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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. $\endgroup$ – Lorenz Mayer Nov 12 '18 at 11:45
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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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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.

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