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If the event horizon of a black hole is the distance from the center from within which light cannot escape, imagine a person with a flashlight falls into the black hole.

He points his flashlight in a precisely radial direction and turns it on. Now there is a light ray moving outward at the speed of light. If it now cannot move in an arc, but rather is constrained to radial motion, it must, at some point before the horizon, switch directions and fall back into the black hole.

If the speed of light is constant, how does it suddenly change directions, without either decelerating, or requiring an infinite amount of energy?

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This question has an open bounty worth +50 reputation from Sir Cumference - Pies ending in 5 days.

This question has not received enough attention.

Looking for an answer in relatively simple terms. More of a hands-on description than multiple equations.

I don't know if this is the correct answer to your question, but light moving in the direction of increasing gravitational potential will be redshifted, thus losing energy without losing speed. I can imagine it being redshifted all the way through zero into "negative" frequencies, which (I guess) amounts to light propagating in the opposite direction. – Nathaniel Jun 11 '13 at 7:41
But the redshift formula doesn't result in negatives, does it? – user25642 Jun 11 '13 at 7:42
maybe not, but why wouldn't it? Consider a photon moving upwards in a constant gravitational field that has energy $E=hf$. As it moves upwards its gravitational potential energy increases, so its $hf$ energy has to decrease in order for the total to be conserved. If it moves far enough its $hf$ energy must reach zero - but what happens then? The photon can't just disappear, and it can't keep moving upwards without getting a negative energy, so the only option is for it to start moving downwards and increasing in frequency. That's my guess, anyway. – Nathaniel Jun 11 '13 at 7:46

9 Answers 9

There are many extra subtle effects that you neglect – the red shift, perhaps the Hawking radiation that makes the black hole shrink, and so forth (one could recommend you to learn the Penrose causal diagrams) – but if one tries to be cooperative, he must say that it is indeed the case that some "velocity of the light from the flashlight" expressed in appropriate variables indeed switches the sign when the flashlight crosses the horizon.

For the neutral black holes, this is manifest in the Schwarzschild coordinates where the metric is $$ c^2 {d \tau}^{2} = \left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \left(1-\frac{r_s}{r}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)$$ Light has to propagate to the future (no one can travel in the past) and indeed, in these coordinates it means $dt\gt 0$. And it must propagate over null geodesics which means $$c\,dt (1-\frac{r_s}{r}) = dr $$ for $ds^2$ to vanish. The angular variables contribute nothing.

You see that $dr/dt$ which is a sort of "coordinate velocity", measuring how much the radial coordinate $r$ changes as a function of the Schwarzschild time $t$, is equal to $c(1-r_s/r)$ and indeed, as we switch from $r\gt r_s$ to $r\lt r_s$, this quantity $dr/dt$ switches the sign ($dr/dt$ is equal to zero when the light is just crossing the event horizon – well, such light is "confined" at a fixed value of $r$).

I must emphasize that this sign flip is an artifact of the chosen (Schwarzschild) coordinates. There exist other coordinates that, in the vicinity of the event horizon of a large black hole and in the reference frame of an infalling observer or the flashlight, resemble the Minkowski space. In these coordinates, the light always propagates along $x=ct$ trajectories with a fixed slope and sign. And in these coordinates, the event horizon is a plane that is moving by approximately the speed of light in the "outward" direction (the event horizon is not static in these coordinates!) which is the explanation in these coordinates why the flashlight's light can't catch up with the event horizon.

One may choose many different coordinates and they may have advantages. A black hole is a static object so one may choose coordinates in which the metric tensor is independent of time $t$; Schwarzschild coordinates are an example. And one may choose coordinates that describe the vicinity of the event horizon "smoothly", without singularities and confusing sign flips of the velocity. But there are no coordinates that would have both properties at the same moment.

