With Compton scattering the total momentum and the total energy of the photon electron system is conserved, so it has to be ELASTIC Collision/Scattering.

But an argument is that just because the kinetic energy of the incoming photon is different than the outgoing photon, it is INELASTIC Collision/Scattering.

However, I thought we had to look at the system, which is the photon and electron interaction; not just the scattered photon itself.

So, what is the real issue here?

  • $\begingroup$ "Compton scattering is an example of elastic scattering. Energy of the incident photon is transferred to the electron (recoil) but only as kinetic energy in the laboratory frame. The electron gains no internal energy, respective masses remain the same, the mark of an elastic collision." en.wikipedia.org/wiki/Compton_scattering#Introduction $\endgroup$
    – anna v
    Jun 5 at 19:55

1 Answer 1


The definition of inelasticity differs between dynamics and particle physics.

In the former, inelasticity refers to the loss of kinetic energy in the system as a whole, as some is lost (e.g. heat).

In the latter, it refers explicitly to the loss of kinetic energy of the incident particle. In the case of Compton scattering, that means the loss of photon energy.

  • $\begingroup$ Thank you this explains a lot. A lot of people confusing it. But how do we know what are they talking about?? How do we know if they referring to former or latter?? $\endgroup$ Jun 5 at 14:59
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    $\begingroup$ I believe in general you just need to know from the context of the problem. As a general rule, if it's macroscopic, refer to the former, and if it's microscopic, refer to the latter. $\endgroup$ Jun 5 at 15:00
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    $\begingroup$ so if it is macroscopic they use the term "COLLISION" which refers to the former, if it is microscopic they use the term "SCATTERING" which refers to the latter. $\endgroup$ Jun 5 at 15:03
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    $\begingroup$ Yes, that would be right. $\endgroup$ Jun 5 at 15:07
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    $\begingroup$ The Wikipedia page for Compton Scattering is in bad shape regarding this issue. Somebody should clean it up. $\endgroup$
    – Xerxes
    Jun 6 at 15:13

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