<p>We use the Schwarzschild radius <span class="math-container">$r_s$</span> (rather than a diameter) because it's convenient. We're want to describe what happens in the vicinity of a black hole, so it's natural to talk about the distance from the black hole.</p>
<p>For example, there's the <a href="https://en.wikipedia.org/wiki/Photon_sphere" rel="nofollow noreferrer">photon sphere</a> at <span class="math-container">$3r_s/2$</span> (for a Schwarzschild BH), and the <a href="https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit" rel="nofollow noreferrer">innermost stable circular orbit</a> or ISCO, the smallest circular orbit in which a test particle can stably orbit a massive object, at <span class="math-container">$3r_s$</span></p>
<blockquote>
<p>The ISCO plays an important role in black hole accretion disks since it marks the inner edge of the disk. [...] Circular orbits are still possible between the ISCO and the photon sphere, but they are unstable.
For a massless test particle like a photon, the only possible circular orbit is exactly at the photon sphere, and is unstable. Inside the photon sphere, no circular orbits exist. [...] The case for rotating black holes is somewhat more complicated. </p>
</blockquote>
<p>Another phenomenon which involves the distance from a BH is <a href="https://en.wikipedia.org/wiki/Gravitational_lens" rel="nofollow noreferrer">gravitational lensing</a>. The amount of deflection of a light beam passing near a BH (or indeed any source of gravity) depends on the distance <span class="math-container">$r$</span> of the beam's path from the centre of mass. The deflection angle is given by <span class="math-container">$\theta=2r_s/r$</span>. I guess you <em>could</em> use diameter in that formula, but it's more natural to think in terms of radius. It's a bit like shooting at a target (with light beams). When you shoot at a circular target, you're concerned with how far your shots are from the centre of the target.</p>
<hr />
<p>As I said in the comments, our usual notions of geometry are not very useful in the highly curved spacetime near a BH. And if you want to get close to a BH, you need to be in an orbit travelling near the speed of light, and / or your spacecraft needs to be capable of the extremely high acceleration required to hover near a BH (and you &amp; your ship need impossible strength to cope with the insanely huge <em>g</em> force resulting from such acceleration). Travelling at high speed adds additional relativistic effects on top of those GR effects due to the BH's gravity.</p>
<p>To describe the locations and times near the black hole, you need to choose a coordinate system. Schwarzschild coordinates are often used to describe a non-spinning black hole, but they have a coordinate singularity at the EH (event horizon), so they're annoying to work with when you want to talk about objects crossing the EH. It's a lot like how the latitude &amp; longitude on Earth break down near the poles. You can't travel further south than the South Pole at 90°S, and the pole itself doesn't have a well-defined longitude. If I tell you that I'm 1 km north of the South Pole, I could be anywhere on a circle of 1 km radius.</p>
<p>Fortunately (as I mentioned in the comments), GR is very flexible regarding coordinates, and there are quite a few standard coordinate systems used in various circumstances, you can see a list of some of them on Wikipedia's page <a href="https://en.wikipedia.org/wiki/Category:Coordinate_charts_in_general_relativity" rel="nofollow noreferrer">Category:Coordinate charts in general relativity</a>. However, that flexibility makes things complicated, and difficult to describe properly without using advanced mathematics. So popular treatments of GR gloss over those difficulties (or avoid them completely), and that has led to a popular understanding of GR elements that is somewhat distorted.</p>

