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JustinLevy
Mar29-10, 07:39 AM
Starting with the Lagrangian for EM, it looks like there are four degrees of freedom for the four-vector potential. But one term is not physical in that it can be expressed completely in terms of the other degrees of freedom (so it is not a freedom itself), and there is another "freedom" that is not physical because it doesn't effect the equations of motion (the "gauge" freedom).

For interactions with higher symmetries (like the weak force SU(2), or the strong force SU(3)), is there an easy "symmetry argument" for how many of the components of their "potentials" will actually be physical freedoms?

For example, there are 8 gluons. How many physical degrees of freedom are there actually amongst these 8?

Physics Monkey
Mar29-10, 10:06 AM
The counting is essentially the same at the linearized level. For example, the energy density of a gas of photons is \epsilon/T^4 = \pi^2/15 while the energy density for a gas of free SU(N) gluons (at high temperature, say) is \epsilon/T^4 = (N^2 - 1) \pi^2/15 . In other words, you get one photon contribution for each gauge boson. In general, you would have 2 times the dimension of the adjoint representation degrees of freedom.