Electromagnetic Plane Waves

Short Topical Videos

• Plane Waves (kridnix, Bucknell U.)

Reference Material

• Plane Waves (Wikipedia)
• Poynting Flux (Wikipedia)
• Wave equation & electromagnetic radiation (Scott Hughes, MIT)

Prerequisites

• Maxwell Equations for E&M waves
• Fourier Transform

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:

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