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mshr
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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?
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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?
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?