Everyone thinks of Kepler's laws as being static with the period of orbital motion enters into his third and not time increments. That is, it is the period that is proportional to the area to the power $3/2$, and not the time increments themselves. Yet, time does enter into his law, and there are two different time depending on which focus is used as a reference. We will show that Kepler's laws are fully compatible with the laws of geometrical optics. It will also help us to think of the eccentricity of an elliptical orbit, and its sign, as a relative velocity for all the laws related to Kepler's theory obey the laws of refraction and aberration.

The energy integral is

$$\mu\left(\frac{2}{r}-\frac{1}{a}\right)r^2-r^2\dot{r}^2=\mu a(1-\varepsilon^2),$$

where $\mu=GM$, $a$ the semi-major axis and $\epsilon<1$, the eccentricity. By considering,

$$n_K=\sqrt{\frac{2a}{r}-1},$$

as the index of refraction, the above equation can be recast as the equation for a light ray, where $r\dot{r}=Kz$, where $K$ is the conserved angular momentum and $z=\frac{1}{r}dr/d\theta$ is the slope of the curve. The analogy with geometrical optics will turn out to be more than a mere analogy.

The eccentric anomaly, $E$ is defined by

$$r=a(1\pm\varepsilon\cos E),$$

depending on which focus, $f^{\prime}$ or $f$ in the figure below, we are considering,

$$1\mp\varepsilon\cos E=\frac{a(1-\varepsilon^2)}{1\pm\varepsilon\cos\theta},$$

which relates the eccentric anomaly to the true anomaly, $\theta$.

Solving for cosine and sine of the true anomaly, we find

$$\cos\theta_{+/-}=\frac{1\pm\varepsilon}{1\pm\varepsilon\cos E},$$

and

$$\sin\theta_{+/-}=\frac{\sqrt{1-\varepsilon^2}\sin E}{1\pm\varepsilon\cos E}.$$

These relations bear an extraordinary similarity to aberration when we regard the eccentricity as a constant relative velocity. For the far focus, $f^{\prime}$, the area of the relevant triangle is added to the area of the sector, $(1/2)Ea^2$ instead of subtracting it as we would due for the near focus $f$.

As a preliminary remark, consider the lineal ratio-property

$$\sin\theta=\frac{b}{r}\sin E,$$

where $b=a\sqrt{1-\varepsilon^2}.$ Employing the above aberration formula, it follows that $d\theta/dE=b/r$, so that

$$\frac{d\theta}{\sin\theta}=\frac{dE}{\sin E},$$

can be integrated to obtain

$$\ln\tan\theta/2=\ln\tan E/2+\mbox{const}.$$

Again, by the use of the above formula it can be shown that the constant is

$$\frac{1}{2}\ln\left(\frac{1+\varepsilon}{1-\varepsilon}\right)=\tanh^{-1}\varepsilon=\bar{\varepsilon},$$

the hyperbolic measure of eccentricity, so that the solution is

$$\tan\frac{\theta}{2}=\sqrt{\frac{1+\varepsilon}{1-\varepsilon}}\tan\frac{E}{2}.$$

As an aside, if we set $\theta=\pi/2$, the side of the triangle corresponds to the semi-latus rectum, shown below,

the above expression becomes the definition of the angle of parallelism,

$$\tan\frac{E}{2}=\exp(-\overline{\varepsilon}),$$

where the hyperbolic eccentricity $\overline{\varepsilon}=\tanh^{-1}\varepsilon$. As $\varepsilon$ varies between $0$ and $1$, its hyperbolic counterpart varies from $0$ to $\infty$. The eccentric anomaly give above is the largest angle that $CP$ touches $Pf$. For all larger angles there are two parallels through the given point $C$, and, moreover, the angle of parallelism is a sole function of the eccentricity, $\varepsilon$, independent of whatever radius $a$ we may choose.

