Can someone explain how Compton got his equation: h/mc^2(1-cos(x)) from energy and momentum of photons and electrons from these two relations which I am not getting: (1) E_y+E_e=E_y'+E_e' (2) p_y=p_y'+p_e' and y did he put it in the E+m=E'+(p_e^2+m^2)
by 
dishelle
ANSWERS: 1
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First, photons act like protons, thus energy= E=hf and momentum = p=hc/l..(l=lambada) We can see the latter is correct because; E=(1-v^2/c^2)^-1/2mc^2 and v=c for our photons meaning m=0 if E is finite Therefore all particles with velocity c have a zero rest mass. Since E^2= p^2c^2=(mc^20^2= p^2c^2+0; than pc=E=hf and p= hf/c =h/l (again l= lambada) So...p=hf/c=h/l l2-l1=dl=h/mc^2(1-cos@) Compton Wavelength is defined as lc=h/mc=.0024nm **the mc in the last two lines would be seen with e as the subscript
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