Consider the resonant cavity produced by closing off the two ends of a rectangular wave guide, at z = 0 and at z = d, making a perfectly conducting empty box. Show that the resonant frequencies for both TE and TM modes are given by
\omega_{l m n}=c \pi \sqrt{(l / d)^{2}+(m / a)^{2}+(n / b)^{2}}, (9.204)
for integers l, m, and n. Find the associated electric and magnetic fields.
Look for solutions of the form E = E _{0}(x, y, z) e^{-i \omega t}, B = B _{0}(x, y, z) e^{-i \omega t} , subject to the boundary conditions E ^{\|}= 0 , B^{\perp}=0 at all surfaces. Maxwell’s equations, in the form of Eq. 9.177, give
\left. \begin{matrix} \text { (i) } \nabla \cdot E =0, & \text { (iii) } \nabla \times E =-\frac{\partial B }{\partial t} \text {, } \\ \text { (ii) } \nabla \cdot B =0, & \text { (iv) } \nabla \times B =\frac{1}{c^{2}} \frac{\partial E }{\partial t}. \end{matrix} \right\} (9.177)
\left\{\begin{array}{l} \nabla \cdot E =0 \Rightarrow \nabla \cdot E _{0}=0 ; \nabla \times E =-\frac{\partial B }{\partial t} \Rightarrow \nabla \times E _{0}=i \omega B _{0} \\\nabla \cdot B =0 \Rightarrow \nabla \cdot B _{0}=0 ; \nabla \times B =\frac{1}{c^{2}} \frac{\partial E }{\partial t} \Rightarrow \nabla \times B _{0}=-\frac{i \omega}{c^{2}} E _{0}\end{array}\right\}From now on I’ll leave off the subscript (0). The problem is to solve the (time independent) equations
\left\{\begin{array}{c} \nabla \cdot E =0 ; \nabla \times E =i \omega B ; \\\nabla \cdot B =0 ; \nabla \times B =-\frac{i \omega}{c^{2}} E\end{array}\right\}From \nabla \times E =i \omega B it follows that I can get B once I know E, so I’ll concentrate on the latter for the moment.
\nabla \times( \nabla \times E )= \nabla ( \nabla \cdot E )-\nabla^{2} E =-\nabla^{2} E = \nabla \times(i \omega B )=i \omega\left(-\frac{i \omega}{c^{2}} E \right)=\frac{\omega^{2}}{c^{2}} E . So
\nabla^{2} E_{x}=-\left(\frac{\omega}{c}\right)^{2} E_{x} ; \nabla^{2} E_{y}=-\left(\frac{\omega}{c}\right)^{2} E_{y} ; \nabla^{2} E_{z}=-\left(\frac{\omega}{c}\right)^{2} E_{z} . Solve each of these by separation of variables:
E_{x}(x, y, z)=X(x) Y(y) Z(z) \Rightarrow Y Z \frac{d^{2} X}{d x^{2}}+Z X \frac{d^{2} Y}{d y^{2}}+X Y \frac{d^{2} Z}{d z^{2}}=-\left(\frac{\omega}{c}\right)^{2} X Y Z, \text { or } \frac{1}{X} \frac{d^{2} X}{d x^{2}}+\frac{1}{Y} \frac{d^{2} Y}{d y^{2}}+\frac{1}{Z} \frac{d^{2} Z}{d z^{2}}=-(\omega / c)^{2} . Each term must be a constant, so \frac{d^{2} X}{d x^{2}}=-k_{x}^{2} X, \frac{d^{2} Y}{d y^{2}}=-k_{y}^{2} Y, \frac{d^{2} Z}{d z^{2}}=-k_{z}^{2} Z , with k_{x}^{2}+k_{y}^{2}+k_{z}^{2}=(\omega / c)^{2} . The solution is
E_{x}(x, y, z)=\left[A \sin \left(k_{x} x\right)+B \cos \left(k_{x} x\right)\right]\left[C \sin \left(k_{y} y\right)+D \cos \left(k_{y} y\right)\right]\left[E \sin \left(k_{z} z\right)+F \cos \left(k_{z} z\right)\right].
