SR: Rapidity of two particles in uniform motion

[masudr]

Homework Statement

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.

Homework 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)

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!

 

[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.

 

[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 realize this was equivalent to \tanh(\phi_B/2)

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

 

[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}

 

[masudr]

One easy way to work with hyperbolic trig functions is to use the change of variables y=e^x.

So it does. Thanks again.