# "One-parameter" gauge transformation

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?

"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 $$\mathfrak{g} = \mathfrak{u}(1) = \mathbb{R}$$.

• note that your "two-parameter" example is still a one-parameter transformation (with parameter $\chi_1+\chi_2$). Even if we had a gauge field that transformed like that, and even if we lived in two dimensions, you could still not set $A_\mu=0$ using that transformation. Dec 30, 2021 at 2:50
• @AccidentalFourierTransform You're right, of course. I've edited the answer, though I'm not sure it now still makes the point so succinctly as I intended it to... Dec 30, 2021 at 2:59
• Yes, I understand why you wanted to use that example. The simplest alternative I could come up with is, in 2d, to use $\delta A_\mu=\partial_\mu \chi_1+\epsilon_{\mu\nu}\partial^\nu\chi_2$ (or in higher dimensions, $\delta A=\mathrm d\chi_1+\star\mathrm d\chi_2$, although again it is unclear whether this will help OP or not) Dec 30, 2021 at 17:20

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 $$\chi$$.

• It is my understanding that OP was already aware of this argument (if possibly not formulated as precisely), and the actual question is, what is meant by "one-parameter". So I am not sure if this actually addresses the confusion in the OP. I might be wrong, of course. Dec 30, 2021 at 17:22