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The gauge fields in Yang Mills theory are

by Lapidus
Tags: fields, gauge, mills, theory, yang
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Sep2-11, 02:30 PM
P: 283
The gauge fields in Yang Mills theory are matrices:

A[itex]_{\mu}[/itex] = A[itex]^{a}_{\mu}[/itex] T[itex]^{a}[/itex]

But A[itex]^{a}_{\mu}[/itex] are vector fields, i.e. a=1,..,n four-vectors. Should not there be a U(1) gauge symmetry for each of them in addition to the non-abelian gauge symmetry?

In Lagrangian for the strong force, does not each of these four vectors correspond to a gluon? Gluons or weak bosons are spin-1 particles, so they most be described by four vectors. How do they follow from matrices??

And how can a vector field/ a four-vector be non-abelian??

help, please!
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Sep2-11, 06:32 PM
P: 343
In 4-dimensions each 4-vector corresponds to a spin-one particle(not 4 spin one particles).
In SU(N) there are N^2-1 generators so a,b goes from 1 to N^2 -1(not N). In SU(3) that makes 3^2 -1 = 8 gluons [itex] A^a_\mu [/itex]
Sep3-11, 06:29 PM
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Bill_K's Avatar
P: 4,160
Lapidus, as you say, the gauge fields Aaμ in Yang-Mills theory are a set of four-vectors, a=1,..,n. The Ta are not fields and not matrices, they are the group generators. They will be represented by matrices if you consider their action on particles making up a particular group representation. Aaμ and Ta occur together in the covariant derivative, Dμ = ∂μ + ig AaμTa.

For example for QCD there are 8 gauge fields and 8 generators, a=1,..,8. Quarks belong to a 3-d representation labeled by color, i=1,2,3, and in the term of the Lagrangian where Dμ acts on them, the Ta will be represented by eight 3x3 matrices. Elsewhere in the Lagrangian, Dμ acts on the eight gauge fields themselves, and in that term the Ta will be represented by eight 8x8 matrices.

Sep3-11, 06:40 PM
P: 313
The gauge fields in Yang Mills theory are

Each generator of the group does lead to a U(1) gauge symmetry, but since the generators have non-trivial commutation relations, these U(1)s are all linked together to form part of a larger group.

Think of the case of rotations in 3-space. There are three basis elements that together generate SO(3). But each generator alone makes rotations in the plane, which is SO(2)~U(1).
Sep3-11, 07:02 PM
P: 283
thank so much you, guys! Got it
Sep4-11, 03:51 AM
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P: 5,366
Theoretically there could very well be an U(1) generator as well.

In the case of QCD there could be a U(3) = U(1) * SU(3) symmetry which would result a 9th generator represented by the 3*3 identity matrix. But this U(1) symmetry would result in a new color-force similar to an el.-mag. like long range force. b/c we do not observe this long range force in nature this extra U(1) factor has to be excluded.

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