<p>In theory, it's possible, but in practice it's <em>extremely</em> unlikely. The deflection angle is small unless the light ray passes very close to the centre of mass of a compact lensing body.</p>
<p>From Wikipedia,</p>
<blockquote>
<p><a href="https://en.wikipedia.org/wiki/Strong_gravitational_lensing" rel="nofollow noreferrer"><strong>Strong Lensing</strong></a></p>
<p>Despite being considered &quot;strong&quot;, 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. <br />
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.</p>
</blockquote>
<p>Here's a simple approximation for the deflection angle which is valid when the deflection angle is small:</p>
<p><span class="math-container">$$\theta = \frac{2r_s}b$$</span>
where <span class="math-container">$\theta$</span> is in radians, <span class="math-container">$r_s$</span> is the <a href="https://en.wikipedia.org/wiki/Schwarzschild_radius" rel="nofollow noreferrer">Schwarzschild radius</a> of the lensing body, and <span class="math-container">$b$</span> is the <a href="https://en.wikipedia.org/wiki/Impact_parameter" rel="nofollow noreferrer">impact parameter</a> 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.</p>
<p>The Schwarzschild radius is proportional to the mass. For our Sun, <span class="math-container">$r_s \approx 2.953250$</span> km, so for a million solar mass black hole it's <span class="math-container">$2953250$</span> km.</p>
<p>When <span class="math-container">$b$</span> is a small multiple of <span class="math-container">$r_s$</span>, the above formula severely under-estimates the deflection.</p>
<p>Here's the deflection by a black hole with <span class="math-container">$b=4r_s$</span>, where <span class="math-container">$\theta$</span> is slightly more than <span class="math-container">$49.2°$</span>. The diagram uses distance units where <span class="math-container">$r_s=1$</span>, so it applies to any Schwarzschild black hole (rotating black holes are more complicated).</p>
<p><a href="https://i.sstatic.net/rEdEt.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/rEdEt.png" alt="Photon trajectory" /></a></p>
<p>The radius <span class="math-container">$r=1$</span> circle is of course the event horizon, the <span class="math-container">$r=1.5$</span> circle is the photon sphere, the dotted circle at <span class="math-container">$r\approx 2.598$</span> shows the critical impact parameter: a ray with this <span class="math-container">$b$</span> feeds into the photon sphere.</p>
<p>The dotted purple line shows where the trajectory is at its minimum distance to the black hole. If we call that distance <span class="math-container">$r$</span> then</p>
<p><span class="math-container">$$b^2 = \frac{r^3}{r-1}$$</span></p>
<p>in units where <span class="math-container">$r_s=1$</span>. For large <span class="math-container">$b$</span>, <span class="math-container">$r\approx b-\frac12$</span>.</p>
<p>To get <span class="math-container">$\theta=90°$</span>, as in the diagram in the question, we need <span class="math-container">$b\approx
3.083783382385$</span>. So you'd need a few bare black holes without pesky stars in the way.</p>
<hr />
<p>Here are a few graphs of <span class="math-container">$\theta$</span> vs <span class="math-container">$b$</span>. The True value was computed using a high-precision evaluation of an elliptic integral, the Simple value uses the equation given above.</p>
<p><a href="https://i.sstatic.net/wocZb.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wocZb.png" alt="deflection 100 - 200" /></a>
As you can see, the simple approximation is still reasonably good in this range.</p>
<p><a href="https://i.sstatic.net/6bkjM.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/6bkjM.png" alt="deflection 20 - 100" /></a></p>
<p>The simple approximation is starting to fall behind.</p>
<p><a href="https://i.sstatic.net/sP5HZ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/sP5HZ.png" alt="deflection 10 - 20" /></a></p>
<p>Now Simple is noticeably lagging.</p>
<p><a href="https://i.sstatic.net/yTVgf.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/yTVgf.png" alt="deflection 5 - 10" /></a></p>
<p>Simple is now way behind. I won't bother including it in the next plot.</p>
<p><a href="https://i.sstatic.net/EJMfi.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/EJMfi.png" alt="deflection ~b0 - 5" /></a></p>
<hr />
<p>Here are some Padé approximants of the elliptic integral that give better results than the simple approximation.</p>
<h3>Padé 1</h3>
<p><span class="math-container">$$\theta = \frac{72}{36r - 35}$$</span></p>
<h3>Padé 3</h3>
<p><span class="math-container">$$\theta = \frac{158331364 (1732790r^2 - 2653829r + 748305)}{
866395
(158331364r^3 - 396486557r^2 + 284743350r - 51433991)}$$</span></p>
<p><a href="https://i.sstatic.net/QLCnz.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/QLCnz.png" alt="Padé 1" /></a></p>
<p><a href="https://i.sstatic.net/IU7pE.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/IU7pE.png" alt="Padé 3" /></a></p>

