Frame where two velocities are the opposite of each other

\[\newcommand{\vini}[1]{\mathbf{#1}_{\mathrm{I}}} \newcommand{\vmid}[1]{\mathbf{#1}_{\mathrm{M}}} \newcommand{\vfin}[1]{\mathbf{#1}_{\mathrm{F}}} \newcommand{\ini}[1]{#1_{\mathrm{I}}} \newcommand{\fin}[1]{#1_{\mathrm{F}}} \newcommand{\vvr}{\mathbf{r}} \newcommand{\vvr}{\mathbf{r}} \newcommand{\vvv}{\mathbf{v}} \newcommand{\vva}{\mathbf{a}} \newcommand{\vvu}{\mathbf{u}} \newcommand{\vvw}{\mathbf{w}}\]

Introduction

Consider one point-like particle. The particle has position \(\vvr_1\) and velocity \(\vvv_1\) at time \(t_1\). It has position \(\vvr_2\) and velocity \(\vvv_2\) at time \(t_2\). In classical mechanics, it is possible to choose an intertial reference system so that \(\vvr_2' = -\vvr_1'\) and \(\vvv_2' = -\vvv_1'\). What about Special Relativity?

Motivation

Even simple calculations get pretty messy in Special Relativity. Being able to choose a reference frame where formulas are mathematically simpler can help a great deal with the caculations. It also gives a big help to the poor human mind that is trying to understand what is going on behind all the math.

This simplification helps solving the analytical positioning problem.

Classical case

Let’s first take a look at the classical case. The following reference frame transformation:

\[\begin{equation} \mathbf{r}' = \mathbf{r} - \frac{\vvr_1 + \vvr_2}{2} - \frac{\vvv_1 + \vvv_2}{2} \, \left(t - \frac{t_1 + t_2}{2}\right), \label{eq:classical} \end{equation}\]

brings us in the reference frame where the positions and the velocities of the particle at the two instants \(t_1\) and \(t_2\) are the opposite of each other. In particular, applying this transformation to \(\vvr_1\), \(\vvv_1\), \(\vvr_2\), and \(\vvv_2\), we get:

\[\begin{eqnarray*} \vvr_1' & = & \frac{\vvr_1 - \vvr_2}{2} - \frac{\vvv_1 + \vvv_2}{2} \, \frac{t_1 - t_2}{2}, \\ \vvr_2' & = & \frac{\vvr_2 - \vvr_1}{2} - \frac{\vvv_1 + \vvv_2}{2} \, \frac{t_2 - t_1}{2} = -\vvr_1', \\ \vvv_1' & = & \frac{\vvv_1 - \vvv_2}{2}, \\ \vvv_2' & = & \frac{\vvv_2 - \vvv_1}{2} = -\vvv_1'. \end{eqnarray*}\]

Relativistic case

Let’s now try to find a reference frame transformation analogous to Eq. \eqref{eq:classical}, but within the framework of Special Relativity.

As derived in this other post, a four-vector \((\vvw,\,w_0)\) seen from a reference frame that moves with speed \(\vvv\) can be written as follows:

\[\begin{equation} \vvw' = P_\perp \vvw + \gamma P_\parallel \vvw - \gamma \frac{\vvv}{c} \, w_0. \label{eq:boost} \end{equation}\]

Let’s see whether we can find a velocity \(\vvv\) which transforms \(\vvv_1\) and \(\vvv_2\) so that \(\vvv_2' = -\vvv_1'\). We then use Eq. \eqref{eq:boost} with \(\vvw = \gamma_1 \vvv_1\), \(w_0 = \gamma_1 c\) and with \(\vvw = \gamma_2 \vvv_2\), \(w_0 = \gamma_2 c\):

\[\begin{eqnarray*} \gamma_1' \vvv_1' & = & P_\perp \gamma_1 \vvv_1 + \gamma P_\parallel \gamma_1 \vvv_1 - \frac{\vvv}{c} \, \gamma \gamma_1 c, \\ \gamma_2' \vvv_2' & = & P_\perp \gamma_2 \vvv_2 + \gamma P_\parallel \gamma_2 \vvv_2 - \frac{\vvv}{c} \, \gamma \gamma_2 c. \\ \end{eqnarray*}\]

We, however, have \(\gamma_2' \vvv_2' = -\gamma_1' \vvv_1'\), because \(\vvv_2' = -\vvv_1'\) and thus \(\gamma_1' = \gamma_2'\). Therefore we can combine the two equations above, obtaining:

\[\begin{equation} P_\perp (\gamma_1 \vvv_1 + \gamma_2 \vvv_2) + \gamma P_\parallel (\gamma_1 \vvv_1 + \gamma_2 \vvv_2) - \vvv \, \gamma(\gamma_1 + \gamma_2) = 0. \\ \label{eq:projs} \end{equation}\]

This equation can be projected into the direction parallel to \(\vvv\) (via \(P_\parallel\)) and the orthogonal directions (via \(P_\perp\)), obtaining two independent equations. In particular, for the orthogonal directions we have \(P_\perp (\gamma_1 \vvv_1 + \gamma_2 \vvv_2) = 0\), which implies that \(\vvv\) is parrallel to \(\gamma_1 \vvv_1 + \gamma_2 \vvv_2\), i.e. \(P_\parallel (\gamma_1 \vvv_1 + \gamma_2 \vvv_2) = \gamma_1 \vvv_1 + \gamma_2 \vvv_2\). Eq. \eqref{eq:projs} can thus be rewritten as:

\[\gamma (\gamma_1 \vvv_1 + \gamma_2 \vvv_2) - \vvv \, \gamma (\gamma_1 + \gamma_2) = 0.\]

Finally, we have:

\[\bbox[lightyellow, 10px, border: 2px solid orange]{ \begin{equation*} \vvv = \frac{\gamma_1 \vvv_1 + \gamma_2 \vvv_2}{\gamma_1 + \gamma_2}. \end{equation*} }\]

A Lorentz boost with this velocity transforms the two velocities \(\vvv_1\) and \(\vvv_2\) so that they are the opposite of each other, i.e. \(\vvv_2' = -\vvv_1'\).

After the boost is applied, the reference frame can be translated by a four-vector \((\vvr_1 + \vvr_2, t_1 + t_2)/2\) to ensure that \(\vvr_2' = -\vvr_1'\) and \(t_2 = -t_1\), similarly to the classical case.