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Meson decay into two pions

  1. Oct 31, 2016 #1
    1. The problem statement, all variables and given/known data
    90af5e5f88.png
    2. Relevant equations
    E^2=m^2~p^2
    k mass = 497.7Mev/c^2 pion mass = 139.6 Mev/c^2

    3. The attempt at a solution
    Using the rest frame i figured that E_pi=m_k/2
    so
    P_pi = sqrt(E_pi^2 - m_pi^2)
    the pion energy is half of the k meson mass so substituting that i end up with 206Mev/c for each pion.

    So if momentum is conserved then surely 2*P_pi should be P_k in the frame where the k meson is moving?

    so using E=sqrt(497.7^2 + 412^2) I got 646Mev for the k meson energy.

    Is this a valid method where I used the rest frame to find the momenta of the pions first?
     
  2. jcsd
  3. Oct 31, 2016 #2

    mfb

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    In the frame where the kaon is moving, what is P_pi? The two pions won't have the same momentum (not even as magnitude).

    There is an easier approach if you stay in the kaon rest frame for one more step. What is the speed of a 206 MeV/c momentum pion?
    You don't even need the pion momentum, you can directly use the pion energy to find the speed.
     
  4. Nov 1, 2016 #3

    vela

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    No, this isn't correct. Your reasoning would work in Newtonian mechanics, but it doesn't work in special relativity. You need to use the Lorentz transformation to calculate how the four-momentum in one frame becomes the four-momentum in another frame.
     
  5. Nov 1, 2016 #4
    Im on my phone at the moment so no pen or paper but do I need to rearrange mv/sqrt(1-v^2/c^2) to find v? If I find that speed in the kaon rest frame then where do I go from here

    also in the frame where the kaon is moving and one pion is at rest the pion momentum will be equal to the kaon momentum
     
    Last edited: Nov 1, 2016
  6. Nov 1, 2016 #5

    mfb

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    You can do that, but there are easier approaches, especially with the energy.
    Once you have that, you know the speed of the pion in the kaon rest frame. You also know the speed of one pion in the lab frame (0). You can find the relative speed between the frames that way.
    3-momentum: yes.
     
  7. Nov 1, 2016 #6
    So i calculated the pion velocity by rearranging for v and got 0.829c.

    I tried to use the lorentz transform for velocity to find the frame velocity but since the lab frame pion velocity is 0 it just ended up cancelling to 0.829c anyway.

    v=(u-u')/(1-u*u'/c^2)
     
  8. Nov 1, 2016 #7

    mfb

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    You don't need fancy transformations. In the kaon frame, the pion moves at 0.829 c. The pion does not move relative to the lab, which means in the kaon frame, the lab also moves at 0.829 c (technically you can add the velocities 0.829 c and 0 c, but the result is obvious). Due to symmetry, this also has to be the speed of the kaon in the lab frame.
     
  9. Nov 1, 2016 #8
    that makes sense. thanks.
    so to find the energy of the kaon in the lab frame do i just do gamma*m? (c=1)
     
  10. Nov 1, 2016 #9

    mfb

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    Sure.

    And if you look at the whole derivation, you'll note that you do not even need the velocity, the gamma factor is sufficient and easy to calculate.


    Beyond the scope of this problem, but related: If you reverse the time direction, you get a fixed-target collision between two identical particles, and the kaon becomes the center of mass energy.
     
  11. Nov 1, 2016 #10

    vela

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    Now that you've solved the problem, it wouldn't hurt to Lorentz transform the four-momenta in the kaon rest frame to see that one of the pions is indeed at rest, the kaon energy is just ##\gamma m_K##, and that the momentum of the other pion in the lab frame isn't simply ##2P_\pi##.
     
  12. Nov 1, 2016 #11
    I'll do that. Thanks a lot guys
     
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