Electron-Proton Elastic Collision

[Magister]

Homework Statement

Consider the electron-proton elastic collision. Prove that the expression for the energy of the outgoing electron is

E_f= \frac{E}{1+(2E/Mc^2)sin^2(\theta /2)}

where E_f is the energy of the outgoing electron, E is the energy of the incident electron, M is the proton mass and \theta is the scattering angle of the electron.

Homework Equations

E = E_f + E_p
P = P_f cos(\theta) + P_p cos(\theta_p)
0 = P_f sin(\theta) + P_p sin(\theta_p)

E^2 = P^2 c^2 + M^2 c^4

The Attempt at a Solution

Well, I have tried the following
(P - P_f cos(\theta))^2 = (P_p cos(\theta_p))^2

(P_f sin(\theta))^2 = (P_p sin(\theta_p))^2

Adding both we get

P_{p}^2 = (P-P_f cos(\theta))^2 + P_f^2 sin^2(\theta) = P^2 + P_i^2 - 2PP_f cos(\theta)

Putting this into the energy equation for the proton we get

E_p^2 = (P^2 + P_{f}^2 - 2PP_f cos(\theta)) c^2 + M^2 c^4

Assuming that E, E_f>>mc^2, so that we can ignore the electron mass, we get

E^2 = P^2 c^2

E_p^2 = E^2 + E_{f}^2 - 2EE_f cos(\theta)+ M^2 c^4

Using the fact that

E_p^2 = (E - E_f)^2 = E^2 + E_f^2 - 2EE_f

we get
0 = 2EE_f - 2EE_f cos(\theta) + M^2 c^2 = 2EE_f (1-cos(\theta)) + M^2 c^4

E_f = \frac{- M^2 c^4 }{2E(1-cos(\theta))} = \frac{- M^2 c^4 }{4E sin^2(\theta/2)}

well, this is obvious wrong… it even gives a negative energy. What am I doing wrong? I have spent a lot of time looking for a mistake.

Thanks a lot for any suggestion

 

[Meir Achuz]

It should be E+M=E_p+E_f

 

[Magister]

That's right. Thanks a lot for the suggestion it was very useful.