# If the speed of light is constant, why can't it escape a black hole?

When speed is the path traveled in a given time and the path is constant, as it is for $c$, why can't light escape a black hole?

It may take a long time to happen but shouldn't there be some light escaping every so often?

I'm guessing that because time is infinite inside a black hole, that this would be one possible reason but wouldn't that mean that we would require infinite mass?

What is contradicting with measuring black holes in solar masses, what means they don't contain infinite mass.

So how can this be?

• This is inaccurate in certain ways, but it'll help your intuition: from the perspective of any object (including a ray of light) inside a black hole, the distance outward to the event horizon is infinite. – zwol Jun 9 '16 at 15:09
• There is no path forward in time the "light ray" could take that would carry it outside of the event horizon. The spacetime is bent that much. – Luaan Jun 9 '16 at 15:17
• Black holes distort spacetime to the extent that spacetime is effectively wrapped around itself. Any light emitted will follow the curve. The required escape velocity exceeds the speed of light - anything less and it shines. – PCARR Jun 9 '16 at 15:22
• If, after you turn off its thrusters, the speed of spacecraft is constant, why can't it escape Earth? Just because it is travelling at a particular speed, doesn't mean it is travelling fast enough. – Shane Jun 9 '16 at 18:45
• Usain Bolt is the world's fastest man. That doesn't mean he is fast enough to outrun a bullet. In a sense, in a black hole space is more warped than light is fast. The speed of light is the fastest possible speed. It is the speed limit of the universe. Gravity has no limits. If you keep piling matter into one spot, gravity gets stronger (space gets more curved). The how/why one limit is higher than the other will involve a lot of maths that I don't know and should probably be a different question. – Shane Jun 9 '16 at 19:41

The speed $c$ that is constant is so when measured locally relative to a freefalling frame (i.e. one for which all points follow spacefime geodesics wrt to the metric $g$). Local means that the frame's extent must be "small" enough that it can be thought of as flat: think of this as zooming in on the spacetime manifold, which is a smooth object, with enough magnification that you can't see any appreciable deviation from Minkowski spacetime (which is the spacetime analogue of flat Euclidean space, which you've probably encountered). In contrast, the speed of light as measured by a distant observer can vary in generally curved spacetime.

The wording of your question suggests that you imagine sitting at some point within the horizon, and since your laser pointer's output must squirt out at the everconstant $c$, and the horizon is only a finite distace above you, it must reach the horizon and leave.

But the geometry is not like this everyday thought picture. The point about an event horizon is that it is not in the future of any event inside the horizon. The spacetime distortion from flatness is so severe that even the future branch of lightlike geodesics will not intersect it. You can only reach the horizon from an event within it by travelling backwards in time.

Some Q and A from Comments

"the speed of light as measured by a distant observer can vary in generally curved spacetime": That's the first time I have heard that. You sure? (Considering that essentially all of space time is curved.)

which question User Jan Dvorak eloquently answers:

don't worry, it will regain the speed of c once it gets close enough to you - if it does. Its wavelength when it meets you might differ drastically from its wavelength when it left its source, however.

and I'd like to explain Jan's answer a bit more fully. You infer something's speed by comparing the changes in your spatial and temporal co-ordinates for that object. Let's begin in special relativity, where at first both observers chart the Universe by Minkowski co-ordinates. The fact that your clock and rulers measure the same intervals differently from what the distant one does doesn't lead to any surprises (at least to someone who has studied SR thoroughly) because there is a unique, well defined transformation that will map your co-ordinates for events to the distant observer's co-ordinates, and contrariwise. That transformation is the (proper, orthochronous) Lorentz transformation, which has the property that $c$ is measured to be the same from both observers' standpoints.

In general curved spacetime it is impossible to define a unique transformation between two local frames that would allow us to directly compare measured speeds of things in this way. Let's look at why this is so.