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So if you took a picture of the two space-dimensional slice of the black hole that intersects the light ray's world line, and plotted it, what would the world line look like? – user25642 Jun 11 '13 at 8:30
Sorry, this question makes no sense. Which slice are you talking about? There are infinitely many slices - even infinitely many slices that intersect a light ray. – Luboš Motl Jun 12 '13 at 5:54

One of the problems with describing any situation in general relativity is choosing an appropriate set of coordinates. Far from the black hole we use the Schwarzschild coordinates $t$ and $r$ (we'll ignore the angular coordinates). These are just the radial distance as measured with your rulers and the time measured on your clock so they have a nice simple interpretation. The trouble is that as you approach the event horizon the time gets dilated by a factor:

$$ \frac{d\tau}{dt} = 1-\frac{2M}{r} $$

and at the event horizon, where $r = 2M$, this factor goes to zero. This means time slows to a stop at the event horizon, and it's the source of the common claim that nothing can ever cross the event horizon.

Obviously we're going to struggle to describe what happens to light from a torch inside the event horizon if it takes an infinite time for the torch to even reach the event horizon, let alone cross it. So we need to look for a better coordinate system. We could try using the coordinate system of the astronaut falling in. The trouble with this is that for any freely falling observer spacetime is locally flat, so (ignoring tidal forces) the astronaut thinks they are motionless in flat space. When they turn on the torch the light just speeds off at $c$ as usual.

Outside the event horizon we can use shell observers i.e. observers hovering at a fized distance from the horizon. The trouble is that inside the event horizon it's impossible to hover at fixed $r$, so we can't use shell coordinates either.

So what to do? Well in cases like this we have to choose a set of coordinates that don't correspond directly to anything seen by an observer. This allows us to describe what happens inside the horizon, but at the expense of simplicity. In particular it becomes hard to match the description with our intuitive feel for what happens. Sadly this is a price we have to pay.

The very best coordinates to use are the Kruskal-Szekeres coordinates $u$ and $v$ because they make the causal structure immediately obvious. However these are prohibitively complicated for the non-specialist. The coordinate $u$ is spacelike but isn't simply radial distance, while the coordinate $v$ is timelike but isn't simply time. So I'm not going to use the KS coordinates to answer this question. However, if you're feeling brave have a look at my answers to Would the inside of a black hole be like a giant mirror? and Taking selfies while falling, would you be able to notice a horizon before hitting a singularity? where I use the KS coordinates to answer related questions.

In this case I'm going to use the Gullstrand-Painlevé coordinates. sometimes known as rainfall coordinates or the river model. In these coordinates $r$ is the same radial distance as in the Schwarzschild coordinates, so that's easy to understand. However the time coordinate $t_r$ is the time recorded by a clock carried by a freely falling observer, and because of the time dilation mentioned above this is not the same as the time recorded by the Schwarzschild observer far from the event horizon. Bear this in mind as you consider what follows.

I've already used the GP coordinates to calculate the speed of light heading to or from the black hole in my answer to Why is a black hole black?. The result is:

$$ \frac{dr}{dt_r} = -\sqrt{\frac{2M}{r}} \pm 1 \tag{1} $$

where the $+$ gives the speed of an outgoing ray and the $-$ gives the speed of an ingoing ray. Note that this uses geometric units where $c = 1$. In these units the event horizon is at $r_s = 2GM$. If we use equation (1) to calculate the speed of an outgoing light ray at the event horizon $r = 2M$ we get:

$$ \frac{dr}{dt_r} = -\sqrt{\frac{2M}{2M}} + 1 = 0 $$

and we find that at the event horizon a light ray doesn't shine out, stop and fall back. Instead its velocity is zero so it is fixed motionless and doesn't go anywhere. Inside the event horizon, where $r < 2M $, the velocity of an outgoing ray is negative. So inside the horizon even a light ray directed outwards actually moves inwards not outwards. This is the key result we need to answer the question.

