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Mutilated Maxwell's Equations (MMEs)

We are now prepared to look at the propagation of waves in volumes of space bounded in some way by conducting surfaces. We'll generally assume that the conductors in question are ``perfect'' as far as boundary conditions on the dimensions of the volume in question are concerned. The place where this will lead to error is in the gradual attenuation of a propagating wave as it loses energy to the Joule heating of the surface of the bounding conductor, but this process will be slow relative to a wavelength and using the results of the previous section we can add this attenuation in by hand afterwards if necessary.

Since we are going to have to solve boundary value problems for the wave equations for the coupled field components, we'd better select a relatively simple geometry or we'll be here all semester. The two geometries we will examine are cylindrical waveguides where propagation is along the $ z$ axis of the cylinder and rectangular waveguides where the propagation is along the $ z$ axis of a waveguide with a rectangular cross-section in the $ x-y$ plane of dimension $ a \times b$ . The transverse coordinates are therefore $ (\rho,\phi)$ or $ (x,y)$ , respectively.

As usual, we will start by assuming that we're dealing with a harmonic wave with time dependence $ e^{-i\omega t}$ , write down Maxwell's equations in free space (the cavity volume), turn them into wave equations for the field separately, note that the fields are coupled by Maxwell's equations themselves, and impose boundary conditions. The only thing that is ``special'' about a cylinder is the form of the Laplacian and how we separate the laplacian to respect the boundary conditions. Let's skip ahead to the wave equation since by now everybody should be able to do this in their sleep:

(&nabla#nabla;^2 + &mu#mu;&epsi#epsilon;&omega#omega;^2){\Vec{E} or \Vec{B} } = 0 We look at propagation along $ z$ , making it ``plane-wave-like'': \Vec{E} (\Vec{x} ,t) & = & \Vec{E} (&rho#rho;,&phis#phi;)e^±ikz - i&omega#omega;t
\Vec{B} (\Vec{x} ,t) & = & \Vec{B} (&rho#rho;,&phis#phi;)e^±ikz - i&omega#omega;t so that the wave equation becomes: (&nabla#nabla;_&perp#perp;^2 + (&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2)){\Vec{E} or \Vec{B} } = 0 (Note that $ \nabla_\perp^2 = \nabla^2 - \frac{\partial^2 }{\partial
z^2}$ ).

Resolve fields into components $ \perp$ and $ \vert\vert$ to $ z$ : \Vec{E} & = & E_z\Hat{z} + (\Hat{z} ×\Vec{E}\Hat{z}
& = & \Vec{E} _z + \Vec{E} _&perp#perp;
\Vec{B} & = & B_z\Hat{z} + (\Hat{z} ×\Vec{B}\Hat{z}
& = & \Vec{B} _z + \Vec{B} _&perp#perp;
(defining $ \Vec{E}_z$ and $ \Vec{E}_\perp$ etc. in fairly obvious ways). Now we try to write Maxwell's equations in terms of these field components, assuming that the only $ z$ -dependence permitted is $ e^{\pm ikz}$ .

This isn't trivial to do - let's start with Faraday's law, for example: \Vec{E} = -\Vec{B} t = i&omega#omega;\Vec{B}

If we project out the $ z$ component of both sides we get: \Hat{z} ·(\Vec{E} ) & = & i&omega#omega;B_z
\Hat{z} ·{ (E_zy - E_yz)\Hat{x} + (E_xz - E_zx)\Hat{y} .& + &
. (E_yx - E_xy)\Hat{z} } & = & i&omega#omega;B_z
(E_yx - E_xy) & = & i&omega#omega;B_z
\Hat{z} ·( \Vec{\nabla} _&perp#perp;×\Vec{E} _&perp#perp;) & = & i&omega#omega;B_z as only the $ \perp$ components of the curl contribute to the $ z$ direction. Similarly: \Hat{z} ×(\Vec{E} )& = & i&omega#omega;(\Hat{z} ×\Vec{B} )
\Hat{z} ×{(E_zy - E_yz)\Hat{x} + (E_xz - E_zx)\Hat{y} . & + &
. (E_yx - E_xy)\Hat{z} } & = & i&omega#omega;(\Hat{z} ×\Vec{B} _&perp#perp;)
(E_zy - E_yz)\Hat{y} - (E_xz - E_zx)\Hat{x} & = & i&omega#omega;(\Hat{z} ×\Vec{B} _&perp#perp;)
\Vec{E} _&perp#perp;z + i&omega#omega;(\Hat{z} ×\Vec{B} _&perp#perp;) & = & _&perp#perp;E_z (where $ \Hat{z} \times \Vec{B} = \Hat{z}\times\Vec{B}_\perp$ , of course).

Ouch! Looks like working through the curl termwise is a certain amount of pain! However, now that we've done it once (and see how it goes) Ampere's law should be straightforward: \Hat{z} ·(\Vec{H} ) & = & -i&omega#omega;D_z
\Hat{z} ·( \Vec{\nabla} _&perp#perp;×\Vec{B} _&perp#perp;) & = & -i&omega#omega;&mu#mu;&epsi#epsilon;E_z and \Hat{z} ×(\Vec{H} )& = & -i&omega#omega;(\Hat{z} ×\Vec{D} )
\Vec{B} _&perp#perp;z - i&omega#omega;&mu#mu;&epsi#epsilon; (\Hat{z} ×\Vec{E} _&perp#perp;) & = & _&perp#perp;B_z

Finally, we have Gauss's Law(s): \Vec{E} & = & 0
_&perp#perp;·\Vec{E} _&perp#perp;+ E_zz & = & 0
_&perp#perp;·\Vec{E} _&perp#perp;& = & -E_zz and identically, _&perp#perp;·\Vec{B} _&perp#perp;= -B_zz

Let's collect all of these in just one place now: _&perp#perp;·\Vec{E} _&perp#perp;& = & -E_zz
_&perp#perp;·\Vec{B} _&perp#perp;& = & -B_zz
\Hat{z} ·( \Vec{\nabla} _&perp#perp;×\Vec{B} _&perp#perp;) & = & -i&omega#omega;&mu#mu;&epsi#epsilon;E_z
\Hat{z} ·( \Vec{\nabla} _&perp#perp;×\Vec{E} _&perp#perp;) & = & i&omega#omega;B_z
\Vec{B} _&perp#perp;z - i&omega#omega;&mu#mu;&epsi#epsilon; (\Hat{z} ×\Vec{E} _&perp#perp;) & = & _&perp#perp;B_z
\Vec{E} _&perp#perp;z + i&omega#omega;(\Hat{z} ×\Vec{B} _&perp#perp;) & = & _&perp#perp;E_z

Gee, only a few pages of algebra to obtain in a shortened way what Jackson just puts down in three short lines. Hopefully the point is clear - to ``get'' a lot of this you have to sooner or later work it all out, however long it may take you, or you'll end up memorizing (or trying to) all of Jackson's results. Something that most normal humans could never do in a lifetime of trying...

Back to work, as there is still plenty to do.

next up previous contents
Next: TEM Waves Up: Wave Guides Previous: Boundary Conditions at a   Contents
Robert G. Brown 2017-07-11