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Axially Symmetric Cavities

The rectangular cavity that we have just discussed has many features in common with axially symmetric cavities of arbitrary cross-section. In every axially symmetric cavity, the allowed values of the wave vector, $ {\bf k}$ , and thus the allowed frequencies, are determined by the cavity geometry. We have seen that for each set of mode numbers, $ k_1$ , $ k_2$ , $ k_3$ , in a rectangular cavity, there are, in general, two linearly independent modes: that is, the polarization remains arbitrary. We can take advantage of this fact to classify modes into two types, 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 such 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, all wave quantities satisfy

$\displaystyle \left(\frac{\partial^{\,2}}{\partial z^{\,2}} + k_3^{\,2}\right)\psi =0,$ (1327)

as well as

$\displaystyle (\nabla^{\,2} + k^{\,2})\, \psi = 0,$ (1328)

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

$\displaystyle \nabla\times {\bf E}$ $\displaystyle = {\rm i}\,\omega\, \mu_0\,{\bf H},$ (1329)
$\displaystyle \nabla\times{\bf H}$ $\displaystyle = -{\rm i}\,\omega\, \epsilon_0 \,{\bf E}$ (1330)

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

$\displaystyle {\rm i}\,\omega\, \mu_0\,{\bf H}_s$ $\displaystyle = \nabla_s\times{\bf E}_z +\nabla_z\times{\bf E}_s,$ (1331)
$\displaystyle -{\rm i}\,\omega \,\epsilon_0 \,{\bf E}_s$ $\displaystyle = \nabla_s\times{\bf H}_z + \nabla_z\times{\bf H}_s.$ (1332)

When one of Equations (1333)-(1334) is used to substitute for the transverse field on the right-hand side of the other, and use is made of Equation (1329), we obtain

$\displaystyle {\bf E}_s$ $\displaystyle =\frac{\nabla_s(\partial E_z/\partial z)}{k^{\,2} - k_3^{\,2}} + \frac{{\rm i}\,\omega\,\mu_0}{k^{\,2}-k_3^{\,2}} \,\nabla_s\times{\bf H}_z,$ (1333)
$\displaystyle {\bf H}_s$ $\displaystyle = \frac{\nabla_s(\partial H_z/\partial z)}{k^{\,2} - k_3^{\,2}} - \frac{{\rm i}\,\omega\,\epsilon_0}{k^{\,2}-k_3^{\,2}}\, \nabla_s\times{\bf E}_z.$ (1334)

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

$\displaystyle \left[\nabla_s^{\,2} + (k^{\,2}-k_3^{\,2})\right] A_z = 0,$ (1335)

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

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 to ensure that $ {\bf H}_s$ in Equation (1336) satisfies the appropriate boundary condition. If the cross-section is a rectangle then 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 sets of eigenvalues, 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 Equations (1335)-(1336).

The mode frequencies are determined by the eigenvalues of Equations (1329) and (1337). 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 Equation (1337) as

$\displaystyle \nabla_s^{\,2} f = - k_s^{\,2} \,f.$ (1336)

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

$\displaystyle f\,\nabla_s^{\,2} f = \nabla_s\cdot(f\,\nabla_s f) - (\nabla_s f)^{\,2}.$ (1337)

It follows that

$\displaystyle -k_s^{\,2} \int_V f^{\,2} \,dV + \int_V (\nabla_s f)^{\,2}\,dV = \int_S f\,\nabla f \cdot d{\bf S},$ (1338)

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

$\displaystyle k_s^{\,2} = \frac{\int_V (\nabla_s f)^{\,2}\,dV} {\int_V f^{\,2}\,dV} > 0.$ (1339)

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 (1338) 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 then given by special cases of Equations (1335)-(1336):

$\displaystyle {\bf E}_s$ $\displaystyle = \frac{1}{k_s^{\,2}} \,\nabla_s \!\frac{\partial E_z}{\partial z},$ (1340)
$\displaystyle {\bf H}$ $\displaystyle =\frac{{\rm i}\,\omega\,\epsilon_0}{k_s^{\,2}}\, {\bf e}_z\times \nabla_s E_z.$ (1341)

For TE modes, in which $ E_z=0$ , the condition that $ H_z$ vanish at the ends of the cylinder demands a $ \sin( n\,\pi \,z/L)$ dependence on $ z$ , and a $ k_s$ which is such that the normal derivative of $ H_z$ is zero at the walls. Equations (1335)-(1336), for the transverse fields, then become

$\displaystyle {\bf H}_s$ $\displaystyle = \frac{1}{k_s^{\,2}} \,\nabla_s \frac{\partial H_z}{\partial z} ,$ (1342)
$\displaystyle {\bf E}$ $\displaystyle = - \frac{{\rm i}\, \omega\,\mu_0}{k_s^{\,2}}\,{\bf e}_z\times \nabla_s H_z,$ (1343)

and the mode determination is complete.

next up previous
Next: Cylindrical Cavities Up: Resonant Cavities and Waveguides Previous: Quality Factor of a
Richard Fitzpatrick 2014-06-27