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https://wikimedia.org/api/rest_v1/media/math/render/svg/d0841c7868cedab8de771d4b9c21ec30ed4b8702 --209.204.41.233 (talk)02:08, 21 October 2021 (UTC)[reply]

I'm not an expert, or even a physicist, but (as I understand it) the values describe the chances of a gluon having the combination of those 'colors' (they're not really colors, just convenient labels). If you want the chances of the combination to come out between zero and one (that's required) you have tonormalize them. Hence you device by the sqrt(1^2+1^2+1^2)=sqrt(3). It's basically the Pythagorean theorem in disguise. Ask an actual physicist for a better answer.Kleuske (talk)10:15, 21 October 2021 (UTC)[reply]

The value of the color charge is acomplexunit vector in a three-dimensional complex vector space (a complex three-dimensionalHilbert space). Theeigenvectors that span the space is the three colors of the color charge (i.e. the basis axes X, Y, Z of this space are labeled "red", "green", "blue", or in physics-speak${\displaystyle |\text{red}⟩,|\text{green}⟩,|\text{blue}⟩}$ where the${\displaystyle |\cdot ⟩}$ is called a "ket", from thebra-ket notation.) These color charge vectors define the probabilities formeasurement of the color charge. The sqrt(3) term is for normalization the quantum state, so it remains a unit vector (seeunitarity). The 1/sqrt(3) factor is theprobability amplitude of each of the possible outcomes of this three-levelquantum state. · · ·Omnissiahs hierophant (talk)18:57, 16 December 2021 (UTC)[reply]

Why don't we list gravity under "Interactions" in the infobox? Or maybe more precisely, why do we list gravity as an interaction for some particles (for examplequarks andphotons) but not for all particles? I thought all particles were affected by gravity in the sense that all of them followgeodesics, and that all particles had a gravitational effect on other particles, since all particles have an energy and energy curves space, which gives rise to orbits that look like they are affected by gravity. —Kri (talk)17:58, 16 December 2021 (UTC)[reply]

Maybe no one has seen a gluon interact with gravity. No experimental data. We should avoid making stuff up just because it makes sense.What if gluons don't interact with gravity?— Precedingunsigned comment added by98.128.172.242 (talk)12:01, 17 December 2021 (UTC)[reply]

In the following:

• red–antired(${\displaystyle r\overline{r}}$), red–antigreen(${\displaystyle r\overline{g}}$), red–antiblue(${\displaystyle r\overline{b}}$)
• green–antired(${\displaystyle g\overline{r}}$), green–antigreen(${\displaystyle g\overline{g}}$), green–antiblue(${\displaystyle g\overline{b}}$)
• blue–antired(${\displaystyle b\overline{r}}$), blue–antigreen(${\displaystyle b\overline{g}}$), blue–antiblue(${\displaystyle b\overline{b}}$)

the following appear to be their own antiparticles:

• red–antired(${\displaystyle r\overline{r}}$), green–antigreen(${\displaystyle g\overline{g}}$), blue–antiblue(${\displaystyle b\overline{b}}$)

while the following appear to be particle/antiparticle pairs:

• red–antigreen(${\displaystyle r\overline{g}}$), green–antired(${\displaystyle g\overline{r}}$)
• red–antiblue(${\displaystyle r\overline{b}}$), blue–antired(${\displaystyle b\overline{r}}$)
• green–antiblue(${\displaystyle g\overline{b}}$), blue–antigreen(${\displaystyle b\overline{g}}$)

Is that a correct interpretation? If that's the case for the particle/antiparticle pairs above, then that would be a case of bosons having antiparticles (or is it only fermions that could have antiparticles?)!137.82.118.58 (talk)00:13, 9 June 2023 (UTC)[reply]

Is it a bion?— Precedingunsigned comment added by203.211.104.191 (talk)06:58, 16 October 2023 (UTC)[reply]

wiki is supposed to be a general encyclopedia, accessible to the average person

I don't know what the average person is, but this article is written at way too high a level, it’s more appropriate for a college senior majoring on physics— Precedingunsigned comment added by50.245.17.105 (talk)20:38, 1 November 2023 (UTC)[reply]

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