<p>The Schwarzschild metric by definition can be written in a spherically symmetric form, the Kerr metric is not spherically symmetric, but in  Boyer–Lindquist coordinates the event horizon has a constant radial coordinate.</p>

<p>For most people this will be good enough to say the event horizon of Kerr black holes (including Schwarzschild black holes) are perfect spheres. A physical black hole has a quasi-neutral charge, but the more general case of a charged Kerr-Newman black hole still has a perfectly spherical event horizon.</p>

<p>Now of course no physical situation actually possess the exact symmetries of Kerr black holes. However the outer Kerr metric is stable, which means that the event horizon mantains a near-perfect spherical shape, except in extreme situations such as when two black holes are undergoing a merger.</p>

CC BY-SA 3.0
2015-11-07T01:37:12.833Z
2015-11-07T01:37:12.833Z
John Davis
<p>Because the gravitational pull of a non-rotating black hole in a void space is equal at all sides (as a result of it being a point) the event horizon should in theory be a perfect sphere. Because of the nature of a blackhole not letting light escape you would need a background for it to be seen. Also, in real world physics a blackhole  would need to interact with light to be "seen" which "correct me if I'm wrong" has it's own very small gravitational pull, it would slightly bend the sphere a bit, the observer will give the same effect.</p>

<p>If rotating it would (as pointed out before) form an ergosphere around the event horizon connecting to it at the poles an being the farthest over the equator, thus forming a more oval shape.</p>

<p>Also, you know who to subscribe to right?</p>

CC BY-SA 4.0
2019-01-15T16:20:05.950Z
2019-01-15T16:57:58.203Z
2019-01-15T16:57:58.203Z
Das Stein
<p>As I understand it, general relativity says that there's gravitation everywhere in the universe, and this gravitation creates dips in space, so to speak, often represented in 2D as a weight on a rubber sheet, like the picture below.</p>

<p><a href="https://i.sstatic.net/mmPP2.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/mmPP2.jpg" alt="enter image description here"></a></p>

<p><a href="http://www.daviddarling.info/encyclopedia/G/genrel.html" rel="nofollow noreferrer">Source</a>, </p>

<p>so, a black hole might generate a perfectly spherical event horizon, but it generates it on a not perfectly flat 3 dimensional surface, and I think the gravitation from other objects makes it not quite a perfect sphere.  For example, a star or planet that orbits a black hole would drag around a ripple on the event horizon as it orbits the black hole.</p>

<p>Precisely what shape that ripple would be . . . I'm not sure.    That said, if you had a black hole as the only object in a universe, then I think the event horizon might be a perfect sphere, or as perfect as possible, given quantum fluctuation, hawking radiation and the impossibility of precise observation and all that good stuff.</p>

<p>Non black holes tend to have lumpy/inconsistent gravity - see <a href="http://www.universetoday.com/116801/the-potsdam-gravity-potato-shows-earths-gravity-variations/" rel="nofollow noreferrer">here</a>.  Even Neutron stars have some inconsistency, but Black holes probably avoid that because the matter is condensed to a point, so there's equal gravitational pull from all directions from the singularity.</p>

<p>Now a Kerr black hole,  that's a whole different question.   I'm not smart enough to try to answer that one.   That might not be a sphere at all.</p>

