# Electromagnetic Plane Waves

{\begin{aligned}{\vec {E}}&=E_{0}e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}\\{\vec {B}}&=B_{0}e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}\\\end{aligned}}\,\! where $E_{0}$ is the magnitude of the electric field, $B_{0}$ is the magnitude of the magnetic field, ${\vec {k}}$ is a “wavevector”, and $\omega$ is the angular frequency. Note that ${\vec {k}}$ is a 3D spatial analog of $\omega$ — it encodes a direction-dependent spatial frequency, with units of inverse length. The larger $k\equiv |{\vec {k}}|$ is, the shorter the distance you have to travel to trace out a period of the sine wave. The directionality of ${\vec {k}}$ means that you walk through a period fastest along the direction of ${\vec {k}}$ , and if you walk perpendicular to ${\vec {k}}$ , you stay at the same phase.

Plane waves are useful because they are the 3D analog of sine waves. If you take a 3D Fourier transform, you decompose any function in 3D into a sum of plane waves. Hence, as a basis for functions in 3D, plane waves can be used to describe many things of interest, including light propagation.

## Light Propagation

Speaking of light propagation, plane waves can be useful for teasing a few more properties about electromagnetic plane waves. Firstly, it’s worth pointing out that ${\vec {k}}$ and $\omega$ can’t be independent. We know how fast light travels ($c$ ), so it must be true that as light propagates through the spatial distance that corresponds to a period, the time that elapses must also correspond to a period for the given frequency. Mathematically, the spatial distance corresponding to one period is $2\pi /k$ , and the time corresponding to one period is $2\pi /\omega$ , so we have

{\begin{aligned}{\frac {2\pi }{k}}{\frac {1}{c}}&={\frac {2\pi }{\omega }}\\ck&=\omega \\\end{aligned}}\,\! This is really just another way of saying

$\nu \lambda =c\,\!$ Moreover, light must be propagating in the direction ${\vec {k}}$ .

### Relative Orientation of ${\vec {E}}$ , ${\vec {B}}$ , and ${\vec {k}}$ Plane waves are great for working out the mathematical constraints of Maxwell’s equations in free space, which we only sketched out in the section on that topic. To begin, the ${\vec {E}}$ plane wave that we wrote down must satisfy the following equation:

{\begin{aligned}\nabla \cdot {\vec {E}}&=0\\\nabla \cdot {\vec {E}}_{0}e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}&=0\\{\vec {E}}_{0}\cdot (i{\vec {k}})e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}&=0\\{\vec {E}}_{0}\cdot {\vec {k}}&=0.\\\end{aligned}}\,\! Similarly, $\nabla \cdot {\vec {B}}=0$ implies that ${\vec {B}}_{0}\cdot {\vec {k}}=0$ . Since ${\vec {k}}$ encodes the direction of propagation, this means that the directions of the ${\vec {E}}$ and ${\vec {B}}$ fields are orthogonal to the direction of propogation.

Using another one of Maxwell’s equations, we can figure out the relative orientation of ${\vec {E}}$ and ${\vec {B}}$ :

{\begin{aligned}\nabla \times {\vec {B}}&={\frac {1}{c}}{\frac {\partial {\vec {E}}}{\partial t}}\\\left({\frac {\partial B_{z}}{\partial y}}-{\frac {\partial B_{y}}{\partial z}}\right){\hat {x}}+\left({\frac {\partial B_{x}}{\partial z}}-{\frac {\partial B_{z}}{\partial x}}\right){\hat {y}}+\left({\frac {\partial B_{y}}{\partial z}}-{\frac {\partial B_{z}}{\partial y}}\right){\hat {z}}&=-{\frac {i\omega }{c}}{\vec {E}}_{0}e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}\\\left[(iB_{0,z}k_{y}-iB_{0,y}k_{z}){\hat {x}}+(iB_{0,x}k_{z}-iB_{0,z}k_{x}){\hat {y}}+(iB_{0,y}k_{z}-iB_{0,z}k_{y}){\hat {z}}\right]e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}&=-ik{\vec {E}}_{0}e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}\\i({\vec {k}}\times {\vec {B}}_{0})e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}&=-ik{\vec {E}}_{0}e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}\\{\vec {k}}\times {\vec {B}}_{0}&=-k{\vec {E}}_{0}\\{\vec {\hat {k}}}\times B_{0}&=-{\vec {E}}_{0}\\\end{aligned}}\,\! (This also demonstrates a neat trick–you can always replace $\nabla$ with $i{\vec {k}}$ for a plane wave.)

So that settles it. ${\vec {B}}$ and ${\vec {E}}$ must be orthogonal, of equal magnitude (${\hat {k}}$ is just a unit vector), and reordering some terms, we have

${\vec {E}}\times {\vec {B}}=|{\vec {E}}|^{2}{\hat {k}}\,\!$ This is called the Poynting vector (the most appropriately named vector in physics). Crossing the E and B fields gives the energy density, transferred in the direction of ${\hat {k}}$ . This is often converted into a Poynting Flux by dividing out by $4\pi$ steradians to get the normalized energy density $E^{2}/4\pi$ , and multiplying by the speed of light $c$ to convert a density into a flux. (A density times a speed becomes a flux: if you multiplied by a cross-sectional area, you’d get a volume per second times a density, which just gives you the total number of things passing through your cross section per second). Hence the Poynting Flux is:

${\vec {S}}\equiv {\frac {c}{4\pi }}{\vec {E}}\times {\vec {B}}\,\!$ 