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

AI Thread Summary
The discussion focuses on simplifying algebraic expressions in relativistic elastic collision problems by leveraging conservation laws. Participants suggest avoiding complex variables like p' by using conservation of energy and momentum separately. It is recommended to express momentum in terms of energy and known parameters to streamline calculations. Squaring the relevant 4-vectors can help eliminate square roots, potentially leading to a solvable quadratic equation. Overall, the emphasis is on reducing algebraic complexity while maintaining accuracy in calculations.
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Homework Statement



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Homework 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.

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&#039;E_m-p&#039;p_m)<br /> \implies Em=E&#039;E_m-p&#039;p_m<br />

now, using the conservation of energy and momentum:
E_m=E+m-E&#039; & p_m=p-p&#039;
Substituting these in and using our invarient again gets us:
Em=E&#039;(E+m)-p&#039;p-M^2

I have tried setting p=\sqrt{E^2-M^2}however its gets so algebraically heavy. Is there an easier way??
 
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yup790 said:
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.
 
mfb said:
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-pm is a bit redundant
 
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.
 
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