# ''Invariant mass of the electron and electron antineutrino''

1. Dec 30, 2014

### Coffee_

1. A neutron decays into a proton, an electron and an electron antineutrino. In the frame of the neutron, the proton is in rest after decay. Calculate the ''invariant mass of the electron and electron antineutrino together''. The rest masses of the proton and neutrno are known

2. Conservation of energy, momentum and conservation of four momentum.

3. The problem is that I don't understand what is asked of me, are they asking for $m_{e} + m_{v}$? If so, I start like this in natural units (h bar = 1 , c=1)

$m_{n}=m_{p} + E_{e} + E_{v}$

Plugging in the relativistic formulas for energy, and knowing that since the proton has no momentum the momenta of the electron and antineutrino have to be equal in size:

$\sqrt{p^{2}+m_{e}^{2}}+\sqrt{p^{2}+m_{v}^{2}}=m_{n}-m_{p}$

Seems like not enough info, hence I'm doubting I'm correctly interpreting the question.

2. Dec 30, 2014

### jbriggs444

They are asking for the "invariant mass of the electron and electron antineutrino together". That is, they are asking the invariant mass of a system consisting of those two particles. That is not the same thing as asking for the sum of the invariant mass of the electron of the electron anti-neutrino.

3. Dec 30, 2014

### Coffee_

Is $\sqrt{ (p_{e} + p_{v})^{2}}$ the correct interpretation then? Where $p_{e}$ and $p_{v}$ represent the four momenta of the particles?

Last edited: Dec 30, 2014
4. Dec 30, 2014

### jbriggs444

When it comes to four-momentum, I am out of my depth, and am working from first principles. But no, that does not appear to be the correct interpretation.

Edit: It looks correct now.

Go back to definitions. What is the definition of invariant mass of X? The norm of the four-momentum of X, right? So if X is a system consisting of two particles...

Last edited: Dec 30, 2014
5. Dec 30, 2014

### Coffee_

Oh damn I messed up my squares being too qucik, here I fixed it. What about now?

6. Dec 30, 2014

### jbriggs444

Yes, that works.