Some questions about polarization and intensity

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Discussion Overview

The discussion revolves around the properties of polarized electromagnetic waves, specifically focusing on the intensity of linear and circular polarization, the behavior of these waves when passing through polarizers, and the implications of their Poynting vectors. The scope includes theoretical considerations and mathematical formulations related to wave polarization.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states that the intensity of a linearly polarized electromagnetic wave is given by I=(1/2)(E_0^2)/c, while questioning the intensity of a circularly polarized wave with the same amplitude E_0.
  • Another participant corrects the first by stating that the time-average of the Poynting vector for a linearly polarized wave is actually (ε₀c/2)E_0².
  • A third participant suggests that the Poynting vector definition leads to a different expression for the average intensity of a linearly polarized wave, indicating a potential misunderstanding in notation or definitions among participants.
  • One participant explains that a circularly polarized wave can be viewed as a superposition of two linearly polarized waves that are 90 degrees out of phase, which may account for differences in expected Poynting flux.
  • There is a query regarding the intensity of a circularly polarized wave after passing through a linear polarizer, and whether a linearly polarized wave passing through a circular polarizer retains its intensity but with a rotated polarization.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between the intensities of linearly and circularly polarized waves, as well as the effects of polarizers. The discussion remains unresolved, with multiple competing interpretations of the underlying physics and mathematics.

Contextual Notes

Some participants acknowledge corrections to earlier claims, but the discussion includes unresolved mathematical steps and varying definitions of the Poynting vector and intensity. The implications of these definitions on the perceived intensity of polarized waves are not fully clarified.

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Hello everybody,
I have some questions:

I'm talking only about traveling waves, not standing waves.
If a linear polarized electromagnetic wave has electric field amplitude E_0, i know that its intensity is given by

I=(1/2) (E_0^2 ) /c

that is the Poynting vector averaged over a period.

But if I have the same E_0 amplitude in a circularly polarized wave, the Poynting vector remains constant so i'd expect to have an intensity

I= (E_0^2)/c

this difference sounds strange to me: am I right or am I making some mistake?

and also, if I send a circularly polarized wave through a linear polarizer, what will it be the intensity of the wave exiting from the polarizer?
And if I send a linearly polarized wave through a circular polarizer (circular dichroic filter), am I right if I say that the exiting wave will have the same intensity but with a linear polarization rotated in respect to the ingoing wave?

Thank you in advance and sorry if my english is not very good.
 
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Actually, the time-average of the Poynting vector of a linearly polarised plane wave is equal to:
\frac{\varepsilon_0 c}{2} E_0^2
 
to BruceW:
I was certainly wrong , but maybe you too, or maybe you were just using a notation that I don't know:
if you define Poynting vector as
\vec S= \frac{1}{\mu } \vec E \times \vec B

then you have, for linear polarization,

<S>= \frac{1}{2} \sqrt{\frac{\varepsilon }{\mu}} E_0^{2}

thank you for your correction.
As you see my questions are still valid, correcting the formulas i previously wrote.
 
Last edited:
A circularly polarized wave can be thought of as a superposition of two linearly polarized waves that are 90 degrees out of phase with each other (which counts as "oppositely polarized" for linear polarization). So you shouldn't expect the Poynting flux of a single linearly polarized wave to equal that of a superposition of two. Thus you are right, but there is no problem with it. Also, when you pass through a linear polarizer, it just selects the component with that linear polarization.
 

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