How does Spherical Symmetry constraint the Components of the Vector Potential I am a bit confused as to what form does the 4 vector potential and Electromagnetic field tensor $F_{\mu \nu}$ take in the presence of magnetic charges. I got confused in this while looking at the Reissner Nordstrom black holes. Hobson etc ( they take only the electric charge) state that due to spherical symmetry the vector potential should be :
\begin{equation}
A=(\phi(r), a(r), 0, 0)
\end{equation}
and thus
\begin{equation}
F_{01}= -F_{10}= E(r)
\end{equation}
which seems perfectly sensible because under spherical symmetry a vector shouldn’t have $\theta$ and $\phi$ components.
But then Carrol who does consider magnetic charges as well states that non zero components of the electromagnetic field strength tensor are
\begin{equation}
F_{01}= -F_{10}= E(r) \\
F_{23}=-F_{32} =B(r)
\end{equation}
That makes me wonder what will be the the components of the 4 vector potential $A^\mu$ corresponding the Electromagnetic field tensor Carrol obtains.
Particularly, since $F_{23}\neq 0 \implies A^2 \neq 0$. That is confusing me. I don’t see what the form of the 4 vector potential $A^\mu$ will be when we consider magnetic charges and spherical symmetry.
Edit-
The answer here addresses -

*

*Why can the 4 vector potential have a $\theta$ and $\phi$ component even in spherical symmetry ( in detail)


