B The physics explanation of ~bendy~ rays

[etotheipi]

Take a medium with $n = n(y)$ and define the functional as usual$$T[y] = \int_{\mathcal{C}} \frac{ds}{cn^{-1}} = \frac{1}{c} \int_{x_1}^{x_2} dx \, n(y)\sqrt{1+ y'^2}$$along $\mathcal{C}$ between $\mathcal{P}_1 \overset{.}{=} (x_1, y_1)$ and $\mathcal{P}_2 \overset{.}{=} (x_2, y_2)$. Because the integrand does not depend explicitly on $x$, Beltrami identity gives$$ \frac{y'^2 n(y)}{\sqrt{1+y'^2}} - n(y) \sqrt{1+y'^2} = \frac{n(y)}{\sqrt{1+y'^2}} = K \in \mathbb{R}$$Consider a ray entering the medium horizontally. When $n(y)$ is not a constant function, the trajectory $y = C$ where $C$ is a constant will not make the functional extremal. Instead you must solve $y' = \sqrt{\left(\frac{n(y)}{K}\right)^2 - 1}$ with the IC $(x_1, y_1) = (0,C)$ or something, which gives the nice bendy trajectory.

Physically [without appealing to variational calculus] if a ray enters such a medium horizontally, what causes the ray to bend away from a possible horizontal path $y=C$ with constant refractive index $n(C)$ [which would, at first glance, seem like the more reasonable path]? How does EM theory describe the same behaviour, for instance?

 

[Dale]

Remember that a ray is a shorthand approximation for a wave front. That wave front is spread out in the direction perpendicular to the ray. Thus some of the wave front is moving slower than the rest, and so the wave front angles, and therefore the direction perpendicular to the wave front also turns, which is the ray.

 

[etotheipi]

Remember that a ray is a shorthand approximation for a wave front. That wave front is spread out in the direction perpendicular to the ray. Thus some of the wave front is moving slower than the rest, and so the wave front angles, and therefore the direction perpendicular to the wave front also turns, which is the ray.

Cool, thanks! I understand now, before I was thinking of a ray as something you could make arbitrarily thin so that it would only "see" one refractive index.

 

[vanhees71]

For the relation between wave optics and ray optics, look for "eikonal approximation". You find a very good treatment in Sommferfeld, Lectures on theoretical Physics, vol. 4 (Optics).

 

[sophiecentaur]

Remember that a ray is a shorthand approximation for a wave front. That wave front is spread out in the direction perpendicular to the ray. Thus some of the wave front is moving slower than the rest, and so the wave front angles, and therefore the direction perpendicular to the wave front also turns, which is the ray.

Cool, thanks! I understand now, before I was thinking of a ray as something you could make arbitrarily thin so that it would only "see" one refractive index.

An example of 'bending' rays where wavelengths are bigger than a house, is in the mechanism of Medium Frequency Ground Wave Propagation. The loss in the ground (not a perfect conductor) produces a lag in the wavefront at ground level and will keep 'pulling' the radiated Power (the "rays") down towards the ground and so it follows the curvature of the Earth. There would still be propagation with a perfectly conducting Earth but the rate of drop off with distance would be higher. This tilting downwards of the wave front can be dramatic in the 'signal shadow' of a large town, in which the signal absorption is high. A few km beyond the town, the signal strength rises again as Power is angled downwards to the ground and the shadow is filled in again.