<p>I'd suggest you take a look at <a href="https://www.tau.ac.il/%7Elab3/MICROLENSING/JeruLect.pdf" rel="noreferrer">Narayan and Bartelmann (1996)</a>. They run through all the math to answer exactly this question. I'll add here the punchline from their paper for posterity.</p>
<p>Shown below is Figure 7 from their paper. It shows the scenario of looking straight on at the gravitational lens of mass <span class="math-container">$M$</span> (a black hole in your scenario). The source (a background galaxy in your scenario) is offset from the line of sight of the black hole by some angle. This produces <em>two</em> images of the source, namely <span class="math-container">$I_+$</span> and <span class="math-container">$I_-$</span> that we would observe here on Earth. The other useful concept here is the <a href="https://en.wikipedia.org/wiki/Einstein_radius" rel="noreferrer">Einstein Radius</a> denoted by the angle <span class="math-container">$\theta_E$</span>. The image that appears outside the Einstein Radius involves a positive magnification (that is, that image is brighter than the source). The image inside the Einstein Radius involves a negative magnification (that is, that image is dimmer than the source).</p>
<p><a href="https://i.sstatic.net/mkf7E.png" rel="noreferrer"><img src="https://i.sstatic.net/mkf7E.png" alt="Source and images during magnification" /></a></p>
<p>The Einstein Radius itself is a function of the geometry of the system and the mass of the lens.</p>
<p><span class="math-container">$$\theta_E = \left(\frac{4GM}{c^2}\frac{D_{ds}}{D_sD_d}\right)^{1/2}$$</span></p>
<p>In this equation <span class="math-container">$D_{d}$</span> is the distance between the observer and the lens (aka the &quot;deflector&quot;), <span class="math-container">$D_s$</span> is the distance between the observer and the source, and <span class="math-container">$D_{ds}$</span> is the distance between the lens and the source.</p>
<p>To jump to the punchline, the total magnification of the source is given by the area of the image over the area of the source. Since there are two images, we have to add the flux from both to get the total magnification. The end result is that the magnification, <span class="math-container">$\mu$</span>, is given by</p>
<p><span class="math-container">$$\mu = \frac{u^2+2}{u\sqrt{u^2+4}}$$</span></p>
<p>where <span class="math-container">$u$</span> is the angular separation of the source from the lens, in units of the Einstein Radius. As a specific example, when the source lies exactly on the Einstein Radius, that is, it's angle of separation from the lens equals <span class="math-container">$\theta_E$</span>, then <span class="math-container">$u = 1$</span> and the magnification of the source is <span class="math-container">$\mu = 1.34$</span>. In other words, the source image (or rather, images) is 1.34 times brighter.</p>
<hr />
<blockquote>
<p>Does it magnify across the electromagnetic spectrum or only the visible light?</p>
</blockquote>
<p>Yes, the magnification is independent of wavelength.</p>
<hr />
<p>Note that the above scenario assumes certain simplifications are true. The primary one being the &quot;thin screen approximation&quot;. This basically assumes that the lensing happens all at once with a single deflection. In reality it would be a constant deflect as the light travels from the source to the detector, but that's generally barely distinguishable from the thin screen approximation, so it's useful to use. In addition, once the lens becomes extended, things get more complicated. You restricted your case to a &quot;point lens&quot; by using a black hole so the math was much simpler.</p>

CC BY-SA 4.0
2022-05-04T22:12:03.923Z
For the positve and negative magnification, did you really mean to use the words "brighter" and "dimmer" rather than "larger" and smaller?
Greg Miller
CC BY-SA 4.0
2022-05-05T01:38:49.653Z
Thanks a lot! I probably don't have enough math to grind through the paper but I'll keep it as a bookmark in my notes.
Nicholas Humphrey
CC BY-SA 4.0
2022-05-05T02:54:21.187Z
@gregmiller No, I meant brighter and dimmer. The surface brightness of the source image is preserved but because the $I_-$ image has a smaller surface area than the source, the magnification is $<1$ and thus is dimmer than the source. Similarly the surface area of the $I_+$ image is larger so it is brighter. That total increase in brightness is what allows us to find microlensing events.
zephyr
CC BY-SA 4.0
2022-05-04T13:54:38.630Z
2022-05-04T13:54:38.630Z
zephyr
<p>Given a black hole of mass <span class="math-container">$M$</span> which has a distance <span class="math-container">$D_1$</span> to Earth and <span class="math-container">$D_2$</span> to a galaxy being magnified, what is the magnification factor?</p>
<p>And a follow-up question:</p>
<p>Does it magnify across the electromagnetic spectrum or only the visible light?</p>

