Following the publication of the first edition of *Principia*, Newton set himself to correct and revise it. Newton proposed three distinct methods for producing the solution to the equation of motion of Kepler's problem. In the first edition, curvature played a minor role, but all this changed in the new edition where curvature was elevated to play a major role in determining the law of force. Kepler's problem was dissected into two problems: the 'direct' problem where given the path and the center of force, determine the force that is necessary to determine the motion, and the 'inverse' method whereby given the force and its center of action, determine the path of the motion. Although there is no direct evidence that Newton did the calculation, there appears that he gave an outline on how to do it.

In the early 1670's Newton wrote down an expression for the radius of curvature, ρρ, in polar coordinates:

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

where the slope of the curve, $z=d\ln r/d\theta$, and $z^{\prime}=dz/\theta$. The latter could be expressed in terms of $1+z^2$ as

$$z^{\prime}=\frac{r}{2}\frac{d}{dr}(1+z^2).$$

Why insist on the factor $1+z^2$? Hindsight shows that it is a tour de force since it is given by optical equation,

$$1+z^2=\left(\frac{nr}{\Delta}\right)^2,$$

which determines the light rays in our medium with index of refraction, $n$, impact parameter, $\Delta$. So, theoretically, given the index of refraction we can completely determine the force law. But this is not all!

Introducing the expression for $z^{\prime}$ into Newton's radius of curvature results in

$$\rho=r\frac{(1+z^2)^{3/2}}{1+z^2-\frac{r}{2}d(1+z^2)/dr}.$$

But, we can do better than this by introducing the index of refraction,

$$\rho=-\frac{n^2r}{\Delta dn/dr},$$

Newton makes a distinction between the force, $F_S$, directed toward the center of force, and

$$F_{\theta}=\frac{v^2}{\rho},$$

the force of centripetal acceleration, as required by his Proposition 4 in *Principia*. $v=r\dot{\theta}$ is the tangential acceleration at point $P$ in the diagram below.

The two forces are related by

$$F_{\theta}=F_S\cos\theta=F_S\sin\alpha,$$

as can be seen from the diagram, where $\alpha$ is the complementary angle to $\theta$ that appears in Kepler's law of areas

$$rv\sin\alpha=L,$$

where $L$ is the angular momentum per unit mass. In terms of the index of refraction,

$$\sin\alpha=\frac{\Delta}{nr}.$$

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Now, it transpires that for Keplerian orbits, the index of refraction

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

is none other than the index of refraction of Maxwell's fish eye, and its stereographic projection onto the plane gives an elliptical orbit determined by the index of refraction

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

which specifies Newton's gravitational potential, $V=1/r$! This is shown in the following diagram.

Newton explicitly employs the assumption of uniform circular motion in Lemma 11, Corollary 3, and in Proposition 7, Problem 2, Corollary 5. Solving for radius of curvature gives

$$\rho=\frac{v^2}{F_S\sin\alpha}=\frac{L^2}{F_S r^2\sin^3\alpha}.$$

However, the radius of curvature,

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

is certainly not constant, as it would be for uniform circular motion. But, if we multiply it by $\sin^3\alpha$, there results a pure constant, $\rho=2$ It also implies that

$$F_Sr^2=\mbox{const},$$

which is what Newton set to prove, if $L$ is a constant of the motion. So, $\sin\alpha$ is a correction that is necessary when the radial force and the centripetal force are not lined up in the same direction. But then what allowed Newton to claim that he is dealing with uniform circular motion?

Look to the sphere! On the sphere, the index of refraction is is given by Maxwell's fish-eye, so that the radius of curvature $\rho=1/2$ is constant. There, there is no need for a correction factor and and Newton's law of attraction is satisfied identically.

Notwithstanding the fact that the radii of curvature are different for the orbit on the sphere and its projection onto the plane, what remains invariant is Newton's law of force!

It wouldn't be fitting to close without considering the Schwarzschild metric, whose index of refraction on the hyperboloid

$$n=\frac{1}{1-\mathcal{R}/r},$$

analogous to that of Maxwell's s fish-eye on a sphere, where $\mathcal{R}$ is the Schwarzschild radius. That is, Maxwell's fish eye corresponds to constant density apart from the sign change in the denominator which occurs when we go from a sphere to an hyperboloid. The Schwarzschild metric is not a metric of constant curvature . It has a Gaussian curvature of $-2/\rho$ where the radius of curvature is

$$\rho=\frac{r^3}{\mathcal{R}\Delta},$$

having reinstated the impact parameter on dimensional grounds. Newton's expression,

$$F_S=\frac{L^2\mathcal{R}\Delta}{r^5},$$

determines the force law as

$$F_S r^5=\mbox{const}.$$

This law of force does not warrant the assumption that for large distances and weak fields the potential is Newtonian, for it is far from it!

Using the fact that the same law of force must apply on "earth as it is in heaven", we can determine the index of refraction in the plane. It is easily found to be

$$n^2=\frac{1}{r^4}+C,$$

where $C$ is an arbitrary constant. Since the square of the index of refraction is proportional to the potential, $V=-1/r^4$, we know that it implies a $1/r^5$ force law. In fact, the orbit is

$$r=a\cos\theta$$

which is circular of radius $a$. The criterion that the orbit be stable is [cf. Danby, *Celestial Mechanics*, eqn (4.2.7)]

$$3F_S(a)-a\frac{dF_S(a)}{da}>0,$$

which is not fulfilled in the case of the Schwarzschild metric.

In fact, Cotes' spiral

which has a potential, $V=-1/r^2$, and a law of force, $F_S=1/r^3$ is unstable since the inequality becomes an equality. A small perturbation is sufficient to tip the balance and render the orbit unstable. This is a true "black" hole since the light trajectories spiral into the center of force. There is no "cosmological censorship" to prohibit a "naked" singularity.

We can understand this using Kepler's third law,

$$\frac{GM}{r}=v^2=\omega^2 r^2.$$

For a constant period, $T=2\pi/\omega$, the mass increases as the cube of the velocity, implying that the density is constant. For Cotes' spiral we would need the central mass increasing as the fourth power of the radius leading to an instability.