180 degree Compton Scattering: low versus high energy initial photon

  • Thread starter ltarhab
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Main Question or Discussion Point

So when you calculate the scattered electron energy and the scattered photon energy (for 180 degree deflection) you get roughly the following (in keV).

Photon(in)__Photon(scattered)__Electron(recoil)
27.5_______24.8______________2.7
81_________62_______________19
140________91_______________49
364________150______________214
511________170______________341
1330_______214______________1116
infinity_____255.5_____________~infinity

So as the Photon(in) energy increases, the scattered photon energy approaches 255.5 keV (aka half of an electron's rest energy). And as the Photon(in) energy decreases, the scattered photon energy approaches the input energy. Why is this so?

For the high energy case, the best explanation I could come up with is that this is due to the conservation of the relativistic energy and/or the invariant mass of the system. In sort of the same way that two 255.5 keV traveling in opposite directions would be equal to an electron at rest. But this doesn't necessarily explain what is going on at the lower energies.

I've run through the equations too many times now and am still having a little trouble conceptualizing exactly what is going on in the low energy case. Any input would be appreciated.
 

Answers and Replies

  • #2
mathman
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If you use wavelength as your definition for photon energy with wavelength scaled so that λ = 1 corresponds to the rest mass of an electron, then backscattering will give λ'=λ+2.
 
  • #3
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If you use wavelength as your definition for photon energy with wavelength scaled so that λ = 1 corresponds to the rest mass of an electron, then backscattering will give λ'=λ+2.
Right, you're basically just stating what Compton did in his original paper; the difference in wavelengths is equal to the Compton wavelength of the electron times 1-Cos(theta). So I guess if you want to deal with it in terms of wavelength, what is the significance of 2 times the Compton wavelength? Again, intuitively speaking with respect to photon-electron interaction.
 
  • #4
Ken G
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It seems to me the low-energy limit is actually the easier one, because in that limit, the rest mass of the target electron might as well be virtually infinite. In other words, there is an ignorable parameter in this problem, which is the actual rest mass of the target-- all that matters is the ratio of the rest mass to the photon energy. So instead of lowering the photon energy and keeping the rest mass constant, you could imagine getting the same answer by keeping the photon energy constant, but raising the rest mass. In the limit of large rest mass, the target acts like a brick wall, that simply bounces back the photon at the same energy it came in at.
 
  • #5
mathman
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Right, you're basically just stating what Compton did in his original paper; the difference in wavelengths is equal to the Compton wavelength of the electron times 1-Cos(theta). So I guess if you want to deal with it in terms of wavelength, what is the significance of 2 times the Compton wavelength? Again, intuitively speaking with respect to photon-electron interaction.
Since you are familiar with Compton's derivation, I suggest you look there. It arises from the conservation laws (energy and momentum).
 
  • #6
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Since you are familiar with Compton's derivation, I suggest you look there. It arises from the conservation laws (energy and momentum).
As stated in the OP, I've already ran the math. About 5 pages back to back running the different conservation of energy/momentum/mass equations, and found all kinds of mathematical relations. The answer I was looking for was what Ken G gave (thanks BTW).
 

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