I have three questions regarding the following extract(s), I have marked red the parts for which I do not understand for later reference. The convention followed for the Minkowski metric in these notes is $\eta_{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$:
The most general Lagrangian which is quadratic in vector field $A_\mu$ and its derivatives is $$\mathcal L=aS^2+bF_{\mu\nu}F^{\mu\nu}+cG_{\mu\nu}G^{\mu\nu}+d A_\mu A^\mu\tag{1}$$ where $\mathrm{a}$, $\mathrm{b}$, $\mathrm{c}$ and $\mathrm{d}$ are arbitrary constants.
- $S$ is a Lorentz scalar,
- $F_{\mu\nu}$ is an antisymmetric rank $2$ tensor, which means that it satisfies $F_{\mu\nu}=-F_{\nu\mu}$,
- and $G_{\mu\nu}$ is a symmetric and traceless rank 2 tensor, which means that it satisfies $G_{\mu\nu}=G_{\nu\mu}$ and $\eta^{\mu\nu}G_{\mu\nu}=0$.
These can be written explicitly as $$S=\partial_\mu A^\mu,\quad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu,\quad G_{\mu\nu}=\partial_\mu A_\nu+\partial_\nu A_\mu-\frac12\eta_{\mu\nu}S\tag{2}$$
A concrete example of a vector field theory is electrodynamics written in terms of the four-vector potential $A^\mu=(\phi, \vec A)$, $\phi$ is the scalar potential and $\vec A$ is the vector potential. Electric and magnetic fields are given by $$\vec E=-\frac{\partial \vec A}{\partial t}-\nabla \phi,\quad\vec B=\nabla\times \vec A\tag{3}$$ Alternatively, the dynamics can be described in a Lorentz covariant way using the field strength tensor, $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$, which coincides with the definition of the antisymmetric tensor. $F_{\mu\nu}$ above.
$\quad$ As we know from electrodynamics, there are many different configurations of the potentials that correspond to the same electric and magnetic fields, in other words the same physical configuration. More precisely, $\vec E$ and $\vec B$ are invariant under gauge transformations $$\phi\to \phi-\frac{\partial \lambda}{\partial t},\quad \vec A\to \vec A+\nabla\lambda\tag{4}$$ where $\lambda$ is an arbitrary scalar function of spacetime. In terms of the four-vector potential, $\color{red}{\text{this gauge invariance can be written compactly as}}$ $$A_\mu\to A_\mu\stackrel{\color{red}{?}}{-}\partial_\mu\lambda\tag{5}$$ The different terms in the general quadratic Lagrangian $(1)$, transform under such a gauge transformation as $$S=\partial_\mu A^\mu\to \color{red}{\partial_\mu A^\mu + \partial^2\lambda}=S+\partial^2\lambda\tag{a},$$ $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu\to \color{red}{\partial_\mu A_\nu+\partial_\mu\partial_\nu \lambda-\partial_\nu A_\mu-\partial_\nu\partial_\mu \lambda}$$ $$=\partial_\mu A_\nu-\partial_\nu A_\mu=F_{\mu\nu}\tag{b},$$ $$G_{\mu\nu}=\partial_\mu A_\nu+\partial_\nu A_\mu-\frac12\eta_{\mu\nu}S\to \color{red}{\partial_\mu A_\nu + \partial_\mu\partial_\nu\lambda+\partial_\nu A_\mu+\partial_\nu\partial_\mu\lambda-\frac12\eta_{\mu\nu}S-\frac12\eta_{\mu\nu}\partial^2 \lambda}$$ $$=G_{\mu\nu}+2\partial_\mu\partial_\nu \lambda-\frac12\eta_{\mu\nu}\partial^2 \lambda\tag{c}$$
My questions are as follows
- How were the gauge transformations in $(4)$ "written compactly" as one four vector potential gauge transformation in $(5)$ (assuming either sign of $(5)$ for now)?
- I think the four vector gauge, $(5)$ should be written as $A_\mu\to A_\mu\color{blue}{+}\partial_\mu\lambda$, ie. there is a sign error. I think this because if the minus sign is assumed true in $(5)$ then the parts marked red in eqns. $(\mathrm{a})-(\mathrm{c})$ is all incorrect. Taking eqn. $(\mathrm{b})$ for instance, by my logic this should be $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu\to \partial_\mu\left( A_\nu+\partial_\nu\lambda\right)-\partial_\nu \left(A_\mu+\partial_\mu\lambda\right)$$ $$=\partial_\mu A_\nu+\partial_\mu\partial_\nu \lambda-\partial_\nu A_\mu-\partial_\nu\partial_\mu \lambda=\partial_\mu A_\nu-\partial_\nu A_\mu=F_{\mu\nu}$$
- There is another problem however, two pages later, the notes go on to discuss Local Symmetry where it is stated that the following is a
position-dependent phase rotation $$\phi(x)\to e^{i\theta(x)}\phi(x)\tag{6}$$ The derivative transforms as $$\partial_\mu\phi\to e^{i\theta}\partial_\mu\phi+i\left(\partial_{\mu}\theta\right)e^{i\theta}\phi\tag{7}$$ [$\cdots$], let us introduce a vector field that transforms according to eqn. $(5)$, that is $$A_\mu\to A_\mu-\partial_\mu\lambda$$ The combination $$D_\mu\equiv \partial_\mu\phi + ieA_\mu\phi\tag{8}$$ where $e$ is an arbitrary constant. Then $(8)$ would then transform as $$D_\mu\phi\to e^{i\theta}\partial_\mu\phi+i\left(\partial_\mu\theta\right)e^{i\theta}\phi+ie A_\mu e^{i\theta}\phi-ie\left(\partial_\nu\lambda\right)e^{i\theta}\phi$$ $$=e^{i\theta}D_\mu\phi+i\left(\partial_\mu\theta-e\partial_\mu\lambda\right)e^{i\theta}\phi\tag{9}$$
$\quad$ and this equation $(9)$ is actually correct (algebraically speaking), so the negative sign in $(5)$ was needed instead of the positive sign. So this leads me to the final question, I have shown an example of where the positive sign is required in the gauge transformation and when the negative sign is required, so does this mean that we actually have $$A_\mu\to A_\mu\color{darkgreen}{\pm}\partial_\mu\lambda,$$ $\quad$ where the sign is case dependent? (I checked other university websites such as this one from Durham university which uses a negative sign and this page from UC San Diego which uses the positive sign).
Closing remarks -
The notes used in the quotations of this post are from ICL physics department. I decided to typeset everything in this post instead of inserting images of the notes (as I think this sort of thing is frowned upon on the S.E websites).
I did my best to make the quotes as concise as possible while maintaining enough relevant information for the reader - despite particular extracts from the notes being several pages apart. For this reason, if more information is required please let me know and I can upload images of the notes.
Update and response to a comment -
Rather clumsily I managed to pick the only gauge transformation out of $(\mathrm{a})-(\mathrm{c})$ that does not depend on the sign of $\partial_\mu \lambda$. To address the question posed in the comment by @KurtG. $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu\to \partial_\mu\left( A_\nu-\partial_\nu\lambda\right)-\partial_\nu \left(A_\mu-\partial_\mu\lambda\right)$$ $$=\partial_\mu A_\nu-\partial_\mu\partial_\nu \lambda-\partial_\nu A_\mu+\partial_\nu\partial_\mu \lambda=\partial_\mu A_\nu-\partial_\nu A_\mu=F_{\mu\nu}$$ which indeed shows that $(\mathrm{b})$ does not depend on the sign of $\partial_\mu\lambda$.
