Composition of Velocities
In classical mechanics, a velocity \({\vb u’}\) in the moving frame will be measured in the stationary frame as \({\vb u}={\vb v}+{\vb u’}\), which is the well-known law of the parallelograms.
In relativity, the parallelogram formula holds only for velocities well below the speed of light, and the composition of velocities needs to account for the fact that no speed can be measured, which is faster than the speed of light.
To derive the more general parallelogram formula, we consider the general motion of a body in \(F’\) described by the four-vector
\begin{equation} \ket{\vb{x’},0} +c t’ \ket{\vb{0},1} \end{equation}The motion of the body observed in \(F\) is given by the Lorentz transformation,
\begin{equation} \vb \Lambda({\vb v}) \left(\ket{\vb{x},0} +c t \ket{\vb{0},1}\right) = \ket{\vb{x’},0} +c t’ \ket{\vb{0},1} \label{lk} \end{equation} \begin{equation} \vb \Lambda({\vb v})= \beta \ket{\vb{e}_1,0}\bra{\vb{e}_1,0} -\ds\beta\frac{v}{c}\ket{\vb{e}_1,0}\bra{\vb{0},1}+ (\vb{I’}-\ket{\vb{e}_1,0}\bra{\vb{e}_1,0}) -\beta\frac{ v} {c} \ket{\vb{0},1}\bra{\vb{e}_1,0} +\beta \ket{\vb{0},1}\bra{\vb{0},1}\\ \label{lambda_1} \end{equation}and the velocity of the body in \(F\) is computed from the time derivative of equation (\ref{lk}) with respect to the time \(t\),
\begin{equation} \vb \Lambda({\vb v}) (\ket{\vb{u},0} +c \ket{\vb{0},1}) = \beta\left(-\frac{ v} {c^2}\bra{{\vb e}_1}\ket{\vb u}+1 \right) (\ket{\vb{u’},0} +c \ket{\vb{0},1}) \label{nm} \end{equation}where the time \(t’\) is a function of the time \(t\) given by the time component of the Lorentz transformation
\begin{equation} t’=\beta\left(-\frac{ v} {c^2}\bra{{\vb e}_1}\ket{\vb x}+t \right) \end{equation}from which we easily compute the derivative with respect to \(t\)
\begin{equation} \frac{dt’}{dt}=\beta\left(-\frac{ v} {c^2}\bra{{\vb e}_1}\ket{\vb u}+1 \right) \label{pkl} \end{equation}We write equation (\ref{nm}) explicitly,
\begin{gather*} (\beta \ket{\vb{e}_1,0}\bra{\vb{e}_1,0} -\ds\frac{v}{c}\beta \ket{\vb{e}_1,0}\bra{\vb{0},1} + \ket{\vb{e}_2,0}\bra{\vb{e}_2,0} + \ket{\vb{e}_3,0}\bra{\vb{e}_3,0} + -\beta\ds\frac{v}{c} \ket{\vb{0},1}\bra{\vb{e}_1,0}+ \beta\ket{\vb{0},1}\bra{\vb{0},1} )\cdot (\ket{\vb{u},0} +c \ket{\vb{0},1} ) \\ = \\ \beta\left(-\frac{ v} {c^2}\bra{{\vb e}_1}\ket{\vb u}+1 \right) (\ket{\vb{u’},0} +c \ket{\vb{0},1}) \end{gather*}which simplifies to,
\begin{gather*} \beta \ket{\vb{e}_1,0}\bra{\vb{e}_1,0} \ket{\vb{u},0} -{v}\beta \ket{\vb{e}_1,0} +\ket{\vb{e}_2,0}\bra{\vb{e}_2,0} \ket{\vb{u},0} +\ket{\vb{e}_3,0}\bra{\vb{e}_3,0} \ket{\vb{u},0} -\beta\ds\frac{v}{c} \ket{\vb{0},1}\bra{\vb{e}_1,0} \ket{\vb{u},0} +c\beta\ket{\vb{0},1} \\ = \\ -\beta\frac{ v} {c^2}\ket{\vb{u’},0}\bra{{\vb e}_1}\ket{\vb u} -\beta\frac{ v} {c}\ket{\vb{0},1}\bra{{\vb e}_1}\ket{\vb u} +\beta\ket{\vb{u’},0} +\beta c\ket{\vb{0},1} \end{gather*}and finally,
\begin{equation} \beta \ket{\vb{e}_1 +\frac{ v} {c^2}\vb{u’},0}\bra{{\vb e}_1,0}\ket{\vb u,0} +\ket{\vb{e}_2,0}\bra{\vb{e}_2,0} \ket{\vb{u},0} +\ket{\vb{e}_3,0}\bra{\vb{e}_3,0} \ket{\vb{u},0} = {v}\beta \ket{\vb{e}_1,0} +\beta\ket{\vb{u’},0} \label{df} \end{equation}The problem with the above equation is that \(\vb u’\) appears both on the left and right sides, and, in order to separate the two velocities, we have to write the above expression as the product of an operator depending only on \(\vb u’\), times \(\vb u\).
