SR: Rapidity of two particles in uniform motion

1. Jun 3, 2007

masudr

1. The problem statement, all variables and given/known data
A particle leaves the spatial origin P of O at time $t=0$ and constant velocity. After a time t as measured by O, a second particle B leaves P at a different constant velocity and in pursuit of A. B catches A after proper-time t as measured by B. Show that the rapidity of B with respect to O is twice the rapidity of A with respect to O.

2. Relevant equations
I assign the coordinates $(t', x')$ within this frame (slightly counter-intuitive I know, but the problem has already denoted some variables as unprimed, which I might otherwise want to use). I'm not adept at drawing diagrams in LaTeX, so I won't bother. However, as I envisaged the problem, the worldlines of particles A and B are as follows:

$$t'_A(x')=x'/v_A$$
$$t'_B(x')=t + x'/v_B$$

I label the event where they meet as $(t'_1,x'_1)$. So we have the equality

$$x'_1 \left(\frac{1}{v_A}-\frac{1}{v_B}\right)=t\,\,\,\,\,\,\,(1)$$

We have the fact that the proper-time as measured by the particle B also happens to be t. I could think of two ways to write that:

$$t=\gamma_B (t'_1-t)\,\,\,\,\,\,\,(2)$$

and also as (which I think is correct)

$$c^2(t'_1-t)^2-x'_1^2=c^2 t^2\,\,\,\,\,\,\,(3)$$

I list some handy rapidity based relations (where the rapidity, $\phi,$ is related to the velocity, v,):

$$\beta = \frac{v}{c} = \tanh(\phi)$$
$$\gamma = (1-\beta^2)^{-1/2} = \cosh(\phi)$$
$$\beta \gamma = \sinh(\phi)$$

3. The attempt at a solution
My method of attack was to use the equation (1) and either (2)/(3) and obtain some relation between the velocities of both particles, which I could then take the artanh of to get the rapidities.

Processing equation (1) gave me

$$t=x'_1\left(\frac{v_B - v_A}{v_A v_B}\right).$$

Substituting this into (2) gave me

$$\left(\frac{v_B - v_A}{v_A v_B}\right) (1+\gamma_B)=\frac{t'_1}{x'_1}=\frac{1}{v_A}$$

Playing around with fractions ended up giving me

$$v_A = v_B \left( \frac{\gamma_B}{1+\gamma_B} \right)$$

Using the definitions/relations regarding rapidity above, I still couldn't get to the answer I wanted, which is

$$\phi_B = 2 \phi_A$$

Does anyone have any ideas? Thanks for reading this long question!

Last edited: Jun 3, 2007
2. Jun 4, 2007

StatusX

You're second to last equation is right. To derive the last one, note that since tanh is one to one, it's equivalent to show tanh(\phi_B)=tanh(2\phi_A), or:

$$v_A = \tanh(\phi_A) = \tanh(\phi_B/2)$$

Try rewriting the last term using hyperbolic trig identities.

3. Jun 4, 2007

masudr

StatusX:

Thank you for your reply. Huh! The answer was right under my nose, all I needed to do was look up the trig. identities.

This problem was the last part of a question in a previous undergrad. SR paper. How was I meant to know the identity? Is there an obvious route for its derivation? It's just that I'd never seen it before, and so didn't recognise it, even though I came across the expression

$$\tanh(\phi_A) = v_A = \frac{\sinh(\phi_B)}{1+\cosh(\phi_B)}$$

I just didn't realise this was equivalent to $\tanh(\phi_B/2)$

Anyway, many thanks for pointing me in the right direction.

4. Jun 4, 2007

StatusX

One easy way to work with hyperbolic trig functions is to use the change of variables y=e^x. Then you get, eg, sinh(x)=1/2(y-1/y), and:

$$\tanh(x) = \frac{y^2-1}{y^2+1}$$

And since $e^{x/2}=\sqrt{y}$, we get:

$$\tanh(x/2) = \frac{y-1}{y+1} = \frac{y^2-1}{(y+1)^2}$$

$$= \frac{y-1/y}{y+1/y+2} =\frac{\sinh(x)}{\cosh(x)+1}$$

5. Jun 4, 2007

masudr

So it does. Thanks again.