My name is Alexander FufaeV and here I write about:

Linearly and Circularly Polarized Electromagnetic Waves

Table of contents

Consider a plane, periodic electromagnetic wave in vacuum. It has an electric field $\boldsymbol{E}$ and a magnetic field $\boldsymbol{B}$.

For polarization only the E-field $\boldsymbol{E} = (E_{\text x}, E_{\text y}, E_{\text z}) $ is relevant. The individual E-field components of a plane wave are:

$$ \begin{align} \begin{bmatrix} \class{purple}{E_{\text x}} \\ \class{purple}{E_{\text y}} \\ \class{purple}{E_{\text z}} \end{bmatrix} ~=~ \begin{bmatrix} \class{purple}{E_{0 \text x}} \, \cos(\class{red}{\omega} \, t - \class{green}{k}\,z + \alpha) \\ \class{purple}{E_{0 \text y}} \, \, \cos(\class{red}{\omega} \, t - \class{green}{k}\,z + \beta) \\ \class{purple}{E_{0 \text z}} \, \, \cos(\class{red}{\omega} \, t - \class{green}{k}\,z + \gamma) \end{bmatrix} \end{align} $$

Here $\boldsymbol{E}_0 = (E_{0 \text x}, E_{0 \text y}, E_{0 \text z}) $ is the amplitude of the E-field, $\omega$ the angular frequency, $k$ the wave number and $\alpha, \beta, \gamma$ are phases to get the possible phase shift between the vector components.

Since the electromagnetic wave is plane, Eq. 1 depends only on one position coordinate (here it is the $z$ coordinate). And, since the wave is periodic, it is described by a sine or cosine function (here it is cosine). Furthermore, the wave propagates in the $z$ direction.

Light, i.e. an electromagnetic wave, can be polarized with a polarization filter, for example. Mathematically, this means that the E-field components 1 are linked to certain conditions, depending on the type of polarization. For this purpose, let's look at two important types of polarization and their conditions, namely linear and circular polarization.

One condition that both types of polarization must meet is:

Condition #1 - for both polarization types

The amplitude vector $\boldsymbol{E}_0 $ is always orthogonal to the propagation direction $z$.

Thus the $E_{\text z}$ component of the E-field is zero:

$$ \begin{align} \begin{bmatrix}E_{\text x} \\ E_{\text y} \\ E_{\text z} \end{bmatrix} ~=~ \begin{bmatrix}E_{0 \text x} \, \cos(\omega \, t - k\,z + \alpha) \\ E_{0 \text y} \, \, \cos(\omega \, t - k\,z + \beta) \\ 0 \end{bmatrix} \end{align} $$

Linear polarized plane wave

A linearly polarized E-field wave.

A linearly polarized electric wave must also satisfy the following condition besides condition #1:

Condition #2 - for a linearly polarized wave

The field components in $x$- and $y$-direction have no phase shift.

You wonder why it has to be that way? Because this is a definition! If the conditions #1 and #2 are fulfilled, then we speak of linear polarized plane waves.

According to condition #2, the phases $\omega \, t - k\,z + \alpha$ and $\omega \, t - k\,z + \beta$ must be equal. For this, $ \alpha = \beta $ must be satisfied. For simplicity, let's set $ \alpha $ and $\beta$ equal to zero (the important part is that they are BOTH equal to zero):

$$ \begin{align} \begin{bmatrix}E_{\text x} \\ E_{\text y} \\ E_{\text z} \end{bmatrix} ~=~ \begin{bmatrix}E_{0 \text x} \, \cos(\omega \, t - k\,z) \\ E_{0 \text y} \, \, \cos(\omega \, t - k\,z) \\ 0 \end{bmatrix} \end{align} $$

Of course, we can write down this E-field vector compactly and get:

$$ \begin{align} \boldsymbol{E} ~=~ \boldsymbol{E}_0 \, \cos(\omega \, t - k\,z) \end{align} $$

Circularly polarized wave

For a circularly polarized wave, the phase shift $ \beta - \alpha $ between the two E-field components is not zero, as it is for a linearly polarized wave, but $\pm \pi/2$ (i.e., 90 degrees).

