Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I was trying to solve an exercise in classical optics, consisting in finding the type polarization of certain field profiles. And, by analyzing this one: $$E_x(z,t) = |E|sin(kz-\omega t)\\ E_y(z,t)=|E|sin(kz-\omega t +\frac{\pi}{4})$$ I first found that it was the superposition of a elliptical polarization and a linear one by splitting the $sin$ in two, then I plotted it and discovered that in fact it's an inclined ellipse. It brought me to the conclusion that superposing a defined longitudinal and an elliptical gives another elliptical one. Now, I recall hearing that the elementary polarization are the two circular ones, which should imply that any elliptical or longitudinal can be decomposed into a sum of circular polarization, am I right? Because formally it's not evident to me.

ps: Also, what do the terminology TM and TE labeling the polarization of single photons refer to? I encountered them a few times, but they were never defined.

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

To answer your first question - Basically, you can use any two orthogonal polarizations as your 'basis' to describe any other polarizations. Like light polarized along $x$ abd $y$, or light that is circularly polarized in opposite directions.

First, the following is a general expression for two waves that are out of phase : $$E_{1} = E_{0x} \cos(kz - \omega t)$$ $$E_{2} = E_{0y} \cos(kz - \omega t + \phi)$$

where $\phi$ is the phase difference between the waves. These would be two linearly polarized waves. If they are $\pi / 2$ out of phase with each other, and you superpose them, the resultant will be circularly polarized light. If $E_{0x} = E_{0y}$ then,

$$E_\text{circ} = E_{0x}(\cos(kz-\omega t) + \sin(kz-\omega t)) $$

Conversely, you can just as easily see how you can get linearly polarized light from two circular polarizations moving in opposite directions.

$$E_\text{lin} = E_{0}(\cos(kz - \omega t) + \sin(kz - \omega t)) + E_{0}(\cos(kz - \omega t) - \sin(kz - \omega t))$$ $$\implies E_\text{lin} = 2E_{0} \cos(kz - \omega t) $$

You could try this with two orthogonal elliptically polarized waves to convince yourself further.

share|improve this answer
    
TM and TE: not quite. If light is incident on a surface, then no transverse polarization is ever perpendicular to that surface. –  ptomato Nov 2 '12 at 13:06
    
@ptomato - I'm not really clear on how TM and TE are defined, I'll probably edit that out of my answer. –  Kitchi Nov 2 '12 at 13:30
    
As far as I know, TM and TE are defined in terms of waveguides. For more info, see Wikipedia: en.wikipedia.org/wiki/Waveguide_%28electromagnetism%29 –  ptomato Nov 2 '12 at 15:47
    
I think if you see people using TM and TE to mean s and p polarization, it's probably wrong (although I admit to having done this myself in the past.) –  ptomato Nov 2 '12 at 15:48
1  
S is for German 'senkrecht'(perpendicular) and P for German 'parallel' (parallel), both referring to the plane of incidence. TM and TE (transverse magnetic and transverse electric) are equivalents to that. en.wikipedia.org/wiki/Polarization_(waves)#S_and_P_Polarization –  roadrunner66 Mar 12 '13 at 8:31
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.