View Full Version : Fundamental process of a photon elastically scattering from an electon
actionintegral
Oct2-06, 09:58 AM
I learned a lot from the discussion of the converse problem, photon absorption. What about this case? Perfectly elastic.
Is there something specific that you have in mind?
actionintegral
Oct2-06, 11:08 AM
Yes, the thread I refer to (inelastic scattering) generated three pages of discussion. Is elastic scattering so well understood that it prompts no discussion?
Meir Achuz
Oct2-06, 02:44 PM
Elastic scattering of a photon by an electron is called Compton scattering.
It is well understood in QED, but that won't stop discussion in this forum.
Elastic scattering of a photon by an electron is called Compton scattering.
Compton scattering is inelastic.
Compton scattering is inelastic.
How so? The total kinetic energy before equals the total kinetic energy after the scattering. (A photon's total energy equals its kinetic energy because it has no rest energy.)
There's momentum transfer in Compton scattering. The total momentum and energy are of course conserved but Compton scattering spectroscopy for example is always referred to as a type of inelastic scattering techinque.
There's momentum transfer in Compton scattering.
Most any collision, elastic or inelastic, has momentum transfer. If there's no momentum transfer, the incoming particles each maintain their original momenta, which is the same result as no collision at all!
Or are you thinking of something different than I am, by "momentum transfer"? Conservation of momentum says
{\vec p}_1 + {\vec p}_2 = {\vec p}_1 \prime + {\vec p}_2 \prime
-({\vec p}_1 \prime - {\vec p}_1) = {\vec p}_2 \prime - {\vec p}_2
To me, either side of the second equation gives the momentum transfer, depending on which particle you think of the momentum as "coming from."
Yeah it seems like we're using different terminology maybe due to coming from different backgrounds. By momentum transfer I mean that the outgoing photon has different magnitude of momentum than the incoming one had. That's what differentiates between inelastic and elastic light scattering techniques in condensed matter research.
actionintegral
Oct3-06, 10:53 AM
That's what differentiates between inelastic and elastic light scattering techniques in condensed matter research.
Please expand on this
RandallB
Oct3-06, 11:56 AM
By momentum transfer I mean that the outgoing photon has different magnitude of momentum than the incoming one had.
That's what differentiates between inelastic and elastic light scattering techniques in condensed matter research.This sounds like “elastic light scattering” means no scattering of the light at all; as it just continues as before unaffected by the electron “absorbing and re-emitting” it.
Is this the type of “Absorption and reemission” that is used to account for light going though water as an example completely unchanged accept by speed while in the water?
The speed change of course accounting for refraction of light at interfaces of media imposing different speeds on the photon.
It has always seemed to me this “absorb & reemit” explanation was a too convenient, and incomplete, explanation of how the speed of light slows down in a media like water.
Is there a detailed understanding of such elastic “Absorption and reemission” effects on a photon; and what is happening to the electron as it does this?
Since the change to the photon is not completely zero as it is “slowed down” while in the media; is there any affect on the water or the electrons there in any measurable way?
actionintegral
Oct3-06, 12:46 PM
So Condensed Matter people refer to elastic scattering as emission and reabsorption? And they refer to inelastic scattering as a loss of energy by the photon?
RandallB
Oct3-06, 03:32 PM
So Condensed Matter people refer to elastic scattering as emission and reabsorption? And they refer to inelastic scattering as a loss of energy by the photon?I don't see any post here that says either of those things.
Meir Achuz
Oct3-06, 05:55 PM
Compton scattering is elastic. If initially at rest, the electron recoils so that the final photon has less energy than the incident photon (by the amount of energy given to the electron. In physics, this is elastic scattering. To the extent that optics is not physics, it could be called inelastic.
actionintegral
Oct4-06, 06:19 AM
This sounds like “elastic light scattering” means no scattering of the light at all; as it just continues as before unaffected by the electron “absorbing and re-emitting” it.
That's what I took away from this statement
Compton scattering is elastic. If initially at rest, the electron recoils so that the final photon has less energy than the incident photon (by the amount of energy given to the electron. In physics, this is elastic scattering. To the extent that optics is not physics, it could be called inelastic.
Well then there's a lot of really bad terminology used in peer reviewed journals. That's how people who scatter x-rays off stuff just talk about this. You are correct about this but I still wouldn't call RIXS an elastic scattering method although total energy and momentum are of course conserved.
RandallB
Oct4-06, 08:11 AM
Compton scattering is elastic. If initially at rest, the electron recoils so that the final photon has less energy than the incident photon (by the amount of energy given to the electron. In physics, this is elastic scattering. The only way the final photon could have less energy is if it changed Frequency (Wavelength). An interaction where the light changed wavelength such as one photon being absorbed pushing an electron high energy “orbit” then emitting two photons by dropping to an intermediate energy level and then back to the original would not be considered “scattering”.
You say Compton is elastic inha calls it inelastic?
Seems to me inelastic is the better description.
actionintegral
Oct4-06, 08:58 AM
You say Compton is elastic inha calls it inelastic?
Seems to me inelastic is the better description.
How so? Energy is conserved. It seems inelastic would be when the photon "merges" with the electron as in pushing it to a higher energy level.
Of course energy is still conserved in this case, but not in terms of the participants of the collision
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