# The Unapologetic Mathematician

## Polarization of Electromagnetic Waves

Let’s look at another property of our plane wave solutions of Maxwell’s equations. Specifically, we’ll assume that the electric and magnetic fields are each plane waves in the directions $k_E$ and $k_B$, repectively:

\displaystyle\begin{aligned}E(r,t)&=\hat{E}(k_E\cdot r-ct)\\B(r,t)&=\hat{B}(k_B\cdot r-ct)\end{aligned}

We can take these and plug them into the vacuum version of Maxwell’s equations, and evaluate them at $(r,t)=(0,0)$:

\displaystyle\begin{aligned}k_E\cdot\hat{E}'(0)&=0\\k_E\times\hat{E}'(0)&=c\hat{B}'(0)\\k_B\cdot\hat{B}'(0)&=0\\k_B\times\hat{B}'(0)&=-\frac{1}{c}\hat{E}'(0)\end{aligned}

The first equation says that $\hat{E}'(0)$ is perpendicular to $k_E$, but the second equation implies, in part, that $\hat{B}'(0)$ is also perpendicular to $k_E$. Similarly, the third and fourth equations say that both $\hat{E}'(0)$ and $\hat{B}'(0)$ are perpendicular to $k_B$, meaning that $k_E$ and $k_B$ either point in the same direction or in opposite directions. We can always pick our coordinates so that $k_E$ points in the direction of the $z$-axis and $\hat{E}'(0)$ points in the direction of the $x$-axis; then $\hat{B}'(0)$ points in the direction of the $y$-axis. It’s then straightforward to check that $k_B=k_E$ rather than $k_B=-k_E$. Of course, it’s possible that $\hat{E}'(0)$ — and thus $\hat{B}'(0)$ also — is zero; in this case we can just pick some different time at which to evaluate the equations. There must be some time for which these values are nonzero, or else $\hat{E}$ and $\hat{B}$ are simply constants, which is a pretty vacuous solution that we’ll just subtract off and ignore.

The upshot of this is that $E$ and $B$ must be plane waves traveling in the same direction. We put this back into our assumption:

\displaystyle\begin{aligned}E(r,t)&=\hat{E}(k\cdot r-ct)\\B(r,t)&=\hat{B}(k\cdot r-ct)\end{aligned}

and then Maxwell’s equations imply

\displaystyle\begin{aligned}k\cdot\hat{E}'&=0\\k\times\hat{E}'&=c\hat{B}'\\k\cdot\hat{B}'&=0\\k\times\hat{B}'&=-\frac{1}{c}\hat{E}'\end{aligned}

where these are now full functions and not just evaluations at some conveniently-chosen point. And, incidentally, the second and fourth equations are completely equivalent. Now we can see that $\hat{E}'$ and $\hat{B}'$ are perpendicular at every point. Further, whatever component either vector has in the $k$ direction is constant, and again we will just subtract it off and ignore it.

As the wave propagates in the direction of $k$, the electric and magnetic fields move around in the plane perpendicular to $k$. If we pick our $z$-axis in the direction of $k$, we can write $\hat{E}=\hat{E}_x\hat{i}+\hat{E}_y\hat{j}$ and $\hat{B}=\hat{B}_x\hat{i}+\hat{B}_y\hat{j}$. Then the second (and fourth) equation tells us

$\displaystyle\hat{E}_x'\hat{j}-\hat{E}_y'\hat{i}=c\hat{B}_x'\hat{i}+c\hat{B}_y'\hat{j}$

That is, we get two decoupled equations:

\displaystyle\begin{aligned}\hat{E}_x'&=c\hat{B}_y'\\\hat{E}_y'&=-c\hat{B}_x'\end{aligned}

This tells us that we can break up our plane wave solution into two different plane wave solutions. In one, the electric field “waves” in the $x$ direction while the magnetic field waves in the $y$ direction; in the other, the electric field waves in the $y$ direction and the magnetic field waves in the $-x$ direction.