However, for $$S=\partial_\mu A^\mu\to \partial_\mu\left(A^\mu-\partial^\mu\lambda\right)=\partial_\mu A^\mu - \partial^2\lambda=S-\partial^2\lambda\ne (\mathrm{a})$$ Similarly for $$G_{\mu\nu}=\partial_\mu A_\nu+\partial_\nu A_\mu-\frac12\eta_{\mu\nu}S$$ $$\to \partial_\mu\left(A_\nu-\partial_\nu\lambda\right)+\partial_\nu\left(A_\mu-\partial_\mu\lambda\right)-\frac12\eta_{\mu\nu}\left(S-\partial^2\lambda\right)$$ $$=\partial_\mu A_\nu - \partial_\mu\partial_\nu\lambda+\partial_\nu A_\mu-\partial_\nu\partial_\mu\lambda-\frac12\eta_{\mu\nu}S+\frac12\eta_{\mu\nu}\partial^2 \lambda$$ $$=\partial_\mu A_\nu+\partial_\nu A_\mu-\frac12\eta_{\mu\nu}S-2\partial_\nu\partial_\mu\lambda+\frac12\eta_{\mu\nu}\partial^2 \lambda$$ $$=G_{\mu\nu}-2\partial_\nu\partial_\mu\lambda+\frac12\eta_{\mu\nu}\partial^2 \lambda\ne (\mathrm{c})$$
So I've just explicitly demonstrated that the gauge transformation, $(5)$: $A_\mu\to A_\mu-\partial_\mu\lambda$ has the wrong sign before the $\partial_\mu\lambda$ term, and should actually be $A_\mu\to A_\mu+\partial_\mu\lambda$ to get agreement with all three equations, $(\mathrm{a})-(\mathrm{c})$ as presented in the extract from the author.
In summary, why is the author using eqn. $(5)$ instead of $A_\mu\to A_\mu+\partial_\mu\lambda$?
So, gauge transformations have nothing to do with physics, and everything to do with Cartan's geometry of principal bundles and connections on principal bundles. This is why gauge symmetry is always referred to as an unphysical symmetry; it's literally an artifact of working with "vector field theories", as physicists just accidentally used the language of principal bundles and connections on said principal bundle to describe electromagnetism, and other field theories.
So here's my take on why there is a difference in the two conventions, and they really shouldn't matter at all. First a primer in mathematical gauge theory for some one who I assume isn't familiar with it. For simplicity, everything will be done over $\mathbb R^{1,3}$ where principal bundles are trivial.
A principal bundle over $\mathbb R^{1,3}$ is a product manifold $\mathbb R^{1,3}\times G$ where $G$ is any Lie group, which we often take to be a compact. There is a right group action of $G$ on $\mathbb R^{1,3}\times G$ on given by $(x,g)\cdot h=(x,gh)$. We can define vector fields on $\mathbb R^{1,3}$ via this action in the following way: choose an vector $X$ in the Lie algebra of $G$, then define the vector field on $\mathbb{R}^{1,3}\times G$ point wise by: $$\tilde{X}_{(x,g)}=\frac{d}{dt}\Big|_{t=0}(x,g)\cdot \exp(tX)$$ A connection on $\mathbb R^{1,3}$ is a one form $A$ valued in the Lie algebra of $G$ such that: $$A(\tilde{X})=X$$ for all fundamental vector fields, and is compatible with the right action in the sense that: $$R_g^*A=\operatorname{Ad}_{g^{-1}}\circ A$$ These notions really aren't that important to understand, but the crux of it is is that first foremost the connection one form (which will be our gauge potential) is defined on the total space.
A (global) gauge is a smooth map $s:\mathbb{R}^{1,3}\rightarrow \mathbb R^{1,3}\times G$ that satisfies $\pi\circ s=\text{Id}_{\mathbb R^{1,3}}$, where $\pi$ is the projection onto $\mathbb R^{1,3}$. We pull back a connection one form by $s$ to get a one form on $\mathbb R^{1,3}$, and denote by $A_s$, and this is precisely what a gauge field is. If we have two gauges, $s$ and $s'$, then a smooth map $\lambda:\mathbb R^{1,3}\rightarrow G$ relates the two, in the sense that $s'=s\cdot h$, and we have that $A$ transforms as: $$A_{s'}=h^{-1}A_sh+h^{-1}dh$$ If our group is $U(1)$, then sections are the of the form $x\mapsto (x,e^{if(x)})$, and gauge transformations are of the form $x\mapsto e^{i\lambda(x)}$, where $f$ and $\lambda$ are smooth functions. We then see that since $U(1)$ is abelian our transformation law becomes: $$A_{s'}=A_s+d\lambda$$ where we have are conveniently leaving out the $i$'s because we can easily identify the Lie algebra $i\mathbb R$ with $\mathbb R$.
