Compton Scattering Angle - CM FRAME

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SUMMARY

The discussion focuses on calculating the scattering angle in the center of mass (CM) frame for a photon with an energy of 10 MeV that scatters at an angle of 25 degrees. The solution involves using Compton scattering formulas, resulting in an electron scattering angle of 35 degrees, an energy of 4.02 MeV for the scattered electron, and an energy of 5.47 MeV for the scattered photon. To transition from the lab frame to the CM frame, participants suggest applying Lorentz transformations to derive the four-vectors, ensuring that total momentum in the CM frame equals zero.

PREREQUISITES
  • Understanding of Compton scattering principles
  • Familiarity with four-vector notation in relativistic physics
  • Knowledge of Lorentz transformations
  • Basic concepts of center of mass frame in particle physics
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  • Study the derivation of Compton scattering formulas
  • Learn about Lorentz transformations in detail
  • Explore the concept of four-momentum in relativistic physics
  • Investigate the implications of scattering angles in different reference frames
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Physics students, particle physicists, and anyone interested in understanding relativistic scattering processes and transformations between different reference frames.

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Homework Statement


Given incoming photon has energy 10Mev and scatters at angle 25 degrees, find the scattering angle in CM frame.
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Homework Equations

The Attempt at a Solution


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In lab frame, let the energy of incoming photon be ##E##. Total 4-vector would be ##P + Q = (\frac{E}{c},\frac{E}{c}) + (mc,0)##.

Using compton scattering formulas, I solved the scattering angle of electron ##\alpha = 35 ^o##, energy of scatted electron ##E_e = 4.02 ~MeV##, energy of scattered photon ##E' = 5.47~MeV##.Having solved this completely in the lab frame before/after scattering, how do I proceed in CM-frame?

In CM frame, total momentum is ##0##, so 4-vector would be ##P_{cm} + Q_{cm} = (\frac{E_{p,cm}+E_{e,cm}}{c},0)##.

Even when I assume that in CM frame they are both scattered along the same line it doesn't help.
 
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Did you mean ##\alpha=35^\circ## or is that a typo?

Since you know everything the lab frame, you can simply use the Lorentz transformation to find the four-vectors in the CM frame. Suppose the CM moves with speed ##\beta=v/c##. Get expressions for the three-momentum of the photon and the electron in the CM frame. As you noted, their sum should be 0, which will allow you to determine ##\beta##.
 
Solved.
 

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