# Unpolarized beam

1. Oct 19, 2006

### asdf1

In diffraction, why when calculating the intensity of an unpolarized X-ray beam do you have to assume that Ey=Ez if the electric field acts perpendicular to the plane of scattering (x-z) and lies in the y-z plane?

2. Oct 20, 2006

### OlderDan

I'm not seeing the geometry. Are you saying the x-ray is travelling in the y-z plane toward the x-z plane, or are you saying E is in the y-z plane. The latter would not make sense because that would mean the x-ray is parallel to the x axis and parallel to the scattering plane. The former has the possibility of the x-ray travelling parallel to the y-axis, in which case Ey is zero. It seems to me the relative average strengths of Ey and Ez for unpolarized light would depend on the angle of incidence to the plane.

3. Oct 20, 2006

### asdf1

Sorry about the confusion. The x-ray is traveling in the x-z plane and the diffrated beam can be divided into the componets of Ey=Ez. Pg.85-86 of Microstructural Characterization of Materials by David Brandon and Wayne D. Kaplan.

4. Oct 20, 2006

### OlderDan

OK, I did misinterpret the geometry, but I'm still bothered by the equality, although it may not be important. I don't have the book, so I can't see what the author is using for a coordinate system.

Assume there is a set of Bragg planes parallel to the x-y plane. The possible polarization components of an incoming x-ray are: 1) perpendicular to the plane of incidence and reflection, i.e., in the y direction, and 2) parallel to the plane of incidence and reflection. The in-plane direction of E is at an angle wrt to the z and x axes. If the incident beam is parallel to the z axis, Ez = 0. In that case you would expect to have on the average in an unpolarized beam Ey = Ex, Ez = 0. For an incoming unpolarized beam you would always expect to have equal average in-plane and out-of-plane components.

If your coordinate system has z perpendicular to the direction of the incident beam, then it makes sense that Ey (out of plane) = Ez (in plane). But then the reflection planes would not be parallel to x-y planes. The point is that whatever the coordinate system, the incoming x-rays have the same average electric field component in the plane of incidence/reflection as the average component perpendicular to that plane.

The reflection coefficients for the parallel and perpendicular polarizations are likely different, and angle dependent, so the diffracted beams are probably partially polarized.

5. Oct 23, 2006

### asdf1

Um... I need to ask a stupid question... Can you explain why the average has to be equal (Ey=Ex)? I'm still a little confused here...

6. Oct 23, 2006

### OlderDan

Unpolarized light means a random distribution of polarizations. If you think of light as photons, each has its own polarization. If you think of it as waves, you have to think of unpolarized light as a mix of light beams with random polarizations. In either case, an individual beam or photon has a definite polarization vector that can be resolved into perpendicular components along any coordinate axes you choose that are perpendicular to the direction of propegation. For reflected light, the plane of incidence/reflection is a natural choice for one of those coordinates, with the direction perpendicular to the plane for the other coordinate.

Any single photon or polarized beam will have its own set of components, but since there is no preferred direction of polarization the average over all beams or photons will be the same for the in-plane components as for the perpendicular components. Since the behavior of any beam or photon can be analyzed by treating its components independently, the average behavior is the same as the behavior of a polarized beam that has equal components. Equating the components is just a way of replacing the unpolarized beam with a polarized beam that will be representative of the aggregate of all the random beams.

Last edited: Oct 24, 2006
7. Oct 24, 2006

### asdf1

Thank you very much for explaning it very thoroughly!!! :)