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In "Principles of Optics" by Max Born and Emil Wolf, the eikonal equation is derived from Maxwell's equations. In there, the authors use the approximation of low wavelengths / high wavenumbers, to neglect some terms in the wave equation and arrive at the Eikonal equation. One of those neglected terms is the term $\nabla \log \epsilon$, and the reason for that is that it is smaller than the "wavenumber" $k$ of the solution. We then proceed to derive the Eikonal equation and fermat's principle and it is happily ever after used in geometric optics.

I may show some steps in the derivation: In general linear inhomogenous media, the wave equations that follow from Maxwell's equations are given as: $$ \Delta \vec{E} - \frac{\epsilon \mu}{c^2} \ddot{\vec{E}} + (\nabla \log \mu ) \times (\nabla \times \vec{E}) + \nabla (\vec{E} \nabla \log \epsilon) = 0 $$ Plugging in $\vec{E} = \vec{E}_0(\vec{r})e^{i(k\chi(\vec{r} - \omega t)}$, looking only at the imaginary part of the equation and neglecting everything that isn't of order $k^2$, you arrive at the Eikonal equation. In a similar way (see my other question on that), you make an approximation to arrive at the commonly used "approximated wave equation" (I will for now call it like that) in a medium:

$$ \Delta \vec{E} - \frac{\epsilon (\vec{x}) \mu (\vec{x})}{c^2} \ddot{\vec{E}} = 0 $$

So here is my question: Think of the case of a sharp material change in surface (that would be a discontinuity of $\epsilon$), and of a plane wave that is refracted at this plane (we all know Snell's law of refraction is applicable here). All in all, the whole solution in this case (incident wave and refracted wave and reflected wave) do have a discontinuity at the plane.

This solution was obtained by solving the wave equation for both materials and then glueing them together, using the right continuity conditions. Similar, one could solve the given general wave equation given at first. Our "approximated wave equation" is not able to describe this solution, it will not yield a discontinuity. So far everything's fine, we made an approximation (neglecting the terms containing derivations of $\epsilon$), we can't expect the equation to still be as powerful as it has been before.

But then (and that really buffles me, and is the core of my question): Making stronger assumptions and neglecting even more terms, we arive at a theory (geometrical optics) that can describe refraction at a surface perfectly well.

Why is it that geometrical optics can describe something like refraction better than the "approximated wave equation", when geometrical optics are even a bigger approximation of said wave equations? And why do geometrical optics describe surfaces of edges of different media, when in there you have discontinuities in the refractive index, and thus can't make those said approximations?

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Quantumwhisp
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    What do you mean by "geometrical optics can describe something like refraction better than the 'approximated wave equation' "? What does geometric optics describe that the 'approximated wave equation' does not? When you solve differential equations you always have to apply boundary conditions to get particular solutions. – ProfRob Dec 07 '16 at 17:45
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    You don't have to apply boundary conditions for the boundary surfaces when deriving the laws of refraction from fermats principle. But you do when solving the wave equations. – Quantumwhisp Dec 07 '16 at 20:16
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    I don't understand. To derive Snell's law from Fermat's principle requires that the speed of light abruptly changes at the boundary. How is that not a boundary condition? The laws of geometric optics also tell you little about the intensity or polarisation state of the refracted light. – ProfRob Dec 07 '16 at 23:51
  • It isn't, it's just a discontinuity. The way you calculate the trace of the lightray doesn't change, no ma tter if you have a continuous change of $n$, or a discontinuity.
    When treating waves, the wave equation (+ knowledge of $n(\vec{x})$ allone isn't sufficient, you need an additional statement about wave at the boundary surface (the boundary-condition). From the wave equation, you get geometric optics, which again DON'T need additional information: Fermat's Principle + knowledge of $n(\vec{x})$ are sufficient here.     – Quantumwhisp Dec 08 '16 at 09:52
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    You used Maxwell's equations to derive the wave equation. You use Maxwell's equations to get the relationships for field at a boundary. I don't think you can separate them. But yes, there must be some trade off that allows you to get directions using Fermat's principle, but not amplitudes. – ProfRob Dec 08 '16 at 11:12
  • I think I'm getting what you tell me: I don't need the boundary conditions at the boundary if I just wan't to derive directions (that would be snell's law). That's why geometric optics which are an even bigger approximation also yield snells law without using the boundary conditions. – Quantumwhisp Dec 08 '16 at 11:30

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