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Photon colliding with stationary mass

  1. Jan 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A photon with energy E collides with stationary mass m. They form a single particle together, what is this new particle's mass and what is its speed?

    2. Relevant equations
    Energy-momentum 4-vector P=(E, px, py, pz)
    Possibly P2=m2

    3. The attempt at a solution
    Using 4- momenta, the particle moving with energy E has a 4-momenta of P1 = (E, p, 0, 0) and the stationary mass is P2 = (m, 0, 0, 0). The total momentum before the collision is then Pbefore = (E+m, p, 0, 0). My question is, what exactly would the Pafter equal to? Could it possibly be: Pafter = (E', p', 0, 0)
    If so that would show that E' = E+m (due to conservation of energy). How can I find the mass of the new particle from that? Does the second equation of m2 have anything to do with it?[/SUB]
     
    Last edited by a moderator: Jan 21, 2016
  2. jcsd
  3. Jan 21, 2016 #2

    Orodruin

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    You have the expression for the mass of a particle right here:

     
  4. Jan 21, 2016 #3
    Great, do I have all the tools to solve this then? Is Pafter = (E', p', 0, 0) correct? How can I find the speed then?
     
  5. Jan 21, 2016 #4

    PeroK

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    For a particle you have three important equations that you must not forget:

    ##E = \gamma mc^2##

    ##p = \gamma mv##

    ##E^2 = p^2 c^2 + m^2 c^4##

    In particular, if you know the energy and momentum of a particle, you can calculate its mass, velocity and gamma factor.

    You should experiment with these equations to see how you can do this.
     
  6. Jan 21, 2016 #5
    These equations are rather useful, but I just want to know if Pafter = (E', p', 0, 0). If so I can substitute E' as E+m and p' as p, using conservation of energy and momentum laws. Thanks to the third equation you've given me, I can say that p = √(E2-m2) and p' = √(E2-m2) and solve for m' since we are saying that c=1 in this example. I can't see why Pafter isn't what I think it is, the new particle doesn't travel on an angle or anything like that.
     
  7. Jan 22, 2016 #6

    Orodruin

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    The 4-momentum is proportional to the 4-velocity. You can use this fact to find the speed. In other words, how would you find the speed if I gave you a 4-velocity?
     
  8. Jan 22, 2016 #7

    PeroK

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    You should satisfy yourself from conservation of momentum that the final particle must move in the same direction as the photon.

    Also, I think you're missing the relationship between the energy and momentum of a photon. Hint: only one of those three equations holds for a photon.
     
  9. Jan 22, 2016 #8

    Orodruin

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    I do not think the problem is what you perceive it is. A student using 4-vector notation should already be familiar with what you have said, which is essentially the statement of the components if the 4-momentum for a massive particle.
     
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