The Unapologetic Mathematician

Mathematics for the interested outsider

Gauss’ Law for Magnetism

Let’s repeat what we did to come up with Gauss law, but this time on the magnetic field.

As a first step, though, I want to finally get a good definition of “current density”: it’s a vector field J that consists of a charge density \rho and a velocity vector v, each of which is a function of space. In our example of an infinite line current, this density was concentrated along the z-axis, where the velocity was vertical. But it could exist along a surface, or throughout space; a single particle of charge q moving with velocity v is a current density concentrated at a single point.

Anyway, so the Biot-Savart law says that the differential contribution to the magnetic field at a point r from the current density at point s is

\displaystyle dB(r)=\frac{\mu_0}{4\pi}J(s)\times\frac{r-s}{\lvert r-s\rvert^3}d^3s

So, as for the electric field, we want to integrate over s:

\displaystyle B(r)=\frac{\mu_0}{4\pi}\int\limits_{\mathbb{R}^3}J(s)\times\frac{r-s}{\lvert r-s\rvert^3}d^3s

Last time we spent a while noting that the fraction here is secretly a closed 2-form in disguise, so its divergence is zero. This time, I say it’s actually a conservative vector field:

\displaystyle\frac{r}{\lvert r\rvert^3}=-\nabla\left(\frac{1}{\lvert r\rvert}\right)

Indeed, this is pretty straightforward to check by rote calculation of derivatives, and I’d rather not get into it. The upshot is we can write:

\displaystyle\begin{aligned}B(r)&=\frac{\mu_0}{4\pi}\int\limits_{\mathbb{R}^3}J(s)\times\left(-\nabla\left(\frac{1}{\lvert r-s\rvert}\right)\right)d^3s\\&=\frac{\mu_0}{4\pi}\int\limits_{\mathbb{R}^3}\nabla\left(\frac{1}{\lvert r-s\rvert}\right)\times J(s)+\frac{1}{\lvert r-s\rvert}\left(\nabla\times J(s)\right)d^3s\end{aligned}

where the extra term on the second line is automatically zero because the curl is in terms of r and the current density J depends only on s. I write it in this form because now it looks like the other end of a product rule:

\displaystyle\begin{aligned}B(r)&=\frac{\mu_0}{4\pi}\int\limits_{\mathbb{R}^3}\nabla\times\left(\frac{1}{\lvert r-s\rvert}J(s)\right)d^3s\\&=\nabla\times\left(\frac{\mu_0}{4\pi}\int\limits_{\mathbb{R}^3}\frac{J(s)}{\lvert r-s\rvert}d^3s\right)\end{aligned}

Indeed, this is clearer if we write it in terms of differential forms; since the exterior derivative is a derivation we can write

\displaystyle d(f\alpha)=df\wedge\alpha+fd\alpha

for a function f and a 1-form \alpha. If we flip \alpha over to a vector field F this looks like

\displaystyle\nabla\times(fF)=(\nabla f)\times F+f\nabla\times F

Okay, so now we see that B is the curl of some vector field, and so the divergence \nabla\cdot B of a curl is automatically zero:

\displaystyle\nabla\cdot B=0

Coupling this with the divergence theorem like last time, we conclude that there is no magnetic equivalent of “charge”, or else the outward flow of B through a closed surface would be the integral on the inside of such a charge. But instead we find

\displaystyle\int\limits_{\partial U}B\cdot dS=\int\limits_U\nabla\cdot B\,d^3s=0

January 12, 2012 Posted by | Electromagnetism, Mathematical Physics | 11 Comments

   

Follow

Get every new post delivered to your Inbox.

Join 389 other followers