CC BY-SA 4.0
2020-11-11T06:37:40.303Z
I believe photons don't exist. Light is an immaterial wave, consisting of itself, not of particles. The photon theory doesn't make any sense to me, and noone ever imaged photons in a microscope.
John
CC BY-SA 4.0
2020-11-12T21:55:54.933Z
@John Modern science doesn't try to say what's really going on in the world, it leaves that job to the philosophers. Instead, it takes an indirect approach. It develops theories that attempt to explain some aspects of the world, and within that theoretical framework it develops models that are consistent with existing empirical measurements /  observations and are hopefully consistent with future observations. So we aren't claiming that photons (or any other particles) truly exist, they're simply an element of our currently most successful theory of electromagnetic phenomena.
PM 2Ring
CC BY-SA 4.0
2020-11-12T21:56:23.053Z
(cont) Photons are not like classical particles. They do have some properties that are reminiscent of such particles, and other properties that are quite unclassical. But all of that isn't really relevant to what I said in my answer. Light can orbit a black hole at the photon sphere, but such orbits are unstable. This "photon sphere theory" is consistent with our observations of actual black holes.
PM 2Ring
CC BY-SA 4.0
2020-11-13T07:41:51.200Z
Perhaps because the spacetime gets curved light behaves like it does around compact stars. I think Einsteinian physics can explain it without photons, in contrast to Newtonian gravity.
John
CC BY-SA 4.0
2020-11-13T10:57:31.597Z
@John Maybe I don't get what you're trying to say. Light gets gravitationally deflected in the vicinity of stars, but it can't orbit a normal star: their radius is too large compared to their Schwarzschild radius. (As Wikipedia says, it's *possible* that some neutron stars have a photon sphere. I expect that's most likely for the more massive neutron stars when they are new, and still have a high spin).
PM 2Ring
CC BY-SA 4.0
2020-11-13T12:27:00.687Z
I see, by "normal star" you include white dwarves and neutron stars. I also think very dense neutron stars (respectively quark stars) have something like a photon sphere.
John
CC BY-SA 4.0
2020-11-13T12:39:10.563Z
@John Well, white dwarfs aren't very normal, and neutron stars even less so, (both being mostly degenerate matter), but I don't class black holes as stars at all. White dwarfs can still undergo fusion on the surface, through accretion from a companion, and with enough material (including via collision) they can go supernova. A neutron star is a supernova remnant, and it can't go supernova again, although the collision of binary neutron stars leads to a kilonova.
PM 2Ring
CC BY-SA 4.0
2020-11-13T12:45:15.887Z
While you're right that black holes might actually be something different than a singularity (I believe they're the entrance of an Einstein Rosen Bridge and you get ejected by a white hole elsewhere in the universe) I anyway consider black holes stars because spherical bodies orbiting them are called planets, and black holes have a spectral classification and they were stars once. If we consider brown and white dwarfs and neutron stars that are so old to have become dim stars, I think we should consider black holes too ones.
John
CC BY-SA 4.0
2020-11-11T02:02:37.973Z
2020-11-11T02:02:37.973Z
PM 2Ring
<p>The diameter of a black hole (the distance &quot;through&quot; it) or the radius (distance to the center) aren't really meaningful concepts. The singularity is in the future, not in any spatial direction. The <span class="math-container">$r$</span> Schwarzschild coordinate is actually a so-called &quot;reduced circumference&quot; – the metric length of a circle around the hole, divided by <span class="math-container">$2π$</span> – so if you wanted to replace <span class="math-container">$r$</span> by a value with more physical meaning, it would make more sense to multiply it by <span class="math-container">$2π$</span> than by <span class="math-container">$2$</span>. The problem with doing either of these things is that they make the Schwarzschild metric uglier. You end up having to divide by <span class="math-container">$2$</span> or <span class="math-container">$2π$</span> again every time the coordinate appears, so it's just <span class="math-container">$r$</span> with extra steps.</p>
<p>The event horizon of an ideal eternal black hole is exactly spherical, even if the hole is rotating. In fact you can't detect the rotation locally near the event horizon (perfect frame dragging). The event horizon of more realistic holes isn't exactly spherical, but it's very close to spherical most of the time.</p>

CC BY-SA 4.0
2020-11-11T06:35:47.813Z
I have accepted the other answer but yours is good and interesting too, and it answers my 2nd question.
John
CC BY-SA 4.0
2020-11-10T23:57:32.620Z
2020-11-10T23:57:32.620Z
benrg
<p>I never heard of Schwarzschild diameters, only of Schwarzschild radii. Why is it like this? Wouldn't it be better to portray the (possible) size of a black hole in a diameter?</p>
<p>Either way, can we use the terms radius or diameter term precisely? Is an event horizon always perfectly spherical or can it have something like an equatorial bulge even though it's a black hole's immaterial event horizon.</p>

