s p})_{3} \sin ω t$ with $(curl E_{s p})_{3} = \partial_{1} (E_{s p})_{2} - \partial_{2} (E_{s p})_{1} = - \frac{π}{b} \cos \frac{π x_{2}}{b} \sin \frac{π x_{3}}{d}$ , which vanishes at $x_{3} = 0, d$ . Thus all six walls satisfy $n \times E = 0$ and $n \cdot B = 0$ at resonance.

3.1step 1.1step 2.1F2F3algebra

[Identification with the cutoff TM instance] In the setting of [F3] with axis $e_{1}$ , cross-section $(0, b) \times (0, d)$ , indices $(1, 1)$ and $β = 0$ , the dispersion relation $β^{2} = ω^{2} ε μ - k_{11}^{2}$ at $β = 0$ is exactly the resonance $k_{11}^{2} = ω^{2} ε μ$ of step 1.1, $e_{11} = \sin \frac{π x_{2}}{b} \sin \frac{π x_{3}}{d}$ , and the amplitudes become $\tilde{E} = e_{1} e_{11}$ and $\tilde{H} = \frac{i ω ε}{k_{11}^{2}} e_{1} \times \nabla e_{11}$ . Then $θ = - ω t$ , $E = Re (\tilde{E} e^{- i ω t}) = e_{1} e_{11} \cos ω t = E_{s p} \cos ω t$ , and $B = μ Re (\tilde{H} e^{- i ω t}) = μ \frac{ω ε}{k_{11}^{2}} (e_{1} \times \nabla e_{11}) \sin ω t = - \frac{1}{ω} curl E_{s p} \sin ω t$ because $curl E_{s p} = - e_{1} \times \nabla e_{11}$ [F2] and $μ ω ε / k_{11}^{2} = 1 / ω$ at resonance. The amplitudes are independent of $x_{1}$ , which is why the end-cap conditions on $x_{1} = 0, a$ hold for every box length $a > 0$ ; the theorem's clause (d) supplies the resonance condition and clause (b) the guide-wall conditions on the four faces with normals $\pm e_{2}, \pm e_{3}$ .

4.1step 1.1step 2.1step 3.1F3∎

[Scope] The displayed pair is an explicit conditional standing-wave mode of the lossless isotropic scalar branch in the rectangular perfectly conducting cavity, with the exact resonance condition and the full wall check, identified as the cutoff $β = 0$ TM $(1, 1)$ instance with axis $e_{1}$ ; the example asserts a model solution and not an observed resonance, and no measurement, completeness or empirical claim is made.

A rectangular-cavity standing mode at resonance: explicit fields and the PEC end-cap check

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources