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I was looking at this question on Mathematics S.E, as I would like to know the origin of the signs in the gauge transformations of the scalar and vector potentials components, $\phi$ and $\vec A$, of the four-vector potential $A^\mu=\left(\phi, \vec A\right)$: $$\phi\to \phi-\frac{\partial \lambda}{\partial t},\quad \vec A\to \vec A+\nabla\lambda\tag{A}$$ where $\lambda$ is a gauge parameter; an arbitrary function of spacetime.


For the purposes of this question I will follow all the same notation and conventions as the linked question, including the Minkowski sign convention, $\eta= \mathrm{diag}(1,-1,-1,-1)$. I know that this question may seem better placed on Mathematics S.E, but I would like to see a physicists take on this. Moreover, I'm not equipped to understand the answers given by the mathematicians.


The OP in the linked question wanted to know why the transformations in $(\mathrm{A})$ can be written as one compact equation, $$A_\mu\to A_\mu+\partial_\mu\lambda\tag{B}$$

The answer given is way beyond my comprehension, the user that gave the answer was talking about "musical isomorphisms", not something I can relate to.


So instead of starting with the transformations in $(\mathrm{A})$ and trying to understand how they combine compactly into $(\mathrm{B})$, I will try the reverse approach.

Given $A^\mu=\left(\phi, \vec A\right)$ then the covariant version is $A_\mu=\left(\phi, -\vec A\right)$, (this was pointed out in an answer to another question asked by the same OP here).

Now, writing out the components $$A_\mu\to\left(A_0+\partial_0 \lambda, -A_1-\partial_1 \lambda, -A_2-\partial_2 \lambda, -A_3-\partial_3 \lambda\right)$$ $$=\left(\phi+\frac{\partial \lambda}{\partial t}, -\vec A - \nabla \lambda\right)\tag{C}$$ So according to $(\mathrm{C})$ the gauge transformation components of $A_\mu$ are $$\phi\to \phi+\frac{\partial \lambda}{\partial t},\quad \vec A\to -\vec A-\nabla\lambda\tag{D}$$

But these signs in $(\mathrm{D})$ are totally different to that of $(\mathrm{A})$.

Does anyone know how the scalar and vector gauge transformations in $(\mathrm{A})$ can be deduced from $(\mathrm{B})$?

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We already know the 4-position is $x^\mu = (t, \vec{x})$ and the 4-gradient is $$\partial_\mu = \frac{\partial}{\partial x^\mu} = \left(\frac{\partial}{\partial t},\vec{\nabla} \right) \tag{1a}$$ or equivalently $$\partial^\mu = \left(\frac{\partial}{\partial t},-\vec{\nabla} \right) \tag{1b}$$

Starting with the relativistic gauge transformation $$A_\mu \to A_\mu + \partial_\mu \lambda \tag{B}$$ is conceptually problematic, because we do not know yet what is the correct 4-potential $A_\mu$. It is not obvious if the correct choice must be $A_\mu=(\phi,\vec{A})$ or $A_\mu=(\phi,-\vec{A})$. So we need to do it the other way round and start with the classical gauge transformation $$\phi\to \phi-\frac{\partial \lambda}{\partial t}, \quad \vec A\to \vec A+\vec{\nabla}\lambda\tag{A}$$

We can write (A) as $$(\phi,\vec{A}) \to (\phi,\vec{A})- \left(\frac{\partial \lambda}{\partial t},-\vec{\nabla}\lambda \right)$$

Here we see, that because of the $-\vec{\nabla}$ we can use $\partial^\mu$ from (1b), but not $\partial_\mu$ from (1a). We get $$(\phi,\vec{A}) \to (\phi,\vec{A})-\partial^\mu\lambda$$ To make this a valid 4-vector equation (with all upper $\mu$), we need to define $A^\mu=(\phi,\vec{A})$ and get $$A^\mu \to A^\mu-\partial^\mu\lambda$$ or equivalently with $A_\mu=(\phi,-\vec{A})$ $$A_\mu \to A_\mu-\partial_\mu\lambda$$

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Your result is almost correct, you just have two sign errors. Lets start with your transformation (B) and be careful with the signs. The four derivative is defined in such a way, that they have a positive sign if they have a lower index: \begin{equation*} \partial_\mu = \left(\frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \end{equation*} Therefore the right hand side of your equation (C) should be \begin{equation*} \left(\phi + \frac{\partial\lambda}{\partial t}, -\vec A + \nabla\lambda\right) \end{equation*} Your second mistake is that the left hand side of the same equation also has a lowered index, and therefore is \begin{equation*} \left(\phi, -\vec A \right) \end{equation*} The four-potential therefore should transform as \begin{equation*} \phi \rightarrow \phi + \frac{\partial\lambda}{\partial t},\quad \vec A \rightarrow \vec A - \nabla\lambda \end{equation*} This is exactly the opposite of your equation (A). To recover it you either have to apply the transformation with $-\lambda$ instead of $\lambda$ or use \begin{equation*} A_\mu \rightarrow A_\mu - \partial_\mu \lambda \end{equation*} as transformation law.

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This sign comes from the fact that ($g^{00}=+1$ assumed) $$\partial^\mu = \left ( \frac{d~}{dt}, -{\vec \nabla} \right ) \,.$$

You can avoid such issues by using covariant notation throughout.

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The perhaps underwhelming answer is, it doesn't matter, but you need to stick with the convention that you choose (when raising and lowering indices.) To see why, consider the following thought experiment: let $\eta = \lambda(-t, x)$. Then $A^\mu \mapsto A'^\mu := (\phi + \partial_t\eta, \vec A + \vec \nabla \eta )$.

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    $\begingroup$ That sure is underwhelming, if it really doesn't matter then why is it that every single source I've seen quotes these gauge transformations as $\phi\to \phi-\partial_t \lambda$ and $\vec A\to \vec A+\nabla\lambda$? I'm yet to find even one book/webpage that writes the scalar gauge transformation as $\phi\to \phi+\partial_t \lambda$ (and still with $\vec A\to \vec A+\nabla\lambda$, so that both gauge transformations have a positive sign). It's almost like there is some hidden meaning/reasoning in the minus sign for the scalar gauge transformation. $\endgroup$
    – Skynet
    Feb 17 at 3:12
  • $\begingroup$ So, it's slightly more transparent to observe the gauge symmetry $A_\mu\rightarrow A_\mu' = A_\mu + \partial_\mu \lambda$ from the definition of $F_{\mu\nu} \propto \partial_\mu A_\nu - \partial_\nu A_\mu$, and this would require working with $\partial^\mu\lambda$ (rather than the more 'natural' $\partial_\mu\lambda$) when evaluating $A^\mu$ from a vector source $j^\mu$. I would guess that this explains the minus sign (although it's worth mentioning that the + - - - convention is used in Landau vol.. 2, and probably other references that emphasize the role of proper time.) $\endgroup$
    – TLDR
    Feb 18 at 23:28

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