# Troubles verifying $\langle \omega \rangle = \frac{1}{2}\epsilon \epsilon_0 |\vec{E}_0|^2$

I'm trying to prove that the time average of the energy density $$\omega$$ is equivalent to

$$\langle \omega \rangle = \frac{1}{2}\epsilon \epsilon_0 \|\vec{E}_0\|^2$$

for a plane EMW. My approach is to use that

$$\langle f \cdot g \rangle = \frac{1}{2} Re(f \cdot g^*)$$

This comes from Griffiths' Introduction to Electrodynamics:

and also $$\omega = \frac{1}{2} ( \epsilon \epsilon_0 E^2 + \mu\mu_0 H^2)$$

1. The time average distributes linearly so using $$\langle f \cdot g \rangle$$ with $$f=g=E$$ yields $$\langle E^2\rangle = \frac{1}{2}\left(\frac{E_0^2}{1}\right)$$

2. Similarly for $$H$$ I get $$\langle H^2 \rangle = \frac{1}{2}\left(\frac{E_0^2}{c^2}\right)$$ using the fact that $$\vec{H} =\frac{1}{c} (\hat{k} \times \vec{E_0}) e^{i(kx-wt)}$$.

3. When I sum this two quantities I get $$\omega = \frac{1}{2} E_0^2 \left(\frac{\epsilon\epsilon_0}{2} + \frac{\mu\mu_0}{2c^2}\right)$$ but this puzzles me. I know it shouldn't be this way; otherwise I would get $$\mu_0^2 = \mu/\epsilon$$ and that's not true from what I know.

$$\vec{H} = \frac{1}{c}(\hat{k}\times\vec{E}_0)e^{i(kx-\omega t)}$$ is incorrect. Perhaps you have $$\vec{H}$$ and $$\vec{B}$$ confused. The correct formulae are $$\vec{B} = \frac{1}{c}(\hat k\times\vec{E})$$ and $$\vec{H} = \frac{1}{\eta}(\hat k\times\vec{E})$$ where $$\eta=\sqrt{\mu\mu_0/\epsilon\epsilon_0}$$ is the wave impedance of the medium. With either one, the magnetic energy density $$w_m = \frac{1}{2}\mu\mu_0 H^2 = \frac{B^2}{2\mu\mu_0}$$ will give you the correct result.