CC BY-SA 4.0
2020-11-10T07:25:34.873Z
A sphere is most simply described by means of radius, not by diameter: volume = 4/3 pi r^3. Also radius is a good choice in polar coordinates which are suitable to describe the non-spherical yet rotational symmetric deviations from a perfect sphere for rotating black holes
planetmaker
CC BY-SA 4.0
2020-11-10T08:16:20.367Z
@planetmaker So rotating singularities have an event horizon that _do_ have an equatorial bulge?
John
CC BY-SA 4.0
2020-11-10T10:01:30.933Z
Distance (& coordinates) in general relativity is a complicated topic. Especially near black holes! It's better to think of the Schwarzschild radius (of a non-rotating black hole) as a way to measure the circumference or area of the event horizon (using the usual formulae), rather than thinking of it as the distance to the centre of the black hole. There are numerous questions on this topic on the Physics stack, but they tend to get fairly technical very quickly.
PM 2Ring
CC BY-SA 4.0
2020-11-10T12:25:40.617Z
@PM2Ring But the S.-radius is actually the radius at which an object of a certain mass would have an escape velocity at 1c. I wonder why the diameter isn't more important since that's the full (1-dimensional) size (and thus also closer to a circumference, isn't it?), not half of it.
John
CC BY-SA 4.0
2020-11-10T12:36:56.430Z
Please see https://physics.stackexchange.com/a/339424/123208 & https://physics.stackexchange.com/a/222563/123208 & https://physics.stackexchange.com/q/109924/123208 & https://physics.stackexchange.com/a/144458/123208
PM 2Ring
CC BY-SA 4.0
2020-11-10T12:40:47.307Z
As those answers to those linked questions explain, it's rather problematic to define the Schwarzschild radius in reference to the interior of the event horizon. It's much better to define it in terms of the spacetime that's actually observable, and even that has to be done carefully.
PM 2Ring
CC BY-SA 4.0
2020-11-10T13:08:09.363Z
Also, in general relativity, coordinates are simply a systematic way of labeling spacetime points (aka events), they don't necessarily correspond to nice observable measurements of distance or time. This can be rather disconcerting for new students of GR, and it generally takes a while for this concept to sink on. OTOH, this aspect of GR gives it great flexibility and power.
PM 2Ring
CC BY-SA 4.0
2020-11-10T13:10:40.503Z
Once again, there are many questions on that topic on Physics.SE, and Wikipedia has several relevant articles, but they're all fairly technical. This question from Physics.SE *might* be helpful. https://physics.stackexchange.com/q/157010/123208 It focuses on time, but much of the info there also applies to space.
PM 2Ring
CC BY-SA 4.0
2020-11-10T13:17:02.300Z
I don't have the time (or energy) to write a proper answer right now. But briefly, the big problem with defining the diameter of a black hole is that a diameter implies a spacelike path from one side of the event horizon, through the centre, to the "diametrically" opposite point. But there are no such paths inside the event horizon.
PM 2Ring
CC BY-SA 4.0
2020-11-10T13:57:12.150Z
@PM2Ring But if someone asked (just as an example) "Would my hand fit into that black hole?" you would have to tell them the black hole's diameter in order to make clear whether they can put their hand into the black hole or not. E.g. if the hypothetical planet Nine is actually a primordial BH it would have a diameter of about 2 inches. So you _could_ barely put your hand into it. I think this is where 'traditional' size becomes relevant.
John
CC BY-SA 4.0
2020-11-10T15:30:49.553Z
Changing the conventions or units would have no practical effects. Why is it important to you?
D. Halsey
CC BY-SA 4.0
2020-11-10T16:42:36.107Z
@D.Halsey Because one wants to learn the _entire_ black hole's size, not half of it. I don't see a reason to mention the radius only, except for stating how far the singularity is behind the event horizon. But a certain distance is complicated anyway as PM2Ring points out above.
John
CC BY-SA 4.0
2020-11-11T00:44:52.623Z
Our usual notions of geometry are practically useless near a black hole. It's bad enough near a SMBH (super-massive black hole) but near a tiny BH the spacetime curvature changes very rapidly. A rigid body can't get near a small BH without being spaghettified (but you can cross the horizon of a SMBH before getting spaghettified). According to https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator the tidal force from one atom to its neighbour atom in normal matter near a 1" radius BH is ~14 trillion *g* (Earth gravities). No hand or tape measure can cope with that.
PM 2Ring
CC BY-SA 4.0
2020-11-11T00:51:38.893Z
But if you'd like to read about what happens to a measuring tape / rope / chain near a huge BH, please see https://www.gregegan.net/SCIENCE/Rindler/RindlerHorizon.html & https://www.gregegan.net/SCIENCE/Rindler/SimpleElasticity.html Greg Egan also has some great intro articles on special & general relativity, black holes, and quantum mechanics (sorry, I should've linked them earlier): https://www.gregegan.net/FOUNDATIONS/
PM 2Ring
CC BY-SA 4.0
2020-11-13T01:01:01.690Z
I'm voting to **leave open!** because closing is totally counterproductive to the site. It's not an unreasonable question and it has attracted two thoughtful and insightful and therefore valuable answers! Closing just reduces visibility of these and blocks further answers. So instead I'll make an edit to achieve "more focus" without doing the damage that closing does.
uhoh
<black-hole><size><event-horizon>
Why are event horizons described in Schwarzschild radii rather than diameters, and either way can they change equator to pole?
John
# Why are event horizons described in Schwarzschild radii rather than diameters, and either way can they change equator to pole?

