Why protons and neutrons don't have less mass than their constituents?

Question

A system of gravitational attracted objects weight less than the sum of their individual masses because it needs energy to move them apart and overcome the gravitational attraction. Same is true for electromagnetic force, where the attraction between opposite charges must be overcome with additional energy to keep the objects move apart, therefore the system has less energy and less mass.

But why this isn't the case for three quarks bound together to form a proton/neutron?. The color force is (in simple terms) also attractive between the three quarks in a bound system and therefore energy must be applied to move them apart. As a consequence, the bounded system of quarks should have less energy (and less mass) than unbound quarks. Why this isn't the case?

Answer 1

Imagine free quarks being brought together to make a baryon. Your intuition is that, since the end product is bound, (i) the total mass-energy has declined and, though you may not have thought of this implication, (ii) putting the difference back in could break the quarks out. Neither is correct.

Tackling (ii) first, if you try pulling quarks out, your energy is spent on the creation of quark-antiquark pairs, and so what's released is new mesons, not the original valence quarks. This is deeply weird when you first learn of it: it's as if we couldn't create cations, because the attempt to remove an electron from a neutral atom simply released positronium.

The difference comes from gluons carrying their own charge; photons, by contrast, are electrically neutral. So not only do quarks emit gluons, the gluons do too. A baryon's three valence quarks are joined by flux tubes of gluons. These tubes can absorb energy, then turn it into mesons, if you try to pull out the original valence quarks.

A gluon has a color and anticolor charge, and can turn into a quark-antiquark pair. This doesn't just give baryons a meson-making party trick; it also means "sea quarks" populate the forcefields between the valence quarks in a baryon. (Gluons are also believed to clump into glueballs.) So if we turn now to (i), what mass do we reduce by a binding energy's worth to get a baryon's mass? Answer: that of many more than just three particles. Another way to put it is to assign multiple notions of mass to a quark: its "current mass" is what we imagined before the baryon formed, but its being in the baryon means it really has a larger constituent mass.

Answer 2

It's a good question and has a subtle answer, as answer from J.G. has already tried to explain. I think it may help to point out a related fact about something more familiar: the electron.

The mass of an electron is generally stated to be about $9.10938 \times 10^{-31}\,$kg. Now let's consider the electric field around a non-moving electron. This field has an energy density $(1/2) \epsilon_0 E^2$, and a magnitude $$ E = \frac{e}{4 \pi \epsilon_0 r^2}. $$ Suppose we could imagine the electron as a point charge. In that case the electric field would tend to infinity at locations approaching the point charge and so would the energy density. So that suggests the point charge model is questionable. If instead we model an electron as a sphere of some very small but non-zero radius, then we get a finite electric field and a finite energy density. By integrating the field energy density over volume, we get an energy. It is the total energy of the electric field of our electron. This energy has a mass associated with it, and this mass contributes to the $9.10938 \times 10^{-31}\,$kg mentioned above. In fact, if you choose the radius of the electron appropriately, then the whole mass of the electron is accounted for as coming from the field!

Please note, this 'classical charged sphere' model of an electron is not the correct model in the end---one needs quantum theory for that---but it makes the point that an electron has more mass overall than the mass which you might assign to the particle in the absence of its surrounding field.

Similar statements apply to quarks, only more so. In view of the field energy contribution, the term 'the mass of a quark' is rather ambiguous. What mass is it referring to? If it is referring to some notion of what the mass would be if there were no field (no gluons etc) then it could be quite unrelated to the mass which is relevant to things like protons and neutrons. It could even be zero, and then the whole mass of the proton comes from the gluons. Indeed that idea is a pretty good first approximation to what happens.

In view of the above, my answer to the question "why is the proton mass more than the sum of the quark masses, which one does not expect for a bound system?" is "it's because you need to understand more clearly what mass you are referring to when you quote some given value for a quark mass".

Answer 3

A bound system can have greater mass than sum of masses of the components. This does not happen in case of gravitational and electric forces, because the corresponding potential energies are monotonic functions of distance. This is because the forces are attractive irrespective of distance.

However, if there was a conservative force acting on the two bodies that was repulsive for distances greater than some distance $r_0$, and attractive for smaller distances, potential energy would have a maximum for some distance "near" $r_0$. If the system was in such a state or close to it, it would have greater mass than the system in decayed state (where the components are infinitely far from each other).

Simply put, a bound system can be like a "compressed locked spring", storing more energy than system whose parts are far from each other.