# How to identify the types of polarization from EM waveforms

This has always been confusing to me, I have referred online from various sources regarding this yet I come across equations with description that sometimes seems to contradict what I have learned. I would hence like to get a summary of what these polarizations are and how they are represented in equations. So the various polarizations are: Plane polarized, Linearly polarized, Circularly polarized, Unpolarized and elliptically polarized. I will list out some pairs of equations so that it could be useful for someone to explain it.

a) $E_x=\frac {E_0}{\sqrt2}\cos(wt+kz), E_y=E_0\sin(wt+kz)$

b) $E_x=E_0\cos(wt+kz), E_y=E_0\sin(wt+kz)$

c)$E_x=E_1\sin(wt+kz), E_y=E_2\sin(wt+kz)$

d)$E_x=E_0\sin(wt+kz), E_y=E_0\sin(wt+kz+\frac {\pi}{4})$

• The term "wave equation" is restricted to the differential equations of the form $\frac{\partial^2}{\partial x^2}f=\frac{1}{c_2}\frac{\partial^2}{\partial t^2}f$ and related PDEs. The functions you show are waveforms. – Emilio Pisanty Apr 18 '17 at 14:20
• You are adding two sinusoid all functions in the xy plane so have a look at Lissajous figures and one of the many simulations with the x and y frequencies the same and see what happens when you change the amplitudes and the phase. – Farcher Apr 18 '17 at 14:24
• @EmilioPisanty Oh yes indeed. It's just that I have been used to calling them equations. – rahul rj Apr 18 '17 at 14:35
• @Farcher I remember learning some criteria regarding phase and amplitude to identify it So I am basically looking for that which I could refer to. – rahul rj Apr 18 '17 at 14:35
• @rahulrj Well, get un-used to that, then. It is decidedly off-standard terminology and it will only serve to confuse your communications with others. – Emilio Pisanty Apr 18 '17 at 14:46

I am not sure if there is a precise algorithm for determining the type of polarization. After all, there is an infinite number of possible polarizations. However, you can usually gather a lot about the polarization in question by looking at the waveforms of the electric field. Namely, you want to look at the shape of the waveform, the amplitude accompanying that shape, as well as the phase of the waveform.

What's more, it helps to conduct this analysis after imagining the oscillations as projected on the plane perpendicular to the directions of oscillation of the given electric field components. In your case, it will be the z-axis. Of course, this is the same as saying that you want to look at Lissajous curves

Now, lets consider the waveforms case by case.

b) $E_x=E_0\cos(wt+kz), E_y=E_0\sin(wt+kz)$ Suppose you only consider $E_x=E_0\cos(wt), E_y=E_0\sin(wt)$ since all $kz$ is telling you is that the wave is propagating along z-axis. Once you remove that, you see that these components will make the tip of the $\vec{E}$ vector trace out a circle in the plane perpendicular to the direction of propagation. We know it is a circle because the amplitudes of the components are equal.

a) $E_x=\frac {E_0}{\sqrt2}\cos(wt+kz), E_y=E_0\sin(wt+kz)$ Here, the idea is the same, except the amplitudes are not equal. This means we have a distorted circle! That is, of course, an ellipse!

c)$E_x=E_1\sin(wt+kz), E_y=E_2\sin(wt+kz)$ This is a linearly polarized field because $sin(wt)$ can be treated as a variable $\theta = sin(wt)$ for both components. That is, in-plane, the tip of this vector goes back and forth along a straight line. By tweaking the values of $E_1$ and $E_2$ one can change the direction of this line. For example, if $E_1 = 0$ then the light is polarized along the y-axis.

d)$E_x=E_0\sin(wt+kz), E_y=E_0\sin(wt+kz+\frac {\pi}{4})$ This one looks quite similar to the previous case, except for the phase. There is a pesky phase added to the y-component. What does that do? It turns out that this is also an elliptic polarization. This technique applies a quarter of a wavelength shift to one of the components, which causes the polarization to become elliptic. See Polarization types for details. To be more precise, in this case we have an eigth of a wavelength added. The effect is still the same. That is, an elliptical polarization is produced. The difference between this ellipse and that of a) is that the ellipse in a) will have its major and minor axes aligned with the directions of x and y axes (as seen on the plane perpendicular to the direction of propagation), whereas the one in d) will have its axes rotated relative to x and y axes by some angle.

• for the case in c, why is it not an ellipse since the amplitude $E_1 and E_2$ are not equal? and for the case in d) if the axes are simply rotated how does it make it an ellipse? – rahul rj Apr 18 '17 at 16:33
• @rahulrj The differing amplitudes in part (c) change the angle of the plane of polarization. I'm not quite sure what you are asking about (d). Part (a) describes an ellipse, and part (d) describes a different ellipse ... one that's rotated. – garyp Apr 18 '17 at 17:26
• @rahulrj in d) We do not rotate the axes. We say that the axes (major and minor) of the ellipse are rotated relative to the stationary coordinate axes. In part c) the axes of the ellipse and the coordinate axes coincide pairwise. – MadPhysicist Apr 18 '17 at 19:00
• @MadPhysicist so then you say case c) is an ellipse? – rahul rj Apr 19 '17 at 0:17
• @rahulrj c) is a linear polarization. – MadPhysicist Apr 19 '17 at 12:34