Polarization and Poincare circle.


by yungman
Tags: circle, poincare, polarization
yungman
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#1
Feb7-13, 04:28 AM
P: 3,843
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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Andy Resnick
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#2
Feb7-13, 10:35 AM
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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.
yungman
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#3
Feb8-13, 01:55 PM
P: 3,843
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

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


If I understand your question, it's a sign convention.
yungman
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#5
Feb8-13, 04:00 PM
P: 3,843
Quote 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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