A Question About Gauge Symmetry and the Interaction Term

I spent the week seeing if I could give the good old college try and find a "way that worked" for the gauge games I have been playing. Nothing panned out. Bummer, I had a blog to write. While collecting crumpled paper, there was an old issue I wasn't able to figure out. The answer was probably known a hundred years ago. The blog will explain where I got stuck, and who knows, maybe someone will provide the answer.

In my previous blog, I focused on the fields, the derivatives of the potential. In the comments, CuriousReader pointed out that careful thought had to be devoted to the interaction term, the contraction of the current with the potential. The potential A will change for both the interaction and the fields due to the gauge. Having never thought about the interaction term in this context, I was the proverbial deer in the headlights, not doing too much but worrying about shotguns.

Here is how gauge symmetry is presented in the wiki way:

$\\ \phi \rightarrow \phi' = \phi - \frac{\partial f}{\partial t} \\ \\ A \rightarrow A' = A + \nabla f$

Drop those into the definition of the electric field E and the magnetic field B [corrected sign errors]:

$\begin{align*} E \rightarrow E' &= -\frac{\partial A}{\partial t} - \frac{\partial \nabla f}{\partial t} - \nabla \phi + \nabla \frac{\partial f}{\partial t} = E \\ \\ B \rightarrow B' &= \nabla \times A + \nabla \times \nabla f = B \end{align*}$

Fini!

In my limited reading, that was presented as the full story. Yet something is being added to the potential. Shouldn't that mess up the interaction term?

$J_{\mu} A^{\mu} \rightarrow J_{\mu} A'^{\mu} \ne J_{\mu} A^{\mu}$

Look at the interaction with the gauge field close up:

$J_{\mu} A'^{\mu} = \rho \phi - \rho \frac{\partial f}{\partial t} - J \cdot A - J \cdot \nabla f$

This caused me to scratch my head.The derivatives of the scalar function f are free to wander in spacetime. Don't read anything into the plus and minus signs because the derivative of f with respect to the four spacetime variables can be positive in one location, and negative in another.

The way out is to realize how the interaction term is used to generate the field equations. One takes the derivative of the current coupling with respect to the potential with or without the gauge field:

$\frac{\partial J_{\mu} A^{\mu}}{\partial A^{\nu}} = \frac{\partial J_{\mu} A'^{\mu}}{\partial A'^{\nu}} = J_{\nu}$

No matter what odd things the gauge field does, the derivative with respect to the new potential will still be the 4-current J. One can find solutions to the field equations that are independent of the choice of the gauge field because it doesn't matter for the field term nor the interaction term. Those solutions when put into the standard force equation are consistent with experiments. So far, so good.

Here is something that still doesn't quite make sense to me. I don't see how the derivation of the equations of motion remains unaltered with a gauge field. To be concrete, suppose the function and gauge field were defined like so:

$\begin{align*} f &= x^2 + y^2 \\ \\ f' &= (0, 2 x, 2 y, 0) \end{align*}$

[ERROR? Looking into this in detail, I think f must be a differential function of t, x, y, and z for this to work out.]
By the preceding analysis, the derivation of the field equations are unaltered because the derivatives are with respect to the potential A and changes in the potential. The same solutions to the field equations will be found. No mystery.

To generate the force equation however, one takes the derivative of the action with respect to position and velocity. There will be an extra factor of 2x and 2y for this gauge field. It looks like the form of the force equation is altered by this gauge choice.

Hopefully my misunderstanding can be corrected in the comments.

A tentative simple road out.

Both David and CuriousReader said I should keep real close track of what the coordinates are. This is the world of calculus of variations, not integrate the area under the curve. The goal is to find extremums of functions are variables that are varied. I will need to spend a few days doodling to see if I can track what is going on.

I was wondering if there might be a more direct solution to the force derivation + gauge transformation riddle, which is only to me, not a mystery to physics.

Do the force derivation the "standard way" as I did a number of blogs ago. Notice that the results can be expressed in terms of the fields E and B. I showed above how those two fields are invariant under a gauge transformation. Ego the force will be invariant under the same gauge transformation.

Doug

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