# HiWhen the concept of polarization is introduced, it is usually

1. Nov 20, 2011

### Niles

Hi

When the concept of polarization is introduced, it is usually done with plane waves, I guess because it is easy to visualize how it looks. But lets say we look at spherical waves or Gaussian waves: How does the polarization look in that case?

Best,
Niles.

2. Nov 21, 2011

### chrisbaird

Re: Polarization

If the electromagnetic wave is transverse, as is the case in free space, then the electric field, magnetic field, and propagation vector of the wave will all be mutually perpendicular. The "polarization" of the wave just describes the direction that the electric field vector points. No matter if a transverse wave is planar, cylindrical, spherical, or whatever, the fields are still perpendicular and polarization still means the same thing. The words "planar", "cylindrical", "spherical", etc. just describe the shape of the wave's wavefront, i.e. the surface over which the field strength is constant. This is a separate concept from polarization. Similarly, "Gaussian" refers to the overall waveshape along its wavefront and is a different concept from polarization.

For example, a line antenna (think radio tower) radiates vertically linearly polarized spherical waves. This means that the waves spread out in all directions - in expanding spheres - but that at any one point in the wave, the magnetic field, electric field, and propagation vector still form an orthogonal triple, with the electric field mostly vertical, and the magnetic field horizontal (depending on the direction of propagation).

3. Nov 21, 2011

### Niles

Re: Polarization

Thanks! I think I understand what you are saying, but it would be very nice if you could confirm if I am right in the following. In the case of a plane wave
$$E(z,t)=A\cos (kz-\omega t)$$
it can be visualized as in http://en.wikipedia.org/wiki/File:Linear.Polarization.Linearly.Polarized.Light_plane.wave.svg. On the left we have the E-field, and on the right the corresponding wavefronts and the associated polarization. In the case of e.g. a spherical wave, we have
$$E(r, t)= \text{Re}\frac{A}{r}e^{i(\omega t - kr)}$$
and this harmonic function determines the polarization of the spherical wavefronts at some r?

By the way, I have been reading your EM notes. If you don't mind, I have a clarifying question regarding lecture 3 of EM Theory II. On page 1 of lecture 1 (http://faculty.uml.edu/cbaird/95.658(2011)/Lecture3.pdf) you write the general solution to the wave equation. Is it correct to say that it this Fourier transform is basically a sum of plane waves, and it works because they form a complete set?

Best,
Niles.

Last edited: Nov 21, 2011
4. Nov 22, 2011

### chrisbaird

Re: Polarization

Yes. But you have to remember that in general, the electric field vector has three components, so the total general solution is three sums of plane waves (weighted by coefficients), one in each direction. This is what makes it possible in principle to expand a spherical wave into a sum over plane waves. Plane sinusoidal waves are just used as the particular solutions because they are convenient. We could just as easily express the general solution as a sum over spherical waves, or cylindrical waves, or any complete set of orthogonal functions that obey the wave equation.

5. Nov 22, 2011

### chrisbaird

Re: Polarization

For a spherical wave:

E = E0 (1/r) ei(kr-ωt)

The ei(kr-ωt) part defines it has a sinusoidal traveling wave. Because it contains kr where r is the radial direction of spherical coordinates, instead of the usual kx, that means the peaks of the sine wave are traveling in the r direction, along ever-expanding spheres centered at the origin.

The (1/r) part ensures that energy is conserved. As the wave spreads out in spheres, its strength must diminish because the same energy is smeared over ever-larger areas. For a given radius r, the magnitude of the original expression is constant. This means that the wavefront is spherical

The E0 part is the polarization vector. It has a magnitude and direction. For free waves, the direction is perpendicular to the direction of propagation and depends on the specific polarization (circular, linear, etc.) prepared. Its magnitude is the peak strength of the field.

6. Nov 24, 2011