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Gauge fields  how many physical degrees of freedom? 
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#1
Oct313, 11:48 AM

P: 290

The photon field has two physical degrees of freedom (dof): its two transverse polarization directions.
But what about nonabelian gauge theories? What about N massless spin1 particles that transfom under SU(N), how do I count their degrees of freedom? Gluons, for example, are massless spin1 particles, so I assume that each of them has only two physical dof and the unphysical dof (the longitudal and timelike) can be 'gauged away'. But there are three color gluons that transform in the adjoint rep of SU(N), so that might be too naive, since all three gluons get mixed into each other. But how many physical dof do gluons have? thank you! 


#2
Oct313, 12:18 PM

Sci Advisor
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P: 11,948

2 dof per each gluon color.



#3
Oct313, 12:28 PM

Thanks
P: 1,948

Yes, 2 dof per gluon. What do you mean by



#4
Oct313, 01:39 PM

P: 290

Gauge fields  how many physical degrees of freedom?
Now let's suppose we want analog to the EM case impose gauge conditions on the gluons to reduce the dof down to just the physical dof. I know the EM fourvec potential has four components, i.e. the two physical transverse states and the unphysical longitudinal and timelike state. Do I have to impose 8x2 gauge conditions on the SU(3) gauge symmetry to cut down the 8x2 physical dof of the SU(3) gauge theory? Every gluon is a massless spin1 particle, so each has two dof. OK. But is not the gauge symmetry bigger than for the photons and does not that imply more unphysical dof? Dof other than the longitudinal and timelike states? thanks again! 


#5
Oct313, 05:36 PM

Thanks
P: 1,948

The gauge group has 8 parameters which is exactly the right number needed.



#6
Oct413, 02:47 AM

Sci Advisor
P: 5,464

The most transparent way to eliminate unphysical d.o.f. and to count physical d.o.f. is the temporal gauge plus Gauß law constraint. You start with Ddimensional spacetime
##\mu =0,1,\ldots,D1## and N colors i.e. SU(N) ##a=1,2,\ldots,N^21## Then for ##A_\mu^a## we have ##Z_\text{tot} = D \cdot (N^21)## d.o.f. in total. Due to the antisymmetry of the field strength tensor we have ##F_{00}^a = 0## That means that A_{0} is no dynamical d.o.f. but acts as a Lagrange multipler generating the Gauß law constraint. We chose the temporal gauge ##A_0^a = 0## and keep the Gauß law as a condition for physical states, i.e. ##G^a \text{phys}\rangle = 0## This reduces the d.o.f. to the physical subspace. Each condition (gauge condition, Gauß constraint) removes ##N^21## unphysical d.o.f. We get ##Z_\text{phys} = Z^\prime = Z_\text{tot}  Z_\text{unphys} = D \cdot (N^21)  2 \cdot (N^21) = (D2) \cdot (N^21)## So we find D = 1+1: Z' = 0 D = 2+1: Z' = N^{2}1 D = 3+1: Z' = 2 * (N^{2}1) That means that (up to topological d.o.f.) in 1dim space gauge fields can be eliminated by imposing Gauß law. In 3dim. space each gluon color (there are 91=8) carries 2 polarizations which results in 16 physical d.o.f. You can use the same reasoning for U(1) replacing N^{2}1 by 1. For U(N) ~ SU(N) * U(1) like in the electroweak theory you have to replace N^{2}1 by N^{2}; that means that the additional d.o.f. is just the photon. 


#7
Oct413, 10:14 AM

P: 290

A big thank you, Tom! This was extremely helpful.



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