How Can We Simplify the Algebra in Relativistic Elastic Collision Problems?

[yup790]

Homework Statement

Homework Equations

et E_m and p_m 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.

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:
$$M^2+m^2 + 2Em=m^2+m^2+2(E'E_m-p'p_m)<br /> \implies Em=E'E_m-p'p_m$$

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??

 

[mfb]

Squaring the 4-vectors we get:

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.

 

[yup790]

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.

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

 

[mfb]

p-p_m is fine. Express p_m in terms of E_m, 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.