The first step is to decompose \(\vb u’\) along the versors,
\begin{equation} {\vb u’} = u’_1\ket{{\vb e}_1,0} + u’_2\ket{{\vb e}_2,0} +u’_3\ket{{\vb e}_3,0} \label{hj} \end{equation}so that equation (\ref{df}) assumes the form,
\begin{gather*} \left(\beta \left(1 +\ds\frac{ v} {c^2}u’_1\right)\,\,\ket{{\vb e}_1,0}\bra{{\vb e}_1,0} + \beta\frac{ v} {c^2}u’_2\,\,\ket{{\vb e}_2,0}\bra{{\vb e}_1,0}+ \beta\ds\frac{ v} {c^2}u’_3\,\,\ket{{\vb e}_3,0}\bra{{\vb e}_1,0} +\ket{\vb{e}_2,0}\bra{\vb{e}_2,0} + \ket{\vb{e}_3,0}\bra{\vb{e}_3,0} \right)\cdot\ket{\vb{u},0} \\ = \\ {v}\beta \ket{\vb{e}_1,0} +\beta\ket{\vb{u’}}\label{yu1} \end{gather*}We have on the right side of the equation an operator that only depends on \(\vb u’\),
\begin{equation} \beta \left(1 +\ds\frac{ v} {c^2}u’_1,0\right)\,\,\ket{{\vb e}_1,0}\bra{{\vb e}_1,0} + \beta\ds\frac{ v} {c^2 }u’_2\,\,\ket{{\vb e}_2,0}\bra{{\vb e}_1,0}+\beta\ds\frac{ v} {c^2}u’_3\,\,\ket{{\vb e}_3,0}\bra{{\vb e}_1,0} + \ket{\vb{e}_2,0}\bra{\vb{e}_2,0} +\ket{\vb{e}_3,0}\bra{\vb{e}_3,0} \end{equation}which can be easily inverted as,
\begin{equation} \ds\frac{1}{\beta \left( 1 +\ds\frac{ v} {c^2}u’_1\right)}\,\,\ket{{\vb e}_1,0 }\bra{{\vb e}_1,0} % \ds-\frac{ v} {c^2}\frac{u’_2}{(1 +\ds\frac{ v} {c^2}u’_1)}\,\,\ket{{\vb e}_2,0}\bra{{\vb e}_1,0} % \ds-\frac{ v} {c^2}\frac{u’_3}{(1 +\ds\frac{ v} {c^2}u’_1)}\,\,\ket{{\vb e}_3,0}\bra{{\vb e}_1,0}+ \ket{\vb{e}_2,0}\bra{\vb{e}_2,0} +\ket{\vb{e}_3,0}\bra{\vb{e}_3,0} \end{equation}We can express the velocity measured in \(F\) as,
\begin{equation} \ket{\vb{u},0}= \left( \ds\frac{1}{\beta (1 +\ds\frac{ v} {c^2} u’_1)}\,\,\ket{{\vb e}_1,0}\bra{{\vb e}_1,0} % \ds-\frac{ v} {c^2}\frac{u’_2}{(1 +\ds\frac{ v} {c^2}u’_1)}\,\,\ket{{\vb e}_2,0}\bra{{\vb e}_1,0} % \ds-\frac{ v} {c^2}\frac{u’_3}{(1 +\ds\frac{ v} {c^2}u’_1)}\,\,\ket{{\vb e}_3,0}\bra{{\vb e}_1,0}+ \ket{\vb{e}_2,0}\bra{\vb{e}_2,0} + \ket{\vb{e}_3,0}\bra{\vb{e}_3,0} \right)\cdot \left({v}\beta \ket{\vb{e}_1,0} +\beta\ket{\vb{u’}}\right) \end{equation}which finally simplifies to,
\begin{equation} \ket{\vb{u},0}= \ds \frac{1}{(1 +\ds\frac{ v} {c^2}u’_1)} \left( \ds (v+u’_1) \,\,\ket{{\vb e}_1,0} – % \ds\sqrt{1-\ds\frac{v^2}{c^2}}\; u’_2 \; \ket{\vb{e}_2,0} – \sqrt{1-\ds\frac{v^2}{c^2}} \; u’_3 \; \ket{\vb{e}_3,0} \right) \label{p12} \end{equation}Equation (\ref{p12}) assumes a more elegant form when we drop the explicit dependence on the time component
\begin{equation} \ket{\vb{u}}= \frac{1}{ 1 +\ds\frac{ \bra{\vb v}\ket{\vb{u’} }} {c^2} } \left( \ket{\vb v} + % \alpha_v \ket{\vb u’} +\left(1-\alpha_v\right)\frac{ \bra{\vb u’}\ket{\vb v}}{v^2}\ket{\vb v} \right) \label{nice} \end{equation}where,
