Stock's theorem and calculation of current density

1. Jun 3, 2013

Hi,

I'm trying to use stock's theorem with the following magnetic field -
$B=1/r\hat{\theta}$ on Cylindrical coordinate.

From one side I get -
$\nabla X B=0 = \mu \int\int J \cdot dA$, means that the current density is zero.

From the other side I get -
$\oint B \cdot dl = 2 \pi r \cdot 1/r = \mu I$
and hence
$I = 2 \pi / \mu$

So... How the current density could be zero?

Thanks,

2. Jun 3, 2013

WannabeNewton

Hi adamp. Let me first show you a "simpler" example from electrostatics and hopefully from this example you can see what goes wrong in your calculation. Consider the Coulomb field of a point charge $E = \frac{q}{4\pi \epsilon_0} \frac{\hat{\mathbf r}}{r^{2}}$. If we naively calculate its divergence, we find that $\nabla\cdot E = 0$ hence by the divergence theorem $\int _{\Sigma}(\nabla\cdot E )dV = \int _{\partial \Sigma}E\cdot dA = 0$. But if we calculate the surface integral from the start we find that $\int _{\partial \Sigma}E\cdot dA = \frac{q}{\epsilon_0}$ so what's the deal?

The problem is that our system consists of a localized point source. Recall that $\nabla\cdot E = \frac{\rho}{\epsilon_0}$ so the divergence will vanish everywhere except at the localized point source because this is the only point in space where there is a charge. As such, when we calculated the divergence of the Coulomb field, we naively ignored what happens at $r = 0$ which is where the charge density "spikes up".

What we really have to do is represent the charge density in terms of a Dirac delta distribution i.e. $\rho = q \delta^{3} (\mathbf r)$. Consequently, $\nabla\cdot \frac{\hat{\mathbf r}}{r^{2}} = 4\pi \delta^{3}(\mathbf r)$ i.e. this will respect the statement of the divergence theorem: $\int _{\Sigma}(\nabla\cdot E)dV = \frac{q}{ \epsilon_0}\int _{\Sigma}\delta^{3}(\mathbf r) = \frac{q}{\epsilon_0} = \int _{\partial \Sigma}E\cdot dA$.

Can you now extrapolate from this and see what goes wrong in your calculation?

3. Jun 3, 2013

Hi,

Thank you very much for your response.

I still don't understand why when we are calculate $\nabla\cdot E$ we "naively" ignored what happens at $r=0$.
Is it comes up from the divergence definition?

4. Jun 3, 2013

WannabeNewton

$\nabla\cdot E = \frac{q}{4\pi \epsilon_0}\frac{1}{r^{2}}\partial_{r}(r^{2}\frac{1}{r^{2}})$ we have a problem at $r = 0$ as you can see :)

5. Jun 3, 2013

So, can we say that it impossible to use the divergence for fields which are not defined at $r=0$ and we need to use stokes' theorem for getting the right answer using Path integral?

6. Jun 3, 2013

WannabeNewton

No we can still use the divergence theorem as long as we include the dirac delta distribution, as I have done in post #2. The problem of things "spiking up" at $r = 0$ comes from us having a vanishing charge density everywhere in space except at a single point. The dirac delta distribution takes care of that.

In your case we have a vanishing current density at all points in space except for the infinitely thin current carrying wire along $r = 0$ (now in cylindrical coordinates); we again have a sudden "spiking up" problem. The situation is analogous to what I talked about in post #2.

7. Jun 3, 2013