# Inverse beta decay; energy of anti-neutrino

Assuming that the target protons are at rest, calculate the minimum energy of the anti-neutrino for this reaction to take place: $$\bar{\nu}_e+p\rightarrow e^++n$$

I know the answer is given by $E_{\bar{\nu}}=m_{e^+}+m_n-m_p$ but I can't see why this is the case. How can this conserve momentum? Which frame of reference was this calculated in? It seems that we are assuming that the momentum of the positron and neutron are zero and so their energy is their rest mass. But If the proton's are at rest, how can this conserve momentum? I have tried to do the problem in other ways by using the invariant mass and calculating it in different reference frames but I always end up with some variable I don't know.

• Good going for thinking carefully about this. Now, try to figure the amount by which is violates momentum conservation. How fast would the proton have to be moving (toward the neutrino) to zero that out? How does it compare to the thermal energy of a proton at, say, room temperature? (It helps to know that the threshold energy is far below the nucleon mass but a few time the electron mass, so you can safely use non-relativistic physics for the proton and neutron when making these computations.) May 13, 2016 at 22:57

## 1 Answer

The solution you quote actually doesn't conserve momentum. You can use that $p^2$ is a Lorentz invariant and solve $(p_\nu+p_p)^2=(p_e+p_n)^2$, considering the left hand side in the lab frame and the right hand side in the CM frame. You can find $E_\nu$ and check that now momentum is conserved. Anyway, as mentioned in the previous comment, $E_\nu=m_e+m_n-m_p$ is a very good approximation.