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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} $?

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  • $\begingroup$ 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. $\endgroup$ Commented Nov 12, 2018 at 11:45
  • $\begingroup$ To my knowledge, the integral from your second point is a solution of Poisson's equation with the boundary condition that $\mathbf A$ vanishes at infinity. It cannot be applied to current distributions that extend to infinity. In textbooks it is usually obtained by reading off the equivalent one-dimensional solution for $V$ in electrostatics, which itself is derived from the formula for the potential of a point charge at the origin $(1/4\pi\epsilon_0)(q/r)$, and works only when $\mathcal O = \infty$. $\endgroup$
    – Chris Yang
    Commented Sep 2, 2022 at 15:30

2 Answers 2

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  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.

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 $\phi=0$, so you chose where you want to calculate the vector potential, easy to see that the answer will only depend on $\rho$ due to the symmetry of the problem.

Then you are good to go.

Calculation of the potential vector for the infinite 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:

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

This is the current density of a infinite cylinder made of circular wires ($n$ is the linear density of wires on the z direction), the current flow on the wires, like its flowing on the cylinder around it. This is not exactly a solenoid, but its a good approximation.

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.

EDIT: For those coming from more elementary Electrodynamics, this expansion is a rather canonical procedure. The complexity of the computation of those kind of integrals lies almost always entirely in the 1/r² term, thus different expansions in infinite series for this term are used in order to perform the integration. The expansion is chosen based on the geometry of the problem. This expansion is explained in detail in the book Classical Electrodynamics, by Jackson. If your aren't expected to know that, than you're supposed to use some sort of approximation procedure in order to calculate the vector potential.

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.

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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.

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