"One-parameter" gauge transformation

Question

In my advanced classical physics course, it was stated that the electromagnetic field strength tensor $F_{\mu\nu} = \partial_{\nu}A_{\mu} - \partial_{\mu}A_{\nu}$ is invariant under "one-parameter" gauge transformations of the 4-vector potential: $A_{\mu} \to A_{\mu} + \partial_{\mu}\chi$. In a problem set, I was asked to show that there is no gauge transformation which imposes $A_{\mu} =0$.

My logic was the following: If there was such a transformation, this would require $\partial_{\mu}\chi = -A_{\mu}$, then we would have $F_{\mu\nu} = 0$, which is impossible in general (unless $F_{\mu \nu} = 0$ in EVERY gauge).

However, in the solution, it is stated that there is no such solution "because $\chi$ only carries one parameter". I understand that the condition would require 4 independent equations to be satisfied, but I don't understand immediately why is this not possible. I don't understand what is meant by a "one-parameter" transformation and I haven't found any useful information about this.

I would like to ask for clarification. In particular, I would like to understand what is meant by a "one-parameter" gauge transformation, and whether the given solution is consistent with/equivalent to my logic above.

As an extra, I would appreciate any short comments on whether (how) this "one-parameter" gauge transformation is related to one-parameter Lie groups?

Answer 1

"One-parameter" in this context just means that there is a single $\mathbb{R}$-valued function $\chi$ that parametrizes the gauge transformation - as opposed to e.g. the "two-parameter" transformation for some $\mathbb{R}^2$-valued field $A_\mu = \begin{pmatrix}A^1_\mu \\ A^2_\mu\end{pmatrix}$ $$\begin{pmatrix}A^1_\mu \\ A^2_\mu\end{pmatrix} \mapsto \begin{pmatrix}A^1_\mu \\ A^2_\mu\end{pmatrix} + \partial_\mu \begin{pmatrix}\chi^1\\ 0\end{pmatrix} + \partial_\mu \begin{pmatrix}0 \\ \chi^2\end{pmatrix}$$ for independent functions $\chi^1,\chi^2$ we could imagine instead.

You've just presented a more explicit argument for why we cannot achieve $A=0$ for arbitrary $A$ with just one $\chi$, i.e. your logic is perfectly correct, that's exactly how one shows that a one-parameter transformation isn't enough.

One might contrast this with the case of a general non-Abelian gauge theory where the gauge field $A$ takes values in a Lie algebra $\mathfrak{g}$ and the gauge transformations are parametrized by a $\mathfrak{g}$-valued function $\chi$ as $$ A_\mu \mapsto A_\mu + \partial_\mu \chi + [\chi, A]$$ where $\chi$ is now arguably a "$\mathrm{dim}(\mathfrak{g})$-parameter gauge transformation" (which is still not enough to zero $A$ since $A$ is also $\mathfrak{g}$-valued). The case of electromagnetism is simply $\mathfrak{g} = \mathfrak{u}(1) = \mathbb{R}$.

Answer 2

The answer to this question obviously depends on $F_{\mu\nu}$, so it is not down to just a counting of degrees of freedom. Let us show that if there is such $\chi$ then $F_{\mu\nu}=0$.

Suppose there is $\chi$ obeying $A_\mu+\partial_\mu\chi=0$. Then it obeys $\partial_\nu \chi =-A_\nu$. Taking a derivative this means that it further obeys $\partial_\mu \partial_\nu \chi = -\partial_\mu A_\nu$. Now we anti-symmetrize, we will have $$\partial_{[\mu}\partial_{\nu]}\chi=-\partial_{[\mu}A_{\nu]}\tag{1}.$$

The left-hand side is zero because $\partial_\mu\partial_\nu\chi$ is symmetric under $\mu\leftrightarrow \nu$. The right-hand side is $-2F_{\mu\nu}$. Therefore if such $\chi$ exists we must have $F_{\mu\nu}=0$.

Since the existence of $\chi$ implies $F_{\mu\nu}=0$, by contraposition, $F_{\mu\nu}\neq 0$ implies there is no such $\chi$.