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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.

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  • $\begingroup$ The concept you are looking for is the rapidity, en.wikipedia.org/wiki/Rapidity This describes the rapid change of velocities via angles in the (to us unimaginable) 4 dimensional space. With it you can derive the addition formula (which is WRONG! by the way). $\endgroup$ – StringTheoretican Jan 13 '17 at 19:07
  • $\begingroup$ What is yout lorentz expansion? $\gamma^2=\frac{1}{(1-V_x^2/c^2)}$ $\endgroup$ – JMLCarter Jan 13 '17 at 20:50
  • $\begingroup$ It's $\frac{1}{\sqrt{1-\frac{\beta^2}{c^2}}}$ $\endgroup$ – Sha Vuklia Jan 13 '17 at 20:52
  • $\begingroup$ In the equation you get, after taking the derivative w.r.t. $t$, did you try to solve for $V_x'$ which appears on both sides? $\endgroup$ – user130529 Jan 13 '17 at 21:25
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    $\begingroup$ Hints: your Lorentz factor is wrong; with your notation, it should be $1/\sqrt{1-\beta^2}$ (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 and you are done. $\endgroup$ – user130529 Jan 13 '17 at 22:15
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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:

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

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  • $\begingroup$ Ohhh, now I see my mistake! I didn't turn $1$ into $\frac{1-\beta^2}{1-\beta^2}$, to factor out $\gamma^2$ - which leads to the solution! Thank you so, so, só much! $\endgroup$ – Sha Vuklia Jan 13 '17 at 23:23
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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.

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  • $\begingroup$ I already stated that I'm familiar with several proofs, and I'm specifically interested in the one given by my textbook. I don't like this proof, because it's lousy mathematics, because $\frac{d}{dt'}$ is an operator - or whatever you want to call it - and not a fraction. $\endgroup$ – Sha Vuklia Jan 13 '17 at 21:39
  • $\begingroup$ +1: It works for me. $\endgroup$ – Mozibur Ullah Sep 26 '18 at 12:49

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