Find the Speed of an Electron in the lab frame

[B3NR4Y]

Homework Statement

An electron moves to the right with a speed of 0.90c in the laboratory frame. A proton moves to the left with a speed of 0.77c relative to the electron.

Using the Lorentz coordinate transformations, find the speed of the proton in the laboratory frame

Homework Equations

$$
\begin{align*}
x &= \frac{x'+ut'}{\sqrt(1-(u^2/c^2))} \\
y & = y \\
z & = z \\
t & = \frac{t'+(\frac{u}{c^2})x'}{\sqrt(1-(u^2/c^2))} \\

\end{align*}
$$

The Attempt at a Solution

I used the velocity transformation in the second part of the problem and found the speed of the proton to be 0.42c, but I'm not sure how to find it with the coordinate transforms.

I tried dividing x and t, but terms that I don't know remain and I can't get rid of them.

 

[Geofleur]

Are you allowed to derive the velocity transform from the coordinate transform and then use the velocity transform? That's how I would want to do it...

 

[B3NR4Y]

Are you allowed to derive the velocity transform from the coordinate transform and then use the velocity transform? That's how I would want to do it...

Haha, I don't think so cause that's exactly what I thought to do, the next part specifically says to use the velocity transformations to find the velocity and that they should be equal to the velocity I found in part A.

 

[Geofleur]

You know, I think you do some of the same steps as if you were deriving the velocity transform, but don't take the limit as $\Delta t$ goes to zero. Then you can calculate $\Delta x$ for the proton relative to the lab frame and $\Delta x'$ for the proton relative to the electron, etc. Know what I mean?

 

[Geofleur]

It might also help to arrange things so that the proton and the electron are both at the origin of the lab frame at t = t' = 0. Then instead of $\Delta x$ and $\Delta t$ you can just deal with $x$ and $t$, for example.

 

[B3NR4Y]

I think I see what you mean, I will try that.

 

[B3NR4Y]

Hm, it's not quite working or I'm doing something wrong. Not sure I see why I shouldn't divide by Δt, otherwise I don't get a velocity.

$\delta x = \frac{x_{f}' + ut_{f}'}{\sqrt{1-\frac{u^2}{c^2}}}$ I tried this with some dummy numbers, like t_f = 1s later, thus x_f = -0.77 c.

 

[B3NR4Y]

Nevermind, it worked. I forgot to make x_f negative. Thanks a LOT!

 

[Geofleur]

Yeah, you do need to divide by time to get a velocity, just not to take limits and all that (I edited my advice above a little!).