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?


2 Answers 2


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

  • $\begingroup$ 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. $\endgroup$ Dec 30, 2021 at 2:50
  • $\begingroup$ @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... $\endgroup$
    – ACuriousMind
    Dec 30, 2021 at 2:59
  • $\begingroup$ 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) $\endgroup$ 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 $\chi$.

  • $\begingroup$ 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. $\endgroup$ Dec 30, 2021 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.