Coordinate-free distance in special relativity

We could "define" the square of distance of two points in the Cartesian plane by $$\Delta x^2+ \Delta y^2,$$ but distance "really is" what we measure by ruler and doesn't need any coordinate description.

In special relativity, the square of distance is $$\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$$ but here what distance "really is"? How could we measure the distance without coordinates?

• Short answer: with a clock. Commented Jun 18, 2023 at 13:48

It's probably worth adding to the conversation that "how do I measure the interval" depends on whether it is timelike, spacelike, or null.

For timelike intervals, you find an inertial frame where the two endpoints are at the same spatial point, and then the size of the interval gets measured with a clock.

For spacelike intervals, you find an inertial frame where the entire interval is simultaneous, and then you measure the size of the interval with a ruler.

For null intervals, the size of the interval is zero in every coordinate system. there are various "tricks" to measuring the "distance" traveled along null curves, but they all rely on some sort of arbitrary convention like "count the number of clock ticks leaving a surface that cross the null line". the core geometric thing is that this is a macroscropic path that traces out zero proper length.

The problem is that you're comparing apples with oranges.

The quantity you describe as a distance in "normal" 3D space is exactly that - a distance.

The quantity you're describing as a distance in the 3+1D spacetime of special relativity is strictly speaking an interval. It's the invariant time between two events. It's called the proper time.

If the events were at the same place you'd have the $$x,y$$ and $$z$$ separations zero and it would be the same as the square of the classical time between two events.

Hence Javier's comment that we would measure the quantity with a clock. It's a time.

Because this is relativity different observers will measure the differences in $$x,y,z$$ and $$t$$ differently, but they'll measure the proper time the same. It's invariant.