Which has lower entropy?

1. Dec 1, 2009

Phrak

Which has lower entropy, a beam of unpolarized light, or this same beam split into polarized components?

2. Dec 2, 2009

ericgrau

The more possibilities, for example the more possibilities for the orientation of something, the higher the entropy.

3. Dec 2, 2009

fluidistic

Wow! I've only been introduced the change of entropy as being $$\Delta S = \int \frac{dQ}{T}$$ which obviously cannot be applied here.
I'd love to learn any other definition of it.
My intuition tells me that the unpolarized light is the one with higher entropy, as suggested by ericgrau.

4. Dec 2, 2009

Phrak

Thermodynamics was my worst subject. I don't know how I managed to pass, so I'm very baffled.

If one can split an unpolarized beam into two polarized beams without significant loses, it would seem to violate the second law of thermodynamics in some way. Can someone help me overcome my ignorance?

5. Dec 3, 2009

ericgrau

Polarized light isn't as bright, that may be it. If anything changing part of the light into heat and reducing the amount of light energy (which is relatively low entropy) may increase overall entropy.

Last edited: Dec 3, 2009
6. Dec 3, 2009

cesiumfrog

Neither.

Say you arrange that one polarisation is reflected, the other transmitted, by some optic. Now, if both beams are totally reflected back, won't they perfectly recombine? Aren't the polarisations independent degrees of freedom right from the beginning?

7. Dec 11, 2009

Phrak

Well... I did say it was my worst subject. How are they independent?

8. Dec 11, 2009

Andy Resnick

Recall that monochromatic radiation (of any poalrization state) does not have a thermodynamic temperature associated with it- only blackbody radiation does.

Even so, it seems logical to think a randomly polarized beam has a higher entropy than a fully polarized beam because polarization is a statistical measure of the beam properties. Splitting the randomly polarized beam into two orthogonal components is not the problem, but recombining them is- it is surprisingly tricky to generate a randomly polarized beam (usually moving ground glass surfaces or other scattering processes are involved).