What is the $\,\phi=0\,$ gauge called?

Question

In electromagnetism textbooks, the gauges most often talked about are the Lorenz gauge and Coulomb gauge. Sometimes it's convenient to work in a gauge in which there is only the vector potential $\vec{A}$ but no scalar potential $\phi$. The following gauge transformation transforms a general pair of potentials $(\vec{A},\phi)$ into $(\vec{A}',0)$, such that

$$\vec{A}'=\vec{A}+\int_0^t\nabla\phi\,dt,\quad \phi'=0.$$

Then one could work with only the vector potential $\vec{A}'$ to produce both the electric field

$$\vec{E}=-\frac{\partial\vec{A}'}{\partial t}=-\frac{\partial\vec{A}}{\partial t}-\nabla\phi,$$

and the magnetic field

$$\vec{B}=\nabla\times\vec{A}'=\nabla\times\vec{A}.$$

The above procedure seems to work generally without assuming there being no electric charge (which would produce the retarded scalar potential in Lorenz gauge). Is there a name for this $\,\phi=0\,$ gauge?

Answer 1

The gauge $\phi = A_0 = 0$ is called Weyl gauge or temporal gauge.

This gauge is incomplete, as one can see from the definition of a gauge transformation, $$A_\mu \to A_\mu + \partial_\mu \alpha(x).$$ We can still perform any gauge transformation with gauge parameter $\alpha$ independent of $t$, as this keeps $A_0$ the same. To remove some of the residual gauge freedom we could, e.g. impose $$A_z|_{t = 0} = 0.$$ The proof this gauge can be reached is just the exact same as the proof that Weyl gauge can be reached, except with effectively one less dimension since nothing depends on $t$. At this point we are still not done, because we can still preserve both gauge conditions using any $\alpha$ independent of both $t$ and $z$. So we impose the further condition $$A_y|_{t = z = 0} = 0$$ leaving only $\alpha$ dependent on $x$, which are removed by imposing $$A_x|_{t = z = y = 0} = 0.$$ These four conditions together are a complete gauge fixing. It's quite a mouthful, which is why you won't see it written out in textbooks too often.

Whether or not you want to perform a complete gauge fixing is up to taste. For example, in the standard presentation of the QCD $\theta$-vacua, one takes the incomplete gauge fixing $A_0 = 0$ and then argues there are multiple vacua $|n \rangle$. But there is a completely equivalent presentation where one takes the complete gauge fixing I gave above (also mentioned here), and finds a unique vacuum but the exact same physical effects. This is related to whether one chooses to regard large gauge transformations as "do-nothing" transformations.

Answer 2

Your specified gauge is actually incomplete. In general, though, any gauge with $\phi = 0$ is a Weyl gauge.