- #1
forrealfyziks
- 13
- 0
"basic high school" algebra, with physics
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.
E=[tex]\gamma[/tex]mc2
p=[tex]\gamma[/tex]mu
massless: E=pc
rest mass: E=mc2
E2=(pc)2+(mc2)2
v/c=pc/E
[tex]\gamma[/tex]=1/[tex]\sqrt{1+(v/c)^2}[/tex]
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+[tex]\gamma[/tex]mc2
po+mc=p+[tex]\gamma[/tex]mc
po=p+[tex]\gamma[/tex]mc-mc
conserving momentum: po=p-p=[tex]\gamma[/tex]mu-p
Plugging the result I got in conserving energy into the momentum equation:
p-p=p+[tex]\gamma[/tex]mc-mc
p=2p+mc([tex]\gamma[/tex]-1)
Homework Statement
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.
Homework Equations
E=[tex]\gamma[/tex]mc2
p=[tex]\gamma[/tex]mu
massless: E=pc
rest mass: E=mc2
E2=(pc)2+(mc2)2
v/c=pc/E
[tex]\gamma[/tex]=1/[tex]\sqrt{1+(v/c)^2}[/tex]
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+[tex]\gamma[/tex]mc2
po+mc=p+[tex]\gamma[/tex]mc
po=p+[tex]\gamma[/tex]mc-mc
conserving momentum: po=p-p=[tex]\gamma[/tex]mu-p
Plugging the result I got in conserving energy into the momentum equation:
p-p=p+[tex]\gamma[/tex]mc-mc
p=2p+mc([tex]\gamma[/tex]-1)