Special relativity: particle collision

[Uku]

SOLVED: Special relativity: particle physics

Tomorrow is the exam! My fourth SR question.

Homework Statement

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.

Homework Equations

In next:

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?

 

[Gear.0]

It comes from the general equation for the energy:

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

 

[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:

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