Circular Polarization in Electrical Engineering: Examining the Poincaré Sphere

[ask_LXXXVI]

*I am using the conventions of circular polarisation according to electrical engineering , not the one used in optics*

Let us take a uniform plane TEM wave traveling in +z direction which is composed of two linearly polarised TEM waves , one whose electric field lies in X direction , the other whose electric field lies in Y direction . Let us take the case of circular polarisation
so we take ,

E_x = E₀ cos($$\omega$$ t - $$\beta$$ z) a_x
E_y = E₀ cos($$\omega$$ t - $$\beta$$ z + $$\pi$$\2) a_y

Now the resultant TEM wave has the Electric field vector left handed circularly polarised .
On the Poincare Sphere this will be given by the north pole point.
Suppose we had same wave ,but traveling in -ve z direction.
E_x = E₀ cos($$\omega$$ t + $$\beta$$ z) a_x
E_y = E₀ cos($$\omega$$ t + $$\beta$$ z + $$\pi$$\2) a_y

My doubt is :- is the wave left handed circularly polarised or right handed circularly polarised ? And where on Poincare sphere is it located?

 

[Andy Resnick]

AFAIK, if you've defined the first as right-handed, the second is left handed, which is located at the 'south pole' of the Poincare sphere.

My go-to book (Azzam and Bashara's 'ellipsometry and polarized light') is not here, so I'm going by memory.

 

[ask_LXXXVI]

yes , if one of them is on South Pole the other is on North Pole, if we find out the sense of polarisation manually.

But According to me while plotting on the Poincare sphere analytically , using the properties of Ex and Ey we simply use their ratios and relative phase difference . We don't take into account direction of propagation . So won't we get same point ?

Please can you give step by step directions as to how you will analytically plot the state on the sphere simply using the equations for Ex and Ey , not by drawing the polarisation ellipse and manually finding sense of rotation . Will you take into account direction of propagation.

 

[ask_LXXXVI]

please help in resolving my doubts.