1 Fields in a cavity

Jackson works out in detail the fields in a waveguide, then uses the waveguide solutions to work out the fields in a cavity. His procedure is to break up the cavity fields into the e^{ikz} part and the e^{−ikz} part, work on each separately, and recombine them. This can be cumbersome.

One can obtain formulas that work directly for fields proportional to any combination of sinusoidal or exponential functions. The result shown here is in cartesian coordinates, and thus it is especially suited for box-shaped cavities.

Start from Maxwell’s equations written out explicitly:

∂_{y}H_{z} − ∂_{z}H_{y} = −i (ω/c) ϵE_{x}  (1)
∂_{z}H_{x} − ∂_{x}H_{z} = −i (ω/c) ϵE_{y}  (2)
∂_{x}H_{y} − ∂_{y}H_{x} = −i (ω/c) ϵE_{z}  (3)

∂_{y}E_{z} − ∂_{z}E_{y} = i (ω/c) μH_{x}  (4)
∂_{z}E_{x} − ∂_{x}E_{z} = i (ω/c) μH_{y}  (5)
∂_{x}E_{y} − ∂_{y}E_{x} = i (ω/c) μH_{z}  (6)

Assuming only that the each field component satisfies ∂_{z}^{2}ϕ = −γ^{2}ϕ, we can obtain all the other field components from the “superpotentials” E_{z} and H_{z} :

(ϵμ (ω/c)^{2} − γ^{2}) E_{x} = ∂_{x}∂_{z}E_{z} + i(ω/c)μ∂_{y}H_{z}  (7)
(ϵμ (ω/c)^{2} − γ^{2}) E_{y} = ∂_{y}∂_{z}E_{z} − i(ω/c)μ∂_{x}H_{z}  (8)
(ϵμ (ω/c)^{2} − γ^{2}) H_{x} = ∂_{x}∂_{z}H_{z} − i(ω/c)ϵ∂_{y}E_{z}  (9)
(ϵμ (ω/c)^{2} − γ^{2}) H_{y} = ∂_{y}∂_{z}H_{z} + i(ω/c)ϵ∂_{x}E_{z}  (10)

To derive (7), for example, all we need is Eq. (1) and the z derivative of (5):

∂_{y}H_{z} − ∂_{z}H_{y} = −i (ω/c) ϵE_{x}
∂_{z}^{2}E_{x} − ∂_{z}∂_{x}E_{z} = i (ω/c) μ∂_{z}H_{y}

Eliminate ∂_{z}H_{y} from these two equations and replace ∂_{z}^{2}E_{x} with −γ^{2}E_{x}.

2 Boundary conditions on E and H

Jackson is rather vague on this point. For a cavity with perfectly conducting walls and end walls perpendicular to the z axis:

E_{z} = 0 on the side walls and ∂_{z}E_{z} = 0 on the end walls ∂_{n}H_{z} = 0 on the side walls and H_{z} = 0 on the end walls

To derive these boundary conditions, we note first that E_{∥} must be inversely proportional to the conductivity, otherwise the current would be infinite. This implies that on the side walls E_{z} = 0, and on the end walls E_{x} = 0, E_{y} = 0 and their lateral derivatives also vanish. Then we can use ∂_{x}E_{x} = 0 and ∂_{y}E_{y} = 0 in the divergence equation ∂_{x}E_{x} + ∂_{y}E_{y} + ∂_{z}E_{z} = 0 to conclude that ∂_{z}E_{z} = 0 on the end walls. We can also use ∂_{x}E_{y} = 0 and ∂_{y}E_{x} = 0 in the equation i (ω/c) μH_{z} = ∂_{x}E_{y} − ∂_{y}E_{x}, from −(1/c)∂B/∂t = ∇ × E, to conclude that H_{z} = 0 on the end walls. In general the normal component H_{n} vanishes on a wall, and so do its lateral derivatives.

Finally, we show that the normal derivative ∂_{n}H_{z} vanishes on the side walls. Consider a wall element and orient the x axis normal to it. Use −i (ω/c) ϵE_{y} = ∂_{x}H_{z} − ∂_{z}H_{x}, from (1/c)∂D/∂t = ∇ × H, and note that E_{y} = 0, because it is a parallel component of E, and that ∂_{z}H_{x} = 0, because it is a lateral derivative of the normal component of H. It follows that ∂_{x}H_{z} = 0. In general, ∂_{n}H_{∥} = 0 on a wall.