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Relativistic kinetic energy and proton collisions

  1. May 21, 2008 #1
    [SOLVED] Relativistic kinetic energy and proton collisions

    1. The problem statement, all variables and given/known data

    Find the minimum proton kinetic energy required to produce an
    antiproton in the reaction

    [tex]P+P\rightarrow P+P+P+\overline{P}[/tex]

    for protons:

    (a) Incident on protons of equal and opposite momentum,

    (b) Incident on stationary isolated protons.



    2. Relevant equations

    [tex]K=(\gamma-1)mc^{2}[/tex]
    [tex]E=\gamma mc^{2}[/tex]

    3. The attempt at a solution

    For part (a) equal and opposite momentum would give,

    [tex]2(\gamma mc^{2}=4mc^{2}[/tex]
    [tex]\gamma=2[/tex]
    [tex]v^{2}=(3/4)c^{2}[/tex]

    So,

    [tex]K=(\gamma-1)mc^{2}[/tex]
    [tex]K=1.5*10^{-10} J[/tex]

    For part (b) assuming only one proton has all the kinetic energy required to create a proton anti-proton pair would give,

    [tex](\gamma mc^{2}=4mc^{2}[/tex]
    [tex]\gamma=4[/tex]
    [tex]v^{2}=(15/16)c^{2}[/tex]

    So,

    [tex]K=(\gamma-1)mc^{2}[/tex]
    [tex]K=4.5*10^{-10} J[/tex]

    Please can anyone tell me if I hae gone wrong somewhere, I'm new to all this relativistic stuff.
     
  2. jcsd
  3. May 21, 2008 #2

    Hootenanny

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    Gold Member

    Looks good to me :approve:
    This is not quite correct. You need to be careful here, you have not considered conservation of momentum. Note that in the previous question you have assumed that all the decay products are produced at rest, which is plausible and satisfies conservation of momentum since the two protons were collided with equal energies in opposite directions.

    However in this case, a proton is colliding with a stationary proton and therefore to satisfy conservation of momentum, the centre of mass of the decay products must have some non-zero velocity. That is, the decay products must have some kinetic energy.

    Part (b) can be solved quite simply using energy-momentum four-vectors, have you met four-vectors before?
     
  4. May 21, 2008 #3
    Hello Hootenanny,

    Thanks for the pointer, now that i look at it that was a silly error to make:blushing:.

    As for four vectors, I had not studied them yet, so I have done a little research, although I still don't really understand the principle.

    Here is my attempt,

    The four-momentum vector before the collision is [tex][P_{p},(E_{p}+m_{p}c^{2})/c][/tex]
    and the lorentz invariant dot product is,

    [tex]P.P=(E_{p}+m_{p}c^{2})^{2}/c^{2}-p^{2}[/tex]

    The four momentum vector evaluted at the centre of momentum after the collision is

    [tex]p^{'}=(0,4m_{p}c)[/tex]

    Giving [tex]p^{'}.p^{'}=16m_{p}c^{2}[/tex]

    Now equating the two dot products gives,

    [tex](E_{p}+m_{p}c^{2})^{2}/c^{2}-p^{2}=16m_{2}c^{2}[/tex]

    [tex](E_{p}^{2}+2m_{p}E_{p}c^{2}+m_{p}^{2}c^{4})/c^{2}-(E_{p}^{2}-m_{p}^{2}c^{4})/c^{2}=16m_{p}c^{2}[/tex]

    [tex]2m_{p}E_{p}+2m_{p}^{2}c^{2}=16m_{p}c^{2}[/tex]

    So,

    [tex]E_{p}=7m_{p}c^{2}[/tex]

    Giving a total proton energy of [tex]1.05*10^{-9}J[/tex]

    Since the kinetic energy is [tex]K=E_{p}-m_{p}c^{2}[/tex] we have,

    [tex]K=9.02*10^{-10}J[/tex]

    Does this look better?
     
  5. May 22, 2008 #4

    Hootenanny

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    Hey Vuldoraq,

    Looks a lot better to me :approve:. I'm impressed that you have managed to work through this problem using four-vectors if you haven't met them before.
     
  6. May 22, 2008 #5
    Hey again,

    Thanks for the help, I really appreciate it. :smile:. When I get stuck on these problems I find it almost impossible to see where I have gone wrong.

    Vuldoraq
     
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