# Pion decay reaction and the invariance of momenta

Let's take the following weak decay reaction:
$$\pi^+ → \mu^+ + \nu_\mu$$ where masses and the pion energy are known.

Considering the relations among momenta of the involved particles, where neutrino mass is considered to be zero
$$P(E_\pi,p_\pi), P(E_\mu, p_\mu), P(p_\nu, p_\nu)$$ to calculate the neutrino energy one would say
$$P_\pi = P_\mu + P_\nu$$ Now, by leveraging the invariance of squared momenta we obtain $$P_\pi - P_\nu = P_\mu$$ $$E_\pi^2+p_\pi^2+E_\nu^2+p_\nu^2-2E_\pi E_\nu - 2p_\pi p_\nu cos(\theta) = E_\mu^2+p_\mu^2$$ However, the latter is actually wrong. Why?

• In frame where pion is static, for 4-momenta it can be shown that $P_{\nu}P_{\mu}=P_{\nu}\cdot(P_{\pi}-P_{\nu})=P_{\nu}P_{\pi}=m_{\pi}E_{\nu}$. It seems useful for your question Commented Feb 5, 2020 at 17:42
• @ArtemAlexandrov Any reference to that relation? Commented Feb 5, 2020 at 18:38
• Due to 4-momenta conservation, you have $P_{\pi}=P_{\mu}+P_{\nu}$, so $P_{\mu}=P_{\pi}-P_{\nu}$, then you assume that $\nu$ is massless, which means $P_{\nu}^2=0$ and finally in static pion frame $P_{\pi}=(m_{\pi},{\bf 0})$. Commented Feb 5, 2020 at 21:25
• For the sake of completeness $P^2_\nu = p^2_\nu - E^2_\nu$ which is 0 because, for the $\nu$, $E=p$ Commented Feb 5, 2020 at 22:41
• maybe this will help hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html Commented Feb 6, 2020 at 5:54

For the decay of a pion into a muon and a neutrino, 4-momentum must be conserved:

$$p_\pi^\mu = p_\mu^\mu + p_\nu^\mu \tag{1}$$

where the individual 4-momenta are given by

$$\begin{pmatrix} E_\pi \\\textbf p_\pi \end{pmatrix} = \begin{pmatrix} E_\mu \\\textbf p_\mu \end{pmatrix} + \begin{pmatrix} E_\nu \\\textbf p_\nu \end{pmatrix} \tag{2}$$

and the square of a 4-momentum gives the particles mass: $$p_x^2 = m_x^2$$.

# Option 1: Rest frame

Let us choose the rest frame of the pion, where its spatial 3-momentum is zero:

$$\begin{pmatrix} E_\pi \\\textbf 0 \end{pmatrix} = \begin{pmatrix} E_\mu \\\textbf p_\mu \end{pmatrix} + \begin{pmatrix} E_\nu \\\textbf p_\nu \end{pmatrix} \tag{3}$$

and since $$E_\pi = \sqrt{\textbf p^2+m_\pi^2}$$, the energy simplifies to

$$\begin{pmatrix} m_\pi \\\textbf 0 \end{pmatrix} = \begin{pmatrix} E_\mu \\\textbf p_\mu \end{pmatrix} + \begin{pmatrix} E_\nu \\\textbf p_\nu \end{pmatrix} \tag{4}$$

Since we know the masses and want to know $$E_\nu$$, it is most convenient to subtract the neutrino momentum to the left, because if we square the equation afterwards, we don't have to deal with $$E_\mu$$ or $$p_\mu$$, since the right-hand side will only be $$p_\mu^2 = m_\mu$$ and we know the mass.

$$\begin{pmatrix} m_\pi \\\textbf 0 \end{pmatrix} - \begin{pmatrix} E_\nu \\\textbf p_\nu \end{pmatrix} = \begin{pmatrix} E_\mu \\\textbf p_\mu \end{pmatrix} \quad\leftrightarrow\quad p_\pi^\mu - p_\nu^\mu = p_\mu^\mu \tag{5}$$

Now, we square the equation:

$$(p_\pi^\mu - p_\nu^\mu)^2 = p_\mu^2 \tag{6}$$

This gives us

$$\underbrace{p_\pi^2}_{m_\pi^2} - 2p_\pi \cdot p_\nu + \underbrace{p_\nu^2}_{m_\nu^2 = 0} = \underbrace{p_\mu^2}_{m_\mu^2} \tag{7}$$

Finally, we have to evaluate the product of two 4-vectors here, which according to $$a\cdot b = a^0 b^0 - \textbf a\cdot \textbf b$$ gives us,

$$m_\pi^2 - 2(m_\pi E_\nu - \textbf 0\cdot \textbf p_\nu) = m_\mu^2 \tag{8}$$

$$\textbf 0\cdot \textbf p_\nu = 0$$ and then you can solve for the neutrino energy $$E_\nu$$!

# Option 2: Any frame

For the pion and the muon, $$E=\sqrt{\textbf p^2+m^2}$$, we will consider the neutrino later. Since we know the masses and want to know $$E_\nu$$, it is most convenient to subtract the neutrino momentum to the left, because if we square the equation afterwards, we don't have to deal with $$E_\mu$$ or $$p_\mu$$, since the right-hand side will only be $$p_\mu^2 = m_\mu$$ and we know the mass.

$$\begin{pmatrix} E_\pi \\\textbf p_\pi \end{pmatrix} - \begin{pmatrix} E_\nu \\\textbf p_\nu \end{pmatrix} = \begin{pmatrix} E_\mu \\\textbf p_\mu \end{pmatrix} \quad\leftrightarrow\quad p_\pi^\mu - p_\nu^\mu = p_\mu^\mu \tag{9}$$

Now, we square the equation:

$$(p_\pi^\mu - p_\nu^\mu)^2 = p_\mu^2 \tag{10}$$

This gives us

$$\underbrace{p_\pi^2}_{m_\pi^2} - 2p_\pi \cdot p_\nu + \underbrace{p_\nu^2}_{m_\nu^2 = 0} = \underbrace{p_\mu^2}_{m_\mu^2} \tag{11}$$

Finally, we have to evaluate the product of two 4-vectors here, which according to $$a\cdot b = a^0 b^0 - \textbf a\cdot \textbf b$$ gives us,

$$m_\pi^2 - 2(E_\pi E_\nu - \textbf p_\pi\cdot \textbf p_\nu) = m_\mu^2 \tag{12}$$

$$\textbf p_\pi\cdot \textbf p_\nu = |\textbf p_\pi||\textbf p_\nu|\cos\theta$$. Since the neutrino is massless, $$p_\nu^2=E_\nu^2-\textbf p_\nu^2=E_\nu^2-|\textbf p_\nu|^2=0$$, the length of the neutrino 3-momentum vector is equal to $$E_\nu$$.

$$m_\pi^2 - 2 E_\pi E_\nu + 2|\textbf p_\pi|\underbrace{|\textbf p_\nu|}_{E_\nu}\cos\theta = m_\mu^2 \tag{13}$$

Finally, we can solve for the neutrino energy $$E_\nu$$!

• Actually you haven't taken into consideration the angle between emitted particles. The solution seems to be $m_\pi-2E_\pi E_\nu + 2p_\pi E_\nu cos(\theta) = m_\mu$ Commented Feb 6, 2020 at 8:47
• I have taken it into consideration, however, in the rest frame of the pion it does not play a role. Nevertheless, I have edited my answer to include the angle dependence. Commented Feb 6, 2020 at 9:00