Deriving relativistic velocity addition formula

Question

I am familiar with multiple ways of deriving the relativistic velocity addition formula. However, I'm interested in the following proof, by taking the derivative $V_x'=\frac{dx'}{dt'}$.

What my textbook does is the following;

$x'=\gamma(x-\beta ct) \\ \frac{dt}{dt'}=\gamma + \frac{\beta \gamma}{c}V_x'$

So we get:

$\frac{dx'}{dt'}=\frac{d}{dt}\left(\gamma(x-\beta ct)\right)\frac{dt}{dt'}=\frac{d}{dt}\left(\gamma(x-\beta ct)\right)(\gamma + \frac{\beta \gamma}{c}V_x')$.

And then they simply say we get $V_x'=\frac{V_x-\beta c}{1-\frac{\beta}{c}V_x}$. I don't get this last step. I've tried several things, like writing out $\gamma$ or simplifying things... but I don't get their result. Could someone help me out? Thanks a lot in advance.

Answer 1

It's just a matter of doing the algebra right. We can take out a factor $\gamma$ from both factors in the expression:

$${V_{x}}'=\frac{d}{dt}\left(\gamma(x-\beta ct)\right)\left(\gamma + \frac{\beta \gamma}{c}{V_x}'\right)$$

because it's constant under differentiation w.r.t. $t$. This yields:

$${V_{x}}'=\frac{1}{1-\beta^2}\frac{d}{dt}\left(x-\beta ct\right)\left(1 + \frac{\beta }{c}{V_x}'\right)$$

The derivative can be evaluated:

$${V_{x}}'=\frac{1}{1-\beta^2}\left(V_x-\beta c\right)\left(1 + \frac{\beta }{c}{V_x}'\right)$$

All we need to do is solve this equation for ${V_{x}}'$. When working with large equations you need to be careful to prevent errors. The best way is to concentrate on the relevant parts of the equation, instead of trying to do everything at once. So, if we want to collect all the ${V_{x}}'$ terms then just concentrate on doing just that. On the left hand side ${V_{x}}'$ is present with a coefficient of $1$, on the right hand side it has a coefficient of:

$$A = \frac{1}{1-\beta^2}\left(V_x-\beta c\right)\frac{\beta }{c}$$

So, if we bring all the ${V_{x}}'$ terms to the left, it will get a coefficient of $1 - A$. The remaining term on the right hand side is:

$$B = \frac{1}{1-\beta^2}\left(V_x-\beta c\right)$$

So, let's see if we can simplify $1 - A$:

$$1-A = \frac{1}{1-\beta^2}\left(1-\beta^2 -V_x\frac{\beta}{c}+\beta^2\right) = \frac{1}{1-\beta^2}\left(1-V_x\frac{\beta}{c}\right)$$

Dividing both sides by $1-A$ yields:

$${V_{x}}'= \frac{V_x-\beta c}{1-\frac{\beta}{c} V_x}$$

Answer 2

It seems your textbook is taking a complicate path. There is a shorter path which I much prefer: $$\frac{dx'}{dt'}=\frac{d(\gamma(x-vt))}{d(\gamma(t-vx/c^2))}=\frac{dx-vdt}{dt-vdx/c^2}=\frac{dx/dt-v}{1-v(dx/dt)/c^2}$$ where $v = \beta c$.

EDIT: your textbook method works too. But your Lorentz factor that you mentioned in your comment is wrong; with your notation, it should be $\gamma=1/\sqrt{1-\beta^2}$ (deduced from your $x'=\gamma(x-\beta ct)$, hence $v=\beta c$ and $v^2/c^2=\beta^2$). Then use $1+\gamma^2\beta^2=\gamma^2$ in the steps I mentioned in my previous comment (compute the derivative w.r.t. $t$ and solve for $V′x$ which appears linearly on both sides) and you are done.