Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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?

share|improve this question
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

4 Answers 4

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.

share|improve this answer
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

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

share|improve this answer
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

You might be interested in my answer to Why is a black hole black?, which illustrates how to calculate the speed of light near a Schwarzschild black hole.

I can't think of a co-ordinate system in which the light beam would move outwards, then stop and fall back. In the Gullstrand–Painlevé coordinates the light beam never moves outwards - it moves inwards from the moment it leaves the torch, but the torch moves inwards faster so your astronaut falling into the black hole sees the light apparently moving outwards. Luboš Motl explains how to do the calculation in Schwarzschild co-ordinates, but note that these co-ordinates do not have a simple physical meaning for observers outside the event horizon so interpret with care.

share|improve this answer

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 $U < 0$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.