Axially symmetric cavities

The rectangular cavity which we have just considered has many features in common with axially symmetric cavities of arbitrary cross section. In every cavity the allowed values of the wave vector ${\bfm k}$, and thus the allowed frequencies, are determined by the geometry of the cavity. We have seen that for each set of $k_1$, $k_2$, $k_3$ in a rectangular cavity there are, in general, two linearly independent modes; i.e., the polarization remains arbitrary. We can take advantage of this fact to classify modes into two kinds according to the orientation of the field vectors. Let us choose one type of mode such that the electric field vector lies in the cross-sectional plane, and the other so that the magnetic field vector lies in this plane. This classification into transverse electric (TE) and transverse magnetic (TM) modes turns out to be possible for all axially symmetric cavities, although the rectangular cavity is unique in having one mode of each kind corresponding to each allowed frequency.

Suppose that the direction of symmetry is along the $z$-axis, and that the length of the cavity in this direction is $L$. The boundary conditions at $z=0$ and $z=L$ demand that the $z$ dependence of wave quantities be either $\sin k_3 z$ or $\cos k_3 z$, where $k_3 = n\pi/L$. In other words, every field component satisfies

$$\left(\frac{\partial^2}{\partial z^2} + k_3^{~2}\right)\!\psi =0,$$ (1067)

as well as

$$(\nabla^2 + k^2) \psi = 0,$$ (1068)

where $\psi$ stands for any component of ${\bfm E}$ or ${\bfm H}$. The field equations

$$\nabla\wedge{\bfm E} = {\rm i}\,\omega \mu_0\,{\bfm H},$$ (1069)

$$\nabla\wedge{\bfm H} = -{\rm i}\,\omega \epsilon_0 \,{\bfm E}$$ (1070)

must also be satisfied.

Let us write each vector and each operator in the above equations as the sum of a transverse part, designated by the subscript $s$, and a component along $z$. We find that for the transverse fields

$${\rm i}\,\omega \mu_0\,{\bfm H}_s = \nabla_s\wedge{\bfm E}_z +\nabla_z\wedge{\bfm E}_s,$$ (1071)

$$-{\rm i}\,\omega \epsilon_0 \,{\bfm E}_s = \nabla_s\wedge{\bfm H}_z + \nabla_z\wedge{\bfm H}_s.$$ (1072)

When one side of Eqs. (6.27) is substituted for the transverse field on the right-hand side of the other, and use is made of Eq. (6.24), we obtain

$${\bfm E}_s = \frac{\nabla_s(\partial E_z/\partial z)}{k^2 - k_3^{~2}} + \frac{{\rm i}\,\omega\mu_0}{k^2-k_3^{~2}} \,\nabla_s\wedge{\bfm H}_z,$$ (1073)

$${\bfm H}_s = \frac{\nabla_s(\partial H_z/\partial z)}{k^2 - k_3^{~2}} - \frac{{\rm i}\,\omega\epsilon_0}{k^2-k_3^{~2}}\, \nabla_s\wedge{\bfm E}_z.$$ (1074)

Thus, all transverse fields can be expressed in terms of the $z$ components of the fields, each of which satisfies the differential equation

$$\left[\nabla_s^2 + (k^2-k_3^{~2})\right] \!A_z = 0,$$ (1075)

where $A_z$ stands for either $E_z$ or $H_z$, and $\nabla^2_s$ is the two-dimensional Laplacian operator.

The conditions on $E_z$ and $H_z$ at the boundary (in the transverse plane) are quite different: $E_z$ must vanish on the boundary, whereas the normal derivative of $H_z$ must vanish so that ${\bfm H}_s$ in Eq. (6.28)(b) satisfies the appropriate boundary condition. When the cross section is a rectangle, these two conditions lead to the same eigenvalues of $(k^2-k_3^{~2})= k_s^{~2} = k_1^{~2} + k_2^{~2}$, as we have seen. Otherwise, they correspond to two different frequencies, one for which $E_z$ is permitted but $H_z=0$, and the other where the opposite is true. In every case, it is possible to classify the modes as transverse magnetic or transverse electric. Thus, the field components $E_z$ and $H_z$ play the role of independent potentials, from which the other field components of the TE and TM modes, respectively, can be derived using Eqs. (6.28).

The mode frequencies are determined by the eigenvalues of Eqs. (6.24) and (6.29). If we denote the functional dependence of $E_z$ or $H_z$ on the plane cross section coordinates by $f(x,y)$, then we can write Eq. (6.29) as

$$\nabla_s^2 f = - k_s^{~2} f.$$ (1076)

Let us first show that $k_s^{~2}>0$, and hence that $k>k_3$. Now,

$$f\,\nabla_s^2 f = \nabla_s\!\cdot\!(f\,\nabla_s f) - (\nabla_s f)^2.$$ (1077)

It follows that

$$-k_s^{~2} \int f^2 \,dV + \int (\nabla_s f)^2\,dV = \int f\nabla f \!\cdot\! d{\bfm S},$$ (1078)

where the integration is over the transverse cross section. If either $f$ or its normal derivative is to vanish on $S$, the conducting surface, then

$$k_s^{~2} = \frac{\int (\nabla_s f)^2\,dV} {\int f^2\,dV} > 0.$$ (1079)

We have already seen that $k_3 = n\pi/L$. The allowed values of $k_s$ depend both on the geometry of the cross section and the nature of the mode.

For TM modes $H_z=0$, and the $z$ dependence of $E_z$ is given by $\cos (n\pi \,z/L)$. Equation (6.30) must be solved subject to the condition that $f$ vanish on the boundaries of the plane cross section, thus completing the determination of $E_z$ and $k$. The transverse fields are special cases of Eqs. (6.28):

$${\bfm E}_s = \frac{1}{k_s^{~2}} \,\nabla_s \!\frac{\partial E_z}{\partial z},$$ (1080)

$${\bfm H} = \frac{{\rm i}\,\omega\epsilon_0}{k_s^{~2}}\, \hat{\bfm z}\wedge \nabla_s E_z.$$ (1081)

For TE modes, in which $E_z=0$, the condition that $H_z$ vanish at the ends of the cylinder demands the use of $\sin( n\pi \,z/L)$, and $k_s$ must be such that the normal derivative of $H_z$ is zero at the walls. Equations (6.28), giving the transverse fields, then become

$${\bfm H}_s = \frac{1}{k_s^{~2}} \,\nabla_s \frac{\partial H_z}{\partial z} ,$$ (1082)

$${\bfm E} = - \frac{{\rm i}\, \omega\mu_0}{k_s^{~2}}\,\hat{\bfm z}\wedge \nabla_s H_z,$$ (1083)

and the mode determination is completed.