PDA

View Full Version : conservation of energy and momentum


forrealfyziks
Feb9-10, 08:42 PM
1. The problem statement, all variables and given/known data
Consider a head-on, elastic collision between massless photon (momentum Pnot and energy Enot) and a stationary free electron. Assuming that the photon bounces directly back with momentum p (in the direction of -Pnot) and energy E, use conservation of energy and momentum to find p.


2. Relevant equations
massless: E=pc
E=\gammamc2
p=\gammamu
Maybe relevant...but probably not : E=hf

3. The attempt at a solution
I'm assuming that p is the momentum of the electron, because it is the only momentum not denoted in the problem. Note: Pnot is momentum of photon before collision, p is momentum of photon after collision, and m is the mass of the electron. I've set up this:

(Pnot)c + mc2 = pc + \gammamc2
Pnot=pe - p

I remove a c from the top equation and isolate Pnot on the left side.
Pnot= p + \gammamc - mc

I plug this into the second equation

p + \gammamc - mc = pe - p = \gammamu - p

From here I try a couple different things, but my main method seems to be putting p on one side and pulling things out

2p= m(\gammau - \gammac + c)

thanks for any help