Two events occur simultaneously at points A and B as observed in...

Let the lab frame be $S and the moving frame be$S'$.<br><br>The time interval between two events in$S'$is given by the Lorentz transformation:<br><br>$\Delta t' = \gamma \left( \Delta t - \dfrac{\vec{v} \cdot \Delta \vec{r}}{c^2} \right)<br><br>Given that the events are simultaneous in the lab frameS$, we have$\Delta t = 0$.<br><br>For the events to be simultaneous in the moving frame$S'$, we must have$\Delta t' = 0$.<br><br>Substituting$\Delta t = 0$and$\Delta t' = 0$into the equation, we get:<br><br>$0 = \gamma \left( 0 - \dfrac{\vec{v} \cdot \Delta \vec{r}}{c^2} \right)$<br><br>$\Rightarrow \vec{v} \cdot \Delta \vec{r} = 0$<br><br>Here,$\Delta \vec{r}$is the position vector$\overrightarrow{AB}$between the two events. The dot product is zero only if the velocity vector$\vec{v}$is perpendicular to$\overrightarrow{AB}$.<br><br>Thus, the moving frame must move in a direction perpendicular to$AB$.$