Deriving the law of velocity addition from the composition of Lorentz transformations

Question

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?

Answer 1

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.

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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 $\tanh(\eta_1+\eta_2) = \frac{\tanh \eta_1+\tanh\eta_2}{1+\tanh\eta_1\,\tanh\eta_2}$.

Answer 2

\begin{equation*} \left( \begin{array}{c} ct^\prime \\ x^\prime\\ \end{array} \right)=\left[ \begin{array}{cc} \gamma_1 & -\gamma_1 \frac{v_1}{c}\\ -\gamma_1 \frac{v_1}{c} & \gamma_1 \\ \end{array} \right] \left( \begin{array}{c} ct \\ x\\ \end{array} \right)~~~~~~~~~~~~~~~~~~~~~~~~~(1) \end{equation*}

\begin{equation*} \left( \begin{array}{c} ct^{\prime\prime} \\ x^{\prime\prime}\\ \end{array} \right)=\left[ \begin{array}{cc} \gamma_2 & -\gamma_2 \frac{v_2}{c}\\ -\gamma_2 \frac{v_2}{c} & \gamma_2 \\ \end{array} \right] \left( \begin{array}{c} ct^\prime \\ x^\prime\\ \end{array} \right)~~~~~~~~~~~~~~~~~~~~~~~~(2) \end{equation*}

Performing two Lorentz transformations with velocities $v_1$ and $v_2$ in the $x$-direction in succession is equivalent to a single Lorentz transformation. Let us consider the velocity of the third observer with respect to the first observer is $v$ and

\begin{equation} \left( \begin{array}{c} ct^{\prime\prime} \\ x^{\prime\prime}\\ \end{array} \right)=\left[ \begin{array}{cc} \gamma & -\gamma \frac{v}{c}\\ -\gamma \frac{v}{c} & \gamma \\ \end{array} \right] \left( \begin{array}{c} ct \\ x\\ \end{array} \right)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3) \end{equation}

Therefore, by using equations (1)-(2)

\begin{equation*} \left( \begin{array}{c} ct^{\prime\prime} \\ x^{\prime\prime}\\ \end{array} \right)=\left[ \begin{array}{cc} \gamma_2 & -\gamma_2 \frac{v_2}{c}\\ -\gamma_2 \frac{v_2}{c} & \gamma_2 \\ \end{array} \right]\left[ \begin{array}{cc} \gamma_1 & -\gamma_1 \frac{v_1}{c}\\ -\gamma_1 \frac{v_1}{c} & \gamma_1 \\ \end{array} \right] \left( \begin{array}{c} ct \\ x\\ \end{array} \right) \end{equation*}

\begin{equation*} ~~~~~~~~~~~~=\left[ \begin{array}{cc} \gamma_1\gamma_2\left(1+\frac{v_1v_2}{c^2}\right) & -\gamma_1\gamma_2 \frac{(v_1+v_2)}{c}\\ -\gamma_1\gamma_2 \frac{(v_1+v_2)}{c} & \gamma_1\gamma_2\left(1+\frac{v_1v_2}{c^2}\right) \\ \end{array} \right] \left( \begin{array}{c} ct \\ x\\ \end{array} \right)~~~~(4) \end{equation*}

Comparing eq. (4) with eq. (3), one gets \begin{eqnarray} \gamma&=&\gamma_1\gamma_2\left(1+\frac{v_1v_2}{c^2}\right)\\ -\gamma\frac{v}{c}&=&-\gamma_1\gamma_2\frac{(v_1+v_2)}{c} \end{eqnarray}

Substituting $\gamma$ into the second equation, gives

\begin{equation} v=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}. \end{equation}

Answer 3

If $x$ denotes coordinates in the reference frame $O$ then the above equations can be written as

$$ x = \Lambda(u)x' ~~~~~~ x'' = \Lambda(v)x ~~~\mbox{and}~~~ x'' = \Lambda(w)x' $$

Note that the equation in the middle can also be written as

$$ x'' = \Lambda(v)x = \Lambda(v)[\Lambda(u) x'] = [\Lambda(v)\Lambda(u)]x' = \Lambda(w)x' $$

you can conclude from this last equation that

$$ \Lambda(w) = \Lambda(v)\Lambda(u) ~~~\Leftrightarrow~~~ \gamma(w)\left(\begin{array}{cc} 1 & w/c \\ w/c & 1\end{array}\right) = \gamma(v)\gamma(u)\left(\begin{array}{cc} 1 & v/c \\ v/c & 1\end{array}\right) \left(\begin{array}{cc} 1 & u/c \\ u/c & 1\end{array}\right) $$

Take for instance the component $(0,0)$ of this equation, you get

$$ \gamma(w) = \gamma(v)\gamma(u)[1 + uv/c^2] $$