(Asked by: username_0 on 2020-11-10)

I never heard of Schwarzschild diameters, only of Schwarzschild radii. Why is it like this? Wouldn't it be better to portray the (possible) size of a black hole in a diameter?

Either way, can we use the terms radius or diameter term precisely? Is an event horizon always perfectly spherical or can it have something like an equatorial bulge even though it's a black hole's immaterial event horizon.

**Tags:** black-hole, size, event-horizon

> **Comments:**
> **username_1**: A sphere is most simply described by means of radius, not by diameter: volume = 4/3 pi r^3. Also radius is a good choice in polar coordinates which are suitable to describe the non-spherical yet rotational symmetric deviations from a perfect sphere for rotating black holes  
> **username_0**: @username_1 So rotating singularities have an event horizon that _do_ have an equatorial bulge?  
> **username_2**: Distance (& coordinates) in general relativity is a complicated topic. Especially near black holes! It's better to think of the Schwarzschild radius (of a non-rotating black hole) as a way to measure the circumference or area of the event horizon (using the usual formulae), rather than thinking of it as the distance to the centre of the black hole. There are numerous questions on this topic on the Physics stack, but they tend to get fairly technical very quickly.  
> **username_0**: @username_2 But the S.-radius is actually the radius at which an object of a certain mass would have an escape velocity at 1c. I wonder why the diameter isn't more important since that's the full (1-dimensional) size (and thus also closer to a circumference, isn't it?), not half of it.  
> **username_2**: Please see https://physics.stackexchange.com/a/339424/123208 & https://physics.stackexchange.com/a/222563/123208 & https://physics.stackexchange.com/q/109924/123208 & https://physics.stackexchange.com/a/144458/123208  
> **username_2**: As those answers to those linked questions explain, it's rather problematic to define the Schwarzschild radius in reference to the interior of the event horizon. It's much better to define it in terms of the spacetime that's actually observable, and even that has to be done carefully.  
> **username_2**: Also, in general relativity, coordinates are simply a systematic way of labeling spacetime points (aka events), they don't necessarily correspond to nice observable measurements of distance or time. This can be rather disconcerting for new students of GR, and it generally takes a while for this concept to sink on. OTOH, this aspect of GR gives it great flexibility and power.  
> **username_2**: Once again, there are many questions on that topic on Physics.SE, and Wikipedia has several relevant articles, but they're all fairly technical. This question from Physics.SE *might* be helpful. https://physics.stackexchange.com/q/157010/123208 It focuses on time, but much of the info there also applies to space.  
> **username_2**: I don't have the time (or energy) to write a proper answer right now. But briefly, the big problem with defining the diameter of a black hole is that a diameter implies a spacelike path from one side of the event horizon, through the centre, to the "diametrically" opposite point. But there are no such paths inside the event horizon.  
> **username_0**: @username_2 But if someone asked (just as an example) "Would my hand fit into that black hole?" you would have to tell them the black hole's diameter in order to make clear whether they can put their hand into the black hole or not. E.g. if the hypothetical planet Nine is actually a primordial BH it would have a diameter of about 2 inches. So you _could_ barely put your hand into it. I think this is where 'traditional' size becomes relevant.  
> **username_3**: Changing the conventions or units would have no practical effects. Why is it important to you?  
> **username_0**: @D.username_3 Because one wants to learn the _entire_ black hole's size, not half of it. I don't see a reason to mention the radius only, except for stating how far the singularity is behind the event horizon. But a certain distance is complicated anyway as username_2 points out above.  
> **username_2**: Our usual notions of geometry are practically useless near a black hole. It's bad enough near a SMBH (super-massive black hole) but near a tiny BH the spacetime curvature changes very rapidly. A rigid body can't get near a small BH without being spaghettified (but you can cross the horizon of a SMBH before getting spaghettified). According to https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator the tidal force from one atom to its neighbour atom in normal matter near a 1" radius BH is ~14 trillion *g* (Earth gravities). No hand or tape measure can cope with that.  
> **username_2**: But if you'd like to read about what happens to a measuring tape / rope / chain near a huge BH, please see https://www.gregegan.net/SCIENCE/Rindler/RindlerHorizon.html & https://www.gregegan.net/SCIENCE/Rindler/SimpleElasticity.html Greg Egan also has some great intro articles on special & general relativity, black holes, and quantum mechanics (sorry, I should've linked them earlier): https://www.gregegan.net/FOUNDATIONS/  
> **username_4**: I'm voting to **leave open!** because closing is totally counterproductive to the site. It's not an unreasonable question and it has attracted two thoughtful and insightful and therefore valuable answers! Closing just reduces visibility of these and blocks further answers. So instead I'll make an edit to achieve "more focus" without doing the damage that closing does.  

