I need help in finding an expression for the instantaneous electric and magnetic field of unpolarized light in order to write down and evaluate the time-averaged norm of the Poynting vector (i.e. the intensity of unpolarized light). I expect this to be a superposition of some basis functions, but how many and how do they look like? Are these linear polarized waves with random orientation and phase shift?

$$\bar{E}(t,\bar{x})=\sum_{i}\bar{E}_{0,i}e^{i(\bar{k}\cdot\bar{x}-\omega t+\delta_{i})}$$ with $\|\bar{E}_{0,i}\|=\|\bar{E}_{0,j}\|$ ?


Unpolarized light is not a superposition of polarized light (which would again be polarized) but a mixture of polarizations in all directions.

It cannot be described by a single solution of the Maxwell equations. Instead one needs for proper modeling the statistical version of electrodynamics. See the book on quantum optics by Mandel and Wolf (its first half is about classical modeling of light).

Edit: Note that mixtures and superpositions are completely different things. A superposition of modes gives another mode. On the other hand, a mixture is a statistical notion. It means that the mode description contains random fluctuations (often thought to consist of extremely high frequency noise that cannot be resolved at the scale of the wavelengths of interest) over which one must average the (typically quadratic) observable responses. This cannot be modelled by a superposition.

  • $\begingroup$ I'd also recommend Statistical Optics by J. W. Goodman. Goodman is excellent, always! $\endgroup$ – Colin K Jul 13 '12 at 17:37
  • $\begingroup$ But doesn't a "mixture of polarizations in all directions" mean a "superposition of polarized light with polarization in all directions"? $\endgroup$ – Wox Jul 16 '12 at 7:56
  • $\begingroup$ @Wox: No. Such a superposition would cancel the signal completely. (Exercise: Write down the superposition as an integral, and work out the value of the integral.) See the edit to my answer. $\endgroup$ – Arnold Neumaier Jul 16 '12 at 8:09
  • $\begingroup$ So if one would plot the electric field vector in a plane perpendicular to the propagation over a period of time, would you get a uniform distribution in a disk with a certain radius? (Edit: that would just cause the intensity to be zero, wouldn't it? I'm trying to get a hold of one of the mentioned books, but I don't have access to them.) $\endgroup$ – Wox Jul 16 '12 at 9:55
  • $\begingroup$ @Wox: Yes, even when plotting it over a single wavelength. (But one cannot measure that finely.) It causes the superposition to be zero, but not the intensity, as the latter is quadratic in the amplitude. (Just like ordinary scalar noise averages to zero, but its square averages to a positive variance.) The link en.wikipedia.org/wiki/Degree_of_coherence may help; if you replace $E$ by a vector, you get correlation matrices, which in the stationary case describe polarization. $\endgroup$ – Arnold Neumaier Jul 16 '12 at 11:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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