Classical Electrodynamics · Problem 8.9

Radiating Systems, Multipole Fields and Radiation — John David Jackson

Physics — E&M · variational-principle-cavity-eigenvalue

Problem

A hollow volume

V

containing a uniform isotropic linear medium

(\epsilon, \mu)

is bounded by a perfectly conducting closed surface

S

(which may have more than one disconnected part). A harmonic electric field inside the cavity satisfies the vector Helmholtz equation,

\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{E}) = k^2 \mathbf{E} \quad \text{with } k^2 = \omega^2\mu\epsilon

The boundary condition is

\mathbf{n} \times \mathbf{E} = 0

(and

\mathbf{n}\cdot\mathbf{B} = 0

) on

S

. (a) Show that

k^2 = \frac{\displaystyle\int_V \mathbf{E}^* \cdot [\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{E})]\, d^3x}{\displaystyle\int_V \mathbf{E}^* \cdot \mathbf{E}\, d^3x}

is a variational principle for the eigenvalue

k^2

in the sense that a change of

\mathbf{E} \to \mathbf{E} + \delta\mathbf{E}

, where both

\mathbf{E}

and

\delta\mathbf{E}

satisfy the boundary conditions on

S

, leads to only second-order changes in

k^2

. (b) Apply the variational principle to the

\text{TM}_{010}

mode of a right cylindrical cavity of radius

R

and length

d

, using the trial longitudinal electric field

E_z = E_0\cos(\pi\rho/2R)

[no variational parameters]. Show that the estimate of the eigenvalue is

kR = \frac{\pi}{2}\sqrt{\frac{\pi^2 + 4}{\pi^2 - 4}}

Compare numerically with the known eigenvalue, the root

x_{01}

J_0(x)

. (c) Repeat the calculation of part b with

E_z = E_0[1 + \alpha(\rho/R)^2 - (1+\alpha)(\rho/R)^4]

, where

\alpha

is a variational parameter. Show that for this trial function the best estimate is

kR = \left[80\cdot\left(\frac{17 - 2\sqrt{34}}{68 + \sqrt{34}}\right)\right]^{1/2}

How much better is this truly variational calculation than part b?