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Regardless of computational cost, light is a kind of electromagnetic wave, so it can be simulated with Maxwell's equations. If we want to simulate light with Maxwell's equations, we need to express the electric field vector of light source with a formula.

If the light source is polarized, this task won't be hard, but what if I want to simulate natural light: to be precise, unpolarized light?

Is there any approximate formula for it?

If the final result I want to get is just light intensity, can I simply replace the unpolarized light source with a polarized one?

Well, to be honest, this question came to my mind when I read a paper that simulate light in a nanometric optic probe with Maxwell's equations and the incident light in that paper is polarized, I just want to know if unpolarized light is also available for Maxwell's equations.

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3 Answers 3

up vote 6 down vote accepted

To simulate unpolarized light, you need to do two separate simulations using Maxwell's equations.

In the first simulation, assume the incoming light has some polarization (any polarization will do). In the second simulation, assume that the light has the opposite polarization (y is opposite to x, right-circular-polarized is opposite to left-circular-polarized, etc.).

Now imagine that at every moment, randomly, the light is switching back and forth between these two simulations, too fast to measure. So the intensity for unpolarized light is the average of the intensity in the two simulations, the optical force is the average of the forces in the two simulations, etc. etc.

For more details see this answer and comments: http://physics.stackexchange.com/a/31975/3811

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The reason for this, if it isn't clear, is that if you do the simulation for just one polarization, then you get some interesting effects like being able to see through specular reflections [en.wikipedia.org/wiki/Brewster's_angle], which might be unintended. –  lionelbrits Nov 10 '13 at 11:24

To answer your question without getting boggled down in quantum mechanics, we can assume the light is of sufficient intensity(*) or that we are talking about run-of-the-mill radio waves (again, with sufficient intensity). Well, regardless of the nature of the light, at any point in space there will be a definite $\vec E$ and $\vec{B}$ directions, so Maxwell's equations will apply nonetheless.

So what does it mean for the light to be polarized? I take it to mean that there is, over some timescale, and some distance scale, some correlation between the direction in which $\vec{E}$ is aligned. In other words, it stays fixed in some region or changes in a highly symmetrical manner (for circularly polarized light). But with the caveats mentioned earlier, you can always choose your timescale and distance scale small enough that it stays correlated - especially as you start talking about distances close to the minimum wavelength of the light you're talking about (even if it's not monochromatic). So let's amend our definition to mean that it's correlated over some distance scale significantly larger than the wavelength, i.e., $\delta D \gg \lambda$. Roughly, a large coherence length.

But to answer your question, no, you can simulate it just fine.

(*)The extreme case here is laser light. see Speckle.

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Er… sorry for my poor English, but what do you mean by saying "to answer your question, no, you can simulate it just fine"? –  xzczd Nov 9 '13 at 13:06
    
Oh, I see. You mean "you can't simply replace the unpolarized light source with a polarized one", right? –  xzczd Nov 11 '13 at 8:03

Yes, Maxwells equations describe any propagation of any light for virtually all technical applications one can imagine.

They can not be used when it comes to low intensities, i.e. at the single or few photon level.

Also generation of light can not always be described by Maxwells equation (e.g. thermal radiation, spontaneous and stimulated emission). For that quantum mechanics is necessary.

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Sorry, my question seemed to be a little obscure, see my edit. –  xzczd Nov 9 '13 at 2:52

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