Compute relativistic addition of velocities

[Office_Shredder]

Homework Statement

A particle moves to the left with speed u from the laboratory's frame. Another particle moves to the right with speed v. Both particles are related at the same time from the same place in the laboratory. From the perspective of the first particle, what is the speed of the second particle?

Relevant Equations

Spacetime lengths are conserved $\Delta s^2=\Delta t^2-\Delta x^2$

edit: I had a sign error that is now corrected, no further help needed.

Consider in the laboratory frame one second passing. Let $(t,x)$ be the coordinates of various events in a spacetime diagram. Both particles are released at $(0,0)$ Then after one second the left moving particle is at $(1,-u)$. The right moving particle is at $(1,v)$.

Now consider the spacetime diagram the first particle draws. Both particles are released at $(0,0)$ still. The event of the particle reaching $(1,-u)$ in the laboratory frame corresponds to reaching $(t_0,0)$ in the particle frame, where $t_0^2-0^2=1-u^2$.

The event of the rightward particle reaching $(1,v)$ in the laboratory frame corresponds to reaching a point $(t,x)$ in the particle frame. Our goal is to compute $x/t$ which I believe should be $\frac{u+v}{1+uv}$. We know the spacetime distance between $(t,x)$ and the origin is preserved, and this gives us $t^2-x^2=1-v^2$. We also know the spacetime interval to the end of the other particle's path which gives us $(t-t_0)^2-x^2= -(u+v)^2$.

Edit: I wrote this whole thing up with a $+(u+v)^2$ which was the source of all my problems.To try to solve for $t$ and $x$, I tried subtracting the second equation from the second to get
$$(t-t_0)^2-t^2=-(u+v)^2-(1-v^2)$$
$$-2t_0t+t_0^2=-(u+v)^2-(1-v^2)$$
$$t=\frac{(u+v)^2+1-v^2+t_0^2}{2t_0}$$
We can use $t_0^2=1-u^2$ in the numerator and expand the $(u+v)^2$ to get
$$t=\frac{1+uv}{\sqrt{1-u^2}}$$

Ok, cool. Now we also have $x2=t^2-1+v^2$ from which we get

$$x^2/t^2=1-\frac{(1-v^2)(1-u^2)}{(1+uv)^2}$$
Putting the 1 over a common denominator and expanding the numerator gives
$$x^2/t^2= \frac{1+2uv+u^2v^2-1+u^2+v^2-u^2v^2}{(1+uv)^2}$$
$$x^2/t^2= \frac{u^2+2uv+v^2}{(1+uv)^2}$$
$$x^2/t^2= \frac{(u+v)^2}{(1+uv)^2}$$

Success! I guess I just needed to post this thread to figure it out. Thanks for the help.

 

[kuruman]

Posting sometimes helps collect your thoughts. In the standard derivation, one would use the Lorentz transformations to find $dx$ and $dt$ in terms of $dx'$ and $dt'$, then take the ratio.