# Can Someone Explain Quark Binding Energy?

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As a chemist, I am struggling with this.

In, say, a helium nucleus, the rest mass is less than the rest mass of the "free" neutrons and protons from which it is made - the so-called mass defect. That makes perfect sense to me because, to separate the nucleus into its components, you have to do work against the strong nuclear interaction that holds the nucleons together, i.e. an energy input is required, which of course is then reflected in a greater rest mass of the separated nucleons. And hence the converse occurs during fusion, leading to a net output of energy when the nucleons combine and become bound.

But when it comes to the quarks that form a proton, say, the opposite seems to apply. The proton has far more mass than combined mass of the three quarks that make it up are said to have in their "free" state. So apparently the bound quarks are in a higher energy state than free ones. This suggests a proton is thermodynamically unstable with respect to the free constituents - and should spontaneously fly apart, if whatever kinetic barrier there is to it doing so could be overcome.

Can someone with a bit of nuclear physics explains to me how this works? I have not found an internet source that tackles this squarely - or not to my satisfaction.

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My understanding is it’s related to asymptotic freedom. When you add energy to a “typical” bound system (e.g. ionize an electron) you end up with free particles. When they combine, you get a release of energy. But adding energy to bound quarks doesn’t do this - you can’t free a bound quark. Their potential energy at large separation doesn’t go to zero as it does with gravity or Coulomb forces.

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As I understand, there is no "free state" of a quark, and quark mass cannot be interpreted consistently as mass of a "free quark." The quark mass is a coefficient of the quadratic term in the quark's Lagrangian. It is a mass term in the corresponding equation of motion, Dirac equation.

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3 hours ago, exchemist said:

you have to do work against the strong nuclear interaction that holds the nucleons together

Technically that is not the strong nuclear force.
The strong nuclear force  is the color force of QCD.
What holds nucleons together is 'residual' color force, which is more easily modelled as an exchange force ( Yukawa ),

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No way. I wanted to make a contribution here. I was thinking about mentioning 'residual QCD forces' to complete the picture (similar to mesonic states flying to and fro), and @MigL beats me to the punch.

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Sorry Joigus.
You snooze, you lose.

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As I understand, there is no "free state" of a quark, and quark mass cannot be interpreted consistently as mass of a "free quark." The quark mass is a coefficient of the quadratic term in the quark's Lagrangian. It is a mass term in the corresponding equation of motion, Dirac equation.

10 hours ago, swansont said:

My understanding is it’s related to asymptotic freedom. When you add energy to a “typical” bound system (e.g. ionize an electron) you end up with free particles. When they combine, you get a release of energy. But adding energy to bound quarks doesn’t do this - you can’t free a bound quark. Their potential energy at large separation doesn’t go to zero as it does with gravity or Coulomb forces.

OK. Thanks. So one can't pull quarks apart by doing work against the binding interaction. I can also understand that the (theoretical) masses of the quarks themselves in the model may be small compared to the total mass of the proton, if the quarks also have a great deal of potential and/or kinetic energy. But I'm struggling with the way some articles , e.g. Wiki, speak of QCD binding energy as if this is a source of extra energy, whereas in the case of nucleons in the nucleus - or electrons in an atom - the binding energy is the amount by which the energy is reduced as a result of the extra stability conferred by the binding attraction. It would seem that the term "binding energy" is being used in the opposite sense to that in which it used in these other contexts. Can this be right, or have I misunderstood?

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1 hour ago, exchemist said:

OK. Thanks. So one can't pull quarks apart by doing work against the binding interaction. I can also understand that the (theoretical) masses of the quarks themselves in the model may be small compared to the total mass of the proton, if the quarks also have a great deal of potential and/or kinetic energy. But I'm struggling with the way some articles , e.g. Wiki, speak of QCD binding energy as if this is a source of extra energy, whereas in the case of nucleons in the nucleus - or electrons in an atom - the binding energy is the amount by which the energy is reduced as a result of the extra stability conferred by the binding attraction. It would seem that the term "binding energy" is being used in the opposite sense to that in which it used in these other contexts. Can this be right, or have I misunderstood?

I have an issue with this energy being called binding energy, too, and would like to know a justification for that.

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I have an issue with this energy being called binding energy, too, and would like to know a justification for that.

Isn't that what gluons do?

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3 minutes ago, StringJunky said:

Isn't that what gluons do?

You mean, it is a binding energy because it is the energy of gluons which bind the nucleon? OK.