\begin{align} \ket{\vb v}&=v\,\, \ket{\vb{e}_1,0} \nonumber\\ \alpha_v&=\frac{1}{\beta}=\ds\sqrt{1-\ds\frac{v^2}{c^2}}\nonumber \end{align}Considering the components of \(u\) along the vectors \(\ket{\vb{e}_1,0}, \ket{\vb{e}_2,0}\) and \(\ket{\vb{e}_3,0}\)
\begin{equation} {\vb u} = u_1\ket{\vb{e}_1,0} +u_2\ket{\vb{e}_2,0} +u_3\ket{\vb{e}_3,0} \end{equation}equation (\(\ref{p12}\)) assumes a lighter and more familiar form
\begin{align} u_1 &= \frac{u’_1+v}{1+\ds\frac{vu’}{c^2}}\nonumber\\ u_2 & = \frac{\sqrt{1-\ds\frac{v^2}{c^2}}}{1+\ds\frac{vu’_1}{c^2}}u’_2\nonumber\\ u_3 &= \frac{\sqrt{1-\ds\frac{v^2}{c^2}}}{1+\ds\frac{vu’_1}{c^2}}u’_3\label{velComp}\\ \end{align}Composition of Accelerations
The composition of accelerations is computed by performing the time derivative of the velocity (\(\ref{nice}\))
\begin{align*} \ket{\vb{a}}&= -\frac{1}{\left( 1 +\ds\frac{ \bra{\vb v}\ket{\vb{u’} }} {c^2}\right)^2} \left( \ket{\vb v} + % \alpha_v \ket{\vb u’} +\left(1-\alpha_v\right)\frac{ \bra{\vb u’}\ket{\vb v}}{v^2}\ket{\vb v} \right)\frac{ \bra{\vb a’}\ket{\vb{v} }} {c^2}\frac{dt’}{dt} \\ &+\frac{1}{ 1 +\ds\frac{ \bra{\vb v}\ket{\vb{u’} }} {c^2} } \left( % \alpha_v \ket{\vb a’} +\left(1-\alpha_v\right)\frac{ \bra{\vb a’}\ket{\vb v}}{v^2}\ket{\vb v} \right)\frac{dt’}{dt} \end{align*}The derivative \(dt’/dt\) reads, using equation (\ref{pkl}) and (\ref{nice}),
\begin{equation} \frac{dt’}{dt} = \frac{\alpha}{1 +\ds\frac{ \bra{\vb v}\ket{\vb{u’} }} {c^2}} \end{equation}substituted in the acceleration expressions \(\ket{\vb a}\) yields
\begin{align*} \ket{\vb{a}}&= \frac{\alpha}{\left( 1 +\ds\frac{ \bra{\vb v}\ket{\vb{u’} }} {c^2}\right)^3} \left[ -\left( \ket{\vb v} + % \alpha_v \ket{\vb u’} +\left(1-\alpha_v\right)\frac{ \bra{\vb u’}\ket{\vb v}}{v^2}\ket{\vb v} \right)\frac{ \bra{\vb a’}\ket{\vb{v} }} {c^2} \right. \\ &\left.+( 1 +\ds\frac{ \bra{\vb v}\ket{\vb{u’} }} {c^2}) \left( % \alpha_v \ket{\vb a’} +\left(1-\alpha_v\right)\frac{ \bra{\vb a’}\ket{\vb v}}{v^2}\ket{\vb v} \right) \right] \end{align*}which simplifies to the final expression,
\begin{equation} \ket{\vb{a}}= \frac{\alpha^2}{\left( 1 +\ds\frac{ \bra{\vb v}\ket{\vb{u’} }} {c^2}\right)^2} \ket{\vb a’} + \; \frac{\alpha^2(\alpha-1)\bra{\vb a’}\ket{\vb v}}{v^2\left( 1 +\ds\frac{ \bra{\vb v}\ket{\vb{u’} }} {c^2}\right)^3} \ket{\vb v}\: – \frac{\alpha^2 \bra{\vb a’}\ket{\vb v}}{c^2\left( 1 +\ds\frac{ \bra{\vb v}\ket{\vb{u’} }} {c^2}\right)^3}\ket{\vb u’}\label{acc} \end{equation}