# Gauge fixing in the classical $U(1)$ gauge theory

My question concerns the gauge fixing in classical v.s. quantum $$U(1)$$ gauge theory. I will ask about the gauging fixing in quantum $$U(1)$$ gauge theory in a separated Phys-SE post.

## For the classical $$U(1)$$ gauge theory,

we have the electric $$\vec E$$ and magnetic $$\vec B$$ written as the scalar $$\phi$$ and vector $$\vec A$$ potentials: $$\vec E = - \vec \nabla \phi -\frac{\partial}{\partial t} \vec A.$$ $$\vec B = \vec \nabla \times \vec A.$$

My understanding about the gauge fixing for this gauge theory is that we can chose for example, $$(\phi, \vec A)$$ to give a set of $$\vec E$$ and $$\vec B$$ fields.

1. But we can also shift to $$(\phi, \vec A) \mapsto (\phi + C_0, \vec A + \vec C)$$ where arbitrary choices of $$(C_0, \vec C)$$ still give the same solutions of $$\vec E$$ and $$\vec B$$. So a certain but arbitrary choice of $$(C_0, \vec C)$$ can be regarded as a way of gauge fixing? correct?

2. Furthermore, we can also shift to $$(\phi, \vec A) \mapsto (\phi + \phi_0, \vec A + \vec A')$$ such that the followings are satisfied: $$- \vec \nabla \phi_0 -\frac{\partial}{\partial t} \vec A'=0$$ $$\vec \nabla \times \vec A'=0$$ Then we have a choice of $$(\phi, \vec A) \mapsto (\phi + \phi_0, \vec A + \vec A')$$ such that any choice is a way of gauge fixing? correct?

3. In a classical differential equation, we have $$* d * F = J$$ $$dF=0$$ so $$F=dA$$ with $$* d * dA = J$$ is a solution. Say $$F=dA'$$ with $$* d * dA' = J$$ is also a solution. And the gauge fixing implies a different solutions of $$F=dA$$ and $$F=dA'$$ where $$A$$ and $$A'$$ are both valid solutions. What are the differential equation constraints then? Is my understanding complete to include ONLY: $$* d * d (A-A')=0$$ or do we need more constraints to do gauge fixing?

Am I correct to say that $$A$$ and $$A'$$ are in the same gauge profile thus should be regarded as the same gauge equivalent classes. Choose $$A$$ or choose $$A'$$ is simply a choice of gauge fixing?

• A gauge transformation is any transformation on $$A = (\phi, \vec{A})$$ that does not change any physical observable, c.q. $$F$$. So any transformation $$A \to A + d \lambda$$ with $$\lambda$$ any scalar field.
• A gauge fix is, within this gauge freedom, a particular choice for $$A$$. This choice can be either complete ($$A$$ completely fixed, no further gauge transformations allowed) or not (there is still some leftover freedom).

A gauge fix is usually applied to facilitate certain computations (you have to then make sure the end result is nonetheless gauge-invariant). Therefore, one is not actually interested in the exact particular form of $$A$$, but rather in certain gauge-fixing conditions that $$A$$ satisfies. Given any gauge potential $$A$$, one can apply a gauge transformation $$A \to A_g = A + d \lambda$$ so that $$A_g$$ satisfies the gauge-fixing condition.

Some examples are

• In the Lorenz gauge fix, the condition $$d \star A_g = (1/c^2)\partial_t \phi_g + \nabla \cdot \vec{A}_g = 0$$ is satisfied.
• In the Coulomb gauge fix, the condition $$\nabla \cdot \vec{A}_g = 0$$ is satisfied.

The Lorenz gauge fix is not complete, because instead of using $$\lambda$$ to arrive at the gauge fix, we could have used any $$\lambda + \psi$$ as long as $$\psi$$ satisfies $$\Box \psi= 0$$. In other words, instead of $$A_g$$ we can have $$A_g + d \psi$$ and still satisfy the gauge-fixing condition. On the other hand, the Coulomb gauge fix is complete.

In your list, the transformation $$C$$ in item 1. is not entirely free to choose. Instead, it must be of the form $$C = d \lambda$$ for any smooth scalar field $$\lambda$$. Then, items 1. and 2. are actually the same. I would use a different wording than what you have written down: $$C$$ is a gauge transformation. A particular transformation $$C$$ to arrive at $$A$$ satisfying a gauge-fixing condition can be called a gauge-fix.

I don't understand your question item 3. If $$F = d A$$ and $$F = d A'$$ then $$A$$ and $$A'$$ differ by some gauge transformation. A gauge fix is some condition on $$A$$, but it cannot affect any physical observable, in particular not $$F$$. So $$d (A - A') = 0$$ is always satisfied. Moreover, physical observables like $$F$$ and $$J$$ can tell you nothing about the gauge fix. Instead, a gauge-fixing condition (which is a choice) can be for instance $$d \star A = 0$$ (Lorenz gauge).

• Edit in my post: "so $F=dA$ with $* d * dA = J$ is a solution. Say $F=dA'$ with $* d * dA' = J$ is also a solution." Apr 15 at 3:29
• I still don't understand the question, but I've tried to emphasize that gauge transformations do not and cannot (by definition) affect any observable properties. If $\star d \star d(A - A') =0$ but $d(A-A') \neq 0$ then $A$ and $A'$ are not gauge-equivalent. Apr 15 at 3:54