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We have an infinite cylinder of radius $a$ around the $z$ axis.
The current density inside the cylinder is uniform: $\vec J=J_0\hat z$.
We wish to find the magnetic vector potential $\vec A$ (though I am not asking for the whole solution, just the specific step described below).

In cylindrical coordinates, $\vec B \parallel \hat \theta$, and therefore $(\vec\nabla\times\vec A)\parallel \hat \theta$.

My course's notes say that from this and from the symmetry of the problem, we can deduce that $\vec A\parallel \hat z$.
How exactly can we deduce that?

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  • $\begingroup$ On a less sophisticated level can one say that the $\hat \theta$ component of $\vec \nabla \times \vec A$ is $\frac{\partial A_{\rm r}}{\partial z}-\frac{\partial A_{\rm z}}{\partial r} $ with the first term being zero leaving only a $\hat z$ component? $\endgroup$ – Farcher Jan 10 at 17:14
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In general, because of gauge indeterminacy, a single potential vector cannot be derived from the magnetic field. If $\overrightarrow{A}$ is appropriate, then $\overrightarrow{A}+\overrightarrow{\nabla }\varphi $ with "any" $\varphi $ is also suitable.

If we use the Coulomb gauge $\overrightarrow{\nabla }\centerdot \overrightarrow{A}=0$, then $\overrightarrow{A}(M)=\frac{{{\mu }_{0}}}{4\pi }\iiint{\frac{\overrightarrow{j}(P)}{PM}d\tau }$

It is a polar vector, which has the same properties of symmetry as the current density (Here, the current density is not zero at infinity and some problems can occurs ...). Here, the plane $(M, x, y)$ normal to the wire is an antisymmetry plane for the current. The vector potential, polar vector, is therefore normal to this plane.

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  • $\begingroup$ Now that you mention the formula to find $\vec A$, it indeed seems obvious that its direction in our case is $\hat z$. However, I am still curious about the specific deduction in my course's notes, which seemingly doesn't use this formula. $\endgroup$ – Oren Milman Jan 9 at 7:05
  • $\begingroup$ I believe it's a symetry argument : a polar vector is orthogonal to an antisymetry plane in a point of the plane. You have the same argument for the electric field of a dipole $Q$ and $-Q$ . For a point in the plane of antisymetry between the two charges, you can easily check that the electric field is normal to the plane. For an axial vector, as the magnetic field, the conclusion are reversed : the magnetic field is normal to a symetry plane (infinite solenoïd...) $\endgroup$ – Vincent Fraticelli Jan 9 at 7:14
  • $\begingroup$ To complete in a less simple case, the symmetry argument makes it possible to show that the (coulomb jauge) vector potential is zero on the axis of a circular loop : all the planes containing the axis are plane of symmetry and the vector potential, normal to all these plans, is zero. But the magnetic field is still not zero .....! $\endgroup$ – Vincent Fraticelli Jan 9 at 7:39

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