# The Unapologetic Mathematician

## Gauss’ Law

Rather than do any more messy integrals for special cases we will move to a more advanced fact about the electric field. We start with Coulomb’s law:

$\displaystyle E(r)=\frac{1}{4\pi\epsilon_0}\frac{q}{\lvert r\rvert^3}r$

and we replace our point charge $q$ with a charge distribution $\rho$ over some region of $\mathbb{R}^3$. This may be concentrated on some surfaces, or on curves, or at points, or even some combination of the these; it doesn’t matter. What does matter is that we can write the amount contributed to the electric field at $r$ by the charge a point $s$ as

$\displaystyle dE(r)=\frac{1}{4\pi\epsilon_0}\frac{\rho(s)}{\lvert r-s\rvert^3}(r-s)d^3s$

So to get the whole electric field, we integrate over all of space!

$\displaystyle E(r)=\frac{1}{4\pi\epsilon_0}\int\limits_{\mathbb{R}^3}\frac{\rho(s)}{\lvert r-s\rvert^3}(r-s)d^3s$

Now we want to take the divergence of each side with respect to $r$. On the right we can pull the divergence inside the integral, since the integral is over $s$ rather than $r$. But we’ve still got a hangup.

Let’s consider this divergence:

$\displaystyle\nabla\cdot\left(\frac{r}{\lvert r\rvert^3}\right)$

Away from $r=0$ this is pretty straightforward to calculate. In fact, you can do it by hand with partial derivatives, but I know a sneakier way to see it.

If you remember our nontrivial homology classes, this is closely related to the one we built on $\mathbb{R}^3$ — the case where $n=2$. In that case we got a $2$-form, not a vector field, but remember that we’re working in our standard $\mathbb{R}^3$ with the standard metric, which lets us use the Hodge star to flip a $2$-form into a $1$-form, and a $1$-form into a vector field! The result is exactly the field we’re taking the divergence of; and luckily enough the divergence of this vector field is exactly what corresponds to the exterior derivative on the $2$-form, which we spent so much time proving was zero in the first place!

So this divergence is automatically zero for any $r\neq0$, while at zero it’s not really well-defined. Still, in the best tradition of physicists we’ll fail the math and calculate anyway; what if it was well-defined, enough to take the integral inside the unit sphere at least? Then the divergence theorem tells us that the integral of the divergence through the ball is the same as the integral of the vector field itself through the surface of the sphere:

$\displaystyle\int\limits_B\nabla\cdot\left(\frac{r}{\lvert r\rvert^3}\right)=\int\limits_{S^2}\left(\frac{r}{\lvert r\rvert^3}\right)\cdot dS=\int\limits_{S^2}r\cdot dS=4\pi$

since the field is just the unit radial vector field on the sphere, which integrates to give the surface area of the sphere: $4\pi$. Remember that the fact that this is not zero is exactly why we said the $2$-form cannot be exact.

So what we’re saying is that this divergence doesn’t really work in the way we usually think of it, but we can pretend it’s something that integrates to give us $4\pi$ whenever our region of integration contains the point $r=0$. We’ll call this something $4\pi\delta(r)$, where the $\delta$ is known as the “Dirac delta-function”, despite not actually being a function. Incidentally, it’s actually very closely related to the Kronecker delta

So anyway, that means we can calculate

$\displaystyle\nabla\cdot E(r)=\frac{1}{4\pi\epsilon_0}\int\rho(s)\nabla\cdot\left(\frac{r-s}{\lvert r-s\rvert^3}\right)d^3s=\frac{1}{4\pi\epsilon_0}\int\rho(s)4\pi\delta(r-s)d^3s$

This integrand is zero wherever $r\neq s$, so the only point that can contribute at all is $\rho(r)$. We may as well consider it a constant and pull it outside the integral:

$\displaystyle\nabla\cdot E(r)=\frac{\rho(r)}{\epsilon_0}\int\delta(r-s)d^3s=\frac{\rho(r)}{\epsilon_0}$

where we have integrated away the delta function to get $1$. Notice how this is like we usually use the Kronecker delta to sum over one variable and only get a nonzero term where it equals the set value of the other variable.

The result is known as Gauss’ law:

$\displaystyle\nabla\cdot E(r)=\frac{\rho(r)}{\epsilon_0}$

and, incidentally, shows why we wrote the proportionality constant the way we did when defining Coulomb’s law. The meaning is that the divergence of the electric field at a point is proportional to the amount of charge distributed at that point, and the constant of proportionality is exactly $\frac{1}{\epsilon_0}$.

If we integrate both sides over some region $U\subseteq\mathbb{R}^3$ we can rewrite the law in “integral form”:

$\displaystyle\int_U\frac{\rho(r)}{\epsilon_0}d^3r=\int_U\nabla\cdot Ed^3r=\int_{\partial U}E\cdot dS$

That is: the outward flow of the electric field through a closed surface is equal to the integral of the charge contained within the surface. The second step here is, of course, the divergence theorem, but this is such a popular application that people often call this “Gauss’ theorem”. Of course, there are two very different statements here: one is the physical identification of electrical divergence with charge distribution, and the other is the geometric special case of Stokes’ theorem. Properly speaking, only the first is named for Gauss.

## 8 Comments	»

1. [...] repeat what we did to come up with Gauss law, but this time on the magnetic [...]

Pingback by | January 12, 2012 | Reply

2. [...] electromotive force around the circuit by chemical or other means. And, as we saw when discussing Gauss’ law, Coulomb’s law gives rise to an electric field that looks [...]

Pingback by | January 14, 2012 | Reply

3. [...] this is great. We know that the gradient of is , and we also know that the divergence of this function is (basically) the “Dirac delta function”. That [...]

Pingback by | January 30, 2012 | Reply

4. [...] But we might note something interesting if we couple this with Gauss’ law: [...]

Pingback by | February 1, 2012 | Reply

5. [...] first is Gauss’ law and the second is Gauss’ law for magnetism. The third is directly equivalent to [...]

Pingback by | February 1, 2012 | Reply

6. [...] Gauss’ law and Gauss’ law for magnetism, we’ve actually already done this. First, we write them in [...]

Pingback by | February 2, 2012 | Reply

7. [...] Gauss’ law tells us that , so we [...]

Pingback by | February 14, 2012 | Reply

8. [...] When we vary with respect to and insist that the variance of be zero we get Gauss’ law: [...]

Pingback by | July 16, 2012 | Reply

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