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Is there any experiment/measurement that would have a different outcome if one of the following scenarios is applied:

  1. The two color-neutral** gluons would not exist
  2. Those gluons would have very large masses

** we are talking about the 2 gluons not exchanging color, conventionally written as

$(r\bar{r}−b\bar{b})/\sqrt2$

$(r\bar{r}+b\bar{b}−2g\bar{g})/\sqrt6$

EDIT: Since I am now aware that 'color-neutral' does not exist, the question only makes sense if stated as

'What would be the experimental evidence if color-symmetry was broken in some way? (such that 2 of the gluons were singled out)'

NoEscape
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    I think that the linear combinations two by two are not color neutral, it is only the second formula that is and that does not exist because of the matrix algebra of SU(3) color http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/gluons.html . Experiments have not shown color neutral gluons in hadronic interactions, one can use SU(3) color to model the behavior of hadrons well and call it QCD. – anna v Jun 04 '14 at 18:27

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Because the color charge is an exact symmetry (there is no "color mass hierarchy" as there is with neutrino flavor) and because colored objects are confined, I don't know that there is experimental evidence for any particular gluon. The evidence is that hadrons have an internal degree of freedom which can be described by the $SU(3)$ symmetry group; we call it "color" mostly because the other is quite a mouthful.

One feature of color as an exact symmetry is that rotations among the different color charges give the same physics. So for instance we could make the substitution \begin{align*} r' &= \frac{r + b}{\sqrt2} \\ b' &= \frac{r - b}{\sqrt2} \\ g' &= g \end{align*} and similarly for the anticolors. The "colorless" gluons in this basis can be represented in the original basis \begin{align} \frac{r'\bar r' - b'\bar b'}{\sqrt2} &= \frac1{\sqrt2}\left( \frac{r + b}{\sqrt2} \frac{\bar r + \bar b}{\sqrt2} - \frac{r - b}{\sqrt2} \frac{\bar r - \bar b}{\sqrt2} \right) = \frac{r\bar b + b\bar r}{\sqrt2} \end{align} as a coherent superposition of color-changing gluons.

If we had some very massive or missing gluons, color symmetry would no longer be exact; there would probably be a favored color, and I'm not sure that you could still have the complete color confinement that we observe in the real world.

rob
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    Can you necessarily conclude that confinement wouldn't exist if the color symmetry were broken? I'm a bit skeptical of that - not that I believe you're wrong, I just have my doubts without seeing some evidence. I bet there's a paper on it somewhere. – David Z Jun 04 '14 at 22:37
  • @DavidZ As I said, I'm not sure. I'm imagining all of the quarks in a hadron evolving to have the favored color, the way that heavy quarks evolve into light quarks and decay products. I'm aware that's not a great analogy. – rob Jun 04 '14 at 22:44
  • Strangely enough, you are answering in a theoretical way, while in my related theoretical question the guys stated experimental evidence. – NoEscape Jun 05 '14 at 05:01
  • If anti-colors transformed in the opposite way (as I would expect), your calculation (the color rotation) would still result in color-neutral!? – NoEscape Jun 05 '14 at 05:04
  • @NoEscape Give it a try! Remember that your transformation must also take a color-neutral meson to a color-neutral meson. – rob Jun 05 '14 at 06:06
  • the latter is not the case for the transformation you are using – NoEscape Jun 05 '14 at 06:15
  • @NoEscape See Peter Shor's answer to your similar question. I always forget that the way to transform an operator $\mathcal O$ by a matrix $M$ is $\mathcal O' = M\mathcal O M^\dagger$. – rob Jun 05 '14 at 20:31
  • OK, your point was right - in principle. My question was EXPLICITLY not about group theory, though. – NoEscape Jun 23 '14 at 09:26
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    @NoEscape I'm not sure you can discuss gluons without discussing group theory. – rob Jun 23 '14 at 11:26
  • I agree with rob. Gluons are a construct of the SU(3) hypothesis for the strong interaction, and the rounding off of a gauge theory format similar to the electroweak one, that could lead to unification of all forces. All of these depend on the observation that data fit beautifully with an SU(3) for the strong interaction SU(3) of course is a group. – anna v Jun 23 '14 at 14:18