Let's re-imagine our scenario above: we're still in Minkowski spacetime with the same physics and doing SR, but with new co-ordinates. At every point in that spacetime, we rotate and boost the "reference" frames a bit so that nearby points have their reference directions and time intervals slightly different. This is altogether analogous to charting Euclidean 3-space by, say, spherical co-ordinates. Locally, the reference directions (of increasing $r$, $\theta$ and $\phi$) are rotated from the Cartesian ones, and that rotation varies smoothly with position. Now there's a very big infinity of ways to do such a gauge transformation: we can choose directions and unit time intervals any way we like as long as the variation is smooth and that the limiting transformations as the distance between the points shrinks is a Lorentz transformation.

So now, in these new co-ordinates, how do we compare measured speeds if we were given only these co-ordinates? Well we could simply move through space and time along a chosen smooth path, making the little Lorentz transformations between neighboring reference frames and multiplying them all together to get an overall transformation for this path. But we could choose an infinity of smooth paths to do this along. So, if we're given only these co-ordinates, it's not immediately obvious that we wouldn't get a different answer from this procedure if we took a different smooth path between the two points.

But we do, because that's what flat means, by definition.

We can always make a transformation of our weird co-ordinates back to Minkowski spacetime if and only if the result of our calculation does not depend on path. The result of so-called parallel transport of a vector around a loop is always the identity transformation. A corollary of this fact is there is a well defined transformation between the two observers that allows us to compare measured speeds: it doesn't matter whether we compute it along path A or path B between two points: the answer must be the same since the inverse of one transformation must invert the other to achieve the identity transformation around the loop. Thus, in theory, we can still compute what the other observer would observe locally from afar in our weird co-ordinates.

If you've made it through this explanation this far, then General Relativity is now only a small conceptual step away. In curved spacetime, the transformation wrought on vectors by the parallel transport around a loop is in general not the identity transformation. So there is no well defined way of comparing speeds from afar, at least from one's own co-ordinate frame.

That's what "curved" means, by definition: nontrivial "holonomy" in parallel transport around closed paths

And this is what people mean when they say the "co-ordinate speed of light can by anything in GR". But if a distant observer measures the speed of light continually, repeatedly and at regular time intervals as measured bu their clock in a laboratory they carry with them, and then send the result to you, all their reports to you will be that their measurement hasn't changed, even though the interval between reports that are set regularly by their clock may reach us at wildly varying intervals by our clock.

Another analogy that might help you is the $2$-sphere, what we call a "ball" in everyday language, compared with the plane. On the plane, tangent planes to the plane are everywhere the same vector space: there is an unambiguous way to parallel transport the tangent plane at any point to that at any other point. On the ball, not so. Tangent planes at different points are not the same plane. They are isomorphic as vector spaces, but they are not the same. In particular, there is no well defined universal way of comparing them, or of assigning reference bases at all the points in any patch of finite extent, because, on the sphere, parallel transport of vectors around loops always leads to a change to the vector when it arrives back at the beginning point. Indeed, a sphere has constant curvature, which means that the rotation of the vector wrought by loop parallel transport is proportional to the area enclosed by the loop.

• "the speed of light as measured by a distant observer can vary in generally curved spacetime": That's the first time I have heard that. You sure? (Considering that essentially all of space time is curved.) – Peter - Reinstate Monica Jun 9 '16 at 16:41
• @PeterA.Schneider don't worry, it will regain the speed of c once it gets close enough to you - if it does. Its wavelength when it meets you might differ drastically from its wavelength when it left its source, however. – John Dvorak Jun 9 '16 at 16:55
• The study of black holes has brought us great understanding of the universe around us, and also reams and reams of terrible scifi. – corsiKa Jun 9 '16 at 20:38
• Man, black holes are cool – MKII Jun 10 '16 at 7:30
• Minor nitpick (can't edit without changing more letters): You misspelled 'spacetime' as 'spacefime' in the first paragraph. – childofsoong Jun 10 '16 at 17:37

Suppose you are floating in a river, and you have with you a model boat, called the SS Lightray, that can do 3 m/sec through the water. When you set the boat travelling upstream as far as you're concerned it is doing 3 m/sec. But I'm standing on the bank watching the river flowing at 1 m/sec, so when I look at your boat I see it travelling at a net speed of 2 m/sec not 3 m/sec.

