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Thanks, and sorry if this is a stupid question... I couldn't find an answer in the literature.

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- #1

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Thanks, and sorry if this is a stupid question... I couldn't find an answer in the literature.

- #2

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(Btw it's not a stupid question!) ;-)

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I'm not an expert in this field, however... Maybe it should be clearer if I say that the initial and final states of a scattering process must be color neutral; during the scattering process the quarks can mix up, as the are confined pratically in the same place. Then they can split, to the final states, only in color-neutral "groups".

And that's why you happen to see only color-neutral particles. Can you understand now? (If you don't, probably is due to my lack of knowledge in this field ;-) )

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blechman

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So if you kick the quark (confined to the proton) HARD enough (as they do in these particle accelerators), then the strong nuclear force that's holding the quark inside the proton actually gets weaker, and the quark can escape!

Once the quark is out of the proton, however, it realizes that a terrible mistake has been made , and it immediately tries to get back inside a hadron (ANY hadron!), and it does this by pulling quarks out of the vacuum to create the final hadrons that we see.

Once the quarks have buried themselves inside the hadrons, the strong nuclear force is no longer at work (all quarks are now bound up in hadrons, which are "color neutral") and so the hadrons can safely move along on their merry way.

Hope that helps!

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- #8

blechman

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so when the quark gets a strong enough kick, its deBroglie wavelength shortens (that's the "short distance") and the strong force weakens.

- #9

tom.stoer

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- color neutrality

- color confinement

- asymptotic freedom

Color neutrality means that each physical state carries total color charge zero. This is equivalent to the statement that in QED a physical state has zero electric charge. This follows from the Gauss law constraint and is a "kinematic" feature of gauge theories. Total charge zero only means that the total charge must vanish globally, not necessarily locally.

Asymptotic freedom is a high-energy / low-distance phenomenon and means that as energy grows, the coupling strength ("force") between two color charges goes to zero. It has nothing to do with confinement! At high pressure QCD has an unconfined phase (quark-gluon-plasma), nevertheless asymptotic freedom still holds.

Confinement means that it takes an infinite amount of energy to separate two color charges by an infinite distance. There is no simple picture like a potential energy (like Coulomb potential in QED), nevertheless one can think about a potential that grows linearily, meaning that the force netween to color charges stays constant even over infinite distance. As the total enery is something like distance * force, the total energy becomes infinite.

- #10

blechman

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I think what you really mean is at high DENSITY (which then implies high pressure by the equation of state), the QGP is expected to exist. At high density the quarks and gluons are on top of each other and asymptotic freedom says that the force then weakens and confinement goes away.Asymptotic freedom is a high-energy / low-distance phenomenon and means that as energy grows, the coupling strength ("force") between two color charges goes to zero. It has nothing to do with confinement! At high pressure QCD has an unconfined phase (quark-gluon-plasma), nevertheless asymptotic freedom still holds.

Asymptotic freedom is not the same thing as confinement, I agree with that. But the existence of asyptotic freedom might IMPLY confinement, since a corollary is that the force increases as energy decreases (distance increases). The reason it is not a PROOF of confinement is because there could be a fixed point, for example, in which case the strong force just becomes a constant, independent of distance/energy, until you get to higher energy scales. In that case, confinement must come from somewhere else...

The linear potential can actually be derived from QCD lattice calculations, so I take some issue with the idea that this potential just materializes out of thin air! If that's not what you meant, then I apologize.Confinement means that it takes an infinite amount of energy to separate two color charges by an infinite distance. There is no simple picture like a potential energy (like Coulomb potential in QED), nevertheless one can think about a potential that grows linearily, meaning that the force netween to color charges stays constant even over infinite distance. As the total enery is something like distance * force, the total energy becomes infinite.

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Hmm?? I don't think I agree. Any pure non-abelian gauge theory is asymptotically free. As you start adding fermions and scalars to the theory, it becomes less asymptotically free. If you add sufficiently large number of scalars and/or fermions, eventually the theory ceases to be asymptotically free. But I believe that the theory should still be confining (assuming the unbroken gauge group is non-abelian) even after it ceases to be asymptotically free. So, I think asymptotic freedom and confinement are related to each other, but they do not always go hand-in-hand.... But the existence of asymptotic freedom might IMPLY confinement, since a corollary is that the force increases as energy decreases (distance increases)...

