## Relativistic decay of a particle

1. The problem statement, all variables and given/known data

Pions can decay via the reaction π+ → μ+ νμ. Show that the energy of the neu- trino in the rest frame of the pion is given by

$$E_v = \frac{m^2_∏-m^2_μ}{2m_∏}$$

Pions with energy Eπ in the laboratory frame (Eπ >> mπc2) decay via the above reaction. Show that the maximum energy of the neutrinos in the laboratory frame is given by

$$E^{max}_v = E_∏\frac{m^2_∏-m^2_μ}{m^2_∏} c^2$$

2. Relevant equations

Lorentz Transform

$$\left(\begin{array}{cc}E'\\p'c\end{array}\right) = \left(\begin{array}{cc}\gamma&\beta\gamma\\-\beta\gamma&\gamma\end{array}\right) * \left(\begin{array}{cc}E\\pc\end{array}\right)$$

3. The attempt at a solution

I am able to get the first result, the problem I am having is changing frames from the first result to the second result. I take the first result and use the Lorentz transform ont he energy, where prime denotes the lab frame and unprimed is the rest frame of the pion.
this give me:

$$E_v' = \frac{m^2_∏-m^2_μ}{2m_∏} \gamma(1-\beta) c^2$$

i then use the fact that E'_∏ = gamma mc^2 to find gamma, giving me:

$$E_v' = \frac{m^2_∏-m^2_μ}{2m^2_∏} E_∏(1-\beta) c^2$$

However I have done something wrong as this means I need beta =-1 to give me the right result. Any help would be greatly appreciated.
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 You've done something wrong with your LT matrix. Both $\beta \gamma$ terms should have the same sign.
 Ahh yes, however that was only a mistake I made when typing the problem up to pf, my calculations had both with the -ve sign to get the result below it.

 Tags collision, decay, special relativity

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