# Relativistic proton collision scattering angle

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!

By conservation of energy, $$m\gamma=2m{\bar\gamma}$$. Use this to find $$\bar\beta$$.
• Aren't you missing the initial energy of $B$, which is $m$? Mar 7 at 2:50
• So the equation should be $m(\gamma+1)=2m\overline\gamma$ Mar 7 at 2:57