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Direction of particles after decay w/ relativity

  1. Oct 12, 2015 #1
    One of the possible decay modes of the neutral kaon is ## K^0 \rightarrow \pi^+ + \pi^- ## The rest masses of the K0 and pion are 498 MeV/c2 and 135 MeV/c2, respectively.

    In 2-dimensions (xz-plane), if the kaon has an initial momentum of 2000 MeV/c in the z direction, what is the momentum of the two pions, assuming an equal distribution of energy between them, and what is their direction in the laboratory frame?


    Energy should be conserved, ##E_{initial} = E_{final}##, and ##E^2 = (pc)^2 + (mc^2)^2##. ##\gamma = \frac{1}{\sqrt{1-\left ( \beta \right )^2 }}##, ##\beta = \frac{v}{c} = \frac{p}{\sqrt{p^2 + m^2}}##.

    $$\sqrt{(p_Kc)^2 + (m_Kc^2)^2} = 2\sqrt{(p_{\pi}c)^2 + (m_\pi c^2)^2}$$
    $$\sqrt{2000^2 + 498^2} = 2\sqrt{p_{\pi}^2 + 135^2}$$
    $$p_{\pi} = 1021 \mathrm{ MeV/c}$$
    But this is equally divided amongst two pions, so they have 510 MeV/c each.

    Can the above be thought of as energy in the z direction? If so, we can solve for energy in the x direction, with the kaon at rest, so that the momentum per pion is 103 MeV/c (equations as above, with ##p_K=0##), with the pions going off at 180° way from each other.

    We now have two momentum vectors ##p^\pi_x = \pm 103## and ##p^\pi_z = 510##.

    From vector addition, ##p_\pi## is 520 MeV/c.

    I don't know how to transform the momentum vectors from the pion reference frame to the laboratory frame. I know that lengths are contracted by ##\gamma##, but I don't know which ##\gamma## to use, and which direction in which ##\gamma## is to be applied.

    Any tips?
  2. jcsd
  3. Oct 12, 2015 #2


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    Hello mshr, welcome to PF :smile: !

    Something is going wrong with your math: the 1021 is already per pion, so dividing by 2 once more seems strange. But is this really what's happening?

    In the K rest frame the ##\pi## fly off with equal energy in opposite directions, so opposite but equal momenta. Can be calculated exactly. It's the direction vector that needs your attention: the azimuthal dependence has been taken out (by looking in the x,z decay plane) but there still is an angle wrt the direction of motion of the original K that gives you a distribution in the lab frame: If the decay is perpendicular, you can assume equal ##\pi## momenta, but otherwise the forward decaying ##\pi## will have more energy than the backward one ...
  4. Oct 12, 2015 #3


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    No, energy isn't a vector. There's no such thing as energy in the z-direction.
  5. Oct 12, 2015 #4
    You're correct, it already was divided by two.

    I think I've got it.

    $$p_\pi = 1021 \mathrm{MeV/c^2}$$
    $$\beta_K = 0.9704$$
    $$\beta_\pi = 0.9909$$

    In the reference frame of the kaon, the two pions leave in opposite directions with velocity 0.99c. If we assume that the pions are emitted perpendicular to the direction of travel of the kaon, then there is no effect from the Lorentz transformation from the kaon rest frame into the laboratory reference frame.

    The pions have a velocity of 0.99c in the x direction and 0.97c in the z direction in the laboratory frame. The angle that this forms is ##\arctan\left ( \frac{0.9909}{0.9704} \right ) = 45.6^\circ##.
  6. Oct 13, 2015 #5


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    But what reason is there to expect the pi to move off perpendicularly ? None, I would say !
  7. Oct 13, 2015 #6
    There isn't any reason to expect them to come off perpendicularly; that was just a self imposed constraint.

    If I can remember from this morning, if the kaon decays in the direction of travel, then in the laboratory frame one pion is moving at about 0.99c and the other at -0.5c.

    So these represent the two extremes, and reality will be in between the two.
  8. Oct 13, 2015 #7


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    In your case I would assume a ##\theta## and come with a result that depends on the value of that variable.
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