[Physics] How to find the direction of velocity of a reference frame where two events are simultaneous in case of a space-like interval

special-relativity

Suppose in a inertial reference frame $S$, an event $A$ occurs at $(ct_A, x_A, y_A, z_A)$ and event $B$ occurs at $(ct_B, x_B, y_B, z_B)$.

Now the invariant interval of these two events is,

$$I = -c^2 (t_A – t_B)^2 + (x_A – x_B)^2 + (y_A – y_B)^2 + (z_A – z_B)^2 = -c^2 \Delta t^2 + \Delta \bar x^2,$$

where I'm using the $(-, +, +, +)$ metric.

Now there can be $3$ particular cases of interest corresponding to time-like, space-like and light-like events.

For $I = 0 \implies c^2 \Delta t^2 = \Delta \bar x^2$, events are light-like.

For $I < 0 \implies c^2 \Delta t^2 > \Delta \bar x ^2$, events are time-like and a reference-frame $\bar S$ exists(accessible by appropriate Lorentz Transformation) for which these two events occur at the same location. The velocity(magnitude and direction) can be computed.

For $I > 0 \implies c^2 \Delta t^2 < \Delta \bar x^2$, events are space-like and a a reference frame $\bar S$ exists(again accessible by appropriate Lorentz Transformation) for which these two events are simultaneous.

I know how to calculate the velocity(direction and magnitude) of the $\bar S$ frame relative to the $S$ frame in case of a time-like event. I also know how to calculate the magnitude of velocity of the $\bar S$ frame relative to the $S$ frame for a space-like event.

How to find the direction of the $\bar S$ frame relative to $S$ for a space-like event?

Best Answer

This problem would have given me fits when I first learned relativity, but since I've started using the geometric point of view as the tool I reach for first it is almost trivial.

The direction is easy: boost in the direction from the earlier to the later event as measured in your current frame. Why? Because you want the space-like axis to tilt upward in that direction.

Getting the speed is also surprisingly easy. You need the new space-like axis to have a slope (in your current coordinate system) equal to $(\Delta x)/(c \,\Delta t)$, which is exactly the $\beta$ of the boost you need.

Related Question