Snell's Law for Rotational Motion

It has been burned into our minds that the index of refraction must be the inverse of the velocity in units where $c=1$. This has many ramifications like Snell's law here the ratio of the sine of the angle with respect to the normal and the velocity of the ray is constant. And in nonhomogeneous, but spherically symmetric, media where the index of refraction varies with the radial coordinate, the frequency of the wave remains constant making the velocity equal to the wavelength. 

Solving the orbital equation,

$$\frac{d\theta}{dr}=\pm\frac{K}{r\sqrt{(nr)^2-K^2}},$$

for the angular momentum $K$ we find

$$K=\pm n\frac{r^2\theta^{\prime}}{\sqrt{1+r^2\theta^2}}=\pm nr\sin\alpha,$$

where the prime stands for differentiation with respect to $r$, and $\alpha$ is the angle that the ray makes with the tangent to the ray at the point $P$, in the following diagram.

Bouger
Bouguer's formula showing that $n\delta=\mbox{const}$ in a medium with spherical symmetry.

 

The above equation states

$$K=\pm nr\sin\alpha$$

is constant along one and the same ray.  This is commonly referred to as Bouguer's formula, which as Born and Wolf [Principles of Optics, p. 122] tell us, this "is the analogue of a well-known formula in dynamics, which expresses the conservation of angular momentum of a particle moving under a central force."

But Newton tells us that this law is

$$K=vr\sin\alpha,$$

where $v$ is the tangential velocity at point $P$. Since $r\sin\alpha=\delta$, the perpendicular distance from the tangent of the ray to the origin $O$, the product $n\delta+\mbox{const}.$ 

Now a comparison of the two formula for $K$ immediately leads to $n=v$, and not its inverse! There have been almost an infinite number of papers drawing attention to Newton's second law in optics, but always with $n$ the inverse of the velocity, and the claim that optics is the "zero-energy" analogue of mechanical orbits. One even goes so far as to write the angular momentum as $K=n^2r^2\dot{\theta}$ [Evans, Rosenquist, "'$F=ma$' optics", Am J Phys, 54 (1986) 876-883.]

Newton also defined the slope $z$ as

$$z=1/r\theta^{\prime}=\cot\alpha.$$

so that

$$1+z^2=\left(\frac{nr}{K}\right)^2-1=1/\sin^2\alpha,$$

leading to the relation

$$nr\sin\alpha=n_0r_0\sin\alpha_0=K,$$

where $r_0$ and $\alpha_0$  are integration constants that determine the origin of the ray, the constant $K$ at its direction at $P$. 

With $r=r_0=\mbox{const}$, we have what looks like Snell's law:

$$n\sin\alpha=n_0\sin\alpha_0.$$

In ordinary optics,  when light passes from a medium of high index of refraction to a medium with a low index of refraction, the wavelength increases because the leading edge of the wave enters first and speeds up while the trailing edge lags behind. The wave begins to stretch out so that the wavelength and speed of light change but the frequency remains constant. Here, no such thing happens because $n=r\omega$, where $\omega$ is the angular speed, so that under the same condition $r=\mbox{const}$, Snell's law becomes

$$\omega\sin\alpha=\omega_0\sin\alpha_0.$$

Rotary motion gives the impression of a temporal variation where the wavelength doesn't change but the frequency, $\omega$, and speed do. Since the leading edge of the wave never encounters a new value of $n$ before the trailing edge does, the wavelength will not change. A change in speed therefore entails a change in the frequency.

As an example of the equivalence between index of refraction and angular velocity, we can derive the inverse square law without recourse to Kepler's area law. Since

$$n^2=\frac{1}{r}-1,$$

the angular velocity is 

$$r\omega=\sqrt{\frac{1-r}{r}}.$$

The force directed to the center of the force is

$$F_c\sin\alpha=\frac{v^2}{\rho},$$

where $\rho$ is the radius of curvature, and is

$$\rho=\frac{(1-r)^{3/2}r^{3/2}}{K},$$

with $K$ the angular momentum. The angular momentum will cancel out in the final expression so we don't have to use the fact that it is conserved. The reason is that $\sin\alpha$ is the ratio of the angular momentum, $K$, and $r^2\omega$, viz.,

$$\sin\alpha=\frac{K}{r^2\omega}=\frac{K}{\sqrt{r(1-r)}}.$$

The expression for the force directed to its center is

$$F_C=1/r^2$$,

or Newton's inverse square law.