So here's the thing, if you reformulate all of this in terms of left actions on a principal bundle, then signs start flipping everwhere. For example, a connection one form has to satisfy: $$L_g^*A=\operatorname{Ad}_g\circ A$$ notice how now we're conjugating by $g$ instead of $g^{-1}$? That changes our transformation law to be: $$A_{s'}=hA_sh^{-1}-(dh)h^{-1}$$ so in the $U(1)$ case we get: $$A_{s'}=A_s-d\lambda$$ and hence the minus sign.
Basically, the the signs differ because some people prefer the (incorrect in my opinion) convention of left actions over right actions. However, there is $\textbf{no}$ difference mathematically between the two, it is just convention. Indeed, if you have a right action, you can always get a left action by just defining the action to act by inverse multiplication, this would then give us the same exact minus sign.
So to sum up, there is absolutely no physics going on here, it is all just a mathematical convention, and due to the fact that some people prefer left actions over right actions. You have to stay consistent with the convention you use however. If you want to learn more about the mathematics behind gauge theory (which I strongly recommend it is a very a cool field), I recommend Hamilton's Mathematical Gauge Theory.
To answer your two other questions, the transformation is written compactly because it's specifying how each component transforms without writing down four equations. It's even more compact in my opinion to write down things without the use of coordinates as I have done above, but that is largely a mathematicians preference to a physicists one. The notion of "compact" notation is completely suggestive (well maybe not completely).
You may indeed find that there is a typo in that transformation law, however, if you choose one sign convention, and stick with it you will find that the results are consistent up to sign. It doesn't really matter if things are consistent up to sign, as again that is completely convention. You'll see that either gauge transformation leaves the curvature tensor invariant (importantly this only happens in $U(1)$ gauge theories! If your gauge theory is non-abelian then your curvature tensor transforms under the adjoint representation of the group on it's Lie algebra). For the other terms, you'll get things up to a sign.
Edit to answer the follow up:
So first, if we have a vector $(A_t,A_x,A_y,A_z)$, we could write a transformation as: $$(A_t,A_x,A_y,A_z)\mapsto (A_t+\partial_t\lambda,A_x+\partial_x\lambda,A_y+\partial_y\lambda, A_z+\partial_z\lambda)$$ or we could just say where components go: $$A_\mu\mapsto A_\mu \partial_\mu \lambda$$
For you other question, we write a vector four potential as $A=A^\mu\partial_\mu$,
$$A=A^\mu\partial_\mu=(V,M_x,M_y,M_z)$$, then under the musical isomorphism induced by the metric with signature $(-,+,+,+)$, we can write this as the one form: $$A=Vdt-M_xdx-M_ydy-M_zdz$$ Then under a gauge transformation this becomes: $$(V +\partial_t\lambda)dt-(M_x-\partial_x\lambda)dx-(M_y-\partial_y\lambda)dy-(M_x-\partial_x\lambda)dx$$ so using the musical isomorphism again, we get: $$A'=(V+\partial_t\lambda, \mathbf{M}-\nabla \lambda)$$ These have the opposite sign as you want, but this again just a difference in convention. Here I am implicitly using right group actions, and a metric of signature $(1,-1,-1,-1)$, if you use left group actions, and this signature, you'll get the result you want. If you use right group actions, and the metric of signature $(-1,1,1,1)$, you'll get the transformation you want as well. It is all just a matter of what you want to you use, and staying consistent with it.