# Does the rest mass energy include the potential energy of the particle?

The potential energy (as far as I have studied - that is, mainly classical physics) depends on the reference level, since its absolute value cannot be calculated. It can therefore be negative as well. Recently I was taught the mass-energy equivalence in class, in context of the mass defect that arises because of the nuclear binding energy. My question is whether the rest mass energy also includes the potential energy of the particle. If it does, its value should depend on the reference level of the potential energy. Since it is given by $E=mc^2$, the rest mass should depend on the reference level. But this does not seem to make sense. And when the potential energy becomes negative, so should the rest mass. This again seems strange.

If it does not include the potential energy, then why does the nuclear potential energy show up as a mass defect? It should not, if the rest mass energy is not related to the potential energy.

• Related: physics.stackexchange.com/q/48490/2451 , physics.stackexchange.com/q/69080/2451 and links therein. Oct 12 '15 at 10:22
• I do understand now that $E=mc^2$ does not include the potential energy. What about the nuclear binding energy, as asked above? Oct 12 '15 at 10:29
• Keep clear in mind that the m in E=mc^2 is NOT the rest mass. It is the relativistic mass and indicates how the particle would react to an external force, for example. Its rest mass is in the formula in the answer by @Timaeus Mar 20 '16 at 6:14

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.

• I understand what you mean by the mass of the system not being the mass of the components added up. But is there is always a change in mass of a system when its potential energy changes? For example when an electron is pulled apart from its atom, or when the distance between two charges is changed? Oct 12 '15 at 16:36
• @Charles Edited. Oct 12 '15 at 16:50