Stock's theorem and calculation of current density

[adamp121]

Hi,

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

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

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

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

Thanks,
Adam.

 

[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?

 

[adamp121]

Hi,

Thank you very much for your response.

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

 

[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 :)

 

[adamp121]

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

 

[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.

 

[adamp121]

Now it's all clear :-)

Thank you again

 

[WannabeNewton]

Anytime. Good luck with your studies!