If your position vector is , then your velocity vector is (velocity is the time derivative of position).

But since relativity messes with distance and time, it’s important to come up with a better definition of time. This way you can talk about how fast an object is moving through time, as well as how fast it’s moving through space.

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Here’s the idea; energy is distance times force (E=DF), and force is mass times acceleration (E=DMA), and acceleration is velocity over time, which is the same as distance over time squared (E = DMD/T or something like that, and still have the units correct.

It may even be possible to mix together a bunch of other universal constants until you get velocity squared, or there may just be a new, previously unknown physical constant involved.

Notice that when you ignore time (t=0), then this reduces to the usual definition.

This fancy new “spacetime interval” conserves the length of things under ordinary rotations (which just move around the x, y, z part), but also conserves length under “rotations involving time”.

This trick is just something used to get a thumbnail sketch of what “looks” like a correct answer, and in this particular case it’s exactly right.

For a more formal derivation, you’d have to stir the answer gravy: of regular vectors, (which could be distance, momentum, whatever) remains unchanged by rotations.

“Spacetime rotations” (changing your own speed) are often called “Lorentz boosts“, by people who don’t feel like being clearly understood.

You can prove that the spacetime interval is invariant based only on the speed of light being the same to everyone.

You can derive the gamma function by thinking about light clocks, or a couple other things, but I don’t want to side track on that just now. That is, is the ratio of how fast “outside time” passes from the point of view of the object’s “on-board time”. For succinctness (and tradition) I’ll bundle the first three terms together: Now check this out!