# High school algebra -> relativistic conservation of momentum and energy

 P: 13 1. The problem statement, all variables and given/known data Consider a head-on, elastic collision between a massless photon (momentum po and energy Eo) and a stationary free electron. (a) Assuming that the photon bounces directly back with momentum p (in the direction of -po) and energy E, use conservation of energy and momentum to find p. 2. Relevant equations E=$$\gamma$$mc2 p=$$\gamma$$mu massless: E=pc rest mass: E=mc2 E2=(pc)2+(mc2)2 v/c=pc/E $$\gamma$$=1/$$\sqrt{1+(v/c)^2}$$ 3. The attempt at a solution Note:First of all I know that this is relativity, but it boils down to just plain algebra. I can't figure it out and help is hard to find, so if you can help I would really appreciate it. I assume that p is the momentum of the electron. m=mass of the electron u=velocity of the electron c=speed of light conserving energy: poc+mc2=pc+$$\gamma$$mc2 po+mc=p+$$\gamma$$mc po=p+$$\gamma$$mc-mc conserving momentum: po=p-p=$$\gamma$$mu-p Plugging the result I got in conserving energy into the momentum equation: p-p=p+$$\gamma$$mc-mc p=2p+mc($$\gamma$$-1)
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,868 Generally, it's a good idea when solving these types of problems to avoid using $\gamma$ and velocities if you can avoid them. Work with energy, momentum, and mass instead. Doing so usually makes the algebra simpler. I think your best approach is to solve for the electron's energy. Once you have that, you can calculate its momentum. You can rewrite your equations as follows: $$p_0 c - p c = E_e - m c^2$$ $$p_0 + p = p_e$$ Multiply the second equation by c, square both equations, then subtract the second one from the first, and use the relation $E^2 - (pc)^2 = (mc^2)^2$ to simplify what you get. Figure out how to eliminate p from the equation and solve for $E_e$.
 P: 13 I think I'm supposed to find p numerically, because the second part of the question asks you to verify your answer in part a using Compton's formula where $$\Theta$$=$$\pi$$.
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,868 Oh! When they asked you to find p, they were probably referring to the momentum of the recoiling photon, not the electron's. What you want to do is get rid of $E_e$ from your answer.