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Polarization and Poincare circle.

 
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Feb7-13, 04:28 AM   #1
 

Polarization and Poincare circle.


What is the theory behind mapping of the latitude and longitude of the sphere in the Poincare Circle to the polarization of the TEM wave?

That is, why:

1) Linear polarization when ε=0 deg?
2) Circular polarization when ε=+/- 45 deg?
3) Elliptical when ε is not 0 or +/- 45 deg?
4) RH rotation if ε=-ve. and LH rotation if ε=+ve.?

Where 2ε= latitude.

Thanks

Alan
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Feb7-13, 10:35 AM   #2
 
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The first step is to represent the polarization state by a complex number: the general elliptic state with azimuth θ and ellipticity ε is combined into χ= tan(ε +π/4)exp(-i*2θ.) This maps polarization states to the "cartesian complex plane", and recall that -π/2 ≤ θ < π/2 and -π/4 ≤ ε ≤ π/4. χ= 0 refers to left-circularly polarized light, χ=∞ is right-circularly polarized light.

To construct the Poincare sphere, perform a stereographic mapping of the plane to a unit sphere: latitudes on the sphere then correspond to circles of constant ε on the complex plane and longitude corresponds to lines of constant azimuth on the plane.

Azzam and Bashara's "Ellipsometry and Polarized Light" is an excellent resource for this material.
Feb8-13, 01:55 PM   #3
 
Thanks for your reply.

The book is way to expensive as This is only a small part of my study in antenna theory.

Can you show me how the polarity of the ellipticity angle relate to the direction of rotation? That is, why +ve ε gives rise to Left hand rotation and -ve ε gives rise to Right hand rotation.

Thanks
Feb8-13, 02:03 PM   #4
 
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Polarization and Poincare circle.


If I understand your question, it's a sign convention.
Feb8-13, 04:00 PM   #5
 
Quote by Andy Resnick View Post
If I understand your question, it's a sign convention.
No so much about convention, but rather why ε affect the direction of rotation.

[tex]\epsilon=\frac {E_{max}}{E_{min}}[/tex]

How do you justify +ve or -ve of [itex]E_{max},E_{min}[/itex]?

Thanks
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