Lorentz Factor for relative velocities

[Hirdboy]

Homework Statement

Two particles have velocities u, v in some reference frame. The Lorentz factor for their relative velocity w is given by
$\gamma(w)=\gamma(u) \gamma(v) (1-\textbf{u.v})$
Prove this by using the following method:
In the given frame, the worldline of the first particle is $X =(ct,\textbf{u}t)$ Transform
to the rest frame of the other particle to obtain
$t' = \gamma_v t (1-\textbf{u.v}/c^2)$
Obtain $dt'/dt$ and use the result that $dt/d\tau = \gamma$

Homework Equations

$ct' = \gamma (ct-v/c)$
$x' = \gamma (x-vt)$
-Define Lorentz Transform as L
$dt/d\tau = \gamma$

The Attempt at a Solution

Firstly we are in the frame where the two particles velocities are u and v.
The first step comes from applying LX to give: $t' = \gamma_v t (1-\textbf{u.v}/c^2)$

Differentiating the result gives $dt'/dt = \gamma_v (1-\textbf{u.v}/c^2)$
I think that then may be equal to $\gamma_u$ but cannot see how that will help me solve it. Very grateful to all suggestions thank you.

 

[TSny]

Welcome to PF!

$dt'/dt = \gamma_v (1-\textbf{u.v}/c^2)$

This looks good. You'll now need to "use the result that $dt/d\tau = \gamma$".

Can you see a way to conjure the proper time $d\tau$ into $dt'/dt$? Hint: chain rule.

 

[Hirdboy]

Thank you,
This means (I think):

That I'd be right in saying

$\frac{dt'}{d\tau} = \gamma_w$
and $\frac{dt}{d\tau} = \gamma_u$

We know $\frac{dt'}{dt}$ = $\gamma_v (1-\textbf{u.v}/c^2)$
and $dt'/d\tau = \frac{dt'}{dt} \frac{dt}{d\tau}$

Subbing in gives the desired result $\gamma_w=\gamma_u \gamma_v (1-\textbf{u.v}/c^2)$

Finding it quite confusing working out what $\gamma$ relates to which velocity, so thank you so much for all your help!