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If the electromagnetic field of an unpolarized plane wave is written as $$\bar{E}(t,\bar{x})=(\bar{E}_{0x}+\bar{E}_{0y}e^{i\delta(t)})e^{i(\bar{k}\bar{x}-\omega t)}$$ $$\bar{B}(t,\bar{x})=\frac{1}{\omega}\bar{k}\times\bar{E}(t,\bar{x})$$ where $\delta(t)$ is a random phase shift, then the intensity of this light is given by the time-average of the norm of the Poynting vector $$I(\bar{x})=\left<\|\bar{P}(t,\bar{x})\|\right>_{t}$$ $$\begin{split}\bar{P}(t,\bar{x})=&\frac{1}{\mu_{0}}\mathcal{R}e(\bar{E}(t,\bar{x}))\times\mathcal{R}e(\bar{B}(t,\bar{x}))\\ =&\frac{1}{\omega}\bar{E}_{0x}^{2}\cos^{2}(\bar{k}\bar{x}-\omega t)\bar{k}+\frac{1}{\omega}\bar{E}_{0y}^{2}\cos^{2}(\bar{k}\bar{x}-\omega t+\delta(t))\bar{k} \end{split}$$ $$\Leftrightarrow I(\bar{x})=\frac{1}{2}c\epsilon_{0}\bar{E}_{0x}^{2}+c\epsilon_{0}\bar{E}_{0y}^{2}\left<\cos^{2}(\bar{k}\bar{x}-\omega t+\delta(t))\right>_{t}$$ Can we simplify this further? Is the remaining average also $1/2$?

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You've got issues with your definition. – user2963 Jul 6 '12 at 16:08
That light is quite polarized. To get unpolarized light you need to average over an ensemble with random polarization. You've simply described some light with a very pure elliptical polarization – Colin K Jul 6 '12 at 16:27
You are right, I introduced the random phase shift to go from elliptical to unpolarized, but this is indeed not enough. But the question still remains, how would you calculate the intensity of unpolarized light? – Wox Jul 10 '12 at 10:00

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