Admittedly we're using an odd time coordinate, but the $r$ coordinate is our good old Schwarzschild coordinate. So while we may may quibble about the exact value of the calculated velocity the sign is unambiguous. That means when our falling astronaut shines his torch outwards the light does not move out, come to a halt and fall back again. The light is moving inwards from the moment it leaves the torch. The reason the astronaut sees the light move away is because the astronaut is falling inwards even faster than the light.

A comment asks if this means the astronaut is moving faster than light, and yes it does. However this shouldn't surprise you as in GR it's only the local velocity of light that is constant at $c$. At distant locations light can move faster or slower than $c$ (though we'll never observe it moving faster as a horizon will get in the way). For example it's well known (or should be!) that sufficiently distant galaxies are moving faster than light.

There is one last loose end to tie up. I claimed above that the astronaut sees the light move away because the astronaut is falling inwards faster than the light is. Can we prove this? It's actually fairly easy to prove if we start from the well known result that the velocity of an observer falling freely from infinity is (in Schwarzschild coordinates):

$$ \frac{dr}{dt} = -\left( 1 - \frac{2M}{r} \right)\sqrt{\frac{2M}{r}} $$

To convert this to the Gullstrand-Painlevé coordinates we note that the rainfall time $t_r$ is just the proper time $\tau$ along the trajectory of the infalling astronaut, and the proper time is related to the coordinate time by the expression I gave above:

$$ \frac{d\tau}{dt} = 1-\frac{2M}{r} $$

The velocity of the astronaut in GP coordinates is then simply:

$$ \frac{dr}{dt_r} = \frac{dr}{dt_r}\frac{d\tau}{dt} = = -\frac{\left( 1 - \frac{2M}{r} \right)\sqrt{\frac{2M}{r}}}{1-\frac{2M}{r}} = -\sqrt{\frac{2M}{r}} $$

Compare this with equation (1) for the velocity of the light, and you'll see that the velocity of light differs from the velocity of the astronaut by $1$. So the light is always moving at a velocity of $c$ relative to the astronaut.

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The torch moves faster than light?! – Sergey Tachenov yesterday
@SergeyTachenov: yes, sort of. It you choose the right coordinates then yes the torch has a coordinate velocity faster than light. But coordinates are tricky things in GR. – John Rennie yesterday

Now there is a light ray moving outward at the speed of light.

I'm afraid that isn't the case; within the event horizon of a Schwarzschild black hole, the radial coordinate is timelike and so, moving 'outward' toward the horizon is as impossible as moving 'backward' in time.

This plain to see in the Kruskal–Szekeres coordinates:

enter image description here

Image credit

See that, considering the light cone shown inside the black hole, even light emitted in the direction of the horizon always gets closer to the singularity (the $r=0$ hyperbola at the top of the diagram), eventually reaching it, and never gets closer to the horizon (the $r=2M$ line through the origin).

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So, given that the coordinate speed of light at the event horizon is zero, how long does it take for the light to go from there to the r=0 singularity? – John Duffield yesterday
@JohnDuffield The speed of light at the horizon is not zero. – 0celo7 21 hours ago
@0celo7 : the coordinate speed of light as measured by the distant observer is indeed zero. Google it. You'll be able to find articles that say "but the locally-measured speed of light is still 299,792,458 m/s". However there's a problem with that. Gravitational time dilation goes infinite, so as far as we're concerned it takes infinite time to measure it. The gedanken observer at the event horizon hasn't measured it yet, and never ever will. – John Duffield 19 hours ago
@JohnDuffield, will you and Ocelo7 please take your discussion to the chat room; I have no interest in participating in this discussion nor do I want my answer to be its home. – Alfred Centauri 16 hours ago
@JohnDuffield: your question how long does it take for the light to go from there to the r=0 singularity is meaningless unless you specify what time coordinate you're using. The proper time for the light to reach the singularity is of course zero. You cannot do the calculation for the Schwarzschild observer because for them the light never reaches the horizon. – John Rennie 1 hour ago

I think a possible analogy would be to imagine that the singularity is a waterfall. By emitting light, you are trying to send a signal upstream using a tame fish. Outside the event horizon the fish is able to make headway against the current. But the river flows so fast within the event horizon as it approaches the waterfall, that your fish ends up going over the waterfall shortly after you do because the river is flowing faster than the fish can swim. Note that at no point does the fish change direction and it moves with respect to you at "the speed of fish".