CC BY-SA 4.0
2019-01-15T17:22:27.960Z
The rubber sheet analogy isn't wrong, per se, but it can be misleading. Eg, what causes bodies to move down into the depression, some kind of meta-gravity?
PM 2Ring
CC BY-SA 4.0
2019-01-15T17:29:47.387Z
"there's equal gravitational pull from all directions from the singularity". It's a bit more subtle than that. We can't get any information from inside the EH (event horizon), and that includes mapping the shape of the black hole's core; gravity cannot escape from the EH anymore than light can. The gravitational response of bodies outside the EH is completely determined by the spacetime curvature that was created as the BH was formed. For further details, please see http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html
PM 2Ring
CC BY-SA 4.0
2019-01-16T09:42:40.700Z
@PM2Ring Thank you for that.   This is perhaps a question that novices like me shouldn't try to answer.
userLTK
CC BY-SA 3.0
2015-11-06T13:32:38.653Z
I would avoid using the rubber sheet analogy here. It's misleading in a number of aspects, and doesn't help your points too much.
HDE 226868
CC BY-SA 3.0
2015-11-06T21:37:52.687Z
@HDE226868 are you saying the gravitational field doesn't effect the perfect sphere of the event horizon?
userLTK
CC BY-SA 3.0
2015-11-07T06:27:23.127Z
So, the rubber sheet analogy is not ideal, but is it wrong?    Can someone say why it's wrong rather than just voting down?   It's not bad logic.  The Kerr metric is obviously bigger and I said that.   Why is this wrong for a non  rotating black hole?   Inquiring minds want to know.    It might be wrong, but what's the explanation?
userLTK
CC BY-SA 3.0
2015-11-06T07:16:05.330Z
2015-11-06T07:16:05.330Z
userLTK
<p>You cannot see the event horizon. That being said:</p>

<p>A non-rotating black hole, free of external influences, has perfect spherical symmetry. All its properties are exactly the same in any direction, period. This is the Schwarzschild metric.</p>

<p><a href="https://en.wikipedia.org/wiki/Schwarzschild_metric" rel="noreferrer">https://en.wikipedia.org/wiki/Schwarzschild_metric</a></p>

<p>Even if it is electrically charged, if it's non-rotating, and free of external influences, it is still perfectly spherically symmetrical - this is the Reissner–Nordström metric.</p>

<p><a href="https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric" rel="noreferrer">https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric</a></p>

<p>As soon as it begins rotating, the black hole is no longer in perfect spherical symmetry. It acquires an ergosphere, which is shaped like a flattened ball. This is the Kerr metric.</p>

<p><a href="https://en.wikipedia.org/wiki/Kerr_metric" rel="noreferrer">https://en.wikipedia.org/wiki/Kerr_metric</a></p>

<p><a href="https://i.sstatic.net/RxP11.png" rel="noreferrer"><img src="https://i.sstatic.net/RxP11.png" alt="Kerr metric with ergosphere"></a></p>

<p>Significant external factors can distort the shape of the event horizon. Two black holes merging will go through a process where spherical symmetry is lost temporarily:</p>

<p><a href="https://www.youtube.com/watch?v=JOQMvyLYmd4" rel="noreferrer">https://www.youtube.com/watch?v=JOQMvyLYmd4</a></p>

<p>Technically, anything near a BH would slightly distort the metric of spacetime, but in practice this would not be easily measurable unless the external factor is very large (another very massive object). So for most practical purposes, the metrics described above would apply.</p>

CC BY-SA 3.0
2015-11-06T22:23:38.807Z
2015-11-06T22:30:46.130Z
2015-11-06T22:30:46.130Z
Florin Andrei
<p>I know all about how black holes form and why their gravity is so strong. However, is the gravity equally powerful in all directions? Will the event horizon appear as a <strong>perfect</strong> sphere?</p>

