Is the Einstein Energy-Momentum equation $E^2 = p^2c^2 + m_0^2c^4$ valid only for Free Particles?

Question

Is the energy -momentum relation $$E^2 = p^2c^2 + m_0^2c^4$$ satisfied only by free particles or even bound particles?

Does the Energy refer to total Energy(including potential) or only (kinetic +rest mass).

Answer 1

The equation is just the kinetic and rest energy, it does not include potential energy. But potential energy in relativity is not the proper concept.

The linked question has some useful answers, but I think your true question is about how to learn to do things relativistically that you used to do non relativistically. And since the other answer so far takes an entirely quantum answer I'll make this an entirely no quantum answer.

So you used to be able to have potentials exert forces. In reality you knew that the forces came in action reaction pairs so there was something somewhere else that felt an equal and opposite force.

But now you know that simultaneity is relative so if a force acts on you right here right now it doesn't make sense to have some distant thing "right now" feel an equal and opposite force since for the far away thing differently moving observers will disagree about when is right noe for it, so when should it feel the equal and opposite force?

The whole classical idea of potential energy is flawed; seriously, deeply, irredeemably flawed. But if you look at a Lagrangian you might be used to it being written as $L=T-U$ but really all you wanted was a $L=L(x,y,z,\dot x,\dot y, \dot z, t)$ that gives the correct equations of motion as the Euler-Lagrange equations.

Sometimes when you do that you might have a term you call potential energy, but that is not the proper way to think about energy.

Setting up an $L=L(x,y,z,\dot x,\dot y, \dot z, t)$ that gives the correct equations of motion as the Euler-Lagrange equations is something we can do. And we do do it. So for instance with electromagnetism you have to bring in the scalar potential and the vector potential and write an $L$ whose Euler-Lagrange equations are the Lorentz Force Law. But that term is called the scalar potential but it depends on gauge. And it is sad and unfortunate that your introduction to electromagnetism might have made you think qV is a potential energy when really it is a gauge dependant quantity. And thinking of it as an energy when it depends on a gauge is setting yourself up to be confused later.

Because we know how energy actually works in electromagnetism. In electromagnetism the electromagnetic fields have an energy density and a momentum density and energy and density for the fields is conserved in regions without charge or current, and merely flows around. But where there are currents there is a flow of energy and/or momentum between the fields and the charges. And that happens at the same place and the same time, the fields right here right now exchange energy and/or momentum with the charges right here right now so this is sensible.

So the dynamics don't need a potential energy per se, you just need dynamics to have equations of motion and vice versa and you need energy and momentum to be local things that flow from here to there. Shake a charge here and energy and/or momentum flow from it to the field, then flow through empty space until they meet another charge where energy and/or momentum flow from the fields to the charge. Everything happens locally and everything has dynamics.

Finally, as a warning there is something called canonical momentum and it is different than momentum. Canonical momentum is something classically conjugate to position in your classical Lagrangian and that you use to get your classical Hamiltonian (through a Legendre transformation) and that might be constant in time. Momentum is part of the total stress-energy tensor which is a tensor field, so total momentum density is a vector field. And total momentum (e.g. the field momentum and the mechanical momentum of the particles) is conserved in the sense that for any region you can measure the total momentum at time t1 and then measure how much net momentum flowed in through the boundary during the time interval from t1 to t2 and and add that to the momentum in the region at time t1 and the sum is always the total momentum in that region at time t2. There isn't really a relationship between the two. For instance in electromagnetism, the canonical momentum depends on the gauge you pick whereas the momentum does not. So they are pretty different. Momentum is physical and can be measured in a region (though it is frame dependent) whereas canonical momentum is mathematical, can't be measured in a lab, and depends on your coordinates as well as your frame.

And if you combine energy and momentum into four components if a single vector then it becomes a frame independent 4d vector, the different numbers in different frames being no different than getting different components with a different basis. So momentum is real, and canonical momentum is math, classically.

Some people might say I went too far saying that canonical momentum is math, classically. But one of the problems is that only total momentum is conserved (I've seen multiple published papers make mistakes by thinking just one type is conserved, only the total is guaranteed to always be conserved locally flowing from here to there) and when I say total momentum I do not include the canonical momentum. And the other problem is that energy-momentum is real (has that frame independence, coordinate independence, gauge independence, and flow from here to there) and they only called canonical momentum a momentum because it looked like momentum in a few special situations, I might speak differently if it had a different name. But you can't always measure it in the lab and it isn't always conserved, but some people love it so so much (since it is related to symmetries). I just wanted to warn you about the difference.

Answer 2

In classical mechanics there is no distinction between free and bound as far as this relation is concerned.

In relativistic quantum mechanics (i.e. QFT), a particle that satisfies this relations is said to be "on-shell" or a physical observable asymptotically free particle. It is certainly not satisfied for virtual particles, but they are as their name suggests not real and artifacts of our perturbative analysis. In a truely relativistic bound state (i.e. strongly coupled like a light quart hadron and unlike the hydrogen atom), it is not clear when and where the constituents are virtual and when they are real because they are forever interacting (a shortcoming of the perturbative description), in other words you cannot identify the constituents in order to see whether they are on shell of off shell. But rather what is physically measurable is the entire bound state, which is as a whole of course on shell and satisfies this relation (unless it is again interacting with other particles). To be concrete take the proton, at low energies according to QFT there are no quarks, your degrees of freedom is the proton, this is all what you see. So in that sense then yes this relation is always satisfied...