Proton A collides elastically with proton B, which is at rest. The outcome of each individual collision cannot be predicted. However, occasionally there is a “symmetric collision” in which the two protons come off with identical speeds along paths that make identical angles $\theta$ with the forward direction. What is this angle of deflection in a symmetric collision? For Newtonian mechanics the total angle of separation is always $90^{\circ}$. You will see that this angle will be less than $90^{\circ}$ for a relativistic impact.
How high must the velocity of proton A be before the separation angle deviates from $90^{\circ}$ by as much as $1/100$ of a radian?
Comparing the initial and final four momentums, we get to the fact that: $$m\gamma \beta = 2m\overline{\gamma}\overline{\beta}\cos(\theta)$$
Where $\beta=\dfrac v c$ and represents the velocity of A in the initial state, $\overline{\beta}=\dfrac {\overline{v}} c$ and represents the velocity of A/B in the final state. Also, $\gamma = \dfrac 1 {\sqrt{1-\beta^2}}$ and $\overline{\gamma}$ is the same but with $\overline{\beta}$, and all of this is in units where $c=1$.
My issue is that I can't figure out what $\overline{\beta}$ is supposed to be, because if I can write $\overline{\beta}$ in terms of $\beta$, then it'll be only simple algebra to find $\theta$. I was thinking of maybe using a new reference frame $S'$ where the momentum of each state is zero, and somehow getting the final velocity using velocity addition formulas maybe? Then I could maybe try and turn that into a final velocity in $S$ to use, but I'm not sure how. If this wouldn't work, I'm still pretty lost on how to get to the answer.
Thanks for any help!
