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Relativistic decay

  1. Jan 2, 2016 #1
    1. The problem statement, all variables and given/known data
    Hi- i've been doing some questions and came across the following

    A high energy particle, travelling towards an observer with velocity v decays to produce a photon with
    energy E in the rest frame of the particle.
    Find the frequency at which this photon would be detected by the observer

    2. Relevant equations


    3. The attempt at a solution
    so initially i tried:

    conservation of energy in particles frame: mc^2=E
    then transformed this into the lab frame
    E'=gamma(E-vp)
    E'=Esqrt((1-v/c)/1+v/c))

    and then used E'=hc/lambda to get f=E/hcsqrt((1-v/c)/1+v/c))

    but then here is my problem- surely the decay of a single particle into a single photon violates the conservation of momentum as in the particles frame it initially has momentum 0, but after the decay it has momentum E/c? I may be wrong, but I'm just slightly confused. Thanks :)
     
  2. jcsd
  3. Jan 2, 2016 #2

    PeroK

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    Are you sure the particle doesn't decay into two photons? By conservation of momentum a single particle cannot decay into a single particle with a different mass.
     
  4. Jan 2, 2016 #3
    This is my problem.... but that is the exact question
     
  5. Jan 2, 2016 #4

    PeroK

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    It could decay into another unspecified particle and a photon. It doesn't say only a photon!
     
  6. Jan 2, 2016 #5
    But surely the frequency of the photon would depend entirely on the mass of the other particle, so it would be odd not to mention it?
     
  7. Jan 2, 2016 #6

    PeroK

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    It gives you the energy of the photon. The mass of the original particle and the details of the decay are irrelevant.
     
  8. Jan 2, 2016 #7
    ahh okay- so with the exception of my 'conservation of energy' line, my above answer should hold?
     
  9. Jan 2, 2016 #8

    PeroK

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    Are you sure that's right? What would Mr Doppler have to say?

    Also, there's no need to introduce ##\lambda## as ##E' = hf_{obs}##
     
  10. Jan 2, 2016 #9
    I did initially think that, but the expression i've obtained is already that of the doppler shift? Or is that just a coincidence?

    Oh yeah- silly me
     
  11. Jan 2, 2016 #10

    PeroK

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    It's no coincidence. The energy transformation for a photon leads to the relativistic Doppler formula. Is the energy/frequency of the observed photon (in your answer) greater than or less than the emitted photon? What should it be?
     
  12. Jan 2, 2016 #11
    It should be greater, so the other way round. I've probably made an algebraic mistake somewhere. But thank you! thats really helpful :)
     
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