<p>You cannot exceed the speed of light "locally". But you can -see- imagine<sup>*</sup> <strong><em>distances increase quicker than the speed of light</em></strong>.</p>

<p>If by traveling you mean "moving compared to local space time", then light cannot <strong><em>travel</em></strong> quicker than the speed of light. In your example, the distance increases faster than the speed of light, because <strong><em>spacetime is dragged along</em></strong> inside the black hole by its gravity.</p>

<p><sub>*You in fact cannot "see" it, because you would need some information to be transferred somehow to do the same! This is not possible, because of this damn <em>c</em> limitation.</sub></p>

CC BY-SA 3.0
2018-01-22T22:57:41.497Z
@j-chomel your answer is brilliant, i especially like the "distances increase quicker than speed of light" now that i am starting to grasp this concept.
tnt-rox
CC BY-SA 3.0
2018-01-23T14:07:33.170Z
@tnt-rox, I had similar question some time ago, and learned a lot from the answers I got: https://astronomy.stackexchange.com/questions/19909/will-we-start-seeing-galaxies-disappear-due-to-universe-expansion .
J. Chomel
CC BY-SA 3.0
2018-01-23T21:15:25.127Z
This is exactly why the diameter of the visible universe (93 billion light years) can be larger than the number of years since its creation (13.8 billion years).
vsz
CC BY-SA 3.0
2018-01-22T10:25:43.177Z
2018-01-22T23:17:25.943Z
2018-01-22T23:17:25.943Z
J. Chomel
<p>The speed of light will remain constant. Though the way it is perceived near a black hole changes with where and how it is perceived, it will remain constant. The speed of light does not increase or decrease just because it is near a black hole.</p>

CC BY-SA 3.0
2018-02-13T08:07:25.697Z
2018-02-13T08:07:25.697Z
DarkMatter
<p>It doesn't work like that. An observer at the light source (and indeed any observers anywhere else) will always see light travelling (in vacuum) at the speed of light locally.</p>

<p>There is also a major problem with your thought experiment. It is <em>not possible</em> for you to have a stationary light source within the event horizon of a black hole. It, and everything else in its vicinity, <em>must</em> be moving inwards. This is as inexorable and unavoidable as is the passage of time for an observer outside the event horizon.</p>

<p>In my opinion, the best "visual" way of thinking about the situation inside the event horizon is to imagine your photons of light like salmon trying to swimming upstream, whilst you are on a boat flowing with the stream and releasing the salmon into the water. You will always see the salmon swimming at some speed with respect to your boat. Unfortunately if the stream flows fast enough then the salmon will make no progress and you will both be swept over a waterfall (the singularity) a little further downstream.</p>

<p>Likewise, your common-sense fails with the situation of firing light towards a black hole. Light is always measured to have a speed of $c$ <em>locally</em>. It is following through with the consequences of this principle that leads to all the weird behaviour that black holes exhibit.</p>

CC BY-SA 3.0
2018-01-22T11:47:58.863Z
that is a beautiful answer. If i understand you correctly, there is no such thing as a "stationary object", it's a bit like a universal constant, or just a figment of our imagination. )
tnt-rox
CC BY-SA 3.0
2018-01-22T12:16:54.893Z
@tnt-rox There is no such thing as a static observer (sometimes called a shell observer) inside the event horizon.
ProfRob
CC BY-SA 3.0
2018-01-22T15:28:56.777Z
Nice explanation without any complicated terminology.
StephenG - Help Ukraine
CC BY-SA 3.0
2018-01-22T21:35:32.270Z
Isn't the word "vacuum" missing? Light doesn't always travel at $c$.
Eric Duminil
CC BY-SA 3.0
2018-01-22T23:36:56.733Z
Do not try this at home.
Strawberry
CC BY-SA 3.0
2018-01-23T05:46:17.263Z
I remember Leonard Susskind using that same "fish swimming upstream" in his lectures. It does capture the essence well.
TT.
CC BY-SA 3.0
2018-01-23T16:18:32.900Z
@EricDuminil I was under the impression that light DOES always travel at c, but in a non-vacuum it ricochets off particles, causing it to take a longer path and making it SEEM slower.
Feathercrown
CC BY-SA 4.0
2018-11-12T16:32:28.177Z
@Feathercrown actually absorbed and re-emitted (different photon). You might be interested in the "illusion" of galaxies exhibiting "faster than light" motion https://en.wikipedia.org/wiki/Superluminal_motion . It's a similar misconception.
Jack R. Woods
CC BY-SA 4.0
2018-11-15T14:33:15.493Z
@JackR.Woods Oh, that makes sense.
Feathercrown
CC BY-SA 3.0
2018-01-22T10:24:21.857Z
2018-01-22T22:34:56.990Z
2018-01-22T22:34:56.990Z
ProfRob
<p>If I had a directional photon-emitting source and placed it inside a black hole pointing upward and out towards the visible universe, I assume the photons traveling at the speed of light would slow and reverse direction back into the center.</p>

