# The Unapologetic Mathematician

## Maxwell’s Equations

Okay, let’s see where we are. There is such a thing as charge, and there is such a thing as current, which often — but not always — arises from charges moving around.

We will write our charge distribution as a function $\rho$ and our current distribution as a vector-valued function $J$, though these are not always “functions” in the usual sense. Often they will be “distributions” like the Dirac delta; we haven’t really gotten into their formal properties, but this shouldn’t cause us too much trouble since most of the time we’ll use them — like we’ve used the delta — to restrict integrals to smaller spaces.

Anyway, charge and current are “conserved”, in that they obey the conservation law:

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

which states that the mount of current “flowing out of a point” is the rate at which the charge at that point is decreasing. This is justified by experiment.

Coulomb’s law says that electric charges give rise to an electric field. Given the charge distribution $\rho$ we have the differential contribution to the electric field at the point $r$:

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

and we get the whole electric field by integrating this over the charge distribution. This, again, is justified by experiment.

The Biot-Savart law says that electric currents give rise to a magnetic field. Given the current distribution $J$ we have the differential contribution to the magnetic field at the poinf $r$:

$\displaystyle dB(r)=\frac{\mu_0}{4\pi}J\times\frac{r}{\lvert r\rvert^3}dV$

which again we integrate over the current distribution to calculate the full magnetic field at $r$. This, again, is justified by experiment.

The electric and magnetic fields give rise to a force by the Lorentz force law. If a test particle of charge $q$ is moving at velocity $v$ through electric and magnetic fields $E$ and $B$, it feels a force of

$\displaystyle F=q(E+v\times B)$

But we don’t work explicitly with force as much as we do with the fields. We do have an analogue for work, though — electromotive force:

$\displaystyle\mathcal{E}=-\int\limits_CE\cdot dr$

One unexpected source of electromotive force comes from our fourth and final experimentally-justified axiom: Faraday’s law of induction

$\displaystyle\mathcal{E}=\frac{\partial}{\partial t}\int\limits_\Sigma B\cdot dS$

This says that the electromotive force around a circuit is equal to the rate of change of magnetic flux through any surface bounded by the circuit.

Using these four experimental results and definitions, we can derive Maxwell’s equations:

\displaystyle\begin{aligned}\nabla\cdot E&=\frac{1}{\epsilon_0}\rho\\\nabla\cdot B&=0\\\nabla\times E&=-\frac{\partial B}{\partial t}\\\nabla\times B&=\mu_0J+\epsilon_0\mu_0\frac{\partial E}{\partial t}\end{aligned}

The first is Gauss’ law and the second is Gauss’ law for magnetism. The third is directly equivalent to Faraday’s law of induction, while the last is Ampère’s law, with Maxwell’s correction.

February 1, 2012

## Conservation of Charge

When we worked out Ampères law in the case of magnetostatics, we used a certain identity:

$\displaystyle\nabla\cdot J+\frac{\partial\rho}{\partial t}=0$

which we often write as

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

That is, the rate at which the charge at a point is increasing is the negative of the divergence of the current at that point, which measures how much current is “flowing out” from that point. This may be clearer if we integrate this equation over some macroscopic region $V$:

\displaystyle\begin{aligned}\frac{\partial}{\partial t}\int\limits_V\rho\,dV&=\int\limits_V\frac{\partial}{\partial t}\rho\,dV\\&=\int\limits_V-\nabla\cdot J\,dV\\&=-\int\limits_{\partial V}J\,dA\\&=\int\limits_{-\partial V}J\,dA\end{aligned}

The rate of change of the total amount of the charge within $V$ is equal to the amount of current flowing inwards across the boundary of $V$, so this flow of current is the only way that the charge in a region can change. This is another physical law, borne out by experiment, and we take it as another axiom.

But we might note something interesting if we couple this with Gauss’ law:

$\displaystyle0=\nabla\cdot J+\frac{\partial\rho}{\partial t}=\nabla\cdot J+\epsilon_0\frac{\partial}{\partial t}(\nabla\cdot E)$

Or, to put it slightly differently:

$\displaystyle\nabla\cdot\left(J+\epsilon_0\frac{\partial E}{\partial t}\right)=0$

Recall that in deriving Ampère’s law we had to assume that $J$ was divergence-free; when things are not static, the above equation shows that the composite quantity

$\displaystyle J+\epsilon_0\frac{\partial E}{\partial t}$

is always divergence-free. The derivative term isn’t associated with any electric charge moving around, and yet it still behaves like a current for all intents and purposes. We call it the “displacement current”, and we add it into Ampère’s law to see how things work without the magnetostatic assumption:

$\displaystyle\nabla\times B=\mu_0J+\epsilon_0\mu_0\frac{\partial E}{\partial t}$

This additional term is known as Maxwell’s correction to Ampère’s law.

February 1, 2012