If $S'$ and $S$ are two different inertial reference frames and $S$ moves along the $x$ axis of $S$ in a standard configuration, then the Lorentz transformation will be:

$$x'=\gamma(v)[x-\beta ct]$$ $$y'= y$$ $$z'= z$$ $$ct'=\gamma(v)[ct-\beta x]$$

Where $\beta=\frac{v}{c}$. But deriving an expression for the velocity from here is a little tricky.

Say examining the infinitesimal form, firstly at the $x$ direction:

$$dx' = \gamma(v) [dx - vdt]$$ $$cdt = \gamma(v) [cdt - \beta dx]=\gamma(v) [1 - \beta u_x]cdt$$

Then combining should yield the result $$\frac{dx'}{dt'}=\frac{u_x-v}{1-\frac{u_xv}{c^2}}=u'_{x'}$$

But I am getting something confused because I am at $$\frac{dx'}{dt'}=\frac{[dx - vdt]}{[1 - \beta u_x]cdt}=\frac{[u_x - v]}{[1 - \beta u_x]c}$$

and wondering what happens to the $c$ that is kicking about on the denominator.

  • 1
    $\begingroup$ You've divided by $cdt'$ rather than $dt'$ $\endgroup$ Nov 28, 2012 at 20:28

2 Answers 2


For the sake of being explicit and double checking I do actually get this now I am answering the question fully myself.

I had made two mistakes. Firstly, $dt'$ should have been:

$$cdt'=\gamma(v)[cdt-\beta dx]= \gamma(v) \left[1- \frac{\beta}{c}\frac{dx}{dt}\right]cdt=\left[1- \frac{\beta v_x}{c}\right]cdt$$

The second was forgetting about the c in $c dt$ when dividing through:

$$\frac{dx'}{cdt'}=\frac{[dx - vdt]}{[1 - \beta u_x]cdt}$$

So solution is:

$$\frac{dx'}{dt'}=\frac{[u_x - v]c}{\left[1 - \frac{v u_x}{c^2}\right]c}=\frac{[u_x - v]}{\left[1 - \frac{v u_x}{c^2}\right]}$$


Look at where you expand $c\mathrm{d}t'$ into $\gamma(v)[1 - \beta u_x]c\mathrm{d}t$. The last step of that is wrong, and you can tell because the units don't match up in the final expression: $\beta u_x$ has units of velocity, so you can't subtract it from $1$ (which is unitless). So check your math there.

  • $\begingroup$ Oh yeah! I forgot it was $cdt'$ to start with $\endgroup$
    – Magpie
    Nov 28, 2012 at 20:27

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