CC BY-SA 4.0
2022-07-14T09:38:34.807Z
2022-08-10T22:42:39.427Z
2022-08-10T22:42:39.427Z
PM 2Ring
<p>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.</p>
<p><a href="https://i.sstatic.net/YZA7P.jpg" rel="noreferrer"><img src="https://i.sstatic.net/YZA7P.jpg" alt="enter image description here" /></a></p>
<p>Is it possible somehow that one of the galaxies in such pictures is actually the image of the Milky Way:</p>
<p><a href="https://i.sstatic.net/tKlor.jpg" rel="noreferrer"><img src="https://i.sstatic.net/tKlor.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2022-07-14T03:25:03.217Z
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? :)
paul garrett
<galaxy><gravitational-lensing><deep-sky-observing><james-webb-space-telescope><light-curve>
Gravitational mirroring: Can we theoretically see the Milky Way using a telescope?
user50618
# Gravitational mirroring: Can we theoretically see the Milky Way using a telescope?

(Asked by: username_0 on 2022-07-14)

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.

[![enter image description here](https://i.sstatic.net/YZA7P.jpg)](https://i.sstatic.net/YZA7P.jpg)

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

[![enter image description here](https://i.sstatic.net/tKlor.jpg)](https://i.sstatic.net/tKlor.jpg)

**Tags:** galaxy, gravitational-lensing, deep-sky-observing, james-webb-space-telescope, light-curve
> **Comments:**
> **username_1**: 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? :)  

---

## 1 answers

### Answer by username_2 on 2022-07-14. Score: 9

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**](https://en.wikipedia.org/wiki/Strong_gravitational_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](https://en.wikipedia.org/wiki/Schwarzschild_radius) of the lensing body, and $b$ is the [impact parameter](https://en.wikipedia.org/wiki/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).

[![Photon trajectory](https://i.sstatic.net/rEdEt.png)](https://i.sstatic.net/rEdEt.png)

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.

[![deflection 100 - 200](https://i.sstatic.net/wocZb.png)](https://i.sstatic.net/wocZb.png)
As you can see, the simple approximation is still reasonably good in this range.

[![deflection 20 - 100](https://i.sstatic.net/6bkjM.png)](https://i.sstatic.net/6bkjM.png)

The simple approximation is starting to fall behind.

[![deflection 10 - 20](https://i.sstatic.net/sP5HZ.png)](https://i.sstatic.net/sP5HZ.png)

Now Simple is noticeably lagging.

[![deflection 5 - 10](https://i.sstatic.net/yTVgf.png)](https://i.sstatic.net/yTVgf.png)

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

[![deflection ~b0 - 5](https://i.sstatic.net/EJMfi.png)](https://i.sstatic.net/EJMfi.png)

---

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

### Padé 1

$$\theta = \frac{72}{36r - 35}$$

### Padé 3

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

[![Padé 1](https://i.sstatic.net/QLCnz.png)](https://i.sstatic.net/QLCnz.png)

[![Padé 3](https://i.sstatic.net/IU7pE.png)](https://i.sstatic.net/IU7pE.png)


---

