Angular Deflection Edit

A Horseshoe Einstein Ring from Hubble

Suppose a photon passes near a massive object of mass $ M $ with impact parameter $ b $. We assume that the energy of the photon is much greater than the gravitational energy at closest approach $ c \gg \sqrt{\frac{G M}{b}} $ where $ c $ is the speed of light in vacuo and $ G $ is the universal constant of gravity. We further assume that the trajectory remains a straight line, and the deviations from it are very small. Let us denote by $ x $ the coordinate along the photons trajectory, such that the unperturbed trajectory is $ x = c \cdot t $ (i.e. the point of closest approach occurs at $ t = 0 $). The lion's share of deflection occurs in the range $ x \in \left[-b,b\right] $. The time the photon spends in this range is $ \Delta t = \frac{2b}{c} $, and the average acceleration is $ a = \frac{GM}{b^2} $, so the velocity difference is $ \Delta v = a \Delta t = \frac{2GM}{cb} $. The thrust can also be calculated in a more rigorous way by integrating over the entire path

$ \Delta v = \int_{-\infty}^{\infty} \frac{G M b}{\left(b^2+c^2 t^2 \right)^{3/2}}dt = \frac{2 G M}{b c} $

Interestingly, we get the same result. The deflection angle can be estimated by

$ \alpha = \frac{\Delta v}{c} = \frac{2 G M}{b c^2} = \frac{R_s}{b} $

where $ R_s $ is the Schwarzschild radius. In retrospect, we could have guessed this result from dimensional analysis. We note that this results is smaller by a factor of two from the deflection angle derived from general relativity.

Einstein RingsEdit

Suppose now that a point light source is situated behind a massive object. In other words, the observer, lens and source are aligned. The observer would see a ring, and in this section we will estimate the radius of this ring. Suppose the distance between the source and the massive object is $ d_s $ and the distance between the massive object and the observer is $ d_o $. In order that the photon reach the observer, the following equation must hold

$ \alpha + \tan^{-1}\frac{d_o}{b} + \tan^{-1}\frac{d_s}{b} = \pi $

This equation can be simplified with the assumption $ d_o, d_s \gg b $

$ 4\frac{G M}{b c^2} - \frac{b}{d_s} - \frac{b}{d_o} = 0 $

Solving for the impact parameter yields the radius of the ring

$ b = \sqrt{\frac{4 G M}{c^2} \frac{d_s d_o}{d_s + d_o}} $

the angular magnitude would be

$ \frac{b}{d_o} = \sqrt{\frac{4 G M}{c^2} \frac{d_s/d_o}{d_s + d_o}} $

Critical Surface DensityEdit

In this section we consider a slightly different problem. Instead of passing near a point mass, our photon now passes through the plane of an infinite flat disk of surface density $ \Sigma $. Therefore, the deflecting mass depends on the impact parameter

$ M = \pi \Sigma b^2 $

The deflection angle is therefore

$ \alpha = 4 \pi \frac{G \Sigma b}{c^2} $

Let us suppose that this disk occupies the y axis, and that we have some photon source at $ \left(-d_s,0\right) $ and an observer at $ \left(d_o,0\right) $. It turns out that at some critical surface density the observer may see multiple images of the source. In this section we will determine this surface density. In what follows we will always assume $ d_s, d_o \gg b $. In order for a photon to reach to observer, the following equality must hold due to simple geometry

$ \alpha + \tan^{-1}\frac{d_o}{b} + \tan^{-1}\frac{d_s}{b} = \pi $

In the limit of a very small impact parameter

$ 4 \pi \frac{G \Sigma b}{c^2} - \frac{b}{d_o} - \frac{b}{d_s} = 0 $

Solving for $ \Sigma $

$ \Sigma = \frac{c^2}{4 \pi G} \left( \frac{1}{d_o} + \frac{1}{d_s} \right) $

At this surface density the light from the source would seem to the observer as though it is coming from every point on the disk.