A particle in the Dirac field can be described with the following equation $$i\gamma^\mu\partial_\mu\psi-m\psi=0$$ This is if the particle is non-interacting. However, if we impose a local $U(1)$ symmetry such that $\psi'=\psi e^{i\Lambda(x)}$ we should expect that the equations of motion should be invariant $$i\gamma^\mu\partial_\mu\psi'-m\psi'=0$$ However this equation is not invariant under a $U(1)$ local symmetry instead the equations of motion should be $$i\gamma^\mu(\partial_\mu +iqA_\mu)\psi-m\psi=0$$ To impose a local U(1) gauge invariance. My question is if the gauge covariant derivative stays invariant under local $U(1)$ transformation, what does this really represent. In the context of General relativity the covariant derivative $$\nabla_\mu A^\alpha = \partial_\mu A^\alpha + \Gamma_{\gamma\alpha}^{\mu} A^\gamma$$ describes the change of the basis vectors when moving through space-time. Does the covariant derivative in the Dirac equation represent the change of phase moving through spacetime? With this covariant derivative can we come up with a geodesic equation, the curvature of the field, and also can we derive a metric from the gauge covariant derivative. Also in general relativity, the manifold that is curved is describing the manifold of space-time, what manifold does the gauge covariant derivative explain? Thanks
1 Answer
As you can see, the covariant derivative has two components: $$\nabla_{\mu} =\partial_{\mu} +\Gamma_{\gamma \alpha}^{\mu}, $$ in which the first describes the change of the vector field and the second is a linear transformation of the vector space, related with the possibility of making different choices for the basis of the vector space in each point of the spacetime (manifold).
In the context of Yang-Mills, the covariant derivative does exactly the same: $$\nabla_{\mu} = \partial_{\mu}+i q A_{\mu},$$ in which the first term is the change of your filed and the second term is a linear transformation in the gauge group instead of the vector space.
This similarity arises from the fact that, "gluing" a vector space or a gauge group to each point on your manifold is very similar (from the point of view of fiber bundles).
From the point of view of fiber bundles, these two situations are equal (there might be some technical differences with which I am not too familiar, I have just started studying it myself). In particular, you can see that if you try to define a derivative for these kinds of objects, you will always have two terms in each one relates to the change of the field and the other is associated with the freedom you have to parameterize the object that you "glued" to each point of the manifold (for a vector space you have the freedom to choose the basis, as you also have for the gauge groups, as you can see from the existence of gauge transformations, which, in this case, reduces to changing the phase).
In fact, and do not quote me on this, the difference between GR and Yang-Mills (from the perspective of fiber bundles) might only be the details of the theories (e.g. in one you have vector spaces and in the other you have groups) and the definition of the action (i.e the mathematical structure of both theories is the same).
(Feel free to comment if you notice any mistake or imprecision).
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$\begingroup$ Obligatory links to physics.stackexchange.com/q/8686, physics.stackexchange.com/q/71476 and physics.stackexchange.com/q/4359 $\endgroup$ Jan 16, 2021 at 11:04
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1$\begingroup$ But does this mean that the geodesic equation with the gauge derivatives results in the equations of motion for a particle moving in the electromagnetic field, also is it possible to define a metric over this manifold such that $iqA_\mu$ is composed of this metric describing the geometry of the gauge group over this manifold? $\endgroup$ Jan 16, 2021 at 11:18
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$\begingroup$ The equations of motion depend on the action, and the action is different, hence the dynamics will be different (meaning the geodesics may be irrelevant). Regarding the metric, even if you define one, it will be in the space of the groups (it will take two elements of the group into a scalar). I think that can be done by taking the integral of the trace of the product (but I am not very familiar with those details). In Chapter 7 of David Skinner's notes (the notes are public) on advanced quantum field theory, from Cambridge, he does a nice introduction to this. I may want to check them out. $\endgroup$– JGBMJan 16, 2021 at 11:34
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1$\begingroup$ It may be irrelevant for the equations of motion, however, what is the interpretation of curvature, parallel transport, etc... on the gauge theory model? $\endgroup$ Jan 16, 2021 at 11:57
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$\begingroup$ I do not have a good physical intuition for those. However, these concepts are important. In fact, the Yang-Mills action is the integral of the curvature. Regarding parallel transport, the only gauge invariant quantities that you can write, the Wilson loops, are the parallel transport of the gauge field around closed paths. Those concepts continue to exist in the same manner (as long as they can be generically defined for fiber bundles). When you parallel transport the field (e.g. $U(1)$) you obtain a change in the value of $\Lambda(x)$, in Yang-Mills, and a rotation of a vector in GR. $\endgroup$– JGBMJan 16, 2021 at 13:15