---

## 2 answers

### Answer by username_2 on 2020-11-11. Score: 3 | Accepted answer

We use the Schwarzschild radius $r\_s$ (rather than a diameter) because it's convenient. We're want to describe what happens in the vicinity of a black hole, so it's natural to talk about the distance from the black hole.

For example, there's the [photon sphere](https://en.wikipedia.org/wiki/Photon_sphere) at $3r\_s/2$ (for a Schwarzschild BH), and the [innermost stable circular orbit](https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit) or ISCO, the smallest circular orbit in which a test particle can stably orbit a massive object, at $3r\_s$

> The ISCO plays an important role in black hole accretion disks since it marks the inner edge of the disk. [...] Circular orbits are still possible between the ISCO and the photon sphere, but they are unstable.
> For a massless test particle like a photon, the only possible circular orbit is exactly at the photon sphere, and is unstable. Inside the photon sphere, no circular orbits exist. [...] The case for rotating black holes is somewhat more complicated.

Another phenomenon which involves the distance from a BH is [gravitational lensing](https://en.wikipedia.org/wiki/Gravitational_lens). The amount of deflection of a light beam passing near a BH (or indeed any source of gravity) depends on the distance $r$ of the beam's path from the centre of mass. The deflection angle is given by $\theta=2r\_s/r$. I guess you *could* use diameter in that formula, but it's more natural to think in terms of radius. It's a bit like shooting at a target (with light beams). When you shoot at a circular target, you're concerned with how far your shots are from the centre of the target.

---

As I said in the comments, our usual notions of geometry are not very useful in the highly curved spacetime near a BH. And if you want to get close to a BH, you need to be in an orbit travelling near the speed of light, and / or your spacecraft needs to be capable of the extremely high acceleration required to hover near a BH (and you & your ship need impossible strength to cope with the insanely huge *g* force resulting from such acceleration). Travelling at high speed adds additional relativistic effects on top of those GR effects due to the BH's gravity.

To describe the locations and times near the black hole, you need to choose a coordinate system. Schwarzschild coordinates are often used to describe a non-spinning black hole, but they have a coordinate singularity at the EH (event horizon), so they're annoying to work with when you want to talk about objects crossing the EH. It's a lot like how the latitude & longitude on Earth break down near the poles. You can't travel further south than the South Pole at 90°S, and the pole itself doesn't have a well-defined longitude. If I tell you that I'm 1 km north of the South Pole, I could be anywhere on a circle of 1 km radius.