Now the river narrows and speeds up to 4 m/sec. As far as you're concerned you're sitting motionless in the water and when you send the boat out upstream again it is still travelling at 3 m/sec. However from the river bank I see the boat is now travelling downstream i.e. its speed upstream is -1 m/sec. The boat can't travel fast enough to make headway against the flow of the river.

Rather surprisingly an argument very like this applies to motion of light away from a black hole. It's called the River Model, and here is a link to a scientific paper giving the details. More formally this technique is an analysis of the motion of the light using the Gullstrand–Painlevé coordinates. I used this technique to explain why light can't escape from a black hole in my answer to Why is a black hole black?.

The local speed of light is always $c$, but if you use the Gullstrand–Painlevé coordinates to analyse what happens at the event horizon you find that:

1. at the horizon you are falling inwards at the speed of light

2. relative to you the light is travelling outwards at the speed of light

3. so the net speed of the light away from the event horizon is zero

And that's why the light can't escape from the black hole.

• The net speed of light never changes it is always constant, what changes is just its trajectory. Inside a black hole the trajectory is circular, that's why it don't escape. – CoffeDeveloper Jun 9 '16 at 15:34
• @DarioOO: In the context the phrase net speed means coordinate speed, and that does change. See for example my answer to Speed of light in a gravitational field?. What is constant is the local speed i.e. the speed measured by an observer at their position. This is always $c$ even if the observer is in a non-inertial frame. – John Rennie Jun 9 '16 at 15:37
• Then sorry, if you can make any change in the answer I would turn my -1 into a +1 :) – CoffeDeveloper Jun 9 '16 at 15:42
• (sorry again eh XD) Isn't anyway the light speed indipendent by the falling speed of the object? It seems you state that final light speed is C - falling speed, while I think final light speed is independent of falling speed (or am I wrong)? – CoffeDeveloper Jun 9 '16 at 16:37
• @DarioOO: In the GP coordinates the infalling observer is stationary - just like the chap in the river. The speed of light is $c$ relative to this infalling observer. If you go back to my analogy of the river, the GP coordinates are baically the coordinates that are stationary with respect to the water. – John Rennie Jun 9 '16 at 16:56

To explain it in layman's terms, without using advanced concepts:

Space is warped in the "inside" of the black hole (that is, under the event horizon) so much, that it behaves completely different than what we perceive hear on Earth. The "outward direction" simply does not exist.

For example, here on Earth, we can go in the three spacial dimensions in both directions, but in time we can move only forward. Imagine that on the "surface" of the black hole, aka at the event horizon, the spatial dimensions have only one direction: inward. This last paragraph is not intended as an accurate description of how black holes work. It's only to conceptualize how there are cases where only one direction of a coordinate exists.

• I realise you're trying to give a non-technical description, but what you've said is actively misleading. It is not true to say that the outward direction does not exist. What happens is that all outward trajectories become spacelike i.e. they can only be followed by something moving faster than light. – John Rennie Jun 9 '16 at 14:50
• @JohnRennie : do you have an idea for a better phrasing? I'm happy to update my answer in that case. Similarly, it would be wrong to say that the backwards direction in time doesn't exist, because a hypothetical particle moving faster than light would be moving backwards in time (from at least one frame of reference). – vsz Jun 9 '16 at 14:53
• @JohnRennie Assuming FTL is impossible, is there a difference between "there are no outward trajectories" and "to follow an outward trajectory, you need to be going FTL"? If I can't go fast enough to get onto an outward path, there are no outward paths, no? – Shane Jun 9 '16 at 18:53
• @Shane: to put it into your styl. If there is a unpenetrable window blocking each posdible way to get further on the path... does this mean, there is no path?:p – Zaibis Jun 9 '16 at 19:56

The explanation I like is thus:

In GR, all things, from planets to photons, travel in straight lines through curved space bent by mass. Black holes bend and distort spacetime so severely that the curvature captures the photon.

