# Question from Rindler's Introduction to Special Relativity

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## Main Question or Discussion Point

I'm stuck on this problem in the "Relativistic Particle Mechanics" section, number 26. I had no trouble with the first part... but the second part I'm stuck.

"Two identical particles move with velocities +-u along the parallel lines z=0, y=+-a in a frame S, passing x=0 simultaneously. Prove that all centroids determined by observers moving collinearly with these particles lie on the open line-segment x=z=0, |y|<ua/c"...

I had no trouble here. But now:

"Also prove that, keeping the same total (relativistic) mass and angular momentum, two such particles cannot move along lines closer than 2ua/c without breaking the relativistic speed limit."

My basic idea was to use the equation for conservation of relativistic mass leading to:

$$\gamma (v_1) + \gamma (v_2) = 2*\gamma (u)$$

And conservation of 3-angular momentum which leads to:

$$\gamma (v_1)*v_1*r_1 + \gamma (v_2)*v_2*r_2 = 2*\gamma (u) * u * a$$

to try and show the required inequality, but haven't been successful. I'd appreciate any help. Thanks.

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CarlB
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You've solved the problem, but you're stuck on the algebra. The problem is all those nasty gamma functions. To get rid of them, try dividing your second formula by your first formula. That is, divide left side by left side, right side by right side.

Carl

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Hi Carl. I still don't see it. After dividing out, I can cancel the $$\gamma (u)$$, but still have the other gamma functions.

So I get:

$$\frac{\gamma (v_1) * v_1 * r_1 + \gamma (v_2) * v_2 * r_2}{\gamma (v_1) + \gamma (v_2) } = u * a$$

but I'm not sure what to do here. I'm probably missing something very simple.

robphy
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Gold Member
Often, gamma $$\gamma$$ factors may be more easily manipulated if you use rapidities $$\theta$$.

$$v{\color{red}/c}=\tanh\theta$$
$$\gamma=\displaystyle\frac{1}{\sqrt{1-(v/c)^2}}=\cosh\theta$$
$$v\gamma{\color{red}/c}=\displaystyle\frac{v{\color{red}/c}}{\sqrt{1-(v/c)^2}}=\sinh\theta$$

Once written in terms of rapidities, apply hyperbolic-trig identities and then do algebra.

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CarlB
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learningphysics said:
I'm not sure what to do here. I'm probably missing something very simple.
Try showing that $$\gamma (v_1) = \gamma (v_2)$$.

Carl

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Thanks Carl and robphy. I see that I was over-generalizing the problem. I thought that r1 and r2 could take any values... however this isn't the case since we could take both particles having the same velocity (u) and position (a), and keep the same total relativistic mass and 3-angular momentum (about the origin) while their separation is 0.

So instead assume that the particles are moving in opposite directions along opposite parallel lines each at an equal distance r from the origin (which is what I believe the problem expects).

So we have:
$$\gamma (v_1) + \gamma (v_2) = 2 * \gamma (u)$$

and

$$(\gamma (v_1) * v_1 + \gamma (v_2) * v_2)*r = 2 * \gamma (u) * u * a$$

Divide the two equations:

$$\frac{\gamma (v_1) * v_1 + \gamma (v_2) * v_2 }{\gamma (v_1) + \gamma (v_2) } * r = u * a$$

Using sinh and cosh and solving for r:

$$r = \frac{cosh(\theta_1) + cosh(\theta_2)}{sinh(\theta_1) + sinh(\theta_2)} * u * a/c$$

Using the hyperbolic identites for sum of 2 cosh, and sum of two sinh...

$$r = coth((\theta_1 + \theta_2)/2) * u * a/c$$

$$coth((\theta_1 + \theta_2)/2) > 1$$ so

$$r > ua/c$$

so the separation of the two particles is > than 2ua/c.