Would higher-mass quarks result in smaller baryons?

In summary, the stable octet baryons, such as the Delta, Lambda, and Xi, have smaller radii than the nucleon due to the spin-spin interaction of quarks. This is also seen in heavy flavor mesons like the D and B, which have smaller radii than the nucleon due to the lack of large spin-spin interactions. The difference between the mean charged radius, mass radius, and other radii are being studied and calculated through various peer-reviewed papers. The force between two nucleons is modeled by meson exchange, not gluon exchange, and is spin-dependent. The EM force becomes important again at moderate distances within the hadron, affecting the mass through the Coulombic 'Cornell potential'.
  • #1
bbbl67
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Would baryons not made with standard Up or Down quarks exhibit smaller radii than neutrons and protons? I'm thinking like for example how muons have much smaller orbitals than electrons, on the lepton side of things.
 
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I will talk mostly about the stable octet baryons, because excited baryons can have vastly different radial states. The distance between two quarks in a spin zero state is smaller than the distance in a spin one state because of the spin-spin interaction of quarks. This is why the Delta has a larger radius than the nucleon, and the Lambda has a smaller radius. There is a longer range, Coulomb-like, force between quarks that would give a smaller radius to heavier quarks. That is why the Xis have a smaller radius than the nucleon. The spin-spin interaction is not large for charmed and bottom baryons. That is why they tend to have a smaller radius than the nucleon.
 
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  • #5
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Meir Achuz said:
I will talk mostly about the stable octet baryons, because excited baryons can have vastly different radial states. The distance between two quarks in a spin zero state is smaller than the distance in a spin one state because of the spin-spin interaction of quarks. This is why the Delta has a larger radius than the nucleon, and the Lambda has a smaller radius. There is a longer range, Coulomb-like, force between quarks that would give a smaller radius to heavier quarks. That is why the Xis have a smaller radius than the nucleon. The spin-spin interaction is not large for charmed and bottom baryons. That is why they tend to have a smaller radius than the nucleon.
There is also a repulsive force from the gluons when quarks get too close. Do higher-mass quarks run into that limit if they get too close? Or is the gluon interaction distances different for these other quarks?
 
  • #7
bbbl67 said:
I suppose all of those radii would do. What's the difference between them?
the mass radii is a bit "speculative" though there are some peer-reviewed papers trying to calculate it.
it is basically about gravitational interactions.
https://journals.aps.org/prd/pdf/10.1103/PhysRevD.104.054015
https://journals.aps.org/prd/pdf/10.1103/PhysRevD.105.096033
bbbl67 said:
There is also a repulsive force from the gluons when quarks get too close
Source? The gluon exchange is asymptotically free at short rage https://en.wikipedia.org/wiki/Asymptotic_freedom
 
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  • #8
malawi_glenn said:
the mass radii is a bit "speculative" though there are some peer-reviewed papers trying to calculate it.
it is basically about gravitational interactions.
https://journals.aps.org/prd/pdf/10.1103/PhysRevD.104.054015
https://journals.aps.org/prd/pdf/10.1103/PhysRevD.105.096033
Would this be to figure out if they would be stable inside super-neutron stars?

malawi_glenn said:
Source? The gluon exchange is asymptotically free at short rage https://en.wikipedia.org/wiki/Asymptotic_freedom
This graph:
strongForce.png
 
  • #9
bbbl67 said:
This graph:
That is the force between two nucleons, not between two quarks. The force between two nucleons is modeled with meson exchange, not gluon exchange. The repulsive part of the nucleon-nucleon force is due to vector-meson exchange (like ω) iirc.

The force between two quarks become weaker at shorter distances.
In Baryons, you would have quarks with same sign of their charge, so due to their electric replusive force you will at some point not be able to push two quarks closer (in some kind of classical analogy).

bbbl67 said:
Would this be to figure out if they would be stable inside super-neutron stars?
No.
 
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  • #10
Btw, speaking of the nucleon-nucleon interaction, it is isopsin dependent. This is why there is only one bound state of two nucleons, the deuteron.
 
  • #11
It's nearly isospin symmetric (the explicit violation is due to the mass difference between u and d quarks of a few MeV, i.e., of the same order of magnitude as the violation of chiral symmetry, which is due to the masses of the u and d quarks themselves, also at the order of a few MeV). The fact that there are no bound pp and nn states and that there is no S=0 bound state of np rather indicates that the two-body nucleon-nucleon interaction is spin dependent (but not isospin dependent).
 
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  • #12
Yeah I f*cked up :)
Of course it is due to spin-dependency, neutrons are identical fermions so their total wavefunction must be antisymmetric (has to form a S=0 state, but there are no bound states with S=0 in the NN-potential)
 
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  • #13
malawi_glenn said:
That is the force between two nucleons, not between two quarks. The force between two nucleons is modeled with meson exchange, not gluon exchange. The repulsive part of the nucleon-nucleon force is due to vector-meson exchange (like ω) iirc.
Ah, I didn't notice that! I had assumed it was the force within a nucleon, i.e. between quarks.

malawi_glenn said:
The force between two quarks become weaker at shorter distances.
In Baryons, you would have quarks with same sign of their charge, so due to their electric replusive force you will at some point not be able to push two quarks closer (in some kind of classical analogy).No.
Okay, so the EM force becomes important again even within the hadron?
 
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  • #14
malawi_glenn said:
The force between two quarks become weaker at shorter distances.
The force at moderate distances that affects the mass is Coulombic, like the so called 'Cornell potential'.
The force "becomes weaker" because, for very short distances (or high Q^2), there is 'asymptotic freedom' which softens the $1/r^2$ singularity. The EM force is part of the explanation of mass differences within isomultiplets, but not for the masses asked about in the original question.
 
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FAQ: Would higher-mass quarks result in smaller baryons?

How do higher-mass quarks affect the size of baryons?

Higher-mass quarks do not directly affect the size of baryons. The size of a baryon is primarily determined by the strong nuclear force between the quarks, which remains constant regardless of the quark's mass.

Can higher-mass quarks lead to the creation of smaller baryons?

No, the size of a baryon is determined by the number and type of quarks it contains, not their individual masses. Therefore, higher-mass quarks would not result in smaller baryons.

Would increasing the mass of all quarks result in a decrease in baryon size?

No, increasing the mass of all quarks would not necessarily result in a decrease in baryon size. The size of a baryon is determined by the strong nuclear force between the quarks, not their individual masses.

Are there any factors other than quark mass that can affect the size of a baryon?

Yes, there are other factors that can affect the size of a baryon, such as the number and type of quarks it contains, as well as the energy of the system. These factors can influence the strength of the strong nuclear force and therefore impact the size of the baryon.

Would lower-mass quarks result in larger baryons?

No, the size of a baryon is primarily determined by the strong nuclear force between the quarks, not their individual masses. Therefore, lower-mass quarks would not necessarily result in larger baryons.

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