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My 1 cent worth.........Are not Quarks bound together by the strong force, which is weaker when the Quarks are closer, but increase in strength as you try to separate them, making it impossible to isolate a single Quark...unless one could go back in time to t+ 10-12 seconds, when pressures and temperatures were such that they could only exist singularly?
Edited by beecee
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15 hours ago, exchemist said:

It would seem that the term "binding energy" is being used in the opposite sense to that in which it used in these other contexts. Can this be right, or have I misunderstood?

The result is different, but it’s because the binding is different. I think perhaps it’s a mistake to try to use the same terminology.

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You mean, it is a binding energy because it is the energy of gluons which bind the nucleon? OK.

But this is exactly the language I find confusing. Binding implies achieving a lower energy state, so that work has to be done to free the bound entities from what binds them.  Whereas what we seem to have here is a higher energy state than the quarks would theoretically have if it were possible to observe them separated (and at rest).

But I'm getting a sense from you and @swansont that the term "binding energy" is best avoided in this context. The mechanisms are clearly quite different, probably related to this asymptotic freedom idea that I have not fully got my head around.

Edited by exchemist
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16 minutes ago, exchemist said:

But this is exactly the language I find confusing. Binding implies achieving a lower energy state, so that work has to be done to free the bound entities from what binds them.  Whereas what we seem to have here is a higher energy state than the quarks would theoretically have if it were possible to observe them separated (and at rest).

But I'm getting a sense from you and @swansont that the term "binding energy" is best avoided in this context. The mechanisms are clearly quite different, probably related to this asymptotic freedom idea that I have not fully got my head around.

But quarks cannot be free from what binds them. If we try to separate them, the energy of what binds them will increase. Thus, it would be a higher energy state.

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While 'binding energy' may be a bit misleading, a sum of the masses of the neutron's three valence quarks would be about 12 MeV, as opposed to the total mass of the neutron of 939 MeV.
The difference is due to the QCD strong interaction, as 'naked' quarks cannot exist in hadrons, and you always have to consider gluons, virtual quarks, and their kintic energies.

Quantum chromodynamics binding energy - Wikipedia

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But quarks cannot be free from what binds them. If we try to separate them, the energy of what binds them will increase. Thus, it would be a higher energy state.

Is it right that when you apply energy to separate quarks, new quark pairs are formed from each one once a certain threshold is reached? This is why they can never exist in the free state? I read a some Quora posts yesterday from a couple of physics professors that mentioned this.

Edited by StringJunky
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21 minutes ago, StringJunky said:

Is it right that when you apply energy to separate quarks, new quark pairs are formed from each one once a certain threshold is reached?

Essentially correct.
You have a difference of 927 MeV acting as binding energy for the three quarks in a neutron.
If you pull at one of the quarks with more than the energy of their 'combined' masses, 12 to 939 MeV, you have enough energy to 'create'  three new quarks; two to associate with the one you were pulling on, and one to 'replace' the one you were pulling on in the original neutron.
IOW, you can never  get to the threshold value of 939 MeV which will allow for dissociation.
That is a property of the color force, and why the 927 MeV difference is termed 'binding energy'.

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2 hours ago, MigL said:

Essentially correct.
You have a difference of 927 MeV acting as binding energy for the three quarks in a neutron.
If you pull at one of the quarks with more than the energy of their 'combined' masses, 12 to 939 MeV, you have enough energy to 'create'  three new quarks; two to associate with the one you were pulling on, and one to 'replace' the one you were pulling on in the original neutron.
IOW, you can never  get to the threshold value of 939 MeV which will allow for dissociation.
That is a property of the color force, and why the 927 MeV difference is termed 'binding energy'.

Energetically so, except you cannot create three quarks, but only quark-antiquark pairs. These pairs will combine between themselves and with the original quarks in different ways. Actually, even 9 MeV could be enough. It could create one quark-antiquark pair, then an original quark with a new antiquark would make a meson, and a new quark would replace an original quark in the baryon.

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Yeah. The process and energies I described were used as an example ofthe futility of trying to dissociate quarks; not necessarily the actual process.

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I think, my example above needs correction, because it doesn't work energetically. For a new hadron to be created at the end, its binding energy needs to be supplied as well, not only the mass of new quarks and antiquarks. It still works, but the required energy to be added to the original baryon is not 9 MeV but rather at least 135 MeV, the lightest pion.

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