CC BY-SA 3.0
2015-11-06T01:28:30.190Z
One can not ***directly*** see the event horizon, but one can see radiation from the infall of material, which might give visual cues to the event horizon. The event horizon would ***be*** a perfect sphere, but it would not ***appear*** as a perfect sphere because of gravitational lensing. If you were stationary or moving directly towards or away from the black hole, it would appear as a perfect circle that was apparently larger than the actual black hole, also because of lensing.
Eubie Drew
CC BY-SA 3.0
2015-11-06T02:35:12.790Z
@Aabaakawad What about a spinning black hole? Surely that would have some effect on the event horizon, maybe making it more oval-shaped?
Sir Cumference
CC BY-SA 3.0
2015-11-06T03:33:12.477Z
you might think that because you are imagining the mass of the black hole to be distributed throughout the volume within the event horizon. It is not. All the mass is either concentrated into a point in the center of the black (a singularity), or an ultracompact object being held up from being a singularity by forces not yet known, (not likely, but who knows). Yes, a singularity can have spin, and I do not know how that works. It can also have an electric charge and/or a magnetic moment.
Eubie Drew
CC BY-SA 3.0
2015-11-06T07:46:18.830Z
See [this Physic.SE Q&A](http://physics.stackexchange.com/questions/72824/what-is-the-shape-of-a-black-hole), or even [the wikipedia page](https://en.wikipedia.org/wiki/Black_hole#Event_horizon).
zibadawa timmy
<gravity><black-hole><light><stellar-evolution>
Is a black hole a perfect sphere?
Sir Cumference
# Is a black hole a perfect sphere?

(Asked by: username_0 on 2015-11-05)

I know all about how black holes form and why their gravity is so strong. However, is the gravity equally powerful in all directions? Will the event horizon appear as a **perfect** sphere?

**Tags:** gravity, black-hole, light, stellar-evolution

> **Comments:**
> **username_1**: One can not ***directly*** see the event horizon, but one can see radiation from the infall of material, which might give visual cues to the event horizon. The event horizon would ***be*** a perfect sphere, but it would not ***appear*** as a perfect sphere because of gravitational lensing. If you were stationary or moving directly towards or away from the black hole, it would appear as a perfect circle that was apparently larger than the actual black hole, also because of lensing.  
> **username_0**: @Aabaakawad What about a spinning black hole? Surely that would have some effect on the event horizon, maybe making it more oval-shaped?  
> **username_1**: you might think that because you are imagining the mass of the black hole to be distributed throughout the volume within the event horizon. It is not. All the mass is either concentrated into a point in the center of the black (a singularity), or an ultracompact object being held up from being a singularity by forces not yet known, (not likely, but who knows). Yes, a singularity can have spin, and I do not know how that works. It can also have an electric charge and/or a magnetic moment.  
> **username_2**: See [this Physic.SE Q&A](http://physics.stackexchange.com/questions/72824/what-is-the-shape-of-a-black-hole), or even [the wikipedia page](https://en.wikipedia.org/wiki/Black_hole#Event_horizon).  

---

## 4 answers

### Answer by username_8 on 2015-11-06. Score: 7 | Accepted answer

You cannot see the event horizon. That being said:

A non-rotating black hole, free of external influences, has perfect spherical symmetry. All its properties are exactly the same in any direction, period. This is the Schwarzschild metric.

<https://en.wikipedia.org/wiki/Schwarzschild_metric>

Even if it is electrically charged, if it's non-rotating, and free of external influences, it is still perfectly spherically symmetrical - this is the Reissner–Nordström metric.

<https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric>

As soon as it begins rotating, the black hole is no longer in perfect spherical symmetry. It acquires an ergosphere, which is shaped like a flattened ball. This is the Kerr metric.

<https://en.wikipedia.org/wiki/Kerr_metric>

[![Kerr metric with ergosphere](https://i.sstatic.net/RxP11.png)](https://i.sstatic.net/RxP11.png)

Significant external factors can distort the shape of the event horizon. Two black holes merging will go through a process where spherical symmetry is lost temporarily:

<https://www.youtube.com/watch?v=JOQMvyLYmd4>

Technically, anything near a BH would slightly distort the metric of spacetime, but in practice this would not be easily measurable unless the external factor is very large (another very massive object). So for most practical purposes, the metrics described above would apply.


---

### Answer by username_3 on 2015-11-07. Score: 4

The Schwarzschild metric by definition can be written in a spherically symmetric form, the Kerr metric is not spherically symmetric, but in Boyer–Lindquist coordinates the event horizon has a constant radial coordinate.

