A Lorentz transformation for a boost in the $x$ direction ($S'$ moves in $+x$, $v>0$) is given by: $$ t'=\gamma\left(t-v\frac{x}{c^2}\right),~x'=\gamma(x-vt)$$ In the derivation of the addition of velocities (see eg Wikipedia), they simply divide the differentials: $$dt'=\gamma\left(dt-v\frac{dx}{c^2}\right),~dx'=\gamma(dx-v\,dt)$$ yielding $$dx'/dt'=\frac{dx-v\,dt}{dt-v\frac{dx}{c^2}}=\frac{dx/dt-v}{1-v\,dx/dt}$$ How do these differentials follow from the Lorentz transformations mathematically? And is there a chain rule involved?
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3$\begingroup$ You could say there is indeed some chain rule involved, e.g. $t'=t'(t,x)$, so $dt' = (dt'/dt)dt + (dt'/dx)dx$. However, since the equations are linear you can in practice just replace $t$ by $dt$ (and $x$ by $dx$). $\endgroup$– Robrecht KeijzerCommented Jun 14 at 15:03
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How do these differentials follow from the Lorentz transformations mathematically? And is there a chain rule involved?
You simply take the total differentials of the Lorentz transform formula you listed. This is for a Lorentz transform, not for some object, so $v$ and $\gamma$ are constant. If you forget that they are constant then indeed the chain rule applies and you would get $$c\ dt'=c\gamma \ dt + ct \ d\gamma - \frac{x\gamma}{c}dv-\frac{v\gamma}{c}dx-\frac{vx}{c}d\gamma$$ $$dx'=\gamma \ dx + x \ d\gamma - \gamma v \ dt - t\gamma \ dv - t v \ d\gamma$$ And as soon as you remember that they are constant you can substitute $dv=0$ and $d\gamma=0$ to get the usual formula.
