# Homework Help: Why is scalar product of momentum in electron scattering conserved?

1. Apr 4, 2013

### gboff21

1. The problem statement, all variables and given/known data
Reading a textbook, I come across a situation where an electron is scattered off a nucleus. The book says p.P = p'.P', where p is the momentum of the electron and P is the momentum of the nucleus.
I don't understand how it gets the conservation of scalar product.

It's steps are:
four-momentum $p = (E/c , \vec{p})$
$p.p = \frac{E^{2}}{c^{2}} - \vec{p}^{2}$
It says: there is always a rest frame to be found
therefore, $m = \sqrt{p^{2}}/c$
and obtain: $E^{2} = p^{2}c^{2} + m^{2}c^{4}$

now conservation of four momentum: $p + P = p' + P'$
square it to make it invariant: $p^{2} + P^{2} + 2pP = p'^{2} + P'^{2} + 2p'P'$

invariant masses are unchanged. Therefore, $p^{2} = p'^{2} = m_{e}c^{2}$
and $P^{2} = P'^{2} = Mc^{2}$

Agreed so far...But now it says:
it follows that p.P=p'.P'

I understand all of that except that last line.
Why does it follow?

Solved now :/
You just plug p.p=p'.p' and P.P=P'.P' back into $p^{2} + P^{2} + 2pP = p'^{2} + P'^{2} + 2p'P'$
they cancel and you get p.P=p'.P'.
Sorry!

Last edited: Apr 4, 2013