This decomposition is the basis of polarized light. We can create filters that only allow waves with the electric field oriented in one direction to pass; generic waves can be decomposed into a component waving in the chosen direction and a component waving in the perpendicular direction, and the latter component gets destroyed as the wave passes through the Polaroid filter — yes, that’s where the company got its name — leaving only the light oriented in the “right” way.

As a quick, familiar application, we can make glasses with a film over the left eye that polarizes light vertically, and one over the right eye that polarizes light horizontally. Then if we show a quickly-alternating series of images, each polarized with the opposite axis, then they will be presented to each eye separately. This is the basis of the earliest modern stereoscopic — or “3-D” — glasses, which had the problem that if you tilted your head the effect was first lost, and then reversed as your neck’s angle increased. If you’ve been paying attention, you should be able to see why.

February 10, 2012

## The Propagation Velocity of Electromagnetic Waves

Now we’ve derived the wave equation from Maxwell’s equations, and we have worked out the plane-wave solutions. But there’s more to Maxwell’s equations than just the wave equation. Still, let’s take some plane-waves and see what we get.

First and foremost, what’s the propagation velocity of our plane-wave solutions? Well, it’s $c$ for the generic wave equation

$\displaystyle\frac{\partial^2F}{\partial t^2}-c^2\nabla^2F=0$

while our electromagnetic wave equation is

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

so we find the propagation velocity of waves in both electric and magnetic fields is

$\displaystyle c=\frac{1}{\sqrt{\epsilon_0\mu_0}}$

Hm.

Conveniently, I already gave values for both $\epsilon_0$ and $\mu_0$:

\displaystyle\begin{aligned}\epsilon_0&=8.85418782\times10^{-12}\frac{\mathrm{F}}{\mathrm{m}}&=8.85418782\times10^{-12}\frac{\mathrm{s}^2\cdot\mathrm{C}^2}{\mathrm{m}^3\cdot\mathrm{kg}}\\\mu_0&=1.2566370614\times10^{-6}\frac{\mathrm{H}}{\mathrm{m}}&=1.2566370614\times10^{-6}\frac{\mathrm{m}\cdot\mathrm{kg}}{\mathrm{C}^2}\end{aligned}

Multiplying, we find:

$\displaystyle\epsilon_0\mu_0=8.85418782\times1.2566370614\times10^{-18}\frac{\mathrm{s}^2}{\mathrm{m}^2}=11.1265006\times10^{-18}\frac{\mathrm{s}^2}{\mathrm{m}^2}$

which means that

$\displaystyle c=\frac{1}{\sqrt{\epsilon_0\mu_0}}=0.299792457\times10^9\frac{\mathrm{m}}{\mathrm{s}}=299\,792\,457\frac{\mathrm{m}}{\mathrm{s}}$

And this is a number which should look very familiar: it’s the speed of light. In an 1864 paper, Maxwell himself noted:

The agreement of the results seems to show that light and magnetism are affections of the same substance, and that light is an electromagnetic disturbance propagated through the field according to electromagnetic laws.

Indeed, this supposition has been borne out in experiment after experiment over the last century and a half: light is an electromagnetic wave.

February 9, 2012

## 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

## 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.

January 30, 2012

Okay, so let’s say we have a closed circuit composed of a simple loop of wire following a closed path $C$. There’s no battery or anything that might normally induce an 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 like

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

As we saw when discussing Gauss’ law for magnetism, we can rewrite the fraction in the integrand:

\displaystyle\begin{aligned}E(r)&=-\frac{1}{4\pi\epsilon_0}\int\rho(s)\nabla\left(\frac{1}{\lvert r-s\rvert}\right)\,d^3s\\&=-\nabla\left(\frac{1}{4\pi\epsilon_0}\int\rho(s)\frac{1}{\lvert r-s\rvert}\,d^3s\right)\end{aligned}

So this electric field is conservative, and so its integral around the closed circuit is automatically zero. Thus there is no electromotive force around the circuit, and no current flows.

And yet, that’s not actually what we see. Specifically, if we wave a magnet around near such a circuit, a current will indeed flow! Indeed, this is exactly how the simplest electric generators and motors work.

To put some quantitative meat on these qualitative observational bones, we have Faraday’s law of induction. This says that the electromotive force around a circuit is equal to the rate of change of the magnetic flux through any surface bounded by that circuit. What? maybe a formula will help:

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

where $\Sigma$ is any surface with $\partial\Sigma=C$. Why can we pick any such surface? Because if $\Sigma'$ is another one then:

$\displaystyle\int\limits_\Sigma B\cdot dS-\int\limits_{\Sigma'}B\cdot dS=\int\limits_{\Sigma-\Sigma'}B\cdot dS$

We can calculate the boundary of this combined surface:

$\displaystyle\partial(\Sigma-\Sigma')=\partial\Sigma-\partial\Sigma'=C-C=0$

Since our space is contractible, this means that our surface is itself the boundary of some region $E$.

$\displaystyle\int\limits_{\partial E}B\cdot dS=\int\limits_E\nabla\cdot B\,dV$

But Gauss’ law for magnetism tells us that this is automatically zero. That is, every surface has the same flux, and so it doesn’t matter which one we use in Faraday’s law.

Now, we can couple this with our original definition of electromotive force:

\displaystyle\begin{aligned}\int\limits_\Sigma\frac{\partial B}{\partial t}\cdot dS&=-\int\limits_{\partial\Sigma}E\cdot dr\\&=-\int\limits_\Sigma\nabla\times E\cdot dS\end{aligned}

But this works no matter what surface $\Sigma$ we consider, so we come up with the differential form of Faraday’s law:

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

January 14, 2012

## Electromotive Force

When we think of electricity, we don’t usually think of charged particles pushing and pulling on each other, mediated by vector fields. Usually we think of electrons flowing along wires. But what makes them flow?

The answer is summed up in something that is named — very confusingly — “electromotive force”. The word “force” is just a word here, so try to keep from thinking of it as a force. In fact, it’s more analogous to work, in the same way the electric field is analogous to force.

We calculate the work done by a force $F$ in moving a particle along a path $C$ is given by the line integral

$\displaystyle W=\int\limits_CF\cdot dr$

If $F$ is conservative, this amounts to the difference in “potential energy” between the start and end of the path. We often interpret a work integral as an energy difference even in a more general setting. Colloquially, we sometimes say that particles “want to move” from high-energy states to low-energy ones.

Similarly, we define the electromotive force along a path to be the line integral

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

If the electric field $E$ is conservative — the gradient of some potential function $\phi$ — then the electromotive force over a path is the difference between the potential at the end and at the start of the path. But, we may ask, why the negative sign? Well, it’s conventional, but I like to think of it in terms of electrons, which have negative charge; given electric field “pushes” an electron in the opposite direction, thus we should take the negative to point the other way.

In any event, just like the electric field measures force per unit of charge, electromotive force measures work per unit of charge, and is measured in units of energy per unit of charge. In the SI system, there is is a unit called a volt — with symbol $\mathrm{V}$ — which is given by

$\displaystyle\mathrm{V}=\frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{C}\cdot\mathrm{s}^2}$

And often electromotive force is called “voltage”. For example, a battery — through chemical processes — induces a certain difference in the electric potential between its two terminals. In a nine-volt battery this difference is, predictably enough, $9\mathrm{V}$, and the same difference is induced along the path of a wire leading from one terminal, through some electric appliance, and back to the other terminal. Just like the potential energy difference “pushes” particles along from high-energy states to low-energy ones, so the voltage difference “pushes” charged particles along the wire.

January 13, 2012