Mass is a number that relates quantitatively to how energy and momentum are balanced, e.g.
$$E=\sqrt{(c|\vec p|)^2+(mc^2)^2}$$
And the mass of a system is not the sum of the masses of the parts.
And particles don't have potential energy, systems do. Consider two massive particles of mass $m$ a distance d apart, there is some potential energy $U=-Gm^2/d^2$ but it only makes sense to associate it with the system not to randomly pick a particle and associate it with that. And potential energy is just for Newtonian physics anyway.
So why does a ground state hydrogen atom have less mass than the sum of the mass of an electron and of a proton? Because in its own frame it has no momentum but it has less energy. How do we know it has less energy? Because it gave off energy when it became bound. And it requires additional energy to become unbound.
So the mass of a system isn't the mass of the parts added up. Its just the parameter that makes $E=\sqrt{(c|\vec p|)^2+(mc^2)^2}$ work, you can even write it as $E^2-(c|\vec p|)^2=(mc^2)^2.$ And then for the mass of a system you use the energy of the system and the momentum of the system and it isn't the mass of the parts added up.
I understand what you mean by the mass of the system not being the mass of the components added up.
And keep in mind this is for special relativity.
But is there is always a change in mass of a system when its potential energy changes?
And in special relativity, potential energy doesn't make sense. When energy is exchanged it has to happen at the exact same when-where because otherwise if energy disappeared at one place and reappeared elsewhere then in some frame it wouldn't be conserved.
For example when an electron is pulled apart from its atom, or when the distance between two charges is changed?
So for instance the way work is done on electrically charged objects is that charged objects have kinetic energy, momentum, and rest energy and electromagnetic fields have their own field energy and field momentum. And what happens to charged objects is they move around. And what happens to fields is they evolve. And in regions where there is no charge the field energy just moves around in a conserved way. And in regions where there is neither charge nor current the field momentum just moves around in a conserved way. But in places with charge the field loses field energy when the charge gains kinetic energy and the field gains field energy when the charge loses kinetic energy. And in places with charge and/or current the field and the charges can exchanged momentum so that total momentum is conserved.
So there isn't potential energy in classical electromagnetism. If you want to see what is going on in the Schrödinger equation you need to learn the difference between a canonical momentum (which is a third type of momentum not really related to mechanical momentum of charges or field momentum of fields) and kinetic momentum, and how gauge choices affect the phase of the wavefunction of a charged object. All the physics is about the phase and the gauge together in the sense that it is their relationship that carries all the information.
But my description earlier is fine. The atom has less energy because it takes energy to ionize it and I gave up energy when it dropped down to the ground state.
In a Lagrangian, there is something that might look similar to a potential, but it is gauge dependent and the term in the Lagrangian have units of energy but are really just what they need to be to give Euler-Lagrange equations that match what we see.
Best Answer
It did. Or rather, the composite system of the Earth, the ball, and their mutual gravitational field lost energy, which is equivalent to mass.
The question is how much energy. Suppose the combined mass of the ball-Earth system is $m_\text{earth} + m_\text{ball} \approx 10^{24}\,\mathrm{kg}$, the mass of the Earth. The ball falls from the table and liberates one joule of energy from the gravitational field. Since the rest energy of the Earth-ball system is $m_\text{earth} c^2 \gg 1\,\mathrm{J}$, the constant-mass approximation is fine.
(Homework for you: compute the total gravitational binding energy of the Earth, and compare to Earth’s mass in energy units. The scale is $-G m_\text{earth}^2 / r_\text{earth}$ from dimensional analysis, but there’s a half or something out front from doing the integral of bringing all the masses together. The constant-mass approximation is good even if you’re assembling the entire Earth from asteroids and dust.)
In chemistry you come closer to caring about the mass stored in the electric field. For example, the ionization energy for ground-state hydrogen atoms, $13\,\mathrm{eV}$, is a part-per-billion correction to the mass of a hydrogen atom, $m_\text{H} \approx m_\text{proton} \approx 1\,\mathrm{GeV}/c^2$. The innermost electrons in heavy atoms can have binding energies of hundreds of keV.
It’s not that only nuclear interactions have mass changes associated with potential energy changes. But in nuclear interactions the energies involved are so large that the mass effects really can’t be ignored. A uranium fission liberates about $180\,\mathrm{MeV}$. A dozen uranium fissions is enough to produce a proton-antiproton pair from nothing. Nuclear binding energies are a part-per-thousand correction to the nuclear mass.