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.


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 $T_{muon} = E_{muon} -m_{muon}c^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.