4
$\begingroup$

Suppose that I shine light on a surface and none of it will be absorbed or transmitted, and the spectrum doesn't change (so that all that's left to determine is the power reflected from the surface as a function of solid angle).

Is there a sense in which diffuse reflection (following Lambert's cosine law) is maximizing the entropy change of the light from before the reflection to after it?

Improve this question
$\endgroup$
1
  • $\begingroup$ If you are looking for an optical entropy maximization machine, look no further than a black body. $\endgroup$
    – CuriousOne
    Commented Apr 20, 2016 at 16:53

1 Answer 1

Reset to default
2
$\begingroup$

It's maximizing the angular entropy in the sense that far more "available" angle vectors are inhabited. I suspect you need to be careful with the word "entropy" here. For example, the photons are not down-converted into a larger number of photons of longer wavelength (aka 'heat death of the universe').

Improve this answer
$\endgroup$
2
  • $\begingroup$ Hi Carl, is it possible to be more precise? We do not spread the light out evenly among all solid angles; instead the intensity per unit solid angle obeys Lambert's law. $\endgroup$
    – Mark Eichenlaub
    Commented Apr 20, 2016 at 15:51
  • $\begingroup$ Well, that's true too, so you may not really be maximizing even the solid-angle-entropy. $\endgroup$
    – Carl Witthoft
    Commented Apr 20, 2016 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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