Gauge fields - how many physical degrees of freedom?

[Lapidus]

The photon field has two physical degrees of freedom (dof): its two transverse polarization directions.

But what about non-abelian gauge theories? What about N massless spin-1 particles that transfom under SU(N), how do I count their degrees of freedom?

Gluons, for example, are massless spin-1 particles, so I assume that each of them has only two physical dof and the unphysical dof (the longitudal and time-like) 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!

 

[dextercioby]

2 dof per each gluon color.

 

[dauto]

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

But there are three color gluons that transform in the adjoint rep of SU(N)

That's not right. there are N^2-1 gauge bosons in the SU(N) group. That's eight (not three) gluons for QCD's SU(3) adjoint representation

 

[Lapidus]

Yes, 2 dof per gluon. What do you mean by
That's not right. there are N^2-1 gauge bosons in the SU(N) group. That's eight (not three) gluons for QCD's SU(3) adjoint representation

Of course! I messed that up.

Now let's suppose we want analog to the E-M case impose gauge conditions on the gluons to reduce the dof down to just the physical dof. I know the E-M four-vec potential has four components, i.e. the two physical transverse states and the unphysical longitudinal and time-like 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 spin-1 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 time-like states?


thanks again!

 

[dauto]

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

 

[tom.stoer]

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 D-dimensional space-time

$\mu =0,1,\ldots,D-1$

and N colors i.e. SU(N)

$a=1,2,\ldots,N^2-1$

Then for

$A_\mu^a$

we have

$Z_\text{tot} = D \cdot (N^2-1)$

d.o.f. in total.

Due to the anti-symmetry of the field strength tensor we have

$F_{00}^a = 0$

That means that A₀ 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

$N^2-1$

unphysical d.o.f. We get

$Z_\text{phys} = Z^\prime = Z_\text{tot} - Z_\text{unphys} = D \cdot (N^2-1) - 2 \cdot (N^2-1) = (D-2) \cdot (N^2-1)$

So we find
D = 1+1: Z' = 0
D = 2+1: Z' = N²-1
D = 3+1: Z' = 2 * (N²-1)

That means that (up to topological d.o.f.) in 1-dim space gauge fields can be eliminated by imposing Gauß law. In 3-dim. space each gluon color (there are 9-1=8) carries 2 polarizations which results in 16 physical d.o.f.

You can use the same reasoning for U(1) replacing N²-1 by 1.
For U(N) ~ SU(N) * U(1) like in the electro-weak theory you have to replace N²-1 by N²; that means that the additional d.o.f. is just the photon.

 

[Lapidus]

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