<p>So if I took the same source and placed it outside of the black hole pointing inward towards the black hole center, can I assume that an emitted photon would travel towards the center faster than the speed of light it is already traveling at?</p>

CC BY-SA 3.0
2018-01-22T08:51:55.653Z
I voted you up because I love the imagery, but light doesn't behave like that.  Light travels in a straight line, only curving when space is curved which it is, significantly, inside the event horizon of the black hole.  All the possible paths of light from your source curve towards the singularity, there isn't any up, slowing or reverse, there's only all directions point at the singularity.  That said, I'd rather not make this an answer as I still find space-time diagrams confusing.
userLTK
CC BY-SA 3.0
2018-01-22T09:07:31.003Z
@userLTK thanks. but if the source was slightly above the singularity, and a single photon was released perpendicular to the singularity, surely the photon would need to reverse direction, and that would imply the speed of light would slow down and stop before it was drawn to the singularity. or am i a complete idiot )
tnt-rox
CC BY-SA 3.0
2018-01-22T13:51:55.397Z
Small warning: All answers here rely on the assumption that our understanding of black holes (i.e. general relativity, along with our models of a black hole, most notably Schwarzschild and Kerr), keeps describing the inside of the event horizon with the same accuracy that it describes the outside. We have no reason to believe anything else, but assuming these models are indeed (close to) the truth, we can _never_ know the actual answer.
Arthur
CC BY-SA 3.0
2018-01-22T17:53:29.177Z
I was under the impression that the concepts "up" and "out" do not make sense in the context of inside the event horizon of a black hole.
emory
CC BY-SA 3.0
2018-01-22T20:32:15.347Z
It's not a an answer to this question but one idea I was told was that a photon can loose energy, by red-shifting, and does as it leaves a gravity well, while moving at the same speed (mostly). So if you emit a photon close to the event horizon it will red-shift down to almost no energy while traveling at the speed. This is similar to a ball of string unrolling to nothing without slowing down. Ref: https://physics.stackexchange.com/questions/11726/can-a-photon-exiting-from-a-gravity-well-ever-reach-a-frequency-of-zero-wavele. Also given curved s/t the photon could return without ever slowing.
P Chapman
CC BY-SA 3.0
2018-01-22T21:21:44.147Z
@tnt-rox the answer is apparenetly c) both of the above see https://en.wikipedia.org/wiki/Redshift second paragraph "Some redshifts are an example of the Doppler effect ... Finally, gravitational redshift is a relativistic effect observed in electromagnetic radiation moving out of gravitational fields." And red light has less energy than blue light https://van.physics.illinois.edu/qa/listing.php?id=2629 so there is an energy loss, probably "relative to the viewer"
P Chapman
CC BY-SA 3.0
2018-01-23T02:43:16.913Z
Why do we assume black-holes are round?
Fandango68
CC BY-SA 3.0
2018-01-23T09:47:39.540Z
@Fandango68 We don't. When you see "round non-rotating black hole", imagine "a black hole in which the event horizon is a sphere's surface". It doesn't say much about the *internal* structure of the black hole, which is usually imaged as a sort of a "funnel" (taking on crazy shapes with rotation and electromagnetism involved) of heavily distorted space-time that may or may not "end" in a singularity (which cannot quite be described by general relativity). And why round event horizon? It's easier on the math. We also have models for rotating and charged black holes, which aren't spherical.
Luaan
CC BY-SA 3.0
2018-01-23T14:42:41.930Z
@tnt-rox You cannot be "slightly above the singularity" because inside the event horizon, the singularity is not a point in space but a point in *time*. It's like saying the source is "slightly above" next Tuesday.
JohannesD
CC BY-SA 3.0
2018-01-23T17:24:19.033Z
@Luaan it sounds like the same reason we have spherical cows :)
emory
CC BY-SA 3.0
2018-01-23T17:25:21.610Z
@Fandango68 I think for some distance metric, black holes are perfectly spherical.  Actually we can say that the event horizon is infinite distance from the center of the black hole in most reasonable distance metrics - so it is by definition a sphere.
emory
CC BY-SA 3.0
2018-01-25T00:14:41.343Z
@Luaan so you're saying that the closer we get to a black-hole, the less and less round it appears, but from a massive distance it looks round, because everything in the universe is "round"?
Fandango68
CC BY-SA 3.0
2018-02-13T22:09:35.957Z
There's isn't such a thing as straight towards the center, of a black hole, because time space is twisted so the photons you shoot forwards and backwards may travel faster sideways and have arrested path compared to an outside radius, which perhaps doesnt exist and is more of a spiral inside.
bandybabboon
<black-hole><photons>
Speed of light in a black hole
tnt-rox
# Speed of light in a black hole

