Re: How does guage symmetry work?

Date: Mon Jul 2 14:13:14 2001
Posted By: Michael Wohlgenannt, Grad student, Ph.D. student, Department of theoretical physics , university of munich
Area of science: Physics
ID: 992822192.Ph

Message:

Hi Gashib,

that is a good question. The concept of gauge symmetry is a fundamental one in the theory of elementary particles (maybe even in all theoretical physics). In an astonishing way the gauge principle introduces interaction into a theory of otherwise free particles. By the gauge principle I mean that physics does not depend on which "coordinate system" one chooses. Different people may conduct the same experiment at different places choosing different "coordinate systems", nevertheless the result of the experiment has to be the same. So you can fix any coordinate system (i.e., choose a gauge). The first to introduce a gauge theory was Maxwell, that is classical electromagnetism. The vector potential A_m=(phi, vec{A}) is the fundamental quantity of this theory. phi is the electrostatic potential, vec{A} is the 3 dimensional vector potential. The electric field E and the magnetic field B are unaltered under the transformation

A_m ---> A_m+1/e d_m P

where d_m is the partial derivative with respect to the m-th coordinate, P is some real function, m=0,1,2,3. This transformation does not change the electromagnetic field strength, i.e. it does not change neither the magnetic nor the electric field. This is called a gauge transformation. A_m can only be determined up to a function P, which does not influence physics. So one can fix P, choose a gauge. Two common ways of fixing the gauge are the Lorentz gauge and the Coulomb gauge. The Lagrange function for electromagnetism is

 L = F_mn F^mn,
 F_mn = d_m A_n - d_n A_m.

Variation with respect to A will lead to Maxwell's eqns. Now we want to introduce also electrons. Free electrons (psi) are given by the Lagrangian

L_e = barpsi (i\gamma^m d_m - m) psi,

where barpsi means the conjugate of psi. \gamma are the usual gamma matrices, m is the mass. Variation of this equation will lead to the equation of motion for free electrons, the dirac equation.
For the total action we have to add L and L_e. Sofar we have introduced no interaction! Further we see that the function L_e is invariant if we replace

psi ---> e^(iP) psi,

if P is constant. But P had not to be constant in the above considerations. It does not matter what we choose for P. P can be chosen differently in Rome and in Paris, and experiments performed in Rome and in Paris yield the same result, independent of P. So we demand, that L+L_e has to be invariant under the transformation

A_m(x) ---> A_m(x) + 1/e d_m P(x),
psi(x) ---> e^(iP(x)) psi(x).

This transformation is called a local gauge transformation, since the gauge parameter depends on x. But we have seen that L+L_e is not unvariant under above transformations! Therefore it cannot be the right Lagrange function. If we want to render L+L_e invariant we have to add terms. We have to replace the derivative of the electron field psi d_m psi by

D_m psi = (d_m + i e A_m) psi,

where e is the electric charge (coupling parameter). This is called the minimal coupling of the gauge field A. The resulting Lagrange function is invariant under this local gauge transformation.

Properly, we have to start at L_e, a theory of free electrons. We demand that the theory is invariant under a local gauge transformation psi --> e^{iP(x)} psi. Then we have to replace derivatives d_m by D_m, covariant derivatives. That means we have to introduce the gauge field A. Finally we add a kinetic term for the gauge field, L.
We start at a free theory and invariance of physics under a certain (local) symmetry introduces gauge fields and interactions. The difference between electromagnetism and (classical) QCD is in the gauge transformation. In case of electromagnetism we had P(x) a function, therefore e^{iP(x)} build a group called U(1), 1 because there is only one parameter, U because of unitarity, e^{iP(x)}* e^{iP(x)}=1. In case of QCD P(x) is a matrix valued function. P(x)=sum_i P_i(x) T^i, where the sum is over i=1,2,3,4,5,6,7,8. T^i antisymmetric 3x3 matrices which are said to generate SU(3). They so to say build a basis of SU(3). It further means that e^{iP(x)} is a 3x3 matrix and psi has to be a 3-vector, in order we can multiply e^{iP(x)} with psi and compute the transformation

psi --> e^{iP(x)}.psi

Now we return to your question for the colours. The components of psi are called colours, because they are not components in space-time but in an internal space. The gauge transformation above is merely a change of coordinates in this internal space, in colour space. psi has 3 components, and there are 3 complementary colours red, blue and green. We fix a coordinate system and call the components of psi red, blue and green. The analogy is that you can compose any colour using the colours red, blue and green. But you can also use 3 different complementary colours, eg. yellow, indigo and violet. This corresponds to a coordinate transformation in the internal space, a gauge transformation in colour space.
I hope I could help you in understanding gauge interactions. otherwise do not hesitate to ask further questions.
Michael

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