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 $\nabla \vec{E} = 0$, but in inhomogenous Media the equation is $$\nabla (\epsilon(\vec{x}) \vec{E}(\vec{x}))=0 $$ 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?

  • $\begingroup$ Yes, you are correct. There is a term $-\nabla (E \cdot \nabla \ln \epsilon)$ on the RHS. $\endgroup$ – Rob Jeffries Nov 30 '16 at 23:17
  • $\begingroup$ I have also faced this annoying problem. I think people implicitly assume $\epsilon$ is slowly varying or something and say $\nabla \epsilon \sim 0$. Check out these notes: opt.zju.edu.cn/zjuopt2/upload/resources/…. -- They were very helpful for me, hopefully they will be for you too. Equation (5.2--14) in particular is what you are looking for. $\endgroup$ – Arthur Suvorov Nov 30 '16 at 23:19
  • $\begingroup$ I have a difficulty to think of this at the edge of 2 media, where $n$ is not continuous, and $\nabla n$ would be some kind of $\delta$ function. $\endgroup$ – Quantumwhisp Nov 30 '16 at 23:31

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.

| cite | improve this answer | |
  • $\begingroup$ What still buffles me is that you use the same approximation in deriving the Eikonal Equation for geometric optics, and geometric optics work out just fine, even it it comes to abrupt changes in the refractive index (Snell's law for refraction at a surface would be a good example for this). Maybe I should post this whole thing in another question. $\endgroup$ – Quantumwhisp Dec 1 '16 at 15:00
  • $\begingroup$ @QuantumWisp I can derive the laws of geometric optics by treating the boundary condition as sharp, with an abrupt change in $\epsilon$ at the boundary and a uniform (but different) $\epsilon$ either side. The problem you have posed would not necessarily enter into it. $\endgroup$ – Rob Jeffries Dec 1 '16 at 16:58
  • $\begingroup$ I don't see why the problem I have posed doesn't enter it: I derive the laws of optics, neglecting the said terms, I use them in a situation were the terms mustn't be neglected, and still everything works out. That's a contradiction to me. $\endgroup$ – Quantumwhisp Dec 1 '16 at 17:29
  • 1
    $\begingroup$ @Quantumwhisp The derivation of the standard laws of geometric optics does not even involve the wave equation in a direct way. You posit a travelling wave either side of a discontinuity and apply the boundary conditions for electromagnetic fields. On either side of the boundary the wave equation is homogeneous (if these are LIH media) with solutions that are standard travelling waves. This does result in discontinuities in the wave properties (e.g. in the wavevector and field amplitudes). Anyway, that goes beyond what you were asking in your question (which I have answered). $\endgroup$ – Rob Jeffries Dec 1 '16 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.