Why is scalar product of momentum in electron scattering conserved?

[gboff21]

Homework Statement

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?

[Edit]
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!

 

[mark726]

Hello,

I can see why this may be confusing. Let me try to explain it in a different way.

The equation p.P = p'.P' is a result of the conservation of four-momentum. In this case, we are considering the conservation of energy and momentum in the interaction between an electron and a nucleus. The four-momentum of a particle is defined as p = (E/c, p), where E is the energy and p is the momentum.

In this case, we have two particles: the electron with four-momentum p and the nucleus with four-momentum P. After the interaction, the electron has a new four-momentum p' and the nucleus has a new four-momentum P'. The conservation of four-momentum states that the sum of the initial four-momenta (p + P) must be equal to the sum of the final four-momenta (p' + P').

Now, when we square both sides of this equation, we get p^{2} + P^{2} + 2pP = p'^{2} + P'^{2} + 2p'P'. As you mentioned, the invariant masses (m_{e} and M) are unchanged, so we can replace p^{2} and P^{2} with their respective invariant masses. This leaves us with p'^{2} and P'^{2}, which are equal to m_{e}c^{2} and Mc^{2}, respectively.

So, our equation becomes m_{e}c^{2} + Mc^{2} + 2pP = m_{e}c^{2} + Mc^{2} + 2p'P'. We can now cancel out the invariant masses on both sides, leaving us with 2pP = 2p'P'. And since p and p' are the four-momenta of the electron, and P and P' are the four-momenta of the nucleus, this can be written as p.P = p'.P'.

I hope this helps clarify why p.P = p'.P' in this situation. Let me know if you have any other questions.