<p>The apparent radius of something residing in a Schwarzschild metric, when viewed from infinity is given by
<span class="math-container">$$ R_{\rm obs} = R \left(1 - \frac{R_s}{R}\right)^{-1/2}\ ,$$</span>
where <span class="math-container">$R_s$</span> is the Schwarzschild radius <span class="math-container">$2GM/c^2$</span>.</p>

<p>This <em>enlargement</em> is due to gravitational lensing and the formula is correct down to the "photon sphere" at <span class="math-container">$R =1.5 R_s$</span>.</p>

<p>Most of the light in the EHT image comes from the photon sphere. It is therefore <em>observed</em> to come from a radius
<span class="math-container">$$ R_{\rm obs} =\frac{3R_s}{2}\left(1 - \frac{2}{3}\right)^{-1/2} = \frac{\sqrt{27}}{2}R_s\ .$$</span></p>

<p>There are small (&lt;10%) corrections to this for a spinning black hole governed by the Kerr metric.</p>

<p>A sketch which illustrates why gravitational lensing leads to an enlarged image can be found in Pela's <a href="https://astronomy.stackexchange.com/a/30360/5264">answer to a related question</a>.</p>

CC BY-SA 4.0
2019-04-12T17:20:45.250Z
2019-04-12T21:38:02.720Z
2019-04-12T21:38:02.720Z
ProfRob
<p>If the light is orbiting the black hole, then it isn't escaping to our telescopes!</p>

<p>What matters is how close to the black hole the plasma of the accretion disk goes, since that's what produces the light. There are lots of papers on this, and the numbers are usually around 5 Schwarzchild radii, together with effects like gravitational redshift relativistic beaming and so on, and the geometry of the paths followed by light, this leads to the image we see.</p>

CC BY-SA 4.0
2019-04-12T14:58:38.633Z
The figure of $5r_s$ is wrong - it is $2.6r_s$ (unless you were referring to a diameter). But the OP knows this; they want to know why?
ProfRob
CC BY-SA 4.0
2019-04-12T15:21:17.603Z
The thing I think is at 5 ish radii is the point at which the accretion disk becomes unstable. This is, I think further out than the innermost stable orbit, because the disk is hot and turbulent.
Steve Linton
CC BY-SA 4.0
2019-04-12T16:42:43.070Z
The ring is seen at a radius of $\sqrt{27} GM/c^2$ for precise geometric reasons.
ProfRob
CC BY-SA 4.0
2019-04-12T14:44:29.943Z
2019-04-12T14:44:29.943Z
Steve Linton
<p>The innermost stable orbit (ISCO) of a massive object in circular orbit around a black hole or any other spherically symmetric nonspinning mass  is located at <span class="math-container">$3\text{R}_s$</span>. From <a href="https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit" rel="nofollow noreferrer">wikipedia</a>:</p>

<p>"The ISCO plays an important role in black hole accretion disks since it marks the inner edge of the disk."</p>

<p>If the image in question depicts radiation emanating from matter in the accretion disk the "ring of light" that is the accretion disk is where it is supposed to be.</p>

CC BY-SA 4.0
2019-04-12T16:40:53.480Z
But it isn't the accretion disk.
ProfRob
CC BY-SA 4.0
2019-04-12T15:46:39.753Z
2019-04-12T15:46:39.753Z
Agerhell
<p>Today we know that a black hole represents a region of space where the gravitational forces are very intense, this consequence of a concentration of matter in a very small region of space. On April 10 of this year, scientists from the EHT project (event Horizont Telescope) presented to the world the image that I believe everyone has seen. The first image of a black hole in the galaxy M87. We know that what we are really seeing is the "shadow" of the black hole.</p>

<p>But the "photon sphere", where light can orbit a black hole, is actually at <span class="math-container">$1.5R_s$</span>, (Schwarzschild radius), whereas the observed ring is a factor of 2 wider than this. Why we not see a ring of light at the radius of the photon sphere?</p>

CC BY-SA 4.0
2019-04-12T14:02:10.137Z
that is my question. why can not you see the ring of light?
jormansandoval
CC BY-SA 4.0
2019-04-12T14:08:27.713Z
no Rob. I think that you think that when I talk about the ring of light it's what the image shows. no no no ... I am talking about the ring of light that is created at a distance from Schwarzschild's radius, which is exactly 1.5 radii. to that ring of light I am referring to.
jormansandoval
CC BY-SA 4.0
2019-04-12T14:09:44.190Z
I remind you that what we are seeing is not the hole itself ... but its shadow amplified as a result of the distortion of space time.
jormansandoval
CC BY-SA 4.0
2019-04-12T14:12:05.087Z
@Pela a chance to include your diagram again.
ProfRob
CC BY-SA 4.0
2019-04-12T15:03:29.010Z
Closely related https://astronomy.stackexchange.com/questions/30317/m87-black-hole-why-can-we-see-the-blackness
ProfRob
CC BY-SA 4.0
2019-04-12T21:05:52.117Z
@RobJeffries' answer is the correct one, and there you have the derivation. For an intuitive drawing (and explanation), see [this answer](https://astronomy.stackexchange.com/a/30360/5264).
pela
<black-hole><m87>
Why is the ring of light around the M87 black hole bigger than the photon sphere?
jormansandoval
# Why is the ring of light around the M87 black hole bigger than the photon sphere?

