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!Remember that the spacetime interval for a spacetime vector with spacial component , and temporal component , is .So whatever that last term is () it’s also conserved (as long as you don’t change your own speed). Could this whole thing (thought Einstein) be the energy of the object in question, divided by c? And, since c is a constant, energy divided by c is also conserved. Notice that the energy and momentum here are not the energy and momentum: and .This only has noticeable effects at extremely high speeds, and at lower speeds they look like: and , which is what you’d hope for.
In relativity you rotate how you see things in spacetime, by running past them (changing your speed with respect to what you’re looking at).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.
New theories should always include the old theories as a special case (or disprove them). If you allow the speed of the object to be zero (v=0), you find that everything other than the first term in that long equation for E vanishes, and you’re left with (drumroll): E=mc of energy.