We have seen that the conformal transformation,

$$z=w^2,$$

converts Newton's nonlinear law of gravity into a linear equation of a harmonic oscillator. Here, we will show that it converts elliptical orbits into hyperbolic ones, and in so doing derive the expression for the angle of deflection of light when it grazes a massive body like the sun.

In a previous blog we have shown that the Schwarzschild metric corresponds to repulsive gravity, a repulsive instead of an attractive force. Using the Maupertuis-Jacobi principle, we saw that the index of refraction for small values of the radial coordinate was given by the Luneburg expression,

$$n^2_L=2-r^2,\;\;\;\;\;\;\;\;\;\; r\le \surd 2.$$

The Luneburg lens is a sphere of radius $r=\surd 2$ which is filled with a material that gives the Hookian profile, $n_L$. Outside the lens, for $r>\surd 2$, light propagates in empty space with an index of refraction $n=1$. However, inside the lens, the rays no longer follow straight trajectories but bend into segments of an ellipse.

Newton's optical law,

$$\frac{d^2r}{d\psi^2}=\frac{\nabla n^2_L}{2}=-r,$$

where $\xi$ is an affine parameter (for if it were time it would make the index of refraction inversely proportional to the speed of light). Using complex variables, the solution is

$$z=\frac{a+b}{2}e^{i\xi}+\frac{a-b}{2}e^{-\xi}=a\cos\xi+ib\sin\xi,$$

where the real quantities $a$ and $b$ have double meanings: $a$ is both the semimajor axis of the ellipse and the point of launch, while $b$ is the semiminor axis and the speed at launch.

An application of $z\mapsto z^2$ to the origin-centered ellipse shifts one focus to the origin,

$$z\mapsto z^2=\frac{(a+b)^2}{4}e^{2i\xi}+\frac{(a-b)^2}{4}e^{-2i\xi}+\frac{a^2-b^2}{2},$$

The first two terms correspond to an origin-centered ellipse with foci at $\pm(a^2-b^2)/2$, while the last term translates the left-handed focus to the origin. One may ask what about hyperbolic orbits?

Attraction and repulsion can be discussed in terms of the harmonic oscillator solution

$$\ddot{z}+z=0,$$

where $z$ is complex, and

$$\ddot{z}-z=0,$$

respectively. For the repulsive field, the solution is

$$z=\frac{a+b}{2}e^{\xi}+\frac{a-b}{2}e^{-\xi}=a\cosh\xi+b\sinh\xi,$$

where $a=|z|$ at the point of launch and $b$ is the speed at launch. The mapping $z\mapsto z^2$ transforms the pair of hyperbolas so they are origin-centered with their foci are symmetrical with respect to the origin. But, the effect has a more profound significance in that the Newtonian field is still inverse square, but now is repulsive.

This has repercussions on the paths of the light rays. There are only two parameters in the problem, the impact parameter $\Delta$ and the magnitude of the gravitational force $\mathcal{R}$, both of which have units of length. We can combine the two into the ratio $\Xi=\Delta/\mathcal{R}$ and render everything dimensionless. The equation of the orbit is

$$\varphi=\int\frac{\Xi dr}{r\sqrt{r^2n_L^2-\Xi^2}}.$$

Transforming to $\varrho=r^2$, the trajectory of the orbit is

$$\varrho=\frac{\Xi^2}{1-\epsilon\sin 2\varphi},$$

where the eccentricity is

$$\epsilon=\sqrt{1+\Xi^2}.$$

The angle of scattering is

$$\cot\frac{\chi}{2}=\sqrt{\epsilon^2-1}=\Xi.$$

For small vales of the scattering angle $\Xi$, $\cot\chi/2\approx 2/\chi$, so that the expression becomes

$$\chi\approx\frac{2}{\Xi}=\frac{2\mathcal{R}}{\Delta},$$

which is identical to the general relativistic result. Hence, the mapping $r\mapsto r^2$ converts elliptical orbits into hyperbolic ones.