Proton e goes to zero, smaller g, , vacuum color polarization.

[Spinnor]

Suppose we consider a proton where we set the electric charge of the three quarks towards zero. At this point I assume we still would have the three quarks bound together?

Now reduce the strength of the strong force by say ten times. I was told in posts that the radius of the proton depends on the strong coupling constant like,

r ≈ exp(c/g) or r ≈ exp(c/g^2)

Can we assume the mass of the proton is dependent on the strength of the strong force?

Quarks polarize the vacuum by two countering effects, virtual quark and virtual gluon production?

As the strength of the color force is reduced is the reduction in the polarization of the vacuum proportional between virtual quark and virtual gluon polarizations? Is there some type of scale invariance here, do they both go to zero at the same rate?

Thanks for any help!

 

[The_Duck]

Suppose we consider a proton where we set the electric charge of the three quarks towards zero. At this point I assume we still would have the three quarks bound together?

Yes.

Can we assume the mass of the proton is dependent on the strength of the strong force?

Yes; the mass of the proton is of order hbar*c/r where r is the proton radius [at least if you can neglect the rest masses of the quarks themselves; that is, if they each have masses much less than hbar*c/r, which they do in the real world].

Quarks polarize the vacuum by two countering effects, virtual quark and virtual gluon production?

This is very slightly mis-stated. There are two countering effects: vacuum polarization by virtual quarks and vacuum polarization by virtual gluons. Quarks have nothing to do with the latter effect.

As the strength of the color force is reduced is the reduction in the polarization of the vacuum proportional between virtual quark and virtual gluon polarizations?

Yes. In QCD there is a function of the coupling constant $g$ called the beta function, which summarizes the effects of the vacuum polarization. The most important term in the QCD beta function is

\beta(g) = -\left(11 - \frac{2}{3}N_f \right) \frac{g^3}{16 \pi^2}

The $11$ comes from the vacuum polarization by gluons. The $-\frac{2}{3}N_f$ comes from the vacuum polarization by quarks--$N_f$ is the number of quark flavors. You can see that both terms are proportional to $g^3/16 \pi^2$ and so both go to zero as $g \to 0$, and the ratio between these terms stays constant as $g \to 0$.

 

[snorkack]

Suppose we consider a proton where we set the electric charge of the three quarks towards zero. At this point I assume we still would have the three quarks bound together?

In which sense? Sum of the three quarks, or each of the quarks individually?

 

[Spinnor]

Yes.
Yes; the mass of the proton is of order hbar*c/r where r is the proton radius [at least if you can neglect the rest masses of the quarks themselves; that is, if they each have masses much less than hbar*c/r, which they do in the real world].
This is very slightly mis-stated. There are two countering effects: vacuum polarization by virtual quarks and vacuum polarization by virtual gluons. Quarks have nothing to do with the latter effect.
Yes. In QCD there is a function of the coupling constant $g$ called the beta function, which summarizes the effects of the vacuum polarization. The most important term in the QCD beta function is

\beta(g) = -\left(11 - \frac{2}{3}N_f \right) \frac{g^3}{16 \pi^2}

The $11$ comes from the vacuum polarization by gluons. The $-\frac{2}{3}N_f$ comes from the vacuum polarization by quarks--$N_f$ is the number of quark flavors. You can see that both terms are proportional to $g^3/16 \pi^2$ and so both go to zero as $g \to 0$, and the ratio between these terms stays constant as $g \to 0$.

Thank you for your help! Looks like I will have to give up on dipole gluon radiation ?

Maybe I just need a space of higher dimensions? Do we still have quark confinement in higher dimensions?

Thanks again!

 

[Spinnor]

In which sense? Sum of the three quarks, or each of the quarks individually?

Three quarks, ignoring quark electric charge does not change the bound state of the three quarks too much?

 

[snorkack]

Three quarks, ignoring quark electric charge does not change the bound state of the three quarks too much?

Total zero charge for three quarks is a property of neutron.

 

[The_Duck]

Thank you for your help! Looks like I will have to give up on dipole gluon radiation ?

Maybe I just need a space of higher dimensions? Do we still have quark confinement in higher dimensions?

I think that if you want to get rid of confinement, you can imagine having $N_f \ge 17$ quark flavors. Then the quantity $(11 - 2N_f/3)$ in the beta function is negative, which means there is no confinement [well, I think you also have to postulate that your quarks are exactly massless, or else confinement reappears at very low energies].

 

[snorkack]

I think that if you want to get rid of confinement, you can imagine having $N_f \ge 17$ quark flavors. Then the quantity $(11 - 2N_f/3)$ in the beta function is negative, which means there is no confinement [well, I think you also have to postulate that your quarks are exactly massless, or else confinement reappears at very low energies].

What happens to confinement if there are over 17 quark flavours, but the quarks are not exactly massless nor close to massless? How low energies would the confinement appear in?

For example suppose that there are in fact 18 flavours of quarks, of which 5 are nearly massless (the known quarks up to beauty), the sixth is 173 GeV (truth) - and the there are 12 more (6 more generations) of quarks, all of which have masses above the observed 173 GeV of truth. But not much more - say, all the 12 quarks have masses between 173 and 200 GeV.

How would large numbers of massive quark flavours affect quark confinement?
What is the currently observed minimum mass of seventh quark?

 

[Spinnor]

I think that if you want to get rid of confinement, you can imagine having $N_f \ge 17$ quark flavors. Then the quantity $(11 - 2N_f/3)$ in the beta function is negative, which means there is no confinement [well, I think you also have to postulate that your quarks are exactly massless, or else confinement reappears at very low energies].

That is a lot of flavors! Is that formula dependent on the dimension of space-time?

Thanks for your help!

 

[mfb]

That is a lot of flavors! Is that formula dependent on the dimension of space-time?

Thanks for your help!

Quantum field theory looks so different in other dimensions I don't know if that formula even exists for a different number of dimensions.

There is confinement for abelian theories in 2 and 3 dimensions according to this, cited at Wikipedia.

 

[The_Duck]

[Deleted since I wrote some stuff I now think is wrong]