Rest energies of individual nucleons considered in isolation

[etotheipi]

I'm conscious I posted a question about a similar topic recently, however I can't seem to resolve an apparent contradiction.

As an example, consider a deuterium atom with 1 proton and 1 neutron. When separated at infinity, each has a certain mass and a certain rest energy differing only by a factor of c². The total rest energy of both nucleons is just the sum of the individual rest energies, and likewise for mass.

When brought together to form a nucleus, the potential energy of the system decreases due to the work done by the strong force, so the rest energy of the system decreases, suppose by amount ΔE. Consequently one usually determines that the mass of the system decreases by ΔE/c², and this principle is the basis of many nuclear calculations, so must be pretty accurate.

However, now let's consider only the proton in the deuterium nucleus. Since it is a single particle considered in isolation, it has no potential energy. The same goes for the neutron. Consequently there is no difference in rest energy compared to the infinite separation state, since the particles had no potential energy to start with either. From this analysis it appears that the rest energies of the individual nucleons have not actually changed from when they were at infinity, and consequently they have the same mass as before!

I wonder if the standard mass defect analysis only applies when considering the whole system of nucleons, which evidently appears to decrease in rest energy and rest mass? If not, then how do we resolve this apparent contradiction between the values of the mass of the nucleus obtained by considering the whole nucleus as opposed to each nucleon in it individually?

Thank you very much in advance!

 

[Nugatory]

Consequently there is no difference in rest energy compared to the infinite separation state... From this analysis it appears that the rest energies of the individual nucleons have not actually changed from when they were at infinity, and consequently they have the same mass as before!

That’s right.
But because we cannot weigh the bound nucleon by itself (to do that we’d have to separate the nucleon from the rest of the nucleus so we’re back to weighing an unbound nucleus which of course gives us the unbound rest mass) this statement is telling us more about how we’ve defined the term “rest mass” than about the mass of anything.

I wonder if the standard mass defect analysis only applies when considering the whole system of nucleons, which evidently appears to decrease in rest energy and rest mass?

Yes, of course. The mass defect is defined as the difference between the rest mass of the nucleus and the sum of the rest masses of its constituent nucleons so you can only use the concept when you have a bunch of nucleons forming a whole system together.

There is no reason to expect the rest mass of any system to be equal to the sum of the rest masses of its constituents.

 

[Steelwolf]

Would the energy loss not be like that of dropping an object to the floor: Potential Kinetic Energy until I let go, then the work of Falling happened, and then they combined into the class of (objects on the ground, lowest energy point).

It seems like much the same happens with black hole collisions, where much of the energy of the 'Work' is lost as gravitational waves and electromagnetic frequencies at the fringes.