Is there a good reason not to pick $U(3)$ as the colour group? Is there any experiment or intrinsic reason that would ruled out $U(3)$ as colour group instead?
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Other Phys.SE questions about the precise form of the standard model gauge group: https://physics.stackexchange.com/q/105816/2451 , https://physics.stackexchange.com/q/116831/2451 – Qmechanic Jun 13 '14 at 19:13
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1$U(3)$ has 9 generators, you would thus get an extra massless gauge boson (resembling very much an extra photon) on top of the eight gluons. – TwoBs Jun 13 '14 at 20:13
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Suppose that $\text{U}(3)$ was the gauge group. We can decompose this as $$\text{U}(3)=\text{U}(1)\times\text{SU}(3),$$
which implies that in addition to the $\text{SU}(3)$ that has eight generators corresponding to eight gluons, there would be an additional generator for $\text{U}(1)$. The latter in principle corresponds to an additional gauge boson, but a theory of the strong interactions containing such a particle is inconsistent with experiment.
Frederic Brünner
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1would the U(1) vector boson carry colour? It would resemble a photon, wouldn't it? – IamZack Jun 13 '14 at 20:31
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No, it would be a singlet under the colour group, i.e. it would be uncharged. In this sense, it would be like a photon. – Frederic Brünner Jun 13 '14 at 20:34
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2Not to be overly mathematically nitpicky, but shouldn't your direct product be a semi-direct product? See, e.g. http://en.wikipedia.org/wiki/Unitary_group#Properties – joshphysics Jun 13 '14 at 21:32
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It would be double counting, since total phase rotations of the quark wave function are already part of the model and the photon that makes them into a gauge symmetry already exists. The total gauge group is SU(3) × SU(2) × U(1), so the question "where has the U(1) gone" has as its answer that it already was included. In a gauge theory you can only use the symmetries that you start with and you can use them only once.
Jos Bergervoet
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