- #12

blechman

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As I said, though: asymptotic freedom is not a SUFFICIENT condition for confinement.

- #13

tom.stoer

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Regarding asymptotic freedom and confinement I do not agree. Asymptotic freedom is valid in the perturbative / UV regime; it allowes scaling of energies like via DGLAP. But already in the running coupling it becomes clear that

[tex]\alpha_S(Q^2) = \frac{12\pi}{(33-2N_f)\ln(Q^2/\Lambda_{QCD}^2)}[/tex]

can no longer valid near (and below) the QCD scale. Another consideration is that if the coupling growths with decreasing energy this does not necessarily mean that the coupling growth w/o upper bound. A last idea is to look at QCD with a large number of flavours. You can see from the above equation that it is no longer asymptotically free.

I agree that one can derive the linear potential U(r) ~ r (for sufficiently large r) from lattice QCD. All what I want to stress is that in QCD you cannot start with such a potential U(r) between two color charges. This U(r) is nothing fundamental but an effective potential extracted from the low energy behavior of (static) quarks. Qualitatively it can be compared with the Coulomb potential.

The basic reason is that in QED the Coulomb potential can be derived easily in a canonical formalism via gauge fixing and solving for the Gauss law. In order to do that you have to invert a certain differential operator D; its Greens function is just 1/k² which can be Fourier transformed to 1/r. In QCD one can do something similar, but unfortunately the differential operator D[A] contains the physical modes of the gluon field; the Greens function is something like 1/(k+A)²; its Fourier transform cannot be calculated analytically - and it depends on the degrees of freedom for the gluon field A(x).

So yes, there is something like a color potential U(r) ~ r; but it is not a "static entity" like in electromagnetism, but a "dynamical entity", even for static quarks.

- #14

blechman

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Historically, however, the reason why people thought QCD was the best thing since sliced bread was that asymptotic freedom might explain confinement. Unfortunately, this was a little too much to hope for, and we are still working at it. So I only claim that asymptotic freedom is theoretical EVIDENCE for confinement, nothing more. But I think your statement that A.F has

I think we would agree on that?

- #15

tom.stoer

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One question: what is the most promising approach towards an understanding of color confinement (besides lattice QCD which does not explain things)

- Coulomb gauge, IR ghost propagators, ...?

- sphalerons and other "-ons"?

- superconductor inspired ideas?

- SUSY (I think Witten found a SUSY for which he was able to prove confinement)

- #16

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But you seem to be implying that asymptotic freedom is a NECESSARY condition for confinement. Why else would you expect the confinement to fail when the asymptotic freedom fails? I on the other hand, believe that asymptotic freedom is neither necessary nor sufficient condition for confinement. You can have either one without the other. This belief comes from the fact that asymptotic freedom is a statement about very high momentum transfer limit, and confinement is a statement about low momentum transfer limit. When you lose asymptotic freedom, the coupling constant does not go to zero any more at the high momentum limit. But it still may become very large at low momentum transfer, maintaining the confined state. The perturbative calculations that tell you when you lose asymptotic freedom after adding too many fermions are valid only for high momentum transfer. They don't tell you anything about the changes in the low momentum region, if indeed anything changes at all. Change of behavior in one limit does not imply the change in the other. If one indeed implies the other, it must be a very deep connection, and I have never seen any good explanation of it, let alone a proof.

As I said, though: asymptotic freedom is not a SUFFICIENT condition for confinement.

- #17

blechman

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Yes we do!

Dem's fightin' words!!!! :grumpy:One question: what is the most promising approach towards an understanding of color confinement (besides lattice QCD which does not explain things)

Well, we still haven't explained confinement to complete satisfaction. My research is in the perturbative part of QCD so I don't worry too much about this. There are arguments you can make using SUSY models and dualities that do well, but of course, we don't live in a supersymmetric world, so nuts to that! Although theoretically they might be of interest. Similarly, we can make some fun arguments from AdS/CFT and other string theory based proposals (strings were originally meant to describe confinement, after all!).- Coulomb gauge, IR ghost propagators, ...?

- sphalerons and other "-ons"?

- superconductor inspired ideas?

