# Rayleigh-scattering and dipole Common characters?

Hello!

This time I've got a question about the "Rayleigh-scattering" in X-Rays.
The Rayleigh-scattering says, that the x-ray changes it direction only without "loosing" energy.
Why does this happen and what has it to do with dipoles? My university professor told me, that the Rayleigh-scattering has something to do with dipoles.
I can't find anything in books about that. I hope you could help me, that I can understand this problem.
Thanks a lot!!

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dextercioby
Homework Helper
There are several types of diffusion of em.radiation on matter particles,even in the classical theory of electromagnetism.

Rayleigh diffusion,treated thoroughly in Jackson's book,indeed assumes the variation of the wave vector only in direction,and not in modulus.

Daniel.

It is beacause Lord Rayleigh considered scattering on small dipoles. The power emmitted by dipole is proportional to the frequency^4, so the scattered wave becames stronger at highier frequencies (that is why the sky is blue)

I am not sure if I understand this... If we've got an atom like H and we firing x-ray at it, does it then oscillate (like a dipole) and scatters the x-ray into another direction? And in WHAT direction???
Am I totally wrong with that? ;)

First of all, the Rayleigh scattering takes places when wavelength is greater then the dipole size. So not any X-ray scattering on atoms can be considered in that approximation.

Second, if a wavelength of the X-ray radiation is bigger then the dipole size, we can assume that the dipole is under an oscillating electric field, thus it oscillates itself. The combination of original wave and the wave due to dipole radiation produces the scattering, because the dipole does not radiate flat wave. So instead of original flat wave we have waves propagating in different directions.

Third, I did not say anything about quantum. That was done deliberately, because it the the nature of the Rayleigh approximation, it is a classical effect. The mechanism will be no longer valid if quantum effects are important.