This "river model" for a black hole is discussed in detail by Hamilton & Lisle (2006).

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It would move at the speed of fish relative to me, yes, but we're talking about light here. Remember, light is constant. Wouldn't it move at lightspeed regardless of a frame of reference? In your analogy, the fish is slowed down and pushed back, so it'd be moving slower relative to the bottom of the waterfall...even though it is moving at "fishspeed" relative to me. Light should move at c from any frame of reference, right? How could it become slower in any sense? – Sir Cumference - Pies 9 hours ago
@SirCumference-Pies Light travels at $c$ in a local inertial frame in GR. Different observers in different (accelerated) frames of reference do not agree on how fast light appears to be travelling. A "free-falling" observer however (travelling with the river) always sees fish travelling at fishspeed. – Rob Jeffries 5 hours ago
@SirCumference-Pies: I've just updated my answer and gone into the gory details of how the river model works. – John Rennie 2 hours ago

First, the speed of light, as measured by a local observer, is always the same, that is $c$.

To correctly pose your problem, you have to use a light-cone version of Kruskal Szekeres coordinates

The metric is given by :

$ds^2 = F(r) dU dV + r^2 d\Omega^2$, where F(r) is some function of r

The black hole interior is given by $U > 0$ and $V > 0$

The outgoing null geodesics are given by $U = Cte$ (with $V$ increasing).

The ingoing null geodesics are given by $V= Cte$. (with $U$ increasing).

The future horizon is given by $U = 0$ and $V > 0$.

The future singularity is in $UV = 1$ with $U > 0$ and $V > 0$.

So, now, imagine, you are in the interior of the black hole, that is $U > 0$ and $V > 0$, you are sending a outgoing radial light signal, but this signal is at $U= Cte$, so the variable $U$ stays $>0$. But the future horizon is $U = 0$ and $V >0$. So your outgoing signal never reachs the (future) horizon, because the value $U=0$ is never reached.

It is better to draw a litte diagram with the coordinates U and V orthogonal, With upwardly directed axes, and with U and V making an angle or 45 degrees from the vertical.

To complete the schema, you have also :

The past singularity is in $UV = 1$ with $U < 0$ and $V < 0$.

The past horizon is given by $V = 0$ and $U < 0$.

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For better clarity, let's define the following:
Axial direction = the direction the person & light beam are drawn into the BH.
Radial direction = the direction perpendicular to the axial direction.

If we, looking in the same direction as the person & light are being drawn into the BH, watch the light beam as it is drawn into the BH, we will see the following:

  1. Whatever reflected light the beam may produce (i.e. due to gas, debris, etc) that it collides with, as it approaches the event horizon. (This assumes we start at a point in which we can still see the reflected light from the light beam).

  2. The force of the light beam will not travel beyond the outer edge of the BH (at some point) as it is drawn further into the BH.

  3. The person & the light beam would disappear from our view as it passed the event horizon (much like a ship going beyond the horizon at sea).

  4. If we change our pov (& acceleration) so that we could look into the light beam as it is drawn into the BH (like a camera that follows a sprinter on a race track), we would see the light beam widen in the axial direction (i.e. parallel to the direction of the moving person) as the forces of the BH overcome the forces of the light beam & distort/weaken it. I think the light would just blink out (from our perspective) when it got too far into the BH.

re "If it now cannot move in an arc, but rather is constrained to radial motion, it must, at some point before the horizon, switch directions and fall back into the black hole."
--Or, perhaps it is just weakened to the point that it cannot exit beyond the BH.
Also, forces near the center of the BH are greater than forces near the rim of the black hole (i.e. like a vortex). Greater forces will affect the light more & lesser forces will affect the light less. So, that probably means the light near the center of the BH is bent/attenuated faster than light near the edge/rim of the BH.

re: "If the speed of light is constant, how does it suddenly change directions, without either decelerating, or requiring an infinite amount of energy?"
--There are forces in the universe greater than light. They don't necessarily need to be infinite to bend/slow, distort, or destroy the force of light.

