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GRE # 37 ( elementary particle question)

  1. Sep 27, 2004 #1
    A Pie zero meson ( rest-mass energy 135 Mev) is moving with velocity .8c K in the laboratory rest frame when it decays into 2 photons, gamma1 and gamma2. In the pie zero rest frame, gamma1 is emitted forward and gamma2 is emitted backward relative to the pie zero direction of flight. The velocity of gamma2 in the laboratory rest frame is
    (A) -c k
    (B) -.2c k
    (C) .8c k
    (D) 1.0c k
    (E) 1.8c k

    I know that it is either (A) or (D), because it is a photon, must travel at the speed of light in any frame, but I narrowed it down to A, in the rest frame of pie zero, and to D in the laboratory frame, However the correct answer is A in the laboratory rest frame. I am not sure why it is different from my intuition :uhh: .

    Thank u so much!
     
  2. jcsd
  3. Sep 27, 2004 #2

    nrqed

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    It's simply conservation of four-momentum. We have in the lab(I set c=1 in all my equations):

    [itex] \vec p_1 + \vec p_2 = \gamma m {\vec v} [/itex]

    and

    [itex] E_1 + E_2 = \gamma m [/itex]


    What's neat about photons is that E is the magnitude of the 3-momentum. So the second equation tells us the sum of the magnitudes. Now, since the sum of the vectors has a magnitude smaller than the sum of the magnitudes (because of the extra factor of v), we see right away that the two vectors must point in opposite directions. And that's it.

    If you plug in v =0.8c, you can also easily find the energies of each photon. You get about 1.5 m c^2 for the forward photon and about 0.16 mc^2 for the backward photon.

    Regards

    Pat
     
  4. Sep 27, 2004 #3
    Pat,
    I think I lost u :(
    I understand that the vector sum is less than the sum of the magnitudes, but doesn't mean that they are opposite in direction, it just means that there is an angle between them, but not necessarily opposite in direction.
    And also, I thought that all photons will travel with the same velocity = c = speed of light, thus my question is how u got different energies of each one of those photons? energy = gamma*m*c^2, if these photons are traveling at the speed of light, means they have the same gamma, thus the same energy...I think I am missing something here :(
    thank u
     
  5. Sep 28, 2004 #4

    nrqed

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    Hi,

    I am rushing to go to class so I'll answer a bit later today, ok?

    We'll need to discuss a bit more the relation between energy, momentum and mass in relativity because it seems like you might have some misconceptions about these concepts.

    Sorry, got to go. Talk to you later today.
    Pat
     
  6. Sep 28, 2004 #5
    I will surely wait, thank u so much! :smile:
     
  7. Sep 28, 2004 #6

    nrqed

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    You are right, for a general case. But the question explicitly stated that in th pion rest frame, the two photons were emitted along the axis of the pion's momentum (observed in th elab frame). Therefore, in the lab frame, the photons could either move along [itex] + {\hat k}[/itex] or along [itex] - {\hat k} [/itex]. That's what I used to say that they had to move in opposite directions.

    First of all, I made a typo in my last line. Sorry about that. I should have written [itex] E_1 \approx 1.5 m c^2 [/itex] and [itex] E_2 \approx 0.16 m c^2 [/itex] i.e. there should be no gamma factors.

    But again, maybe my notation confused you. I should have written things more explicitly, that is

    [itex] E_1 \approx 1.5 m_{\pi} c^2 [/itex]

    [itex] E_1 \approx 0.16 m_{\pi} c^2 [/itex]

    i.e. my mass in these equations is the value associated to the pion. So these are not meant tp be general formula for energies of photons, they are special cases for this specific problem. Plugging in values, you would get [itex] E_1 \approx 203 MeV[/itex] and [itex] E_2 \approx 22 MeV [/itex]. So the sum is 225 MeV, which is about [itex] \gamma_{\pi} m_{\pi} c^2 [/itex] and the difference is 181 MeV which is about [itex] 0.8 \gamma_{\pi} m_{\pi} c^2 [/itex].

    So I apologize if my notation was the source of confusion. All I meant to say is that we can (of course) express the energies of the photons in terms of the energy of the pion, with the formula given above.

    *But* part of your post may indicate a deeper misunderstanding. You
    wrote

    Be very careful! Of course photons may have different energies! The key point is that of course, one cannot use gamma m c^2 for a photon if one means the "mass of the photon times the gamma of the photon". That's not defined. One can may only use [itex] E = c p [/itex] where p is the magnitude of the three-momentum. All photons travel indeed at the speed of light, but they may have different energies and momenta.

    Again, maybe it's my notation which confused you. I was writing the energies of the photons in terms of the pion energy!

    Hope this clarifies things.

    Pat
     
  8. Sep 28, 2004 #7
    yes indeed, thank u Pat, it is clear now :approve: .
    I don't know what I was thinking when I said photon energies couldn't be different. Thank u for clearing that up.
    But for some reason, I thought that a pion traveling in the K direction could decay into 2 photons that arent in the k direction, in such a way that the momentums of the non k directions cancel out. :rolleyes:
    but I guess, I should use the KISS theory. :blushing:
    did u see my point?

    Thanks again Pat! u r so helpful.
     
    Last edited: Sep 28, 2004
  9. Sep 28, 2004 #8

    nrqed

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    You are welcome. One point: you are absolutely right that the photons don't have, a priori, to be in the k direction. It was not a question of applying the KISS principle. It was explicitly stated in the question (by saying that in the frame of the pion, the photons were emitted in opposite directions along the axis of motion of the pion in the lab)! So I did not assume this.

    Take care

    Pat

    PS: just a detail: it's called a "Pi meson" (and not a "Pie meson" :smile: )
     
  10. Sep 28, 2004 #9
    :biggrin: I thought they are the same, because we eat pies at pi day :biggrin: .
     
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