- #1

etotheipi

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 c

^{2}. 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/c

^{2}, 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!