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Apr 10, 2014 at 17:44 answer added stanley dodds timeline score: -1
Apr 10, 2014 at 8:10 vote accept PML
Apr 9, 2014 at 16:12 history edited PML CC BY-SA 3.0
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Apr 9, 2014 at 15:50 comment added Jordan $rgb$ isn't the only colorless combination. $r\bar{r} g \bar{g}$ and other such 4-quark states are colorless as well.
Apr 9, 2014 at 15:28 answer added innisfree timeline score: 11
Apr 9, 2014 at 15:27 comment added dmckee --- ex-moderator kitten Tetraquarks fit just fine in the standard model and have been sought for a long time. They are neither baryon nor meson, but they are still hadrons. (I suppose the definition of "meson" might be extended to cover this case, but pentaquarks---if found for real---would call for a re-think.)
Apr 9, 2014 at 14:20 comment added PML @innisfree That was my doubt, yes. But you answered it in you first comment.
Apr 9, 2014 at 14:17 comment added innisfree ah hang on, I think this question is if $rgb$ is "white/colorless", a regular hadron of 3 quarks could be colourless, but how can four quarks give $rgb$ and end up being "colourless"?
Apr 9, 2014 at 14:15 comment added innisfree Why can't $c \bar c d \bar u$ be color neutral? Doesn't $3\times\bar3\times3\times\bar3$ contain a singlet?
Apr 9, 2014 at 14:07 comment added PML But then it can't be color neutral if it's composed by $c \bar{c}d \bar{u}$. I thought all hadrons had to be neutral.
Apr 9, 2014 at 13:58 comment added John Rennie The Z(4430) is believed to be made up of quarks, or at least no-one is suggesting differently. It's just that it appears to be made up of four quarks not three.
Apr 9, 2014 at 13:58 comment added George G The particle they found is a Tetraquark. The traditional quark model only had particles with 3 quarks or a quark-antiquark pair.
Apr 9, 2014 at 13:50 history asked PML CC BY-SA 3.0