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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?

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  • $\begingroup$ 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 $\endgroup$ Commented Feb 5, 2020 at 17:42
  • $\begingroup$ @ArtemAlexandrov Any reference to that relation? $\endgroup$
    – Anelito
    Commented Feb 5, 2020 at 18:38
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    $\begingroup$ 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 $P_{\pi}=(m_{\pi},{\bf 0})$. $\endgroup$ Commented Feb 5, 2020 at 21:25
  • $\begingroup$ For the sake of completeness $P^2_\nu = p^2_\nu - E^2_\nu$ which is 0 because, for the $\nu$, $E=p$ $\endgroup$
    – Anelito
    Commented Feb 5, 2020 at 22:41
  • $\begingroup$ maybe this will help hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html $\endgroup$
    – anna v
    Commented Feb 6, 2020 at 5:54

1 Answer 1

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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 $E_\nu$!

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  • $\begingroup$ Actually you haven't taken into consideration the angle between emitted particles. The solution seems to be $m_\pi-2E_\pi E_\nu + 2p_\pi E_\nu cos(\theta) = m_\mu$ $\endgroup$
    – Anelito
    Commented Feb 6, 2020 at 8:47
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    $\begingroup$ 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. $\endgroup$
    – ersbygre1
    Commented Feb 6, 2020 at 9:00

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