# Vector potential of a solenoid in the Coulomb gauge

I understand the usual argument for calculating the vector potential outside of a solenoid of radius $R$ with $n$ turns per unit length carrying current $I_0$ using $$\oint \mathbf{A} \cdot d \mathbf{l} = \iint \nabla \times \mathbf{A} \cdot d\mathbf{a} = \iint \mathbf{B} \cdot d\mathbf{a}$$ which gives (in Gaussian units) $$A_{\varphi} = \frac{2\pi}{c} \frac{nI_0 R^2}{r}$$ However, I am asked explicitly to find the vector potential in the Coulomb gauge. I have two main questions:

1) Is showing that this vector potential satisfies $\nabla \cdot \mathbf{A} = 0$ and $\mathbf{B} = \nabla \times \mathbf{A}$ sufficient? That seems a bit too much like a 'physicist proof' to me.

2) How can I compute the vector potential explicitly from the form $$\mathbf{A} = \frac{1}{c} \int \frac{\mathbf{J}(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|} d^3 r$$ I have written $$\mathbf{J}(\mathbf{r}',t) = n I_0 \frac{\delta(r'-R)}{R} \ \hat{\varphi}$$ which gives after some algebra and one integration $$\mathbf{A} = \frac{n I_0}{c} \int_0^{2\pi} \int_{-\infty}^{\infty} \frac{1}{\sqrt{r^2+R^2 - 2rR \cos(\varphi-\varphi') + (z-z')^2 }}dz' d\varphi' \ \hat{\varphi}$$ But doesn't the integral over $z$ diverge? This integral isn't doable by Mathematica. Have I done something wrong?

EDIT:

I suppose I can simplify this integral by (without loss of generality) letting $\phi = 0$ and $z=0$. The integral becomes \begin{align*} \mathbf{A} &= \frac{n I_0}{c} \int_0^{2\pi} \int_{-\infty}^{\infty} \frac{1}{\sqrt{r^2+R^2 - 2rR \cos(\varphi') + (z')^2 }}dz' d\varphi' \ \hat{\varphi} \\ &= \frac{n I_0}{c} \int_{-\infty}^{\infty} \frac{2 K\left(-\frac{4 r R}{(r-R)^2+(z')^2}\right)}{\sqrt{(r-R)^2+(z')^2}}+\frac{2 K\left(\frac{4 r R}{(r+R)^2+(z')^2}\right)}{\sqrt{(r+R)^2+(z')^2}} dz' \end{align*} But this still seems to diverge. How can I show that this reduces to $\frac{2\pi}{c} \frac{nI_0 R^2}{r}$?

• If you have the expression for the vector potential and then show that $\text{div} A =0$ and $\text{rot} A = B$, then this is a mathematician's proof. – Lorenz Mayer Nov 12 '18 at 11:45

1) Yeah it is enough. Coloumb Gauge is defined by $$\nabla \cdot \mathbf{A} = 0$$. Also this expression for the vector potential was obtained through the solution of Poisson`s equation ($$\nabla^2 \mathbf{A} = -\mu_0\mathbf{J}$$), which already required $$\nabla \cdot \mathbf{A} = 0$$ (the full equation is $$-\nabla(\nabla \cdot \mathbf{A})+\nabla^2 \mathbf{A} = -\mu_0\mathbf{J}$$). So calculating through this expression already implies you will obtain an answer in the Coulomb Gauge.

2) First. In the integral at the end it's not "$$\hat{\phi}$$" that should be there, but "$$\hat{\phi^{\prime}}$$" since this is part of the current density and in the current density you put prime coordinates. And yeah, you integrate unitary vectors, as I will explain.

Second "$$\hat{\phi^{\prime}}$$" is not a constant of integration. Remenber that ALWAYS. Very easy to forget.

$$\hat{\phi^{\prime}} = -\sin(\phi^{\prime})\hat{x} + \cos(\phi^{\prime})\hat{y}$$

Unitary vectors are not constants in the integration when using curvilinear coordinates. Only cartesian unitary vectors are constants in the integration.

As you've said, put $$z=0$$ and $$\phi=0$$

Then you are good to go.

## Calculation of the potential vector for the cylindrical solenoid

Now I'm going to show how to calculate the vector potential of a Solenoid. Starting from:

$$\mathbf{A} = \frac{\mu_0}{4\pi} \int \frac{\mathbf{K}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d^2 r^\prime$$

$$d^2 r^\prime= Rdz^\prime d\phi^\prime$$

Where the surface current density is:

$$\mathbf{K}(\phi^\prime) =nI\hat{\phi^\prime} = nI( -sin(\phi^{\prime})\hat{x} + cos(\phi^{\prime})\hat{y})$$

We will use the expansion:

$$\frac{1}{|\mathbf{r}-\mathbf{r}'|}=$$

$$\frac{4}{\pi}\int_0^\infty dk\cos{[k(z-z^\prime)]}(\frac{1}{2}I_0(k\rho_<)K_0(k\rho_>)+\sum_{m=1}^\infty\cos{[m(\phi-\phi^\prime)]}I_m(k\rho_<)K_m(k\rho_>))$$

Where $$\rho_< = min[\rho,R]$$ and $$\rho_>=max[\rho,R]$$. This expression can be obtained by finding an expansion for a Green function in cylindrical coordinates.

Now, using what I said previously is just a matter of simple integration. Using the fact that:

$$\int_0^{2\pi}\sin(\phi^\prime)\cos{[m(\phi-\phi^\prime)]}d\phi^\prime=\pi\delta_{m,1}\sin(\phi)$$

$$\int_0^{2\pi}\cos(\phi^\prime)\cos{[m(\phi-\phi^\prime)]}d\phi^\prime=\pi\delta_{m,1}\cos(\phi)$$

$$\int_{-\infty}^{\infty}\cos{[k(z-z^\prime)]}dz^\prime=\Re{(e^{(ikz)}\int_{-\infty}^{\infty}e^{(-ikz^\prime))}dz^\prime))}=2\pi\delta(k)cos(kz)$$

and that:

$$lim_{k\rightarrow 0}\cos(kz)I_1(k\rho_<)K_1(k\rho_>))=\frac{\rho_<}{2\rho_>}$$

we get:

$$\mathbf{A}=\frac{\mu_0 n I\rho}{2}\hat\phi$$ for $$\rho < R$$

and

$$\mathbf{A}=\frac{\mu_0 n IR^2}{2\rho}\hat\phi$$ for $$\rho > R$$

As expected.

I think you can use an electrostatic analogy of a cylinder with constant surface charge density only, which one can solve via Gauss' law, giving the known potential.

To do the vector potential, one can take for example, its x-component, and one sees that it is a cylinder with current distribution $$-I_0 \sin \phi.$$ Such a distribution can be produced via the superposition of two cylinders with constant current distributions, opposite signs, slightly displaced with respect to each other in the y-direction.

For the y-component, the current is $$I_0 \cos \phi.$$ In this case we have two cylinders slightly displaced in the x direction.

This is one way of getting the vector potential, without doing the integral explicitly.