I How to use geometrical symmetries -- general advice? (Vector potential)

[LeoJakob]

The following is an example from my script. I always have trouble identifying useful symmetries. Can someone explain to me why (for example) the vector potential doesn't have a $z$ dependence? I understand that there is no $\varphi$ dependency.
I don't understand why the field of $\vec{A}$ has to be parallel to the $z$ axis. What about $d^{3} \overrightarrow{r^{\prime}}$??? Is there a way to show mathematically that the vector potential is independent of the two variables $z\varphi$?
In general, I have problems identifying symmetries and using them correctly.

Magnetic flux density of an infinitely long hollow cylinder

The hollow cylinder is homogeneously traversed by the current $I$. Calculate the magnetic flux $\vec{B}(\vec{r})$ as the curl of the vector potential $\vec{A}(\vec{r})$.

$$
\begin{aligned}
\vec{A}(\vec{r}) & =\frac{\mu_{0}}{4 \pi} \int \limits_{V} \frac{\vec{j}\left(\overrightarrow{r^{\prime}}\right)}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} d^{3} \overrightarrow{r^{\prime}} \\
\vec{j}(\vec{r}) & =j \vec{e}_{z}
\end{aligned}
$$

Use cylindrical coordinates $\vec{r}=(\rho, \varphi, z)$. Due to the symmetry, we have:
$$
\vec{A}(\vec{r})=A(\rho, \varphi, z) \vec{e}_{z}=A(\rho) \vec{e}_{z}
$$

 

[Ibix]

If you have an infinitely long cylindrical object then the field can't depend on either $z$ or $\varphi$ because the charge and current densities are symmetrical under rotation around the $z$ axis and translation along it. If the sources are symmetric like that, how can the fields be otherwise? If they were (e.g.) weaker at some $z=z_0$, why there?

So the $\vec B$ field is cylindrically symmetric. What ways can a vector field be cylindrically symmetric? What does that tell you about $\vec A$?

 

[Orodruin]

There is no way to generally prove that the vector potential does not depend on particular coordinates since it is always possible to perform a gauge transformation $\vec A \to \vec A + \nabla \phi$ with the same resulting field for any scalar function $\phi$.

Any symmetry argument must be based on this particular expression for the vector potential.