Fortunately (as I mentioned in the comments), GR is very flexible regarding coordinates, and there are quite a few standard coordinate systems used in various circumstances, you can see a list of some of them on Wikipedia's page [Category:Coordinate charts in general relativity](https://en.wikipedia.org/wiki/Category:Coordinate_charts_in_general_relativity). However, that flexibility makes things complicated, and difficult to describe properly without using advanced mathematics. So popular treatments of GR gloss over those difficulties (or avoid them completely), and that has led to a popular understanding of GR elements that is somewhat distorted.

> **Comments:**
> **username_0**: I believe photons don't exist. Light is an immaterial wave, consisting of itself, not of particles. The photon theory doesn't make any sense to me, and noone ever imaged photons in a microscope.  
> **username_2**: @username_0 Modern science doesn't try to say what's really going on in the world, it leaves that job to the philosophers. Instead, it takes an indirect approach. It develops theories that attempt to explain some aspects of the world, and within that theoretical framework it develops models that are consistent with existing empirical measurements /  observations and are hopefully consistent with future observations. So we aren't claiming that photons (or any other particles) truly exist, they're simply an element of our currently most successful theory of electromagnetic phenomena.  
> **username_2**: (cont) Photons are not like classical particles. They do have some properties that are reminiscent of such particles, and other properties that are quite unclassical. But all of that isn't really relevant to what I said in my answer. Light can orbit a black hole at the photon sphere, but such orbits are unstable. This "photon sphere theory" is consistent with our observations of actual black holes.  
> **username_0**: Perhaps because the spacetime gets curved light behaves like it does around compact stars. I think Einsteinian physics can explain it without photons, in contrast to Newtonian gravity.  
> **username_2**: @username_0 Maybe I don't get what you're trying to say. Light gets gravitationally deflected in the vicinity of stars, but it can't orbit a normal star: their radius is too large compared to their Schwarzschild radius. (As Wikipedia says, it's *possible* that some neutron stars have a photon sphere. I expect that's most likely for the more massive neutron stars when they are new, and still have a high spin).  
> **username_0**: I see, by "normal star" you include white dwarves and neutron stars. I also think very dense neutron stars (respectively quark stars) have something like a photon sphere.  
> **username_2**: @username_0 Well, white dwarfs aren't very normal, and neutron stars even less so, (both being mostly degenerate matter), but I don't class black holes as stars at all. White dwarfs can still undergo fusion on the surface, through accretion from a companion, and with enough material (including via collision) they can go supernova. A neutron star is a supernova remnant, and it can't go supernova again, although the collision of binary neutron stars leads to a kilonova.  
> **username_0**: While you're right that black holes might actually be something different than a singularity (I believe they're the entrance of an Einstein Rosen Bridge and you get ejected by a white hole elsewhere in the universe) I anyway consider black holes stars because spherical bodies orbiting them are called planets, and black holes have a spectral classification and they were stars once. If we consider brown and white dwarfs and neutron stars that are so old to have become dim stars, I think we should consider black holes too ones.  

---

### Answer by username_5 on 2020-11-10. Score: 2

The diameter of a black hole (the distance "through" it) or the radius (distance to the center) aren't really meaningful concepts. The singularity is in the future, not in any spatial direction. The $r$ Schwarzschild coordinate is actually a so-called "reduced circumference" – the metric length of a circle around the hole, divided by $2π$ – so if you wanted to replace $r$ by a value with more physical meaning, it would make more sense to multiply it by $2π$ than by $2$. The problem with doing either of these things is that they make the Schwarzschild metric uglier. You end up having to divide by $2$ or $2π$ again every time the coordinate appears, so it's just $r$ with extra steps.

The event horizon of an ideal eternal black hole is exactly spherical, even if the hole is rotating. In fact you can't detect the rotation locally near the event horizon (perfect frame dragging). The event horizon of more realistic holes isn't exactly spherical, but it's very close to spherical most of the time.

> **Comments:**
> **username_0**: I have accepted the other answer but yours is good and interesting too, and it answers my 2nd question.  

---

