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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!

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1 Answer 1

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By conservation of energy, $m\gamma=2m{\bar\gamma}$. Use this to find $\bar\beta$.

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  • $\begingroup$ Aren't you missing the initial energy of $B$, which is $m$? $\endgroup$
    – mr_e_man
    Mar 7 at 2:50
  • $\begingroup$ So the equation should be $m(\gamma+1)=2m\overline\gamma$ $\endgroup$
    – mr_e_man
    Mar 7 at 2:57
  • $\begingroup$ Thank you. I was just careless. $\endgroup$ Mar 8 at 11:28

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