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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