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As I know, Field Theory, that to what appeals the topic creator cannot explain the very powerful gravitation fields . So trying to understand what happens with a photon there are inside the Black Hole in meaning of Field Theory, or Special Relativity, isn't a good idea.

The Nature has no the alone space , and the alone time , you can abstractly image that, but there aren't created the pure space , or the pure time in the laboratory.

So why the speed of light is constant, how does it suddenly change directions, without either decelerating, or requiring an infinite amount of energy?

Photons don't change their directions, they moving straight on the geodesic lines of the space-time, and inside the Black Hole curvature of the spacetime is so huge that all of this lines, leads the way straight to the singularity.

Like ants on the apple (but imagine that the ants inside an apple and cannot break the border, moving fast as they can , but always traps in the center of the apple). They CANNOT get through the apple , and Photons cannot broke the spacetime and get trough the spacetime CONTINUUM.

You can image that, like hockey players cannot break the ice, and will moving straightforward on the surface of the ice, even if you will make the huge ice surface curvature! :) It will be not a good picture...

Gravitation Wheeler, Misner, Thorne

Photons have no mass (yeah , it's another hard thing to imagine but nevertheless), and you cannot do force to them in the classic meaning of that thing... :)

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What are the dashes around words/phrases (e.g., -space- in the 2nd paragraph) supposed to represent? – Kyle Kanos 1 hour ago
It's a way to pay attention on the word :) Like said it , with slowly deduction . – Frostys Shad 1 hour ago
Well you can replace those dashes with a single asterisks (*space* gives space) to get italics (you seem to be aware of the bold). – Kyle Kanos 1 hour ago
Oke-doki ;) Thanx for a , dont know how it's say , a useful asking. – Frostys Shad 1 hour ago

the problem is that light dissapears by just nearing the event of horizons. eventually it desacelerates (technically it will not) to a point of producing radio waves instead of light just like happens to magnetars which are considered dark objects and very similar in mass to black holes in terms of measuring instruments there's probably no difference at all

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This doesn't really answer my question. Also, magnetars emit gamma rays and X-rays. If you're referring to the gravitational redshift, you could have just talked about neutron stars. This answer, to me, seems to be unrelated and pretentious in its wording. – user25642 Jun 11 '13 at 8:34
lightspeed is just a theoretical constant along the planck constant which defines mostly how space and time deform nearing an event of horizons eventually a neutron star. when physicists say light redshifts or blueshifts it's not a change in the constant since the constant always remains E. the same applies to theoretical physics either the light is blue or red it doesn't make the case. – sphericsf Jun 11 '13 at 8:57
I think you're answering a different question. Either way, I can't understand exactly what you mean. I'm not asking about the constants of nature or what redshift is. I'm asking how light can flip direction without being absorbed and re-emitted quantum mechanically. – user25642 Jun 11 '13 at 9:01

How does light behave within a black hole's event horizon?

It doesn't behave at all.

If the event horizon of a black hole is the distance from the center from within which light cannot escape, imagine a person with a flashlight falls into the black hole.

I've explored this with a variety of relativists, and posed this question. The answer comes as a surprise to most people.

He points his flashlight in a precisely radial direction and turns it on. Now there is a light ray moving outward at the speed of light. If it now cannot move in an arc, but rather is constrained to radial motion, it must, at some point before the horizon, switch directions and fall back into the black hole.

This isn't true. At no point in the scenario does the light ever switch direction and fall back into the black hole.