Condition #2 - for a circularly polarized wave

The $E_{\text x}$ and $E_{\text y}$ components are 90 degrees out of phase with each other.

Let us apply the definition to the E-field 2:

$$ \begin{align} \begin{bmatrix}E_{\text x} \\ E_{\text y} \\ E_{\text z} \end{bmatrix} ~=~ \begin{bmatrix}E_{0 \text x} \, \cos(\omega \, t - k\,z) \\ E_{0 \text y} \, \, \cos\left(\omega \, t - k\,z - \frac{\pi}{2}\right) \\ 0 \end{bmatrix} \end{align} $$

Since cosine and sine are also 90 degrees out of phase, the second E-field component in 5 can be replaced with sine:

$$ \begin{align} \begin{bmatrix}E_{\text x} \\ E_{\text y} \\ E_{\text z} \end{bmatrix} ~=~ \begin{bmatrix}E_{0 \text x} \, \cos(\omega \, t - k\,z) \\ E_{0 \text y} \, \sin(\omega \, t - k\,z) \\ 0 \end{bmatrix} \end{align} $$

Another condition that a circularly polarized wave must meet is:

Condition #3 - for a circularly polarized wave

The amplitudes $E_{0 \text x}$ and $E_{0 \text y}$ must be equal: $ E_{0 \text x} = E_{0 \text y} := E_0$.

With the third condition E-field 6 becomes:

$$ \begin{align} \begin{bmatrix}E_{\text x} \\ E_{\text y} \\ E_{\text z} \end{bmatrix} ~=~ E_0 \, \begin{bmatrix}\cos(\omega \, t - k\,z) \\ \sin(\omega \, t - k\,z) \\ 0 \end{bmatrix} \end{align} $$

The E-field 7 corresponds exactly to the polar representation. Thus, when the time $t$ changes, the E-field vector $\boldsymbol{E}$ rotates in the $x$-$y$ plane (see illustration 2). This is where the term "circular" comes from. Along the $z$-axis the E-field vector thus spirals.

A right-circular polarized E-field wave.

If the circularly polarized plane wave is viewed orthogonal to the $x$-$y$ plane in the propagation direction, the E-field vector rotates leftwards for the observer. Therefore the E-field vector 7 is called a left-circularly polarized wave (or short: $\sigma^{-}$ wave).

$$ \begin{align} \boldsymbol{E} ~=~ E_0 \, \begin{bmatrix} \cos(\omega \, t - k\,z) \\ \sin(\omega \, t - k\,z) \\ 0 \end{bmatrix} \end{align} $$

If cosine and sine are interchanged in 7, the field vector for the described observer turns clockwise, as shown in illustration 2. This wave is called a right-circularly polarized wave (or short: $\sigma^{+}$ wave):

$$ \begin{align} \boldsymbol{E} ~=~ E_0 \, \begin{bmatrix} \sin(\omega \, t - k\,z) \\ \cos(\omega \, t - k\,z) \\ 0 \end{bmatrix} \end{align} $$

Now you should have a theoretical understanding of the definitions of linearly and circularly polarized plane waves.

How can you experimentally check whether a wave is transverse?

In contrast to longitudinal waves, transverse waves can be polarized. You can use this, for example, to prove that an electromagnetic wave propagates transversely. To do this, you send the wave through a polarization filter, which polarizes a wave linearly in the x-direction. If the wave is actually transverse, then it will be linearly polarized in the x-direction after passing through the polarization filter.

After passing through the first polarizer, you send the wave through the second polarizer (also known as the analyzer). This polarizes the wave linearly in the y-direction and only lets through the part of the wave that oscillates in the y-direction.

If the wave can no longer be detected after passing through the analyzer, then it was a transverse wave. After passing through the first polarizer, it only had one component in the x-direction. After passing through the analyzer, it was extinguished.
If the wave can be detected after passing through the analyzer, then it was a longitudinal wave.