# Gravitational mirroring: Can we theoretically see the Milky Way using a telescope?

As I am watching the beautiful deep field picture of the James Webb Space Telescope. I noticed the gravitational lensing. So I asked myself whether it is possible somehow a successive cluster of galaxies or black holes forming a mirror instead of a lens. Which means making light coming from our galaxy doing a U-turn. This way we can maybe observe our galaxy in the past.

Is it possible somehow that one of the galaxies in such pictures is actually the image of the Milky Way:

• And possibly several different ages and versions !?!?! :) An interesting question. Also, not just about gravitational lensing, but about weird geometry of the large-scale universe? Like mirrors in a fitting room at a clothing store? :) Jul 14, 2022 at 3:25

In theory, it's possible, but in practice it's extremely unlikely. The deflection angle is small unless the light ray passes very close to the centre of mass of a compact lensing body.

From Wikipedia,

Strong Lensing

Despite being considered "strong", the effect is in general relatively small, such that even a galaxy with a mass more than 100 billion times that of the Sun will produce multiple images separated by only a few arcseconds.
Galaxy clusters can produce separations of several arcminutes. In both cases the galaxies and sources are quite distant, many hundreds of megaparsecs away from our Galaxy.

Here's a simple approximation for the deflection angle which is valid when the deflection angle is small:

$$\theta = \frac{2r_s}b$$ where $$\theta$$ is in radians, $$r_s$$ is the Schwarzschild radius of the lensing body, and $$b$$ is the impact parameter of the light ray, which is basically the minimum distance between the ray and the centre of the lensing body if you could turn gravity off so that the ray was just a straight line.

The Schwarzschild radius is proportional to the mass. For our Sun, $$r_s \approx 2.953250$$ km, so for a million solar mass black hole it's $$2953250$$ km.

When $$b$$ is a small multiple of $$r_s$$, the above formula severely under-estimates the deflection.

Here's the deflection by a black hole with $$b=4r_s$$, where $$\theta$$ is slightly more than $$49.2°$$. The diagram uses distance units where $$r_s=1$$, so it applies to any Schwarzschild black hole (rotating black holes are more complicated).

The radius $$r=1$$ circle is of course the event horizon, the $$r=1.5$$ circle is the photon sphere, the dotted circle at $$r\approx 2.598$$ shows the critical impact parameter: a ray with this $$b$$ feeds into the photon sphere.

The dotted purple line shows where the trajectory is at its minimum distance to the black hole. If we call that distance $$r$$ then

$$b^2 = \frac{r^3}{r-1}$$

in units where $$r_s=1$$. For large $$b$$, $$r\approx b-\frac12$$.

To get $$\theta=90°$$, as in the diagram in the question, we need $$b\approx 3.083783382385$$. So you'd need a few bare black holes without pesky stars in the way.

Here are a few graphs of $$\theta$$ vs $$b$$. The True value was computed using a high-precision evaluation of an elliptic integral, the Simple value uses the equation given above.

As you can see, the simple approximation is still reasonably good in this range.

The simple approximation is starting to fall behind.

Now Simple is noticeably lagging.

Simple is now way behind. I won't bother including it in the next plot.

Here are some Padé approximants of the elliptic integral that give better results than the simple approximation.

$$\theta = \frac{72}{36r - 35}$$
$$\theta = \frac{158331364 (1732790r^2 - 2653829r + 748305)}{ 866395 (158331364r^3 - 396486557r^2 + 284743350r - 51433991)}$$