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Special Relativity: Pion Decay - Two Photons

  1. Feb 5, 2014 #1
    Apologies, this is going to be a bit disjointed, I dont want to write the full question down as I dont want anyone to give me a solution as its an assignment question.

    1. am i correct in assuming when a pion decays at rest its energy will be given by:

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

    which goes to:

    E^2 = m^2 c^4

    because I get something ridiculous like 1394J. ive cancelled the c^2 of the pion mass with the c^4 in the energy eqn.

    2. to find the energy of the photons i use conservation of energy:

    E(pion) = E(photon1) + E(photon2)

    E(photon1) = GAMMA * m c^2

    i) what do i use for the mass of a photon?
    ii) how do i calculate gamma when no velocity of the photons (or the pion) is given?

    thanks in advance for any suggestions.
     
  2. jcsd
  3. Feb 5, 2014 #2

    phyzguy

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    The photon has a rest mass of zero, so the energy equation reduces to:

    E^2 = p^2 c^2, or E = pc.

    The equation E = gamma m c^2 applies only for a massive particle.

    So take the equation E = pc, then conserve energy and momentum to calculate the energy of the two photons.
     
  4. Feb 5, 2014 #3
    thanks for the reply!

    but using conservation of energy E(pion) = E(photon1) + E(photon2) doesnt seem to help because although we now have E(pion) the energies of each photon may not be equal. also how would you calculate momentum without a velocity?

    thanks again.

    btw the question says to assume relativistic, so im back to the problem of how to find gamma with no velocities given.
     
  5. Feb 5, 2014 #4

    phyzguy

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    What is the momentum of the initial pion at rest? What does this tell you about the momentum of the final system?

    Also on your question about momentum, for the photon E = pc. What is p?
     
    Last edited: Feb 5, 2014
  6. Feb 7, 2014 #5
    i) the momentum of the pion is zero at rest, so it's energy would be zero, so you would have to assume that it had a momentum before it came to rest and that it is the dissipation of this momentum that is 'given' to the photons?

    how else could it impart energy if it had none at the point of decay?

    unless it has an intrinsic energy inherent to the particle itself, would that be in the form of mass?

    so E = GAMMA * mc^2

    so the momentum of the final system would be only the energy its mass imparts

    so E= GAMMA * mc^2 = pc

    but again im stuck with GAMMA when i have no given velocities.
     
  7. Feb 7, 2014 #6
    OK, think im there,

    E(pion) = GAMMA * mc^2

    COE:

    E(photon) = 1/2 E(pion)

    p(photon) = E(photon) / c

    Only thing is how to find GAMMA with no velocities given? should i just write the answer in terms of GAMMA?

    I see that K = (GAMMA - 1) mc^2 is the kinetic energy, can you assume this is equal to its rest mass, hence find GAMMA this way, or does it have to be moving for this to be true?

    Thanks for the direction!
     
    Last edited: Feb 7, 2014
  8. Feb 7, 2014 #7

    phyzguy

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    No. The pion has a mass m. E^2 = p^2c^2 + m^2c^4. If p = 0, then E = mc^2.

    Also, as you wrote, E = gamma mc^2. If p=0, v=0, so gamma = 1, giving again E = mc^2.

    So the initial momentum is zero, but not the initial energy.
     
  9. Feb 7, 2014 #8

    phyzguy

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    You are given the initial velocity. It says at rest.
     
  10. Feb 7, 2014 #9
    ahhh i see and you can calculate the momentum of the photon from COE because we know the energy of the pion.

    thanks for all the pointers, really appreciated!
     
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