Scale things down and it behaves much the same way passing asteroids can be captured by a star. For us, the speed of the asteroid(photon) is only relevant up until the point it crosses the threshold of capture, the point of no return and tips into the star's(black hole) gravity well. The asteroid(photon) will never escape, the "walls" of the well are too high. Does it matter if it is me or the worlds best high jumper who's trapped at the bottom of a pit, if the walls are 50 feet high? Neither one has any chance of escaping, the fact that one of us can jump the highest a person can possibly jump is irrelevant.

The fact the light goes the the fastest anything can go is a similar red herring. What matters is that it has gotten into a situation from which there is no escape. The speed of light is not a get out of jail free card any more than the high jumpers.

The picture I always liked is for an observer free-falling into the black hole, when they're just outside of the event horizon, it looks like the event horizon is propagating outward at nearly the speed of light. After the observer falls just inside, the event horizon now looks like it's propagating outward at greater than the speed of light.

The answer is that it has nothing to do with light, c, black holes, event horizons or relativity. It is simply escape velocity. We know that two bodies attract each other with a force $f = G\frac{M m}{d^2}$ and we know that something is in a stable orbit when its potential energy $PE = -GMm/d$ matches its kinetic energy $KE = m{v^2}/2$. Solve these two for speed and you get the the speed needed for an orbit is ${V_o} = \sqrt{\frac{2GM}{r_o}}$. So any speed greater than $V_o$ will make it leave orbit and move away. Attraction from gravity will slow it down but never stop it so in effect, it has 'escaped' which is why it is sometimes called $V_{esc}$. However, any speed less than $V_o$ means it will come back eventually.

As the distance $d$ gets smaller or the mass $M$ gets bigger, the value $V_o$ increases. For a body starting from the surface of the earth, to get off (and stay off) the surface, $V_o$ is about 11 kilometers a second. On the surface of our sun it is about 620 km/s. If all the mass of our sun were compressed to half its current size, $V_o$ would double. If you kept compressing the sun until its radius was about 3km, then $V_{esc}$ would reach and exceed the speed of light (known as the Schwarzschild radius). If you were on the surface of such a body and shone a light to your friend who was outside this limit, you would see the beam moving away from you at the speed of light towards them but the light is not traveling fast enough to escape gravity it could never reach your friend. So how can this be? As you pointed out, $c$ is constant to all observers and speed is just distance over time and there is nothing you can do to change the distance. That just leaves time. In order for $c$ to remain constant in your frame of reference, your time would have to slow down to the point where it effectively stops. With no time, there can be no distance, and with no distance, there can be no dimensions and this is why we call it a singularity.

The short answer is the speed of light is constant until its not. Its constant until it runs into a wall and its constant until effected by gravity. The extreme gravity of a black hole will deflect the path of the photons more and more until finally at the event horizon all deflection is toward the black hole.

Isn't the saying "The speed of light is constant 'in a vacuum'"?

The 'in a vacuum' bit is commonly missed off (in the same way, "'The love of' money is the root of all evil", is often commonly misquoted).

If a black hole isn't a vacuum (because at some point, particles are close enough together it is no longer a vacuum), then the speed of light slows down. For example, the speed of light through air is around 90 km/s slower than a vacuum (not much, but there it is).

• I'm afraid this is rather wide of the mark. The coordinate speed of light slows down as you approach a black hole due to the spacetime curvature. This is true in a complete vacuum and is not dependent on having particles present. See Speed of light in a gravitational field? for the details. – John Rennie Jun 9 '16 at 14:48
• I thought about this too. but this would if I get it right, still mean to make sure that never ever light could escape, it would require infinite particles (->infinite mass) to be inside. – Zaibis Jun 9 '16 at 14:48
• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – honeste_vivere Jun 9 '16 at 22:56
• @honeste_vivere: Yes, it does, although it's a wrong answer. Please do not abuse the review system to denote answers with which you disagree; downvote, instead. – Lightness Races in Orbit Jun 12 '16 at 19:55