But E ^{\|}=0 at the boundaries \Rightarrow E_{x}=0 at y = 0 and z = 0, so D = F = 0, and E_{x}=0 at y = b and z = d, so k_{y}=n \pi / b \text { and } k_{z}=l \pi / d , where n and l are integers. A similar argument applies to E_{y} \text { and } E_{z} . Conclusion:
E_{x}(x, y, z)=\left[A \sin \left(k_{x} x\right)+B \cos \left(k_{x} x\right)\right] \sin \left(k_{y} y\right) \sin \left(k_{z} z\right),
E_{y}(x, y, z)=\sin \left(k_{x} x\right)\left[C \sin \left(k_{y} y\right)+D \cos \left(k_{y} y\right)\right] \sin \left(k_{z} z\right),
E_{z}(x, y, z)=\sin \left(k_{x} x\right) \sin \left(k_{y} y\right)\left[E \sin \left(k_{z} z\right)+F \cos \left(k_{z} z\right)\right],
where k_{x}=m \pi / a . (Actually, there is no reason at this stage to assume that k_{x}, k_{y}, \text { and } k_{z} , and kz are the same for all three components, and I should really a!x a second subscript (x for E_{x}, y \text { for } E_{y}, \text { and } z \text { for } E_{z} ), but in a moment we shall see that in fact they do have to be the same, so to avoid cumbersome notation I’ll assume they are from the start.)
Now \nabla \cdot E =0 \Rightarrow k_{x}\left[A \cos \left(k_{x} x\right)-B \sin \left(k_{x} x\right)\right] \sin \left(k_{y} y\right) \sin \left(k_{z} z\right)+k_{y} \sin \left(k_{x} x\right)\left[C \cos \left(k_{y} y\right)-D \sin \left(k_{y} y\right)\right] \sin \left(k_{z} z\right)+k_{z} \sin \left(k_{x} x\right) \sin \left(k_{y} y\right)\left[E \cos \left(k_{z} z\right)-F \sin \left(k_{z} z\right)\right]=0 . In particular, putting in x = 0, k_{x} A \sin \left(k_{y} y\right) \sin \left(k_{z} z\right)=0 , and hence A = 0. Likewise y = 0 ⇒ C = 0 and z = 0 ⇒ E = 0. (Moreover, if the k’s were not equal for different components, then by Fourier analysis this equation could not be satisfied (for all x, y, and z) unless the other three constants were also zero, and we’d be left with no field at all.) It follows that -\left(B k_{x}+D k_{y}+F k_{z}\right)=0 (in order that ∇· E = 0), and we are left with
E =B \cos \left(k_{x} x\right) \sin \left(k_{y} y\right) \sin \left(k_{z} z\right) \hat{ x }+D \sin \left(k_{x} x\right) \cos \left(k_{y} y\right) \sin \left(k_{z} z\right) \hat{ y }+F \sin \left(k_{x} x\right) \sin \left(k_{y} y\right) \cos \left(k_{z} z\right) \hat{ z },
\text { with } k_{x}=(m \pi / a), k_{y}=(n \pi / b), k_{z}=(l \pi / d)(l, m, n \text { all integers }), \text { and } B k_{x}+D k_{y}+F k_{z}=0 .