Returning to the relation

$$\sin\theta=\frac{b}{r}\sin E,$$

we may write it as Snell's law by defining the index of refraction to be

$$n=\frac{b}{r}.$$

Since $nr=b$, a constant, the ray equation

$$\left(\frac{dr}{d\theta}\right)^2=\frac{r^2}{K^2}\left(n^2r^2-K^2\right),$$

vanishes This determines where $r(\theta)$ reaches an extreme value. However, returning to the above aberrational expressions, we use

$$\sin\theta_{-}=\frac{\sqrt{1-\varepsilon^2}\sin E}{1+\varepsilon\cos E},$$

to eliminate $\sin E$ from

$$\sin_{+}=\frac{\sqrt{1-\varepsilon^2}\sin E}{1-\varepsilon\cos E},$$

to obtain

$$\sin\theta_{+}=\left(\frac{1+\varepsilon\cos E}{1-\varepsilon\cos E}\right)\sin\theta_{-},$$

which is the *square* of the Keplerian index of refraction, $n_K$, given above. The intermediate angle, $E$, gives a "double dose" to Snell's law. We can "compound" Snell's law, like we do Doppler shifts, when light pases through more than two media of differing index of refraction.

We also have

$$\frac{dt_{+}}{dt_{-}}=n_K^2, $$

since the two mean anomalies, depending on our reference point, are

$$t_{\pm}=\frac{2\pi}{T}\left(1\pm\varepsilon\cos E\right).$$

The two mean anomalies are related to the gravitational field created by the central mass $M$, which gives rise to a non-unitary index of refraction, $n_K$. However, unlike the usual form of the index of refraction, $n_K$ is proportional to the angular velocity rather than be proportional to the inverse of the translational velocity.

Finally, if we divide the left side of Snell's law by $1+\cos\theta_{+}$ and the right side by $1+\cos\theta_{-}$ while multiplying it by $n_K^{-2}$ so as to leave the identity intact, we get

$$\tan\frac{\theta_{+}}{2}=\left(\frac{1+\varepsilon}{1-\varepsilon}\right)\tan\frac{\theta_{-}}{2},$$

This is precisely the square of what we found in the above expression leading to the angle of parallelism. If we consider $\varepsilon$ as a relative velocity, the above relation is what we would get for a Doppler shift upon reflection by a mirror. Because of reflection, the light ray gets a double dose of a Doppler shift.

Parenthetically, I want to correct a typo in eqn (10.1.5b) in my book, *A New Perspective on Relativity: *there should not be an exponent of $1/2$ in that expression.

Now, to go from the square root to its square, we define a new variable,

$$\gamma\equiv\frac{2\varepsilon}{1+\varepsilon^2},$$

and write

$$\frac{1+\gamma}{1-\gamma}=\left(\frac{1+\varepsilon}{1-\varepsilon}\right)^2.$$

The same transformation is used to go from the Poincare' disc model (conformal} to the Klein disc model (non-conformal) of hyperbolic geometry. Here, we see that it is related to the conformal transformation,

$$w=\frac{1}{2}\left(z+\frac{1}{z}\right),$$

known as the Joukowski transform, where $w=1/\gamma$ and $z=1/\varepsilon$ that transforms a one parameter family of ellipses onto a doubly covered family of ellipses. Its square,

$$z^2+\frac{1}{z^2}+2,$$

is still an ellipse, but displaced by two units. For hyperbola, $\varepsilon>1$, we would be dealing with the direct quantities, and not their inverses.

In terms of the conformal transforms we used in earlier blogs, the $z^2$ term gives the dual laws of Hooke and Newton, while the $1/z^2$ gives the new, and physically unknown, laws of the inverse fourth and seventh force laws. We know that $|z\pm f|=\mbox{const}$ are ellipses, since their sum (arithmetic mean)

$$|z+f|+|z-f|=\mbox{const},$$

while their square, $|z^2\pm f^2|$ (geometric mean)

$$|z+f|\cdot|z-f|=R^2=\mbox{const},$$

lead to Cassinian ovals. For $R>|f|$, we get one closed oval, while for $R<|f|$, the Cassinian splits into two separate ovals. At the boundary $R=|f|$ there results an ordinary leminiscate. In our next blog we we relate these to force laws. Here we have given the motivation for writing down a specific conformal transformation.