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The mass of nucleons is bigger than the sum of the masses of its constituent quarks. I understand that it's because you have to take into account the binding energy of these quarks.

What I don't understand is this: In a nuclear physics course I was tought that the mass of the nucleai is less than the sum of the masses of its constituent nucleons because the binding energy between them is negative in order to make it stable.

So, if the binding energy of nucleons is negative because its suppose to be an atractive force that keeps nucleons stable, why does the binding energy of quarks add to the mass of nucleons instead of reducing it if it's also an atractive force keeping quarks stable?

I understand that in the quark regime we have to deal with quantum chromodynamics and the concept of force and energy are not that simple but I was expecting the binding energy of any atractive interaction to be negative anyway...


marked as duplicate by John Rennie, Jon Custer, David Hammen, rob Jan 23 '17 at 19:13

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When you say that the binding energy of, say, a hydrogen atom is negative you are comparing two states

  • A hydrogen atom, where the two particle are in close proximity to one another
  • A free proton and a free electron where the two particles are arbitrarily far away from one another (where they are free).

The same thing applies when one says that a bound nucleus is less massive than the constituents: you compare to a case when they are been separated from one another (i.e. they are free).

However, it doesn't make sense to talk about comparing the mass of a baryon with the masses of the valence quarks taken to a long distance from one another: confinement means that the second state is not a physically realizable situation.

So what do we compare to, then?

I'm glad you asked. We notice that in deeply inelastic collisions, we seem to bounce off partons that act as though they are free. This is called "asymptoptic freedom".

So we compare the mass of the baryon, to sum of the masses of the valence quarks as determined at high interaction energy (where they are free).

So, why does that set the sign of the binding? We are in the habit of defining the zero of potential energy at the situation where the interaction you are worried about has gone to zero. For the Coulomb force that is at infinite separation. For the strong nuclear force you can use anything over around 10 fermis, but we generally also take it to be at infinity. For the real strong interaction between quarks and gluons it has to be set where they are free and that is at zero separation. And it takes energy to pull them apart, so all non-zero separations have positive potential energy giving the bound state more mass than the sum of the constituent masses.

  • $\begingroup$ "For the real strong interaction between quarks and gluons it has to be set where they are free and that is at zero separation. ". Why zero seperation? Do you mean at zero seperation force is zero? $\endgroup$ – Paul Jan 21 '17 at 3:58
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    $\begingroup$ @Theasgardian It is analogous to the potential and kinetic energy of a stretched spring. Have a look at the images here en.wikipedia.org/wiki/Color_confinement $\endgroup$ – anna v Jan 21 '17 at 6:14
  • $\begingroup$ @anna v when a spring is unstretched ,the restoring force is zero.Similarly force between quarks is less when separation between them is less ,and zero when separation is zero..do you mean this? $\endgroup$ – Paul Jan 21 '17 at 6:27
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    $\begingroup$ @Theasgardian yes, the force between quarks and gluons depends on the distance, the larger the distance the stronger the force. Have a look here profmattstrassler.com/articles-and-posts/… $\endgroup$ – anna v Jan 21 '17 at 6:43
  • $\begingroup$ Great answer! I have one more question: if the state where the quarks are separated is not physical, how can we know the mass of the quarks? and, more important, does it make any sense to talk of mass if the particle can't be really free? $\endgroup$ – P. C. Spaniel Jan 23 '17 at 2:18

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