For most people this will be good enough to say the event horizon of Kerr black holes (including Schwarzschild black holes) are perfect spheres. A physical black hole has a quasi-neutral charge, but the more general case of a charged Kerr-Newman black hole still has a perfectly spherical event horizon.

Now of course no physical situation actually possess the exact symmetries of Kerr black holes. However the outer Kerr metric is stable, which means that the event horizon mantains a near-perfect spherical shape, except in extreme situations such as when two black holes are undergoing a merger.


---

### Answer by username_4 on 2019-01-15. Score: 0

Because the gravitational pull of a non-rotating black hole in a void space is equal at all sides (as a result of it being a point) the event horizon should in theory be a perfect sphere. Because of the nature of a blackhole not letting light escape you would need a background for it to be seen. Also, in real world physics a blackhole would need to interact with light to be "seen" which "correct me if I'm wrong" has it's own very small gravitational pull, it would slightly bend the sphere a bit, the observer will give the same effect.

If rotating it would (as pointed out before) form an ergosphere around the event horizon connecting to it at the poles an being the farthest over the equator, thus forming a more oval shape.

Also, you know who to subscribe to right?


---

### Answer by username_5 on 2015-11-06. Score: 0

As I understand it, general relativity says that there's gravitation everywhere in the universe, and this gravitation creates dips in space, so to speak, often represented in 2D as a weight on a rubber sheet, like the picture below.

[![enter image description here](https://i.sstatic.net/mmPP2.jpg)](https://i.sstatic.net/mmPP2.jpg)

[Source](http://www.daviddarling.info/encyclopedia/G/genrel.html),

so, a black hole might generate a perfectly spherical event horizon, but it generates it on a not perfectly flat 3 dimensional surface, and I think the gravitation from other objects makes it not quite a perfect sphere. For example, a star or planet that orbits a black hole would drag around a ripple on the event horizon as it orbits the black hole.

Precisely what shape that ripple would be . . . I'm not sure. That said, if you had a black hole as the only object in a universe, then I think the event horizon might be a perfect sphere, or as perfect as possible, given quantum fluctuation, hawking radiation and the impossibility of precise observation and all that good stuff.

Non black holes tend to have lumpy/inconsistent gravity - see [here](http://www.universetoday.com/116801/the-potsdam-gravity-potato-shows-earths-gravity-variations/). Even Neutron stars have some inconsistency, but Black holes probably avoid that because the matter is condensed to a point, so there's equal gravitational pull from all directions from the singularity.

Now a Kerr black hole, that's a whole different question. I'm not smart enough to try to answer that one. That might not be a sphere at all.

> **Comments:**
> **username_7**: I would avoid using the rubber sheet analogy here. It's misleading in a number of aspects, and doesn't help your points too much.  
> **username_5**: @HDE226868 are you saying the gravitational field doesn't effect the perfect sphere of the event horizon?  
> **username_5**: So, the rubber sheet analogy is not ideal, but is it wrong?    Can someone say why it's wrong rather than just voting down?   It's not bad logic.  The Kerr metric is obviously bigger and I said that.   Why is this wrong for a non  rotating black hole?   Inquiring minds want to know.    It might be wrong, but what's the explanation?  
> **username_6**: The rubber sheet analogy isn't wrong, per se, but it can be misleading. Eg, what causes bodies to move down into the depression, some kind of meta-gravity?  
> **username_6**: "there's equal gravitational pull from all directions from the singularity". It's a bit more subtle than that. We can't get any information from inside the EH (event horizon), and that includes mapping the shape of the black hole's core; gravity cannot escape from the EH anymore than light can. The gravitational response of bodies outside the EH is completely determined by the spacetime curvature that was created as the BH was formed. For further details, please see http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html  
> **username_5**: @PM2Ring Thank you for that.   This is perhaps a question that novices like me shouldn't try to answer.  

---