The corresponding magnetic field is given by B =-(i / \omega) \nabla \times E:
B_{x}=-\frac{i}{\omega}\left(\frac{\partial E_{z}}{\partial y}-\frac{\partial E_{y}}{\partial z}\right)=-\frac{i}{\omega}\left[F k_{y} \sin \left(k_{x} x\right) \cos \left(k_{y} y\right) \cos \left(k_{z} z\right)-D k_{z} \sin \left(k_{x} x\right) \cos \left(k_{y} y\right) \cos \left(k_{z} z\right)\right],
B_{y}=-\frac{i}{\omega}\left(\frac{\partial E_{x}}{\partial z}-\frac{\partial E_{z}}{\partial x}\right)=-\frac{i}{\omega}\left[B k_{z} \cos \left(k_{x} x\right) \sin \left(k_{y} y\right) \cos \left(k_{z} z\right)-F k_{x} \cos \left(k_{x} x\right) \sin \left(k_{y} y\right) \cos \left(k_{z} z\right)\right],
B_{z}=-\frac{i}{\omega}\left(\frac{\partial E_{y}}{\partial x}-\frac{\partial E_{x}}{\partial y}\right)=-\frac{i}{\omega}\left[D k_{x} \cos \left(k_{x} x\right) \cos \left(k_{y} y\right) \sin \left(k_{z} z\right)-B k_{y} \cos \left(k_{x} x\right) \cos \left(k_{y} y\right) \sin \left(k_{z} z\right)\right].
Or:
B =-\frac{i}{\omega}\left(F k_{y}-D k_{z}\right) \sin \left(k_{x} x\right) \cos \left(k_{y} y\right) \cos \left(k_{z} z\right) \hat{ x }-\frac{i}{\omega}\left(B k_{z}-F k_{x}\right) \cos \left(k_{x} x\right) \sin \left(k_{y} y\right) \cos \left(k_{z} z\right) \hat{ y }
-\frac{i}{\omega}\left(D k_{x}-B k_{y}\right) \cos \left(k_{x} x\right) \cos \left(k_{y} y\right) \sin \left(k_{z} z\right) \hat{ z } .
These automatically satisfy the boundary condition B^{\perp}=0\left(B_{x}=0 \text { at } x=0 \text { and } x=a, B_{y}=0\right. at y = 0 and y = b, and B_{z}=0 at z = 0 and z = d).
As a check, let’s see if ∇· B = 0 :
\nabla \cdot B =-\frac{i}{\omega}\left(F k_{y}-D k_{z}\right) k_{x} \cos \left(k_{x} x\right) \cos \left(k_{y} y\right) \cos \left(k_{z} z\right)-\frac{i}{\omega}\left(B k_{z}-F k_{x}\right) k_{y} \cos \left(k_{x} x\right) \cos \left(k_{y} y\right) \cos \left(k_{z} z\right)-\frac{i}{\omega}\left(D k_{x}-B k_{y}\right) k_{z} \cos \left(k_{x} x\right) \cos \left(k_{y} y\right) \cos \left(k_{z} z\right)
=-\frac{i}{\omega}\left(F k_{x} k_{y}-D k_{x} k_{z}+B k_{z} k_{y}-F k_{x} k_{y}+D k_{x} k_{z}-B k_{y} k_{z}\right) \cos \left(k_{x} x\right) \cos \left(k_{y} y\right) \cos \left(k_{z} z\right)=0.
The boxed equations satisfy all of Maxwell’s equations, and they meet the boundary conditions. For TE modes, we pick E_{z}=0 , so F = 0 (and hence B k_{x}+D k_{y}=0 , leaving only the overall amplitude undetermined, for given l, m, and n); for TM modes we want B_{z}=0 \text { (so } D k_{x}-B k_{y}=0 , again leaving only one amplitude undetermined, since B k_{x}+D k_{y}+F k_{z}=0 ). In either case \left( TE _{l m n} \text { or } TM _{l m n}\right) , the frequency is given by
\omega^{2}=c^{2}\left(k_{x}^{2}+k_{y}^{2}+k_{z}^{2}\right)=c^{2}\left[(m \pi / a)^{2}+(n \pi / b)^{2}+(l \pi / d)^{2}\right], \text { or } \omega=c \pi \sqrt{(m / a)^{2}+(n / b)^{2}+(l / d)^{2}}.