# 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 the Maxwell 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 the Maxwell 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:

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

As we already know, an electromagnetic wave in free space undergoes transverse oscillations, meaning that the electric and magnetic vectors of the wave lie in a plane perpendicular to the direction of motion of the wave. The direction of this electric vector determines the polarization state of a given wave. This can be written as

${\vec {E}}_{i}={\hat {\varepsilon _{i}}}E_{i}e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}\,\!$ where the wave is linearly polarized in the ${\hat {\varepsilon _{i}}}$ -direction.

Now let us consider the case of two waves of the same frequency propagating in the same direction (in our case, in the x-direction) whose electric vectors are orthogonal to each other. Since the two waves have orthogonal field components, they are independent of each other and can therefore have independent phases and amplitudes. These can be superimposed to give an equation for the overall electric field component

${\vec {E}}=({\hat {\varepsilon _{z}}}E_{0z}e^{i\varphi _{z}}+{\hat {\varepsilon _{y}}}E_{0y}e^{i\varphi _{y}})e^{i({\vec {k}}\cdot {\vec {x}}-\omega t)}\,\!$ From this more generalized set-up, we are now prepared to understand the various forms of polarized waves. Now let $\varphi _{z}=a$ and $\varphi _{y}=b$ and $a\neq b$ , and let us set $\tau ={\vec {k}}\cdot {\vec {x}}-\omega t$ . We can now rewrite our general plane wave equation from above as

{\begin{aligned}{\vec {E}}_{z}=E_{0z}\cos {\tau +a}\\{\vec {E}}_{y}=E_{0y}\cos {\tau +b}\end{aligned}}\,\! By dividing out the amplitude and applying the double angle formula, you get

{\begin{aligned}{\frac {E_{z}}{E_{0z}}}=\cos {\tau }\cos {a}-\sin {\tau }\sin {a}\\{\frac {E_{y}}{E_{0y}}}=\cos {\tau }\cos {b}-\sin {\tau }\sin {b}\end{aligned}}\,\! With some rearranging, this then gives

{\begin{aligned}{\frac {E_{z}}{E_{0z}}}\sin {b}-{\frac {E_{y}}{E_{0y}}}\sin {a}=\cos {\tau }\sin {(b-a)}\\{\frac {E_{z}}{E_{0z}}}\cos {b}-{\frac {E_{y}}{E_{0y}}}\cos {a}=\sin {\tau }\sin {(b-a)}\end{aligned}}\,\! By squaring and summing these equations, and setting $\delta =b-a$ , we finally get

${\frac {E_{z}^{2}}{E_{0z}^{2}}}+{\frac {E_{y}^{2}}{E_{0y}^{2}}}-2{\frac {E_{z}}{E_{0z}}}{\frac {E_{y}}{E_{0y}}}\cos {\delta }=\sin ^{2}{\delta }\,\!$ which is the equation for an ellipse! We have now defined the polarization ellipse.

For specific values of $E_{0z},~E_{0y}$ and $\delta$ , the polarization ellipse takes on special forms.

If $E_{0z}=0$ , then we will have vertical linear polarization. If $E_{0y}=0$ , then we will have horizontal linear polarization. This maps to the Stokes Q plane.

If $E_{0z}=E_{0y}$ and $\delta =0,~\pi$ , then the light will be linear $\pm 45^{\circ }$ polarization. This maps to the Stokes U plane.

If $E_{0z}=E_{0y}$ and $\delta =\pi /2,~3\pi /2$ , then the light will be circularly polarized. If $E_{z}$ is advanced by $\lambda /4$ , then the light will be right circularly polarized. If $E_{y}$ is advanced by $\lambda /4$ , then the light will be left circularly polarized. This maps to the Stokes V plane.

In all other cases, the light will be elliptically polarized.

### Methods of Polarization

#### Reflection

When light strikes a surface, its vertical elements will get refracted into the surface while its horizontal elements will reflect off of the surface. Therefore, reflection leads to linear polarization of the light.

#### Scattering

During a scattering interaction, the incoming light vibrates the scattering particles. This vibration causes the particle to act as an antenna, and re-radiate light that is polarized perpendicular to the original wave. This is why Earth’s blue sky is partially polarized – even though the original light from the Sun is unpolarized, the scattering interactions in the atmosphere lead to partial linear polarization of the light!

#### Absorption

Polarizing absorptive filters work by absorbing one component of polarization while transmitting the perpendicular component. The intensity of the transmitted light then depends on the the relative orientation of the polarization direction of the incoming light and the polarization axis of the filter. These are made from dichroic materials – materials which have different absorption for perpendicular polarization modes, the most common of which is polaroid material, used in the manufacturing of polarized lenses for sunglasses.

#### Magnetic Fields

Polarization often arises due magnetic fields. For instance, the presence of a magnetic field can cause one component of an electric field to lag. This leads to a rotation in the phase of linearly polarized light. This phenomenon is called Faraday rotation. Also see Cyclotron and Synchrotron polarization and Zeeman splitting.