(Asked by: username_0 on 2018-01-22)

If I had a directional photon-emitting source and placed it inside a black hole pointing upward and out towards the visible universe, I assume the photons traveling at the speed of light would slow and reverse direction back into the center.

So if I took the same source and placed it outside of the black hole pointing inward towards the black hole center, can I assume that an emitted photon would travel towards the center faster than the speed of light it is already traveling at?

**Tags:** black-hole, photons

> **Comments:**
> **username_1**: I voted you up because I love the imagery, but light doesn't behave like that.  Light travels in a straight line, only curving when space is curved which it is, significantly, inside the event horizon of the black hole.  All the possible paths of light from your source curve towards the singularity, there isn't any up, slowing or reverse, there's only all directions point at the singularity.  That said, I'd rather not make this an answer as I still find space-time diagrams confusing.  
> **username_0**: @username_1 thanks. but if the source was slightly above the singularity, and a single photon was released perpendicular to the singularity, surely the photon would need to reverse direction, and that would imply the speed of light would slow down and stop before it was drawn to the singularity. or am i a complete idiot )  
> **username_2**: Small warning: All answers here rely on the assumption that our understanding of black holes (i.e. general relativity, along with our models of a black hole, most notably Schwarzschild and Kerr), keeps describing the inside of the event horizon with the same accuracy that it describes the outside. We have no reason to believe anything else, but assuming these models are indeed (close to) the truth, we can _never_ know the actual answer.  
> **username_3**: I was under the impression that the concepts "up" and "out" do not make sense in the context of inside the event horizon of a black hole.  
> **username_4**: It's not a an answer to this question but one idea I was told was that a photon can loose energy, by red-shifting, and does as it leaves a gravity well, while moving at the same speed (mostly). So if you emit a photon close to the event horizon it will red-shift down to almost no energy while traveling at the speed. This is similar to a ball of string unrolling to nothing without slowing down. Ref: https://physics.stackexchange.com/questions/11726/can-a-photon-exiting-from-a-gravity-well-ever-reach-a-frequency-of-zero-wavele. Also given curved s/t the photon could return without ever slowing.  
> **username_4**: @username_0 the answer is apparenetly c) both of the above see https://en.wikipedia.org/wiki/Redshift second paragraph "Some redshifts are an example of the Doppler effect ... Finally, gravitational redshift is a relativistic effect observed in electromagnetic radiation moving out of gravitational fields." And red light has less energy than blue light https://van.physics.illinois.edu/qa/listing.php?id=2629 so there is an energy loss, probably "relative to the viewer"  
> **username_5**: Why do we assume black-holes are round?  
> **username_6**: @username_5 We don't. When you see "round non-rotating black hole", imagine "a black hole in which the event horizon is a sphere's surface". It doesn't say much about the *internal* structure of the black hole, which is usually imaged as a sort of a "funnel" (taking on crazy shapes with rotation and electromagnetism involved) of heavily distorted space-time that may or may not "end" in a singularity (which cannot quite be described by general relativity). And why round event horizon? It's easier on the math. We also have models for rotating and charged black holes, which aren't spherical.  
> **username_7**: @username_0 You cannot be "slightly above the singularity" because inside the event horizon, the singularity is not a point in space but a point in *time*. It's like saying the source is "slightly above" next Tuesday.  
> **username_3**: @username_6 it sounds like the same reason we have spherical cows :)  
> **username_3**: @username_5 I think for some distance metric, black holes are perfectly spherical.  Actually we can say that the event horizon is infinite distance from the center of the black hole in most reasonable distance metrics - so it is by definition a sphere.  
> **username_5**: @username_6 so you're saying that the closer we get to a black-hole, the less and less round it appears, but from a massive distance it looks round, because everything in the universe is "round"?  
> **username_8**: There's isn't such a thing as straight towards the center, of a black hole, because time space is twisted so the photons you shoot forwards and backwards may travel faster sideways and have arrested path compared to an outside radius, which perhaps doesnt exist and is more of a spiral inside.  