- SUSY (I think Witten found a SUSY for which he was able to prove confinement)

"Sphalerons" have nothing to do with confinement. They give us electroweak baryogenesis. As to other topological curiosities: again, they ended up being something of a dead end. Similar to SUSY: they give us confinement in very contrived models, but nothing like the real world. There are also "models" (like the Skyrme Model, for example) where hadrons are topological objects, but these suffer from some pretty ludicrous problems and are probably no more than an amusing curiosity rather than an actual real-to-life explanation of confinement.

- #18

tom.stoer

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I studied Skyrmions about 20 years ago; they are just low-energy effective objects with the right baryonic quantum numbers; I don't regard them as something fundamental which can explain QCD effects.

We also considered models similar to Anderson localization where quarks scattering off "gauge defects" result in suppression of quark propagators; quark propagation cancels due to "random distribution" of these defects.

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May I ask a simple question regarding this. In a hypothetical universe which initially contains only the vacuum and aOnce the quark is out of the proton, however, it realizes that a terrible mistake has been made , and it immediately tries to get back inside a hadron (ANY hadron!), and it does this by pulling quarks out of the vacuum to create the final hadrons that we see.

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Obviously color is conserved. If you start with a single isolated quark, the process of confinement will propagate in your universe for all eternity.May I ask a simple question regarding this. In a hypothetical universe which initially contains only the vacuum and asingledown quark particle of "red" color, what would happen? Would it grab other quarks from the vacuum to form a hadron? Also assume that other forces such as weak interaction are absent in this universe.

- #21

tom.stoer

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This is forbidden by color-neutrality.In a hypothetical universe which initially contains only the vacuum and asingledown quark particle of "red" color, what would happen?

One can do the following:

- fix the A°=0 gauge

- derive the equations of motions

- observe that one e.o.m is just a constraint ensuring that A°=0 holds

- this constraint is the (generalized) Gauss-law in color space: G(x)|phys> = 0

The last equation means that it does not hold as an operator equation but as a constraint on physical states; the Gauss law generates residual gauge transformations respecting A°=0 in the sense that the gauge function f(x) is t-independent.

Now one proceeds as follows

- integrate G(x)

- and observe that up to a surface term this is just the total color charge Q

- therefore the integrated constraint reads Q|phys> = 0

There is one exception, namely the possibility to have surface degrees of freedom generating global color charges; I do not know if there are additional theorems ruling out these surface charges as well. Of course they do not contribute if there is no boundary at all.

That means that the state you described above is not in the subspace of physical states and is therefore ruled out by the theory. Color neutrality is a "kinematic" property of the theory and must not be confused with color confinement; it simply follows from a single constraint equation. B.t.w.: by the same arguments charge neutrality must hold in QED as well.

- #22

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If I understand correctly, your argument only works for compact spaces. For infinitely extended Minkowski space, you can't make the surface term vanish unless the field configuration is localized. For example, a single electron in QED generates a Coulumb potential which is neither localized in time (because the electron will persist) nor in space (because electromagnetism is long-ranged, and because there's no other stuff in the universe that can provide shielding). In this scenario, the surface term can never go away however large the integration volume is.That means that local gauge symmetry forces physical states via the Gauss law to be color singulet states!

There is one exception, namely the possibility to have surface degrees of freedom generating global color charges; I do not know if there are additional theorems ruling out these surface charges as well. Of course they do not contribute if there is no boundary at all.

- #23

blechman

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it's amusing that you ask this: when i was in graduate school i asked EXACTLY the same question. we decided that the universe would explode! ;-)May I ask a simple question regarding this. In a hypothetical universe which initially contains only the vacuum and asingledown quark particle of "red" color, what would happen? Would it grab other quarks from the vacuum to form a hadron? Also assume that other forces such as weak interaction are absent in this universe.

fortunately, gauge invariance forbids such a "hypothetical universe".

- #24

tom.stoer

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I do not see a convincing argument in a non-compact space.

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Why does gauge invariance forbid a universe containing only a single quark?it's amusing that you ask this: when i was in graduate school i asked EXACTLY the same question. we decided that the universe would explode! ;-)

fortunately, gauge invariance forbids such a "hypothetical universe".

This only proves that the argument cannot be valid. For a single electron can be an asymptotic state of QED, and hence exist by itself (far away from everything else).Color neutrality is a "kinematic" property of the theory and must not be confused with color confinement; it simply follows from a single constraint equation. B.t.w.: by the same arguments charge neutrality must hold in QED as well.

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