If the speed of light is constant, how does it suddenly change directions, without either decelerating, or requiring an infinite amount of energy?

It isn't constant. See the Einstein digital papers. This dates from 1920:

enter image description here

Or see Irwin Shapiro talking about it in 1964:

enter image description here

The speed of light varies with gravitational potential. Light goes slower when it's lower. And what that means, is that the upward vertical light beam speeds up. I know that's not what you've been told, but I'm not making this up. See this PhysicsFAQ article for more. Now, take a look at Wikipedia. Note this: "The coordinate speed of light (both instantaneous and average) is slowed in the presence of gravitational fields". That ties in with what Einstein and Shapiro were saying, and Don Koks and others. Now pay attention to this: "at the event horizon of a black hole the coordinate speed of light is zero". The light doesn't get out because the coordinate speed of light, the speed of light as measured by distant observers, is zero. As your observer descends, the light coming from his torch emerges slower and slower and slower, until in the end, it doesn't emerge at all. That's why the black hole is black.


Einstein dismissed Gullstrand-Painleve coordinates for good reason. A gravitational field is a place where a concentration of energy in the guise of a massive body has "conditioned" the surrounding space, altering its metrical properties, this being modelled as curved spacetime. But it's curved spacetime, not falling-down spacetime. The notion that we live in some Chicken-Little world where space is falling down is popscience pseudoscience.

IMHO Einstein would have similarly dimissed Kruskal-Szekeres coordinates. You'll be aware that at the event horizon gravitational time dilation goes infinite and a clock doesn't tick? What Kruskal-Szekeres coordinates do is effectively place a stopped observer in front of the stopped clock and claim that "in his frame" he sees the clock ticking normally. Even though he's at a place where the clock has stopped, and light has stopped. I'm afraid it's "nonsense in neverland". It's a dead parrot sketch.

As for redshift, an ascending photon doesn't lose energy. In similar vein a descending photon doesn't gain any. You know this because you know that conservation of energy applies: if you send a 511keV photon into a black hole, the black hole mass increases by 511lkeV/c². The redshifted photon is emitted at a lower frequency when the emitting body is at a lower elevation, it doesn't lose any energy as it ascends. As for Hawking radiation, after forty years we have no evidence. Which is hardly surprising since it ignores the infinite time dilation and relies upon virtual particles popping into existence even though they only exist in the mathematics of the model. And it relies on negative-energy particles. If you see any of those, make sure you put in a call to Stockholm. In addition and despite much mathematical handwaving, "a black hole is a static object", so the event horizon isn't moving at the speed of light.

Kevin Brown's mathspages article The Formation and Growth of Black Holes is worth a read. Pay attention to the frozen star interpretation. He doesn't rate it, but he wrote this article before the Einstein digital papers came online. IMHO this relatively unknown interpretation will eventually become the mainstream interpretation. It's just a matter of time.

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Does the light ever reach the singularity? – 0celo7 yesterday
IMHO it reaches the Schwarzschild $r_s$ singularity. Most people will tell you that this is a mere coordinate artefact, but once you've seen Einstein talking about the varying speed of light year after year and looked closely at KS and GP coordinates, it can't be. In similar vein the point singularity in the middle can't exist. The force of gravity at some location relates to the gradient in the coordinate speed of light at that location. And it can't go lower than zero. Of course some people will tell you we know better than Einstein, but I've yet to find anything convincing. – John Duffield yesterday
So the light is going outwards, gets slower and slower, grinds to a halt and somehow makes it to the singularity? – 0celo7 yesterday
No, the light going outwards ascends faster and faster. As the observer with the torch falls down, the light from his torch is emitted slower and slower. When he's at the event horizon, the speed at which it is emitted is zero. Of course, there's always Friedwardt Winterberg's firewall to think about, but that's one for another day. – John Duffield yesterday
No, the question is talking about being inside of the black hole. If you didn't even grasp that, then I'm afraid this is "not an answer". – 0celo7 yesterday

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