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!