# Conservation of four momentum

1. Oct 11, 2011

### *LouLou*

Hi physics people,

This is a past (3rd year university level) exam question, so I hope it's ok that I didn't post this in the homework section even if it's set out like a homework question.

The Question:
Suppose we are observing the collision

Anti-electron-neutrino + electron ---> W-minus boson

What energy has the Anti-Electron-Neutrino, to produce the W particle?

Relavant equations and data:
mass of electron = 0.511 MeV/c^2
mass of W-minus boson = 80.403 GeV/c^2

Four-momentum equation
$$S = -(\overrightarrow{{p_{e}}}+\overrightarrow{{p_{\nu }}})^{2} = -(cp_{e}+cp_{\nu })^{2}+(E_{e}+E_{\nu})^{2}$$

Thoughts so far:
My understanding is that the four-momentum is always conserved so I have to equate the initial four-momentum to the final four-momentum using the assumption that the electron and the W boson have no kinetic energy.

$$S_{initial} = -c^{2}p_{\nu }^{2} + E_{\nu }^{2}$$
$$S_{final} = ((m_{W^{-}}) c^{2})^{2}$$

to the initial four-momentum becomes zero since
cp = E

So I'm not sure how to carry on, the energy of the neutrino can't be zero!?

Thanks to anyone who helps
xoLouLouox

Last edited: Oct 12, 2011
2. Oct 12, 2011

### Parlyne

You'll find that it isn't possible for both the electron and the W to be at rest.

3. Oct 12, 2011

### TrickyDicky

Is there not a frame where they would both be at rest?. That seems to be at odds with the "no preferred rest frame" postulate.

4. Oct 12, 2011

### PAllen

I shoot a green pea at billiard ball and it sticks, moving the billiard ball slightly. Is there *one* frame where both the initial billiard ball and the composite (billiard+pea) are at rest? No. The existence of such a frame would mean you have a frame where momentum isn't conserved. The described situation is equivalent.

5. Oct 12, 2011

### Hepth

I hope its ok to just show how to do this as it isn't a homework question. I just use conservation of momentum in the rest frame of the W.
$$p_{\nu} +p_{e} = p_W$$
$$(E_{\nu},E_{\nu} \hat{z}) + (E_{e}, p_e \hat{z}) = (E_W, \vec{p}_W)$$
Now in this frame we have:

$$(E_{\nu}, E_{\nu} \hat{z}) + (E_{e}, -E_{\nu} \hat{z}) = (M_W, 0)$$

with
$$m_e^2 = E^2_{e} - E^2_{\nu}$$
from the momentum squared of the electron in this frame, we have $E_{e} = \sqrt{m_e^2 + E^2_{\nu}}$

$$(E_{\nu}, E_{\nu} \hat{z}) + (\sqrt{m_e^2 + E^2_{\nu}}, -E_{\nu} \hat{z}) = (M_W, 0)$$

Looking at just the energies:

$$E_{\nu} + \sqrt{m_e^2 + E^2_{\nu}} = M_W$$

Giving us

$$E_{\nu} = \frac{M_W^2 -m_e^2}{2 M_W}$$

This is all assuming all particles are on-shell.

6. Oct 12, 2011

### TrickyDicky

Nice analogy, thanks.

7. Oct 12, 2011

### *LouLou*

Thanks very much, this helps with a lot of other things as well, that aren't so greatly explained in my notes. Cheers!