(Asked by: username_0 on 2019-04-12)

Today we know that a black hole represents a region of space where the gravitational forces are very intense, this consequence of a concentration of matter in a very small region of space. On April 10 of this year, scientists from the EHT project (event Horizont Telescope) presented to the world the image that I believe everyone has seen. The first image of a black hole in the galaxy M87. We know that what we are really seeing is the "shadow" of the black hole.

But the "photon sphere", where light can orbit a black hole, is actually at $1.5R\_s$, (Schwarzschild radius), whereas the observed ring is a factor of 2 wider than this. Why we not see a ring of light at the radius of the photon sphere?

**Tags:** black-hole, m87

> **Comments:**
> **username_0**: that is my question. why can not you see the ring of light?  
> **username_0**: no Rob. I think that you think that when I talk about the ring of light it's what the image shows. no no no ... I am talking about the ring of light that is created at a distance from Schwarzschild's radius, which is exactly 1.5 radii. to that ring of light I am referring to.  
> **username_0**: I remind you that what we are seeing is not the hole itself ... but its shadow amplified as a result of the distortion of space time.  
> **username_1**: @Pela a chance to include your diagram again.  
> **username_1**: Closely related https://astronomy.stackexchange.com/questions/30317/m87-black-hole-why-can-we-see-the-blackness  
> **username_2**: @RobJeffries' answer is the correct one, and there you have the derivation. For an intuitive drawing (and explanation), see [this answer](https://astronomy.stackexchange.com/a/30360/5264).  

---

## 3 answers

### Answer by username_1 on 2019-04-12. Score: 3

The apparent radius of something residing in a Schwarzschild metric, when viewed from infinity is given by
$$ R\_{\rm obs} = R \left(1 - \frac{R\_s}{R}\right)^{-1/2}\ ,$$
where $R\_s$ is the Schwarzschild radius $2GM/c^2$.

This *enlargement* is due to gravitational lensing and the formula is correct down to the "photon sphere" at $R =1.5 R\_s$.

Most of the light in the EHT image comes from the photon sphere. It is therefore *observed* to come from a radius
$$ R\_{\rm obs} =\frac{3R\_s}{2}\left(1 - \frac{2}{3}\right)^{-1/2} = \frac{\sqrt{27}}{2}R\_s\ .$$

There are small (<10%) corrections to this for a spinning black hole governed by the Kerr metric.

A sketch which illustrates why gravitational lensing leads to an enlarged image can be found in Pela's [answer to a related question](https://astronomy.stackexchange.com/a/30360/5264).


---

### Answer by username_3 on 2019-04-12. Score: -1

If the light is orbiting the black hole, then it isn't escaping to our telescopes!

What matters is how close to the black hole the plasma of the accretion disk goes, since that's what produces the light. There are lots of papers on this, and the numbers are usually around 5 Schwarzchild radii, together with effects like gravitational redshift relativistic beaming and so on, and the geometry of the paths followed by light, this leads to the image we see.

> **Comments:**
> **username_1**: The figure of $5r_s$ is wrong - it is $2.6r_s$ (unless you were referring to a diameter). But the OP knows this; they want to know why?  
> **username_3**: The thing I think is at 5 ish radii is the point at which the accretion disk becomes unstable. This is, I think further out than the innermost stable orbit, because the disk is hot and turbulent.  
> **username_1**: The ring is seen at a radius of $\sqrt{27} GM/c^2$ for precise geometric reasons.  

---

### Answer by username_4 on 2019-04-12. Score: -1

The innermost stable orbit (ISCO) of a massive object in circular orbit around a black hole or any other spherically symmetric nonspinning mass is located at $3\text{R}\_s$. From [wikipedia](https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit):

"The ISCO plays an important role in black hole accretion disks since it marks the inner edge of the disk."

If the image in question depicts radiation emanating from matter in the accretion disk the "ring of light" that is the accretion disk is where it is supposed to be.

> **Comments:**
> **username_1**: But it isn't the accretion disk.  

---

