the electric field is described by E=Eo exp (j(wt-k.z)) where w is the angular frequency and k is the wave vector. Z is the distance vector and t the time.

If k is kr-jki then E=Eo exp (jwt-jkr.z-ki.z)=Eo exp (jwt-jkr.z) exp (-ki.z).

exp (-ki.z) is exponential decay in the z direction.

If k is a complex number its real part is the propagation coefficient (or the wave number) and its imaginary part is the attenuation coefficient giwing the wave decay.

What's not been mentioned is that the exponential loss term is plane wave loss through a lossy medium (i.e. the loss tangent of the material). That is different from the spherical wave 1/r field or 1/r^2 power loss in the Friis transmission equation which is simply the spherical expansion of the wave reducing the power density.

The exponential loss term is always there whenever there is imaginary wave vector, because it multiplies everything else.

Cylindrical wave

E=Eo exp (jwt-jkr.z-ki.z)/sqrt(R)=Eo exp (jwt-jkr.z) exp (-ki.z)/sqrt(R)

Spherical wave

E=Eo exp (jwt-jkr.z-ki.z)/R=Eo exp (jwt-jkr.z) exp (-ki.z)/R

One thing I don't yet know for sure is what happens near the origin or focus (caustic) of a cylindrical or spherical wave, or the waist of a gaussian beam, where the real part of the wave-number becomes smaller. I don't know what happens to the imaginary part. I suspect it remains the same.

Malcolm White , it may always be there in the equation, but if the loss tangent (or rather attenuation) is zero (i.e. free space) then the contribution is always 1. No exponential loss. I think that's what you're saying in reverse with "whenever there is imaginary wave vector". However, it's more common to see this written in terms of attenuation alpha (a) and phase constant beta (b) so E = E0 e-az ej(wt-bz). If a = 0 then e-az = 1

BTW, note that Z and R should really be the same variable in your equations, assuming uniform media. Of course for non-uniform, it's k(Z) that varies, not Z , so having them the same still makes sense.

You are right, I added R without changing Z. alpha is ki, but I decided not to complicate things by introducing it, especially as it usually appears in

gamma = alpha + j beta and then you have all the confusion with sign changes (multiplying by j) to get into into real and imaginary parts of wavenumber.

jk = gamma = jbeta+alpha = jkr + ki = j(kr - jki)

Not to mention the i = -j question between physics and engineering! =)

Hi Mr Karkare, It might depend on the ansatz for solution. I guess that if you want to describe polarisation, you can use the imaginary part as a perpendicular direction. The phase and the imaginary part is not related to decaying, in general, as I remember, but there are different types of modeling on this. And EM certainly shows in variety. If deriving back to the Sun, or thunder( that has a name of its own; cosmic rays, but also described with electrons) or the meteors that burns in the atmosphere; they are radiating, or static electricity that sparks.

Dear Mr. Ketan Karkare.

You can consider this situation in the framework of the photon description.

A photon having an imaginary wave vector cannot propagate. But, anyway, it can be detected (according to the Heisenberg principle). The probability of detection depends exponentially on the distance. This is the answer to your question.

You may read the old paper (around 1970) by E.G Neumann on elektromagnetische Oberflächenwellen and there you find rather simple mathematics as well as some field plots and figures to visualize all this.

Michael D. Foegelle

Not to mention the i = -j question between physics and engineering! =)

I am not used to i = - j, and don't think I have ever come across it in 43 years of microwaves, and electromagnetic theory. I use i and j interchangeably, but j much more often.

I have to watch out for traps set by americans, though, who seem usually to have their waves travelling in the -z direction, instead of sensibly in the z direction!