Space versus space-time YouTube

Figure 4.12 - Triangle in space or space-time

Figure 4.12 - Triangle in space or space-time

I've added this section to point out the difference between the separation between two points in space and two events in space-time.

Consider the triangle in figure 4.12. It could be a triangle in space or space-time.

In space, any side of the triangle will be shorter than the sum of the lengths of the other two sides and in particular, $a \lt b + c$.

If we are in space-time, then the lines are all world-lines so that the separations between the events are time-like.

Now, $OA$ is the world-line of a stationary observer, $OBA$ is (could be) the world-line of a moving observer, in which case it follows that $a \gt b + c$.

Figure 4.13 - Triangle in space

Figure 4.13 - Triangle in space

To see this clearly, figure 4.13 shows how we might compare $a$ with $b + c$ as spatial lengths.

First draw a circle of radius $b$ about $O$ (dashed blue) and see where it intersects $OA$, at $B'$.

Then translate the line $BA$ around this circle to $B'A'$ as indicated.

Finally, draw a circle of radius $c$ about $B'$ (dashed green) and see where it intersects $OA$, at $C$.

Clearly, the length $OC = b + c$ is greater than the length $OA = a$

Figure 4.14 - Triangle in space-time

Figure 4.14 - Triangle in space-time

We can do the same thing in space-time, but now we are dealing with proper time, as opposed to spatial distance, in which case we need to use hyperbolae to calibrate times, not circles. This means that we cannot trust our eyes when comparing proper times in a space-time diagram.

Figure 4.14 shows how we might compare the proper times, $a, b + c$.

First draw a hyperbola of radius $b$ about $O$ and see where it intersects $OA$, at $B'$.

Then translate $BA$ to $B'A'$.

Finally, draw a hyperbola of radius $c$ about $B'$ and see where it intersects $OA$, at $C$.

Clearly, the proper time $OC = b + c$ is less than the proper time $OA = a$.