When it comes to the electromagnetic wave equation in linear inhomogenous media, I find many sources who will just use the usual formula: $$ (\Delta - \frac{\epsilon \mu }{c^2} \frac{d^2}{dt^2})\vec{E} = 0 $$ and render $\epsilon$ and $\mu$ being position dependend: $$ (\Delta - \frac{\epsilon (\vec{x}) \mu (\vec{x}) }{c^2} \frac{d^2}{dt^2})\vec{E}(\vec{x}) = 0 $$ Why are we allowed to do that? When I derive the wave equation from Maxwell's equations, I require $\nabla \cdot \vec{E} = 0$, but in inhomogenous Media the equation is $$\nabla\cdot (\epsilon(\vec{x}) \vec{E}(\vec{x}))=0 $$ which leads to $$ \nabla \vec{E} = \nabla( \log{\epsilon} ) $$ This would be an additional term in the wave equation, and it's similar with $\mu$. Can somebody tell me why this additional terms are neglected, what approximation is made and when this is valid?
1 Answers
You are correct that there is a term on the RHS of $-\nabla(\vec{E}\cdot \nabla \ln \epsilon)$.
In order to neglect this you compare it's magnitude with $\nabla^2 \vec{E} = -k^2 \vec{E}$.
So, just looking at magnitudes, we can say that the RHS of your wave equation is $\sim 0$ if $$ |\nabla(\vec{E}\cdot \nabla \ln \epsilon)| \simeq kE \nabla \ln \epsilon \ll k^2 E$$ $$ \nabla \ln \epsilon \ll k $$ $$ \nabla \epsilon \ll k\epsilon =2\pi \frac{\epsilon}{\lambda}$$
So as long as a typical length scale for a change in $\epsilon$ is much longer than the wavelength of light, then this approximation is good. You certainly could not use this approximation (re your comment) where there are abrupt changes in refractive index, where you would have to employ the usual continuity conditions for the electromagnetic fields.
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