*Though the question of actually deriving which components of a vector or tensor vanish in spherical symmetry is not discussed mathematically.
 A: Alright, so there are several things going on here which are each fairly interesting, so let me try and touch on them one by one. First of all, let me comment on the implications of symmetry in general and spherical symmetry in particular.
The usual examples we think about go something like this, using cylindrical symmetry as an example because it's simpler: suppose we have some (scalar!) function $\phi(z,r,\theta)$ which is supposed to have cylindrical symmetry. Cylindrical symmetry is defined by $z\mapsto z+\alpha$ and $\theta\mapsto\theta+\alpha$ for arbitrary $\alpha$. In particular, symmetry of $\phi$ then implies
$$
\phi(z+\alpha,r,\theta+\beta)=\phi(z,r,\theta).
$$
By taking $\alpha$ and $\beta$ derivatives of this expression, we are able to find the conditions
$$
\partial_z\phi=\partial_\theta\phi=0,
$$
which is really what we would expect from cylindrical symmetry.
If we were to think about some vector $\boldsymbol{v}(z,r,\theta)$ instead, it would be very tempting to conclude that it cannot have any $\theta$ component to it in addition to the statements above about $z$ and $\theta$ dependence. However, this is a bit too quick because the components of $\boldsymbol{v}$ will also transform. For example, when we apply a rotation, we not only need to write $\theta\mapsto\theta+\alpha$, but we also need to rotate the basis vectors of $\boldsymbol{v}$. This can make things much more complicated in the case of vectors, and may mess with what we would "intuitively" think symmetry should imply. Yet we are familiar with the magnetic field about an infinitely long wire. This field has cylindrical symmetry, yet it clearly also has a component in the $\theta$-direction. The point is, we always need to be careful about exactly what symmetry is telling us.
In the case of spherical symmetry, things are even more complicated than in the case of cylindrical symmetry. The key thing that makes spherical symmetry harder is that there is no coordinate system which is adapted to spherical symmetry. What I mean by this can be seen if we look back at cylindrical symmetry again. Cylindrical coordinates are "adapted" to cylindrical symmetry in the sense in these coordinates the transformations which define cylindrical symmetry (translations in $z$ and rotations in the $xy$-plane) can be written as translations of the coordinates ($\theta\mapsto\theta+\alpha$).
Once we move to spherical symmetry this is no longer possible. To see this, consider the "obvious" choice of coordinates that might be adapted to spherical symmetry...spherical polar coordinates. Here there are two angles and a radius, $(r,\varphi,\theta)$. From these, we can translate $\varphi$ and $\theta$ to produce two rotations. That's fine, but it actually takes three rotations to form spherical symmetry...rotations about the $x$, $y$, and $z$ axis for example. There are simply too many different kinds of rotations to be able to find a coordinate system which is simultaneously adapted to all of them.
Things get extremely ugly, but the invariance conditions for spherical symmetry can be written out and indeed do imply that only the radial component of the vector survives and may only depend on the radial coordinate. Since this was not the main point of OP's question though, I will spare the details.
$$
{\ast}\,{\ast}\,{\ast}
$$
So, that's all well and good assuming we have spherical symmetry. But we also have to ask ourselves when we do, in fact, have spherical symmetry and when we don't. In physics there are always two components which are needed to answer this question about whether or not we have a particular symmetry. The first is whether it's a symmetry of the theory, and the second is whether it's a symmetry of the boundary conditions and external charge distributions. In this case we have external charge distributions rather than boundary conditions (though magnetic monopoles are best understood as enforced boundary conditions, which will become important).
Maxwell's equations invariant under the entire Poincare group, which in particular includes rotations. In fact, this is why the angular momentum of the electromagnetic field is a conserved quantity (by Noether's theorem). So that's one of the two boxes checked.
What about the symmetry of the charge distribution? In the case of an electric point charge sitting at the origin, we would indeed have spherical symmetry...essentially the coupling of the vector potential to a point charge looks like $qA_t$ (see the Lagrangian for the Lorentz force) for a point charge fixed to sit at the origin. This involves no spatial components of $A_\mu$, and hence is manifestly invariant under rotations.
The tricky thing about the vector potential, however, is that it's not gauge invariant. So while $A_\mu$ for a fixed electric charge might be a solution to Maxwell's equations and the vector potential we obtain after rotating will necessarily also be a solution to Maxwell's equations, they could, in principle, differ by a gauge transformation. If that were the case, we would not be able to find something akin to the $\phi(z+\alpha)=\phi(z)$ mentioned above, but instead would find something like $\phi(z+\alpha)=\phi(z)+(something)$ where the $+(something)$ would, in the case of the vector potential, be a gauge transformation. This kills our ability to draw the same conclusions from the symmetry of the system.
Now, we of course know that the vector potential for a single electric point charge can be rotated without needing to throw in a gauge transformation, so we can demand it satisfy spherical symmetry and go from there. We are able to rotate the vector potential in this case without adding a gauge transformation at the same time because the vector potential for a single electric point charge is well-defined everywhere in space. This is not the case for the magnetic monopole.
So let's think about monopoles. There are always sources that insist monopoles should be thought of as adding a magnetic charge distribution on the RHS of the magnetic Gauss' law, $\nabla\cdot\boldsymbol B=\rho_m$. However, this is nonsense because doing so would break gauge symmetry, which is a big no-no. The correct way to understand monopoles is as boundary conditions and points deleted from space.
That sounds a little strange, so let's think about why for a moment. I'm going to use differential forms here because it makes everything much simpler to say. Gauss' law essentially tells us that $d F=0$, that is, the field strength is a closed 2-form. This is the entire reason we are able to conclude that there exists a potential $A$ such that $F=d A$. But the Poincare lemma only guarantees that $A$ exists locally. This, I will note, is extremely similar to the statement that $\nabla\cdot\boldsymbol B=0$ implies there exists an $\boldsymbol A$ such that $\boldsymbol B=\nabla\times\boldsymbol A$.
The importance of this is that the magnetic charge enclosed within a surface $\Sigma$ is defined to be $q_m=\int_\Sigma F$. Normally, if $\Sigma$ is closed, then the fact that $F=d A$ would imply, by Stokes' theorem, that $q_m=0$. If $A$ exists only locally, however, then essentially we have to define $A$ patch by patch, and on the overlap regions in spacetime, we demand that the $A$'s differ only by a gauge transformation. This is what allows $q_m$ to be non-zero, and also shows that whenever $q_m$ is non-zero, so we have a monopole somewhere, we necessarily have to work in this patch-by-patch manner with $A$. In terms of fancy language, we could say that monopoles are predicated upon space having a nontrivial deRham cohomology (there exists a closed but not exact form).
The reason I mention all this is because it actually matters for our symmetry considerations. In the case of a single monopole (see, for example, David Tong's lecture notes), we can get away with working on two patches. Usually these are taken to be the northern and southern halves of space, so the overlap of the two patches will be in, say, the $xy$-plane. The $A$ we define patch-by-patch will be different on the north and south, but on the $xy$-plane, they only differ by a gauge transformation.
Now imagine we apply a rotation to this. If the rotation maps the $xy$-plane into itself, there are no problems. However, if the rotation were to move the $xy$-plane, we would be changing the location of the overlap region where the $A$'s need to match. This is not a matter of mapping the solution exactly to itself, and so part of our spherical symmetry is broken (but as noted, we still keep some things, namely those which would map the $xy$-plane to itself). Hence we cannot use all the conclusions that full spherical symmetry would normally grant us.
Another way of seeing that we need to use the patch-by-patch procedure is to, by direct calculation, note that the vector potential which would produce a Coulomb magnetic field (which is what a single monopole would produce) has a singularity along half the $z$-axis. So we define two such vector potentials with the above mentioned overlap conditions so in their respective patches, the vector potentials have no singularities. This is usually referred to as the Dirac string.
The Dirac string can be thought of as a boundary condition for the vector potential in that, along half the z-axis, the vector potential is required to have a singularity. If you're willing to think of it as such, then this is a boundary condition that will clearly kill part of our spherical symmetry.
If we only allow ourselves any fewer than the three rotations in full spherical symmetry, it's no longer enough to conclude that only the radial component is allowed or even that the vector potential need only depend on the radial coordinate. I'm fairly certain this is the reasoning behind your observation that, indeed, Carroll allows for a $\theta$ component to the vector potential.
I realize this was a rather long explanation, but hopefully you will find it more satisfying than simply pointing out that, if the vector potential was of the form $\langle V(r),a(r),0,0\rangle$, then we could not have a magnetic field and hence could not have a monopole, so the vector potential clearly cannot take the form dictated by full spherical symmetry.
Edit: There is one last thought I realize I never actually wrote down here. If we were to apply a rotation which moves the $z$-axis then, as mentioned, we would be moving the boundary between the two patches and we would not be mapping the solution back to itself. This would also be equivalent to moving the Dirac string. However, because the gauge potentials on both patches are equal up to gauge transformations, the vector potential maps into itself up to a gauge transformation. This is an explicit realization of the point I made about the possibility of having $\phi(z+\alpha)=\phi(z)+(something)$ where the something is the result of a gauge transformation.
The usual way to put this observation into words is to say that the Dirac string is a gauge artefact: it can be moved around, essentially at will, by gauge transformations. So, if we were to simultaneously apply both a rotation and an appropriate gauge transformation, we could move the string back to the $z$-axis and would have found a transformation under which the vector potential really is invariant, it just wouldn't be a pure rotation by itself.
It's also worth pointing out that this is why the magnetic field, which is gauge invariant, is correctly spherically symmetric (it's a Coulomb field, after all). Essentially, if we were to apply the rotation, the vector potential goes back to itself plus a possible gauge transformation. But the magnetic field is blind to this gauge transformation, so it just returns to itself under the rotation.
