# The Unapologetic Mathematician

## Ampère’s Law

Let’s go back to the way we derived the magnetic version of Gauss’ law. We wrote

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

Back then, we used this expression to show that the divergence of $B$ vanished automatically, but now let’s see what we can tell about its curl.

$\displaystyle\begin{aligned}\nabla\times B&=\frac{\mu_0}{4\pi}\nabla\times\nabla\times\left(\int\limits_{\mathbb{R}^3}\frac{J(s)}{\lvert r-s\rvert}\,d^3s\right)\\&=\frac{\mu_0}{4\pi}\left(\nabla\left(\nabla\cdot\int\limits_{\mathbb{R}^3}\frac{J(s)}{\lvert r-s\rvert}\,d^3s\right)-\nabla^2\int\limits_{\mathbb{R}^3}\frac{J(s)}{\lvert r-s\rvert}\,d^3s\right)\end{aligned}$

Let’s handle the first term first:

$\displaystyle\begin{aligned}\nabla_r\left(\nabla_r\cdot\int\limits_{\mathbb{R}^3}\frac{J(s)}{\lvert r-s\rvert}\,d^3s\right)&=\nabla_r\int\limits_{\mathbb{R}^3}J(s)\cdot\nabla_r\frac{1}{\lvert r-s\rvert}\,d^3s\\&=-\nabla_r\int\limits_{\mathbb{R}^3}J(s)\cdot\nabla_s\frac{1}{\lvert r-s\rvert}\,d^3s\\&=-\nabla_r\int\limits_{\mathbb{R}^3}\nabla_s\frac{J(s)}{\lvert r-s\rvert}-\frac{1}{\lvert r-s\rvert}\nabla_s\cdot J(s)\,d^3s\\&=-\nabla_r\int\limits_{\mathbb{R}^3}\nabla_s\frac{J(s)}{\lvert r-s\rvert}\,d^3s+\nabla_r\int\limits_{\mathbb{R}^3}\frac{\nabla_s\cdot J(s)}{\lvert r-s\rvert}]\,d^3s\end{aligned}$

Now the divergence theorem tells us that the first term is

$\displaystyle-\nabla_r\int\limits_S\frac{J(s)}{\lvert r-s\rvert}\cdot dS$

where $S=\partial V$ is some closed surface whose interior $V$ contains the support of the whole current distribution $J(s)$. But then the integrand is constantly zero on this surface, so the term is zero.

For the other term (and for the moment, no pun intended) we’ll assume that the whole system is in a steady state, so nothing changes with time. The divergence of the current distribution at a point — the amount of charge “moving away from” the point — is the rate at which the charge at that point is decreasing. That is,

$\displaystyle\nabla\cdot J=-\frac{\partial\rho}{\partial t}$

But our steady-state assumption says that charge shouldn’t be changing, and thus this term will be taken as zero.

So we’re left with:

$\displaystyle\nabla\times B(r)=-\frac{\mu_0}{4\pi}\int\limits_{\mathbb{R}^3}J(s)\nabla^2\frac{1}{\lvert r-s\rvert}\,d^3s$

But this is great. We know that the gradient of $\frac{1}{\lvert r\rvert}$ is $\frac{r}{\lvert r\rvert^3}$, and we also know that the divergence of this function is (basically) the “Dirac delta function”. That is:

$\displaystyle\nabla^2\frac{1}{\lvert r\vert}=-4\pi\delta(r)$

So in our case we have

$\displaystyle\nabla\times B(r)=\frac{\mu_0}{4\pi}\int\limits_{\mathbb{R}^3}J(s)4\pi\delta(r-s)=\mu_0J(r)$

This is Ampère’s law, at least in the case of magnetostatics, where nothing changes in time.

## 6 Comments	»

1. [...] we worked out Ampères law in the case of magnetostatics, we used a certain [...]

Pingback by | February 1, 2012 | Reply

2. [...] magnetism. The third is directly equivalent to Faraday’s law of induction, while the last is Ampère’s law, with Maxwell’s correction. Share this:StumbleUponDiggRedditTwitterLike this:LikeBe the first [...]

Pingback by | February 1, 2012 | Reply

3. [...] we can use Ampère’s law to [...]

Pingback by | February 14, 2012 | Reply

Pingback by | February 17, 2012 | Reply

5. [...] the components of we get Ampère’s law with Maxwell’s [...]

Pingback by | July 16, 2012 | Reply

6. How can this be derived without the use of the delta distribution? Just curious

Comment by Matthew Kvalheim | August 28, 2012 | Reply

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