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I was stuck on the following question and I can't see what the relation of this is, The question in regard was

Determine the energy E of the muon in terms of $m0_µ$, $m0_π$ and $c$ when the resting Pion decays into a muon with a smaller mass and a massless neutrino.

For this, it seemed like I could utilize the four-momentum vector so here is what I tried

$P_{pion}=P_{muon}+P_{Neutrino}$ we do not know the four-momentum vector of the neutrino so $P_{Neutrino}=P_{pion}-P_{muon}$

With $P_{pion}$=\begin{pmatrix} m_{pi}*c \\ 0 \\ 0 \\ 0 \end{pmatrix} and $P_{muon}$=\begin{pmatrix} E_{\mu}/c \\ p_x \\ p_y \\ p_z \end{pmatrix}

if we would then square the last equation we would find that $(P_{pion})^2+(P_{muon})^2+2*P_{pion}\cdot P_{muon}$ =$(m_{pi}*c)^2+(m_{\mu}*c)^2-2*m*E_u=0$ since the massless particle has no momentum squared.

I have two main questions.

the first being that I do not see how $(P_{muon})^2=(m_{\mu}*c)^2$. The reason for this is that this is only the rest mass of the particle right, shouldnt there also be a term included for the kinetic energy that this particle receives, it seems so simple but it doesn't click I know it is true and the answer makes sense but why.

The second question in regard is, now that I can find the energy of the muon if we solve for $E_\mu$ but can we then also solve for the kinetic energy, since there is no potential energy in the system it is just $E_{\mu}-m_{\mu}*c^2$=T right?

Anyway I hope you can help me on the right track.

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Since $P_{muon}$ is the 4-momentum, its square magnitude is the mass-squared times $c^2$, which can written in terms of an observer's decomposition of $P_{muon}$ into a temporal part, the relativistic energy $E_{muon}/c$ (which includes the kinetic energy) and the spatial part, the relativistic momentum $\vec p_{muon}$, where $$P_{muon}^2=m_{muon}^2c^2=(E_{muon}/c)^2-\vec p_{muon}\cdot\vec p_{muon}.$$

Yes, the observer determines the relativistic kinetic energy of the muon as $T_{muon} = E_{muon} -m_{muon}c^2$.

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