# Relativity- CM and Lab frame energy

## Homework Statement

Hi, I've been doing some relativity past paper questions and I totally got stuck on 2 of them.
Both regarding an elastic collision of 2 electrons at relativistic speeds.

First question: I am to show that it is true that the total energy of the electron which is in motion in the lab frame can be given by the following formula:
$$E _{1}= \frac{2E* ^{2}-m ^{2} c ^{4} }{mc ^{2} }$$

Where E* is the energy of one of the electrons in the CM frame.
E is energy in lab frame, m is mass, c is speed of light

Second question: According to an observer in CM frame scattering angle is 90 degrees.
Whais is the scattering angle in Lab frame (in terms of E* and m)

2. The attempt at a solution

I tried doing it in various ways and I can't get an answer.

My 2 best bets are noticing that:
$$E _{1}= mc ^{2} + \frac{2p ^{*2} _{1}}{m }$$
where p* is the momentum of an electron in CM frame, but what next??

Or I try this way, don't even know if its valid:
(total energy in Lab frame) squared=(total energy in CM frame) squared
So for lab frame total energy is
$$E _{1} +mc ^{2}$$
and for CM frame i take 2E*
I can then expand any of these E's taking
$$E = \sqrt{m ^{2} c ^{4}+(pc) ^{2} }$$
This gets me really close to the answer but I have an additional term i can't get rid of: $$\frac{p _{1} ^{2}c ^{2} }{2}$$, which is the momentum of moving particle in lab frame times c^2

For second question I don't even know what to start with