# Special relativity: particle collision

1. Jun 27, 2010

### Uku

SOLVED: Special relativity: particle physics

Tomorrow is the exam! My fourth SR question.

1. The problem statement, all variables and given/known data

There is a $$\beta$$ breakdown(?) of a neturon, resulting in
$$n \rightarrow p + e^{-} + \nu^{-}_{e}$$

I have to find the maximum speed of the electron, the decomposing neutron is still.
I'm also given the masses of the proton, the electron and the neutron.

2. Relevant equations

In next:

3. The attempt at a solution
Right, here is what I have written down from the lecture:

We can consider (in simplification) the antineutrino and the proton as one particle and their impulse as:
$$p^{-}=p_{p}+p_{\nu^{-}_{e}}$$

The total impulse is conserved, resulting in:

$$0=p_{p^{-}}+p_{e}$$

This is because the decomposing neutron has no impulse. We also take $$c=1$$

The total energy is conserved:

$$m_{n}=E_{p^{-}}+E_{e}$$ !NB! to the c=1 and stationary neutron

Right, but now comes the thing I don't get;

$$E_{p^{-}}=\sqrt{p^{2}_{p}+m^{2}_{p^{-}}}$$
and
$$E_{e}=\sqrt{p^{2}_{e}+m^{2}_{e}$$

Where does this square root expression come from?

Last edited: Jun 27, 2010
2. Jun 27, 2010

### Gear.0

It comes from the general equation for the energy:

$$E^{2} = (m c^{2})^{2} + p^{2} c^{2}$$

3. Jun 27, 2010

### Uku

Thanks!

I went trough the derivation and I'll put it here for further reference;

We know that:
$$p^{\mu}=(mc,p)$$

$$p_{\mu}=(mc,-p)$$

$$p^{\mu}p_{\mu}=m^{2}_{0}b^{\mu}b_{\mu}=m^{2}_{0}c^{2}$$

But we can write

$$p^{\mu}=(\frac{E}{c},p)$$

Now

$$p^{\mu}p_{\mu}=\frac{E^{2}}{c^{2}}-p^{2}$$

Putting the two impulse squares together:

$$m^{2}_{0}c^{2}=\frac{E^{2}}{c^{2}}-p^{2}$$

From there:

$$E^{2}=p^{2}c^{2}+m^{2}_{0}c^{4}$$