---

## 3 answers

### Answer by username_12 on 2018-01-22. Score: 31 | Accepted answer

It doesn't work like that. An observer at the light source (and indeed any observers anywhere else) will always see light travelling (in vacuum) at the speed of light locally.

There is also a major problem with your thought experiment. It is *not possible* for you to have a stationary light source within the event horizon of a black hole. It, and everything else in its vicinity, *must* be moving inwards. This is as inexorable and unavoidable as is the passage of time for an observer outside the event horizon.

In my opinion, the best "visual" way of thinking about the situation inside the event horizon is to imagine your photons of light like salmon trying to swimming upstream, whilst you are on a boat flowing with the stream and releasing the salmon into the water. You will always see the salmon swimming at some speed with respect to your boat. Unfortunately if the stream flows fast enough then the salmon will make no progress and you will both be swept over a waterfall (the singularity) a little further downstream.

Likewise, your common-sense fails with the situation of firing light towards a black hole. Light is always measured to have a speed of $c$ *locally*. It is following through with the consequences of this principle that leads to all the weird behaviour that black holes exhibit.

> **Comments:**
> **username_0**: that is a beautiful answer. If i understand you correctly, there is no such thing as a "stationary object", it's a bit like a universal constant, or just a figment of our imagination. )  
> **username_12**: @username_0 There is no such thing as a static observer (sometimes called a shell observer) inside the event horizon.  
> **username_13**: Nice explanation without any complicated terminology.  
> **username_14**: Isn't the word "vacuum" missing? Light doesn't always travel at $c$.  
> **username_15**: Do not try this at home.  
> **username_16**: I remember Leonard Susskind using that same "fish swimming upstream" in his lectures. It does capture the essence well.  
> **username_17**: @username_14 I was under the impression that light DOES always travel at c, but in a non-vacuum it ricochets off particles, causing it to take a longer path and making it SEEM slower.  
> **username_18**: @username_17 actually absorbed and re-emitted (different photon). You might be interested in the "illusion" of galaxies exhibiting "faster than light" motion https://en.wikipedia.org/wiki/Superluminal_motion . It's a similar misconception.  
> **username_17**: @JackR.username_18 Oh, that makes sense.  

---

### Answer by username_9 on 2018-01-22. Score: 9

You cannot exceed the speed of light "locally". But you can -see- imagine\* ***distances increase quicker than the speed of light***.

If by traveling you mean "moving compared to local space time", then light cannot ***travel*** quicker than the speed of light. In your example, the distance increases faster than the speed of light, because ***spacetime is dragged along*** inside the black hole by its gravity.

\*You in fact cannot "see" it, because you would need some information to be transferred somehow to do the same! This is not possible, because of this damn *c* limitation.

> **Comments:**
> **username_0**: @j-username_9 your answer is brilliant, i especially like the "distances increase quicker than speed of light" now that i am starting to grasp this concept.  
> **username_9**: @username_0, I had similar question some time ago, and learned a lot from the answers I got: https://astronomy.stackexchange.com/questions/19909/will-we-start-seeing-galaxies-disappear-due-to-universe-expansion .  
> **username_10**: This is exactly why the diameter of the visible universe (93 billion light years) can be larger than the number of years since its creation (13.8 billion years).  

---

### Answer by username_11 on 2018-02-13. Score: 1

The speed of light will remain constant. Though the way it is perceived near a black hole changes with where and how it is perceived, it will remain constant. The speed of light does not increase or decrease just because it is near a black hole.


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