# [Physics] Deriving the law of velocity addition from the composition of Lorentz transformations

special-relativity

I have trouble understanding the derivation of the law of velocity addition from the composition of Lorentz transformations. The proof is from Special Relativity by Nicholas Woodhouse:

The author sets up three reference frames $O,O'$ and $O''$, where $O'$ moves with velocity $u$ relative to $O$, $O$ with $v$ relative to $O''$
and $O'$ with $w$ relative to $O''$:

$\begin{pmatrix} ct\\ x \end{pmatrix} = \gamma(u)\begin{pmatrix} 1 & \frac{u}{c} \\ \frac{u}{c} & 1 \end{pmatrix}\begin{pmatrix} ct'\\x' \end{pmatrix}$

$\begin{pmatrix} ct''\\ x'' \end{pmatrix} = \gamma(v)\begin{pmatrix} 1 & \frac{v}{c} \\ \frac{v}{c} & 1 \end{pmatrix}\begin{pmatrix} ct\\x \end{pmatrix}$

$\begin{pmatrix} ct''\\ x'' \end{pmatrix} = \gamma(w)\begin{pmatrix} 1 & \frac{w}{c} \\ \frac{w}{c} & 1 \end{pmatrix}\begin{pmatrix} ct'\\x' \end{pmatrix}$

From there follows that
$\gamma (w)\begin{pmatrix} 1 &\frac{w}{c} \\ \frac{w}{c} & 1 \end{pmatrix} = \gamma(u) \gamma(v)\begin{pmatrix} 1 & \frac{v}{c} \\ \frac{v}{c} & 1 \end{pmatrix}\begin{pmatrix} 1 & \frac{u}{c} \\ \frac{u}{c} & 1 \end{pmatrix}$.

The the author states that because of that $\gamma(w)=\gamma(u)\gamma(v)(1+\frac{uv}{c^2}$). How does that follow from the above equation? I can see that it holds if $w=\frac{u+v}{1+\frac{uv}{c^2}}$ which is the law of velocity addition. But isn't that circular reasoning? Couldn't you say for example that $\gamma(w)=\gamma(u)\gamma(v)$ if $w=v+u$?

My question is how can you derive the law of velocity addition from the composition of Lorentz transformations without assuming it a priori? Or am I missing something here?

#### Best Answer

Look at your last equation:

$$\gamma (w)\begin{pmatrix} 1 &\frac{w}{c} \\ \frac{w}{c} & 1 \end{pmatrix} = \gamma(u) \gamma(v)\begin{pmatrix} 1 & \frac{v}{c} \\ \frac{v}{c} & 1 \end{pmatrix}\begin{pmatrix} 1 & \frac{u}{c} \\ \frac{u}{c} & 1 \end{pmatrix}$$

It implies, on equating matrix elements, two equations:

$$\gamma(w)=\gamma(u)\,\gamma(v)\,\left(1+\frac{u\,v}{c^2}\right)\quad\text{(equate elements (1,1) and (2,2))}$$ $$\gamma(w)\,w=\gamma(u)\,\gamma(v)\,(u+v)\quad\text{(equate elements (1,2) )}$$

Now eliminate the $\gamma$s (e.g. divide one by the other) and you should make some headway.

My personal favorite way for this problem is to reason with rapidities. Rapidities, defined by $\eta = \operatorname{artanh}\left(\frac{v}{c}\right)$, combine linearly. This assertion is readily proven by writing the boost matrix as $\exp\left(\eta\,\left(\begin{array}{cc}0&1\\1&0\end{array}\right)\right)$. Then the velocity addition law simply follows from the addition identity for hyperbolic tangents $\tanh(\eta_1+\eta_2) = \frac{\tanh \eta_1+\tanh\eta_2}{1+\tanh\eta_1\,\tanh\eta_2}$.