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How can one argue by symmetry that the vector potential $A$ such that $$B = \nabla \times A$$ is along the $z$ axis assuming the wire is also along the $z$ axis and is carrying current in the $+z$ direction?

I know that $B=\frac{\mu_0I}{2\pi r}\hat{e_r}$ and I also know the expression for the curl in cylindrical coordinates, which I eventually want to use to calculate $A$ but I first want to understand why is $A$ along the $z$ axis and in particular why is $A$ oriented along the direction of current?

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  • $\begingroup$ B is not in the r-direction. It is in the $\phi$ direction. $\endgroup$ – Rob Jeffries Apr 10 '18 at 22:47
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The answer is you can't and the A-field is not necessarily directed only along the z-axis. $\vec{A}$ is not uniquely defined. You can add the gradient of any scalar field to it and arrive at the same B-field.

What you can do with symmetry is argue that $\vec{A}$ cannot vary with $z$ (for an infinitely long wire) and cannot vary with $\phi$. So only derivatives with respect to $r$ can be non-zero. However, this still does not uniquely define the A-field.

Taking a curl in cylindrical coordinates, the only possible non-zero terms are $$\nabla \times \vec{A} = -\frac{\partial A_z}{\partial R} \hat{\phi} + \frac{1}{R}\frac{\partial (R\ A_{\phi})}{\partial R} \hat{z}$$

The B-field is $$ \vec{B} = \frac{\mu_0 I}{2\pi R} \ \hat{\phi}$$ so this immediately tells you that there must be an $A_z$ (and gives you a possible functional form for it).

However, the second term in the curl above is zero, suggesting you could also add a cylindrically symmetric $A_{\phi}$ term of the form $$ A_{\phi} = \frac{k}{R}$$ This has a curl of zero (is irrotational) and so does not change the curl of the A-field.

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Up the gauge freedom described by Rob Jeffries's answer, and focusing on "why is $A$ along the $z$":

At any point $\vec{r}$, there are 2 vectors given by the problem:

$$ \vec j, \ \ \vec r . $$

(Where $\vec r$ is in spherical coordinates). From that, all the vector you can create are:

$$ \vec I , \ \ \vec r,\ \ \vec I \times\vec r .$$

The cross product can be ruled out very generally by parity symmetry (it's a pseudo vector), while $\vec r$ breaks the translation symmetry along the $z$-axis.

That leaves you with:

$$ \vec A \propto \vec I$$

(at fix $||r||$).

The remaining vectors can be scaled by the scalars in the problem (or functions of them):

$$1,\ \ ||I||^2,\ \ ||r||^2,\ \ \vec I\cdot\vec r, \ \ ||\vec I \times \vec r|| $$

The constant term gives a field with infinite energy, so throw that out. $||r||^2$ and $\vec I \cdot \vec r$ break the ($z$) translation symmetry, so we're stuck with

$$ ||I||^2, \ \ ||\vec I \times \vec r||.$$

The result is that the constant of proportionailty is:

$$ -\frac{\mu_0}{2\pi} \ln{\big(\frac{||\vec I\times\vec r||}{||I||}\big)}$$

Of course, $\vec A$ is not physical, and the gradient of a scalar field lacking any symmetry can be added. $\vec B$ is physical, and its a axial vector; hence the cross product term.

The sign is a result the order of the cross products. Looking just at the direction, the Biot-Savart law says (emphasis: loosely, we're just looking at the direction):

$$ d\vec B \propto \vec I \times \hat r $$

Working on the x-axis:

$$ dB_y \propto I_zx-I_zx = I_zx $$

So that's a positive $B_y$ for positive $I_z$ and $x$.

Meanwhile the curl of $A$ goes like:

$$ B_y \propto \frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x} = \frac{\partial A_z}{\partial x}$$

But $A_z$ is decreasing (in magnitude) with increasing $x$ (it's further from the wire), so to get a positive result, $A_z$ must be negative, or opposite the positive current $I_z$.

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