The Unapologetic Mathematician

Mathematics for the interested outsider

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.

About these ads

January 30, 2012 - Posted by | Electromagnetism, Mathematical Physics

6 Comments »

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

    Pingback by Conservation of Charge « The Unapologetic Mathematician | 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 Maxwell’s Equations « The Unapologetic Mathematician | February 1, 2012 | Reply

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

    Pingback by Energy and the Magnetic Field « The Unapologetic Mathematician | February 14, 2012 | Reply

  4. [...] start with Ampère’s law, including Maxwell’s [...]

    Pingback by Conservation of Electromagnetic Energy « The Unapologetic Mathematician | February 17, 2012 | Reply

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

    Pingback by The Higgs Mechanism part 1: Lagrangians « The Unapologetic Mathematician | 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


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 366 other followers

%d bloggers like this: