# 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.

About these ads

February 10, 2012 -

## 4 Comments »

1. [...] other important thing to notice is what this tells us about our plane wave solutions. If we take such an electromagnetic wave propagating in the direction and with the electric field [...]

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

2. It’s was great fun (for me, anyway) to take two polarized sun glasses and lay the lens of one over the lens of the other and rotate until they turn dark, although I haven’t kept up with the latest in sun glass technology. I’m not sure if any of them only use polarization. It would seem like a bad way to block UV A or B.

Comment by Hunt | February 18, 2012 | Reply

3. That’s not just polarization. A quick search turns up the ANSI standards being about the amount of transmittance within certain frequency bands. Polarized sunglasses are mainly to reduce glare and offer little UV protection.

Still, the rotation experiment is a neat one for teaching kids. Unfortunately it won’t work with most current 3D glasses.

Comment by John Armstrong | February 18, 2012 | Reply

4. [...] charge, the magnetic field has units of force per unit charge per unit velocity. Further, from our polarized plane-wave solutions to Maxwell’s equations, we see that for these waves the magnitude of the electric field is [...]

Pingback by The Meaning of the Speed of Light « The Unapologetic Mathematician | February 24, 2012 | Reply