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Relativistic momentum and energies

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data
    A meson (elementary particle) decays into two photons, each of energy 150MeV in its rest frame. Find the mesonic kinetic energy in the case of a symmetric decay in flight with an angle of 60degrees between the photon momenta (30 degrees for each particle on each side of the direction of the meson's motion).

    2. Relevant equations
    well e=mc^2 is clearly used
    maybe ke=1/2mv^2
    the momentum of a photon is p=e/c
    E(total)^2 = (pc)^2+E(rest)^2
    and E(total)^2-(pc)^2 is an invariant (i know this equation is the main way to solve the problem but i'm not exactly sure how to use this properly)
    and E(total)=KE+E(rest)

    3. The attempt at a solution
    First I analyzed the case of the rest frame:
    the meson at rest will decay into two photons with 180 degrees between the momenta so the total momentum will be zero, the energies of each photon is 150MeV and the momentum of each individual photon will be p=150/2.99e8.
    I also assumed that the meson had a total rest energy of 300MeV since energy cannot be created/destroyed.
    Now I know I'm supposed to use the invariance equation now to find the energies at the relativistic velocity with 300 as the E(total)^2 but I'm not sure what p^2 I would use. I was thinking I would use the p=e/c but even still I don't know if I need to multiply that by 2 because there's two photons or…
    But on the other side of the invariance equation I think I replace E(total)^2 with (KE+E(rest))^2 which is just (KE+300)^2 and once again I'm not sure which p^2 I would use on that side.
    So I think I'm really close to just solving for KE but I have no idea which momentums to use in each p^2 in the invariance.

    Thank you all in advance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 22, 2010 #2
    ohhhhhh i think for the first invariant i just make p^2=0 cuz in the rest frame there is no momentum…i'm dumb…
    as for the invariant on the right side of the equation i think i use p^2 as p=(e/c)cos30 to find the x component since the y component cancels out anyway, so the p would be (150*2/2.99e8)cos30
    thus the equation becomes:
    E^2=(Eo+KE)^2+(pc)^2
    300^2=(300+KE)^2+((150*2/2.99e8)cos30*2.99e8)^2
    and i solve for ke, i hope that's right :/
     
  4. Mar 22, 2010 #3

    vela

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    It's kind of hard to follow what you're talking about, but your final equation will give you a negative value for the kinetic energy.
     
  5. Mar 23, 2010 #4

    berkeman

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    Staff: Mentor

    (two threads merged. please do not multiple post. Thanks.)
     
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