# Relativistic elastic collision

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1. Jan 6, 2017

### yup790

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

2. Relevant equations
et Em and pm be the energy and momentum of the mass m after the collision. Let p and p' be the momentum of mass M before and after the collision.

From conservation of 4 momentum:
$$\begin{bmatrix}E+m \\ p\end{bmatrix}=\begin{bmatrix}E_m+E' \\ p'+p_m\end{bmatrix}$$

We also have our invariants
E2-p2=M2, etc.

3. The attempt at a solution
Squaring the 4-vectors we get: $$E^2+2Em+m^2-p^2=E'^2+2E'E_m+E_m^2-p'^2-2p_mp'-p_m^2$$

Using our invarients, this becomes:
$$M^2+m^2 + 2Em=m^2+m^2+2(E'E_m-p'p_m) \implies Em=E'E_m-p'p_m$$

now, using the conservation of energy and momentum:
$$E_m=E+m-E'$$ & $$p_m=p-p'$$
Substituting these in and using our invarient again gets us:
$$Em=E'(E+m)-p'p-M^2$$

I have tried setting $$p=\sqrt{E^2-M^2}$$however its gets so algebraically heavy. Is there an easier way??

2. Jan 6, 2017

### Staff: Mentor

You can make a stronger statement. Energy and momentum are both conserved separately. That way you can avoid p' with its ugly influence on the equations.

3. Jan 6, 2017

### yup790

What would I replace it with? p-pm is a bit redundant

4. Jan 6, 2017

### Staff: Mentor

p-pm is fine. Express pm in terms of Em, then replace it by E' and known parameters. Squaring it at the right place will get rid of any square roots, and then you'll probably have to solve a quadratic equation.