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Relativstic energy and momentum.

  1. Aug 3, 2007 #1


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    A neutral pi meson, rest mass 135 MeV, decays symmetrically into two photons while moving at high speed. the energy of each photon in the lab system is 100 MeV.
    find the angle 'a' in the lab system between the momentum of each photon and the initial line of motion.

    the answer in the book is 42 degrees, but i found that it's 39.88 or something like that, here's my solution the initial four momentum is: (E/c,P)=(2E_ph,P1+P2)
    now by using lorentz invariant here:
    and we know that E_ph/c=P1=P2
    so we get (2E_ph^2/c^2)-(E_ph/c)^2*cos(a)=M_pi^2c^2
    from there i calclulated the angle, but didn't get 42 degrees, perhaps the book is outdated and this accounts for the rounding up.

    any advice as to wether i got it confused or the book?
  2. jcsd
  3. Aug 3, 2007 #2


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    i have another question, not quite sure if i got it right.
    A high energy photon (gamma ray) collides with a proton at rest. A neutral pi meson is produced according to the reaction:
    gamma photon+ proton=> proton+neutral pi meson
    What is the minimum energy the gamma ray must have for this reaction to occur? the rest mass of a proton is 938 MeV and that of the neutral pi meson is 135 MeV.
    well again im using here four-momentum:
    before the reaction: P=(E'/c+m_p*c,E'/c) where E' is the energy of the photon.
    now after the reaction we have P'=(E1/c,p_1)+(E2/c,p_2) where 1 is the proton and 2 is the pi meson, now from this we have the equations:
    after rearranging i get the next equation (i reming you that i need to find E'):
    now here is where im stuck, i need to replace p1 and p2 and also E1 and E2, but it has to be some nasy algebraic manipulation, how to get my head over it?

    thanks in advance.
  4. Aug 3, 2007 #3
    I get 42.45°. Perhaps double-check your calculations starting from [tex] m_\pi{}^2 = 2 E_\gamma{}^2 - 2 E_{\gamma}{}^2 \cos \alpha [/tex], which I also got.

    Disclaimer: I omitted all factors of c. Think I should explicitely note it since it often seems a source of confusion.

    EDIT: Got a bit confused over your non-use of tex. You forgot the factor 2 at the cos-term stemming from [tex] (\vec p_1 + \vec p_2)^2 = \| \vec p_1 \| + \| \vec p_2 \| + \mathbf{2} \| \vec p_1 \| \vec p_2 \| \cos \alpha [/tex], either as a typo due to the badly readable ASCII or in your calculation.
    Last edited: Aug 3, 2007
  5. Aug 3, 2007 #4


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    ok, i see where i got my mistake, forgot the two, in the term with the cosine, thanks.
    can you help me with the next question?

    welcome aboard, i see you're new to the forums, it's a great site, hope you'll enjoy your stay here.
  6. Aug 3, 2007 #5
    You should really use tex to display your equations - makes your posts much more readable (especially when considering your relatively condensed posting style) and therefore makes it more likely that someone will take the time reading them and possibly replying, then.

    Two remarks on the 2nd problem:
    - Since you're explicitely keeping factors of c, there's an error in your "P=...": The first E'/c should be an energy, i.e. only E'.
    - I've not worked through the problem but the first thing that I'd try is assuming [tex]\vec p_1= \vec p_2=0[/tex] and [tex]E=m_p+m_\pi[/tex] in the center-of-mass frame (which is not the frame that E' is given/asked!) and see if it gives sensible results.
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