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Special Relativity - energy-momentum conservation

  1. Aug 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A positron of rest mass me, kinetic energy equal to its rest mass-energy, strikes an electron at rest. They annihilate, creating two high energy photons a and b. The photon a is emitted at the angle of 90 degress with respect to the direction of the incident positron.

    (a) Split 4-vector law of energy-momentum conservation into energy and momentum conservation laws in the rest frame of the electron.

    (b) Show that the total energy of the emitted photons Ea+Eb=3me*c^2 and that Eb^2 = Ea^2 + (p^2)*(c^2) where p is 3momentum of positron and Ea, Eb are energies of photons a and b.

    (c) Use these results and identity E^2 = (p^2)*(c^2) + m0^2*c^4 to show Eb = 2me*c^2 and Ea = me*c^2. Find the direction of motion of photon b. In particular show that the angle between its direction and the direction of the positron is theta = arcsin(1/2)



    2. Relevant equations



    3. The attempt at a solution

    Qp, Qe = 4momenta of positron and electron
    Pa, Pb = 4momenta of photon a and b.

    Qp = (2me*c, me*v) v=velocity of positron
    Qe = (me*c, 0)
    Pa = (Ea/c, Ea/c n) n=direction of photon
    Pe = (Eb/c, Eb/c n)

    This is far as I can get! Can anybody please help? Have I split up the conservation law okay?
    Thanks
     
  2. jcsd
  3. Aug 27, 2009 #2

    kuruman

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    A conservation law is usually represented as an equation. Here, a "split" would be to say

    Momentum before = Momentum after
    Energy before = Energy after

    You have not done that.
     
  4. Aug 27, 2009 #3
    Sorry, I didn't put that in.

    Qp + Qe = Pa + Pb
     
  5. Aug 27, 2009 #4

    kuruman

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    That's only one equation. You need two plus you need to put in symbols for the rest masses, speed of positron, etc. in these two equations.
     
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