# Non-Uniform Plane Waves, Multiple values of E and H?

1. Oct 14, 2008

### Tawoos

I have a question in Electromagnetics, precisely about non-uniform plane waves.

In the lecture, the professor made a strange assumption, he first used two of Maxwell's equations to get an expression for E and H (which he called E1 and H1). Then he used the remaining two to derive what he called E2 and H2. He also defined the complex wave number k. Then he stated that if alpha (the attenuation) was zero, the wave would be a uniform plane wave (Okay I understand this, but ..) then he said that if alpha was not zero and alpha was perpendicular (why perpendicular? why not just "at an angle"?) to beta, the wave is a non-uniform plane wave.

Okay, I think I understand the meaning of giving alpha and beta directions. I also understand that if there is an angle between them, then planes of constant phase wouldn't coincide with planes of constant amplitude. Therefor the wave wouldn't be uniform. Meaning, any plane of constant phase would contain points of different amplitudes.

What I don't understand is what he calls E1, E2, H1, H2. How can Maxwell's equations give different values for the same quantity!? I know that the four equations are not independent, but can they contradict? What “is” E1 and E2? And which is the one that's actually there in space?

Another question,

k is a complex vector, I understand. But I can't imagine it's orientation in space when alpha and beta aren't in the same direction. I know that the wave propagates – by definition – in the direction of beta. But what about k? What “is” k??

if k = (3 + 2j) ax + (1 – 5j) ay + (4 + 10j) az for example. How would such a quantity point in space?

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2. Oct 14, 2008

### Ben Niehoff

For an ordinary plane wave, k is the "wave vector". In this case, k represents the direction the wave is traveling, and the space frequency (i.e., reciprocal wavelength).

For a wave traveling through an absorptive medium, there will be attenuation. The simplest case is where the medium is isotropic: then the absorption always happens in the direction the wave is traveling (otherwise, you will have uneven absorption of the wave, which amounts to changing the wave's direction). The imaginary part of k is the absorption coefficient.

So, consider a plane wave traveling through an absorptive, isotropic medium. It will experience exponential attenuation; you might represent the amplitude of the wave as

$$e^{-\alpha x} \cos(\beta x)$$

which is, conveniently, the real part of

$$e^{(-\alpha + i \beta)x} = e^{i(\beta + i \alpha)x}$$

which is just

$$e^{i \vec k \cdot \vec x}$$

for the appropriate k and x. If you visualize the shape of this wave, it is like an infinite sheet, oscillating and attenuating in the x direction.

But if you look in a direction perpendicular to the x direction, the amplitude of the wave does not attenuate: consider, for example, tracing the contour of the wave surface as you travel from (x, 0, 0) to (x, y, 0). In this sense the "wave fronts of constant amplitude" are perpendicular to the direction the wave is propagating.

As for saying what direction k "points", another way to think of it is that

$$\vec k = k\hat n$$

where n is a unit vector pointing in the direction the wave is propagating, and k is some complex number; in this case, we treat k formally and forget about what "direction" an imaginary vector points.

As for E and H, you do all your calculations with complex numbers, and at the end you finally take the real parts. The only reason for all of this is that e^(iwt) is much nicer to deal with than sin(wt).

3. Oct 17, 2008