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Decay into photons

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A mass M moving at speed V going east decays into 2 photons. One moves perpendicular (south) and the other at some angle from the horizontal. Show that if tan(theta) = 1/2, then V/C = [sqrt(5)-1] / 2

    3. The attempt at a solution

    (gamma is the lorentz factor)

    Momentum in x: x momentum of angled photon: gamma*m*v
    Momentum in y: y momentum of angled photon = momentum of perpendicular photon = E1/C

    Solving for E1, I get mc^2/2*gamma

    Since tan(theta) = 1/2 = [P in y]/[P in x]

    then 1/2 = [E1/C]/[gamma*m*v]

    I get V = C/gamma^2...
     
  2. jcsd
  3. Nov 29, 2009 #2

    tiny-tim

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    Hi mathman44! :smile:

    Why gamma? :confused:

    Photons travel at the speed of light. :wink:
     
  4. Nov 29, 2009 #3
    Conservation of momentum says that the x momentum of the angled photon is equal to the x momentum of the mass before decay... right? So to find the momentum of the mass before decay, I need gamma.
     
  5. Nov 29, 2009 #4

    Redbelly98

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    Looks good so far. What is gamma, by definition?
     
  6. Nov 29, 2009 #5
    gamma = 1/sqrt(1 - v^2/c^2)

    :S?
     
  7. Nov 29, 2009 #6

    tiny-tim

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    oh i see, you're using gamma for the mass M …

    but you need the momentum of both photons, and you also need conservation of energy.
     
  8. Nov 29, 2009 #7
    I used cons momentum with both photons, no?

    y: momentum of photon 1 = momentum of photon 2
    x: momentum of photon 2 = momentum of mass before decay

    where photon 2 is the angled one. I have already solved for E1, so what do I need cons. of energy for?
     
  9. Nov 29, 2009 #8

    Redbelly98

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    You can combine these equations to find v
     
  10. Nov 29, 2009 #9
    Then V = C - V^2/C? Doesn't make sense to me...
     
  11. Nov 29, 2009 #10

    tiny-tim

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    actually, V2/c = 1 - V2/c, but still gives the wrong answer :redface:
    mathman44, i don't know how you solved for E1, but you do need cons. of energy :smile:
     
  12. Nov 29, 2009 #11
    Yes, you're right... my teacher gave us E1 in class. Does

    gamma*m*c^2 = p1c + p2c

    look good? And if so, I don't know p2 so how can I solve for p1c (E1)? My brain hurts.
     
  13. Nov 29, 2009 #12

    tiny-tim

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    Yes, but split p2 into x and y components, and use Pythagoras. :wink:

    (going to bed now … goodnight! :zzz:)
     
  14. Nov 29, 2009 #13

    Redbelly98

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    If you solve this quadratic equation in v, you'll get the answer you are supposed to get.

    There is a problem with inconsistent units here :confused:

    Hmmm, I'm also puzzled how E1 was solved for. But mathman44's equation does yield the same answer given in the problem statement.
     
  15. Nov 29, 2009 #14

    Redbelly98

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    In that case, you don't need conservation of energy. Presumably it was already used by your teacher to determine E1.

    Just solve the quadratic equation for v you had earlier.
     
  16. Nov 29, 2009 #15
    Great, thanks, that works perfect!
     
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