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Is the Compton scattering elastic or inelastic? Because the kinetic energy is conserved (in addition to the total energy conservation) so it should be elastic, but the energy of the photon is changed, so from that point of view is inelastic. I have found both definitions online, which one is correct?

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    $\begingroup$ There is some fussing of this in the talk page of wikipedia. A Solomonic answer would call it inelastic scattering underlain by an elastic collision. Don't waste any time on the terms: just try to understand the kinematics and ignore the terms. $\endgroup$ – Cosmas Zachos Mar 16 '17 at 22:30
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When you are considering scattering, you are looking to what happens to the incoming particle. In Compton scattering, the incoming particle, the photon, comes out after the scattering process with less energy than it started with, and so the scattering is called inelastic; whereas, with Thomson scattering, since the energy of the photon does not change, that would be termed as elastic scattering.

It might be that the confusion arises because of the term "elastic" alternatively being used for collisions when the total kinetic energy of the interacting particles is conserved?

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Let me give you an example here.

Consider the following process $\gamma+d\rightarrow\gamma+d$ (Compton scattering off deuteron):

  • In center of mass frame, energy of the photon under scattering doesn't changed (of course the direction does), then according to your definition in your post, it's a $\textbf{elastic}$ process

  • In lab frame, the photon loses some energy after scattering off the deuteron, and again, it's classified as $\textbf{inelastic}$ process by the definition in your post

However, in my field, i.e., [nucl-th/0512064], elastic Compton scattering means that: whatever comes in, and the same comes out; while inelastic Compton scattering means that whatever comes in, something different comes out.

As long as you understand what's going on, you shouldn't worry about too much about the terms they used.

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  • $\begingroup$ The center of mass frame is not the deuteron rest frame, but the one in which the sum of 3-vectors of the photon and the deuteron is zero. $\endgroup$ – HeisenbergCat Dec 29 '20 at 21:25
  • $\begingroup$ In the center of mass frame, the incoming photon has 4-vector (k,0,0,k), and the deuteron has $(sqrt(m^2+k^2),0,0,-k); with scattering angle theta, the outgoing photon has 4-vector (k',k' sin(theta),0,cos(theta)) and the deuteron has (sqrt(m^2+k'^2),-k'sin(theta,0,-cos(theta))). By 4-momentum conservation, one can easily derive that k' is the same as k. $\endgroup$ – HeisenbergCat Dec 30 '20 at 6:00
  • $\begingroup$ Absolutely correct. I withdraw my comments. $\endgroup$ – my2cts Dec 30 '20 at 9:29

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