Thanks for the details.
Should one view ##E^{2}=m^{2}c^{4}+p^{2}c^{2}## then, as describing the energy of a particle without regard of any potential that may be present?
Why is this? Is it because a particle collision occurs at a point, and since only potential energy differences are physical...
That's what I thought. It comes from purely kinematical considerations, without any mention of a potential.
I guess maybe the author is just observing the fact that the energy only contains kinetic energy plus a constant term, and thus can be interpreted as the energy of a free particle...
I mean, not subject to any external forces.
This was my understanding, however, I have read several notes (of which I can't find of hand, but I'm currently looking for an example) in which they refer to this expression for ##E## as giving the energy of a free particle, leaving me confused.
I'm in the process of learning special relativity (SR), and I'm a bit confused as to why the relativistic energy dispersion relation ##E^{2}=m^{2}c^{4}+p^{2}c^{2}## gives the energy for a free particle? I get that it is the sum of (relativistic) kinetic energy plus the rest mass term (a...