"high school" algebra > relativistic conservation of momentum and energyby forrealfyziks Tags: algebra, conservation, energy, high school, momentum, relativistic 

#1
Feb1110, 11:52 PM

P: 13

1. The problem statement, all variables and given/known data
Consider a headon, elastic collision between a massless photon (momentum p_{o} and energy E_{o}) and a stationary free electron. (a) Assuming that the photon bounces directly back with momentum p (in the direction of p_{o}) and energy E, use conservation of energy and momentum to find p. 2. Relevant equations E=[tex]\gamma[/tex]mc^{2} p=[tex]\gamma[/tex]mu massless: E=pc rest mass: E=mc^{2} E^{2}=(pc)^{2}+(mc^{2})^{2} v/c=pc/E [tex]\gamma[/tex]=1/[tex]\sqrt{1+(v/c)^2}[/tex] 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: p_{o}c+mc^{2}=pc+[tex]\gamma[/tex]mc^{2} p_{o}+mc=p+[tex]\gamma[/tex]mc p_{o}=p+[tex]\gamma[/tex]mcmc conserving momentum: p_{o}=pp=[tex]\gamma[/tex]mup Plugging the result I got in conserving energy into the momentum equation: pp=p+[tex]\gamma[/tex]mcmc p=2p+mc([tex]\gamma[/tex]1) 



#2
Feb1210, 01:31 AM

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PF Gold
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Generally, it's a good idea when solving these types of problems to avoid using [itex]\gamma[/itex] 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: [tex]p_0 c  p c = E_e  m c^2[/tex] [tex]p_0 + p = p_e[/tex] Multiply the second equation by c, square both equations, then subtract the second one from the first, and use the relation [itex]E^2  (pc)^2 = (mc^2)^2[/itex] to simplify what you get. Figure out how to eliminate p from the equation and solve for [itex]E_e[/itex]. 



#3
Feb1210, 04:29 AM

P: 13

Thank you for the hints it has really helped. I worked through the things vela posted, and found myself stuck again, though.
If I add the two equations, I end up with (p_{o}^{2}c^{2}+p^{2}c^{2})=E_{e}^{2}E_{e}mc^{2} If I subtract the second from the first 2pp_{o}c^{2}=E_{e}mc^{2}+(mc^{2})^{2} On the first one, it obviously has the momentums on the left, and on the second one I also have momentums on the left. With the first one the left almost looks like one of the conservations, but not quite. I'm going to fiddle around with the information vela has already given me and see if I can't find another way. 



#4
Feb1210, 04:43 AM

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"high school" algebra > relativistic conservation of momentum and energy
It wasn't clear in your original post what terms are allowed in the final answer. I assume you want to get rid of p, the photon's final momentum. You can do that by solving for it in one of the conservation equations in terms of the other variables.




#5
Feb1210, 04:50 AM

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 [tex]\Theta[/tex]=[tex]\pi[/tex].




#6
Feb1210, 05:08 AM

P: 13

I found out from a peer that the momentum you are finding is actually the photon's second momentum, and that you should solve in terms of p_{o}. It became incredibly easy once I knew I wasn't looking for something numerical... Thank you so much for your help!




#7
Feb1210, 05:09 AM

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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 [itex]E_e[/itex] from your answer.




#8
Feb1210, 05:44 AM

P: 13

yes it was supposed to be photon's new momentum. Guess I should have put that second part of the question in, but I didn't think it was necessary. Going back over the derivation of Compton's formula, I noticed it was almost the exact thing I was doing. heh, well thanks again.



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