14.5: Totally Elastic Collision - Compton Scattering
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As a final example of a collision in special relativity, we consider the totally elastic case: a collision in which the total momentum, total kinetic energy, and the mass of all particles are conserved. An example of such a collision is Compton scattering: the collision between a photon and an electron, resulting in a transfer of energy from one to the other, visible in a change of wavelength of the photon. For our example, we’ll take the electron to be initially stationary, and the photon to be coming in along the x-axis; after the collision, both particles have nonzero velocities in both the x and y directions (see Figure 14.4.1).
The four-momenta of the electron and photon before and after the collision are given by:
¯pe,i=(mec000),¯pγ,i=Eic(1100),¯pe,f=(Ee,f/cpe,fcosϕpe,fsinϕ0),¯pγ,f=Efc(1cosθ−sinθ0)
We can now solve for the energy Ef of the outgoing photon (and thus its wavelength) in terms of that of the incoming photon (Ei) and the scattering angle θ. There are again (at least) two ways to do this. One is to compare the components of the initial and final energy-momentum four-vector term by term. The other is to again use the fact that we know about the length of the four-vector to immediately eliminate the scattering angle ϕ of the electron. To do so, we first rewrite the conservation of energy-momentum equation, ¯pe,i+¯pγ,i=¯pe,f+¯pγ,f to isolate the term of the outgoing electron, and then take the square, to get:
(¯pe,i+¯pγ,i−¯pγ,f)2=¯p2e,f
¯p2e,i+¯p2γ,i+¯p2γ,f+¯pe,i⋅¯p2γ,i−2¯pe,i⋅¯pγ,f−2¯pγ,i⋅¯pγ,f=¯p2ef
m2ec2+0+0+2meEi−2meEf−2EiEfc2(1−cosθ)=m2ec2
from which we can solve for Ef. Rewriting to wavelengths (throughE=hf=hc/λ), we get
λf=λi+hmec(1−cosθ)