CC BY-SA 4.0
2022-05-04T13:36:45.697Z
Have you done any preliminary research? What have you tried so far?
WarpPrime
CC BY-SA 4.0
2022-05-04T21:35:12.170Z
different but related: [What exactly is it that is being magnified 50 times in this gravitational lensing observation?](https://astronomy.stackexchange.com/q/20865/7982)
uhoh
CC BY-SA 4.0
2022-05-05T13:32:09.490Z
Does this answer your question? [Can we see the past image of Earth because of curving light by massive objects?](https://astronomy.stackexchange.com/questions/12062/can-we-see-the-past-image-of-earth-because-of-curving-light-by-massive-objects)
Carl Witthoft
<black-hole><gravitational-lensing>
Black hole magnification factor
Nicholas Humphrey
# Black hole magnification factor

(Asked by: username_0 on 2022-05-04)

Given a black hole of mass $M$ which has a distance $D\_1$ to Earth and $D\_2$ to a galaxy being magnified, what is the magnification factor?

And a follow-up question:

Does it magnify across the electromagnetic spectrum or only the visible light?

**Tags:** black-hole, gravitational-lensing
> **Comments:**
> **username_1**: Have you done any preliminary research? What have you tried so far?  
> **username_2**: different but related: [What exactly is it that is being magnified 50 times in this gravitational lensing observation?](https://astronomy.stackexchange.com/q/20865/7982)  
> **username_3**: Does this answer your question? [Can we see the past image of Earth because of curving light by massive objects?](https://astronomy.stackexchange.com/questions/12062/can-we-see-the-past-image-of-earth-because-of-curving-light-by-massive-objects)  

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## 1 answers

### Answer by username_4 on 2022-05-04. Score: 7 | Accepted answer

I'd suggest you take a look at [Narayan and Bartelmann (1996)](https://www.tau.ac.il/%7Elab3/MICROLENSING/JeruLect.pdf). They run through all the math to answer exactly this question. I'll add here the punchline from their paper for posterity.

Shown below is Figure 7 from their paper. It shows the scenario of looking straight on at the gravitational lens of mass $M$ (a black hole in your scenario). The source (a background galaxy in your scenario) is offset from the line of sight of the black hole by some angle. This produces *two* images of the source, namely $I\_+$ and $I\_-$ that we would observe here on Earth. The other useful concept here is the [Einstein Radius](https://en.wikipedia.org/wiki/Einstein_radius) denoted by the angle $\theta\_E$. The image that appears outside the Einstein Radius involves a positive magnification (that is, that image is brighter than the source). The image inside the Einstein Radius involves a negative magnification (that is, that image is dimmer than the source).

[![Source and images during magnification](https://i.sstatic.net/mkf7E.png)](https://i.sstatic.net/mkf7E.png)

The Einstein Radius itself is a function of the geometry of the system and the mass of the lens.

$$\theta\_E = \left(\frac{4GM}{c^2}\frac{D\_{ds}}{D\_sD\_d}\right)^{1/2}$$

In this equation $D\_{d}$ is the distance between the observer and the lens (aka the "deflector"), $D\_s$ is the distance between the observer and the source, and $D\_{ds}$ is the distance between the lens and the source.

To jump to the punchline, the total magnification of the source is given by the area of the image over the area of the source. Since there are two images, we have to add the flux from both to get the total magnification. The end result is that the magnification, $\mu$, is given by

$$\mu = \frac{u^2+2}{u\sqrt{u^2+4}}$$

where $u$ is the angular separation of the source from the lens, in units of the Einstein Radius. As a specific example, when the source lies exactly on the Einstein Radius, that is, it's angle of separation from the lens equals $\theta\_E$, then $u = 1$ and the magnification of the source is $\mu = 1.34$. In other words, the source image (or rather, images) is 1.34 times brighter.

---

> Does it magnify across the electromagnetic spectrum or only the visible light?

Yes, the magnification is independent of wavelength.

---

Note that the above scenario assumes certain simplifications are true. The primary one being the "thin screen approximation". This basically assumes that the lensing happens all at once with a single deflection. In reality it would be a constant deflect as the light travels from the source to the detector, but that's generally barely distinguishable from the thin screen approximation, so it's useful to use. In addition, once the lens becomes extended, things get more complicated. You restricted your case to a "point lens" by using a black hole so the math was much simpler.

> **Comments:**
> **username_5**: For the positve and negative magnification, did you really mean to use the words "brighter" and "dimmer" rather than "larger" and smaller?  
> **username_0**: Thanks a lot! I probably don't have enough math to grind through the paper but I'll keep it as a bookmark in my notes.  
> **username_4**: @gregmiller No, I meant brighter and dimmer. The surface brightness of the source image is preserved but because the $I_-$ image has a smaller surface area than the source, the magnification is $<1$ and thus is dimmer than the source. Similarly the surface area of the $I_+$ image is larger so it is brighter. That total increase in brightness is what allows us to find microlensing events.  

---

