Physics Asked on January 29, 2021
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 $$omega = frac{1}{2} ( epsilon epsilon_0 E^2 + mumu_0 H^2)$$
The time average distributes linearly so using $langle f cdot g rangle $ with $f=g=E$ yields $$langle E^2rangle = frac{1}{2}left(frac{E_0^2}{1}right)$$
Similarly for $H$ I get $$langle H^2 rangle = frac{1}{2}left(frac{E_0^2}{c^2}right)$$ using the fact that $vec{H} =frac{1}{c} (hat{k} times vec{E_0}) e^{i(kx-wt)}$.
When I sum this two quantities I get $$omega = frac{1}{2} E_0^2 left(frac{epsilonepsilon_0}{2} + frac{mumu_0}{2c^2}right)$$ 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.
$$vec{H} = frac{1}{c}(hat{k}timesvec{E}_0)e^{i(kx-omega t)}$$ is incorrect. Perhaps you have $vec{H}$ and $vec{B}$ confused. The correct formulae are $$vec{B} = frac{1}{c}(hat ktimesvec{E})$$ and $$vec{H} = frac{1}{eta}(hat ktimesvec{E})$$ where $eta=sqrt{mumu_0/epsilonepsilon_0}$ is the wave impedance of the medium. With either one, the magnetic energy density $$w_m = frac{1}{2}mumu_0 H^2 = frac{B^2}{2mumu_0}$$ will give you the correct result.
Answered by Puk on January 29, 2021
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