# Electromagnetic Plane Waves

{\displaystyle {\begin{aligned}{\vec {E}}&={\vec {E}}_{0}e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}\\{\vec {B}}&={\vec {B}}_{0}e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}\\\end{aligned}}\,\!}

where ${\displaystyle E_{0}}$ is the magnitude of the electric field, ${\displaystyle B_{0}}$ is the magnitude of the magnetic field, ${\displaystyle {\vec {k}}}$ is a “wavevector”, and ${\displaystyle \omega }$ is the angular frequency. Note that ${\displaystyle {\vec {k}}}$ is a 3D spatial analog of ${\displaystyle \omega }$ — it encodes a direction-dependent spatial frequency, with units of inverse length. The larger ${\displaystyle k\equiv |{\vec {k}}|}$ is, the shorter the distance you have to travel to trace out a period of the sine wave. The directionality of ${\displaystyle {\vec {k}}}$ means that you walk through a period fastest along the direction of ${\displaystyle {\vec {k}}}$, and if you walk perpendicular to ${\displaystyle {\vec {k}}}$, 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 ${\displaystyle {\vec {k}}}$ and ${\displaystyle \omega }$ can’t be independent. We know how fast light travels (${\displaystyle c}$), 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 ${\displaystyle 2\pi /k}$, and the time corresponding to one period is ${\displaystyle 2\pi /\omega }$, so we have

{\displaystyle {\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

${\displaystyle \nu \lambda =c\,\!}$

Moreover, light must be propagating in the direction ${\displaystyle {\vec {k}}}$.

### Relative Orientation of ${\displaystyle {\vec {E}}}$, ${\displaystyle {\vec {B}}}$, and ${\displaystyle {\vec {k}}}$

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 ${\displaystyle {\vec {E}}}$ plane wave that we wrote down must satisfy the following equation:

{\displaystyle {\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, ${\displaystyle \nabla \cdot {\vec {B}}=0}$ implies that ${\displaystyle {\vec {B}}_{0}\cdot {\vec {k}}=0}$. Since ${\displaystyle {\vec {k}}}$ encodes the direction of propagation, this means that the directions of the ${\displaystyle {\vec {E}}}$ and ${\displaystyle {\vec {B}}}$ fields are orthogonal to the direction of propogation.

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

{\displaystyle {\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 {B}}_{0}\times {\vec {k}})e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}&=-ik{\vec {E}}_{0}e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}\\{\vec {B}}_{0}\times {\vec {k}}&=-k{\vec {E}}_{0}\\{\vec {B}}_{0}\times {\hat {k}}&=-{\vec {E}}_{0}\\\end{aligned}}\,\!}

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

${\displaystyle {\vec {E}}\times {\vec {B}}=|{\vec {E}}|^{2}{\hat {k}}\,\!}$

This is called the Poynting vector (the most appropriately named vector in physics). Crossing the E and B fields gives the energy, transferred in the direction of ${\displaystyle {\hat {k}}}$.