# Can someone explain Compton scattering?

1. Apr 14, 2014

### Coop

Hi,

My textbook describes Compton scattering the following way:

"...an x-ray photon...has a collision with a slow moving electron...the photon transfers energy and momentum to the electron [and recoils]..."

Is it not true that by definition when a photon collides with a particle it donates its energy in an all or none fashion depending on the particle's energy levels? I don't get how this x-ray photon is capable of only donating SOME of its energy.

Any help?

Thanks,
Cooper

2. Apr 14, 2014

### Bill_K

The electron absorbs one photon and emits another.

The scattering is elastic, and in the center-of-energy frame the ingoing and outgoing photon have the same energy. But in the lab frame, the collision causes the electron to recoil, imparting some of the photon's energy to it. The process is described by these Feynman diagrams.

3. Apr 14, 2014

### UltrafastPED

Atomic absorption requires an available energy level; in that case an electron "moves" to a higher energy state, and for emission there is a drop. This is what they study in spectroscopy.

Scattering is a different process: a particle arrives, has an interaction with the body (atom, nuclei, proton, etc) - and leaves. Rutherford's atomic model (1911) was inspired by nuclear scattering experiments.

In Compton's case very high energy photons are being scattered, and in doing so energy is transferred from the photon to the electron. This results in a change of wavelength for the photon, and a corresponding increase in energy for the electron.

See http://en.wikipedia.org/wiki/Compton_scattering

4. Apr 14, 2014

### Matterwave

The easier explanation might be that the reason that an electron has to absorb a given amount of energy (discrete spectra) is that the electron is in a bound state (inside an atom) and therefore can only take on a discrete set of energy levels. In Compton scattering or Thompson scattering, the electron is free, and can take on a continuum of energy levels.

5. Apr 14, 2014

### Coop

Thanks guys,

So is absorption actually taking place here or is it a different type of interaction? I am getting mixed answers.

6. Apr 14, 2014

### Matterwave

The photon is scattering off the electron. The electron absorbs some of the energy and momentum. If you want to interpret the Feynman diagram as an actual physical process, then the electron is absorbing the photon, and quickly (immediately) re-emitting another one. But the Feynman diagrams are just a good diagrammatic way of solving for the S-matrix, so whether you take them to mean "physically" what's happening, is really up to you.

You start out with one photon and one electron, you end up with one photon and one electron, with some momentum and energy from the photon transferred to the electron. That's all QED can really say about the matter.

7. Apr 15, 2014

### DParlevliet

Indeed it is not known what physically does happen (see wikipedia), so everyone has his own believe. Probably one must limit to look what is going in and out, like Matterwave tells.

8. Apr 15, 2014

### UltrafastPED

The usual terminology is scattering.

The Feynman diagram is not a literal representation, but rather a guide to setting up the calculations in a systematic way.

9. Apr 15, 2014

### Staff: Mentor

And in between the absorption and re-emission, the electron is virtual. (Cue the arguments about "real" versus "virtual" particles here. )

10. Apr 15, 2014

### DParlevliet

I am not so fond about the word "virtual", because it is always vague what it means. What happens between input and output is real; we only don't know what it is.

11. Apr 15, 2014

### Coop

Thanks a lot everyone :) I appreciate the input.
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