# The Unapologetic Mathematician

## The Electromagnetic Wave Equations

Maxwell’s equations give us a collection of differential equations to describe the behavior of the electric and magnetic fields. Juggling them, we can come up with other differential equations that give us more insight into how these fields interact. And, in particular, we come up with a familiar equation that describes waves.

Specifically, let’s consider Maxwell’s equations in a vacuum, where there are no charges and no currents:

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

Now let’s take the curl of both of the curl equations:

\displaystyle\begin{aligned}\nabla\times(\nabla\times E)&=-\frac{\partial}{\partial t}(\nabla\times B)\\&=-\frac{\partial}{\partial t}\left(\epsilon_0\mu_0\frac{\partial E}{\partial t}\right)\\&=-\epsilon_0\mu_0\frac{\partial^2 E}{\partial t^2}\\\nabla\times(\nabla\times B)&=\epsilon_0\mu_0\frac{\partial}{\partial t}(\nabla\times E)\\&=\epsilon_0\mu_0\frac{\partial}{\partial t}\left(-\frac{\partial B}{\partial t}\right)\\&=-\epsilon_0\mu_0\frac{\partial^2 B}{\partial t^2}\end{aligned}

We also have an identity for the double curl:

$\displaystyle\nabla\times(\nabla\times F)=\nabla(\nabla\cdot F)-\nabla^2F$

But for both of our fields we have $\nabla\cdot F=0$, meaning we can rewrite our equations as

\displaystyle\begin{aligned}\frac{\partial^2 E}{\partial t^2}-\frac{1}{\epsilon_0\mu_0}\nabla^2E&=0\\\frac{\partial^2 B}{\partial t^2}-\frac{1}{\epsilon_0\mu_0}\nabla^2B&=0\end{aligned}

which are the wave equations we were looking for.

February 7, 2012

## Deriving Physics from Maxwell’s Equations

It’s important to note at this point that we didn’t have to start with our experimentally-justified axioms. Maxwell’s equations suffice to derive all the physics we need.

In the case of Faraday’s law, we’re already done, since it’s exactly the third of Maxwell’s equations in integral form. So far, so good.

Coulomb’s law is almost as simple. If we have a point charge $q$ it makes sense that it generate a spherically symmetric, radial electric field. Given this assumption, we just need to calculate its magnitude at the radius $r$. To do this, set up a sphere of that radius around the point; Gauss’ law in integral form tells us that the flow of $E$ out through this sphere is the total charge $q$ inside. But it’s easy to calculate the integral, getting

$\displaystyle4\pi r^2\lvert E(\lvert r\rvert)\rvert=\frac{q}{\epsilon_0}$

or

$\displaystyle\lvert E(\lvert r\rvert)\rvert=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

which is the magnitude given by Coulomb’s law.

To get the Biot-Savart law, we can use Ampère’s law to calculate the magnetic field around an infinitely long straight current $I$. We again argue on geometric grounds that the magnitude of the magnetic field should only depend on the distance from the current and should point directly around the current. If we set up a circle of radius $r$ then, the total circulation around the circle is, by Ampère’s law:

$\displaystyle2\pi r\lvert B(\lvert r\rvert)\rvert=\mu_0I$

or

$\displaystyle\lvert B(\lvert r\rvert)\rvert=\frac{\mu_0}{2\pi}\frac{I}{r}$

Now, we can compare this to the last time we computed the magnetic field of the straight infinite current by integrating the Biot-Savart law directly and got essentially the same answer.

Finally, we can derive conservation of charge from Ampère’s law, with Maxwell’s correction by taking its divergence:

$\displaystyle\nabla\cdot(\nabla\times B)=\mu_0\nabla\cdot J+\epsilon_0\mu_0\frac{\partial}{\partial t}(\nabla\cdot E)$

The quantity on the left is the divergence of a curl, so it automatically vanishes. Meanwhile, Gauss’ law tells us that $\epsilon\nabla\cdot E=\rho$, so we conclude

$\displaystyle0=\mu_0\left(\nabla\cdot J+\frac{\partial\rho}{\partial t}\right)$

or

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

which is the “continuity equation” expressing the conservation of charge.

The importance is that while we originally derived Maxwell’s equations from four experimentally-justified laws, those laws are themselves essentially derivable from Maxwell’s equations. Thus any reformulation of Maxwell’s equations is just as sufficient a basis for all of electromagnetism as our original physical axioms.

February 3, 2012

## Maxwell’s Equations (Integral Form)

It is sometimes easier to understand Maxwell’s equations in their integral form; the version we outlined last time is the differential form.

For Gauss’ law and Gauss’ law for magnetism, we’ve actually already done this. First, we write them in differential form:

\displaystyle\begin{aligned}\nabla\cdot E&=\frac{1}{\epsilon_0}\rho\\\nabla\cdot B&=0\end{aligned}

We pick any region $V$ we want and integrate both sides of each equation over that region:

\displaystyle\begin{aligned}\int\limits_V\nabla\cdot E\,dV&=\int\limits_V\frac{1}{\epsilon_0}\rho\,dV\\\int\limits_V\nabla\cdot B\,dV&=\int\limits_V0\,dV\end{aligned}

On the left-hand sides we can use the divergence theorem, while the right sides can simply be evaluated:

\displaystyle\begin{aligned}\int\limits_{\partial V}E\cdot dS&=\frac{1}{\epsilon_0}Q(V)\\\int\limits_{\partial V}B\cdot dS&=0\end{aligned}

where $Q(V)$ is the total charge contained within the region $V$. Gauss’ law tells us that the flux of the electric field out through a closed surface is (basically) equal to the charge contained inside the surface, while Gauss’ law for magnetism tells us that there is no such thing as a magnetic charge.

Faraday’s law was basically given to us in integral form, but we can get it back from the differential form:

$\displaystyle\nabla\times E=-\frac{\partial B}{\partial t}$

We pick any surface $S$ and integrate the flux of both sides through it:

$\displaystyle\int\limits_S\nabla\times E\cdot dS=\int\limits_S-\frac{\partial B}{\partial t}\cdot dS$

On the left we can use Stokes’ theorem, while on the right we can pull the derivative outside the integral:

$\displaystyle\int\limits_{\partial S}E\cdot dr=-\frac{\partial}{\partial t}\Phi_S(B)$

where $\Phi_S(B)$ is the flux of the magnetic field $B$ through the surface $S$. Faraday’s law tells us that a changing magnetic field induces a current around a circuit.

A similar analysis helps with Ampère’s law:

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

We pick a surface and integrate:

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

Then we simplify each side.

$\displaystyle\int\limits_{\partial S}B\cdot dr=\mu_0I_S+\epsilon_0\mu_0\frac{\partial}{\partial t}\Phi_S(E)$

where $\Phi_S(E)$ is the flux of the electric field $E$ through the surface $S$, and $I_S$ is the total current flowing through the surface $S$. Ampère’s law tells us that a flowing current induces a magnetic field around the current, and Maxwell’s correction tells us that a changing electric field behaves just like a current made of moving charges.

We collect these together into the integral form of Maxwell’s equations:

\displaystyle\begin{aligned}\int\limits_{\partial V}E\cdot dS&=\frac{1}{\epsilon_0}Q(V)\\\int\limits_{\partial V}B\cdot dS&=0\\\int\limits_{\partial S}E\cdot dr&=-\frac{\partial}{\partial t}\Phi_S(B)\\\int\limits_{\partial S}B\cdot dr&=\mu_0I_S+\epsilon_0\mu_0\frac{\partial}{\partial t}\Phi_S(E)\end{aligned}

February 2, 2012

## 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