I have read about an object, that once at the even horizon the object would be seen as stopped to the observer. So my question is in regards to the light reflection off of the object. If the object has stopped and my understanding is the light itself is unable to escape, how does the light reflect off of the object for eternity in order for the observer's eyes to receive. If it is due to the light on the facing side is still able to escape, then my thought is the object surface facing the observer would reduce in area unil only the last surface able to reflect light would be available. Thus the item would result in a dot of light reflection.

I hope this question is clear and is sensible.

  • $\begingroup$ once at the even horizon the object would be seen as stopped to the observer” - In classical GR, this is incorrect. The object gradually slows down, but never fully stops and never reaches the horizon. (This picture may be different once quantum gravity is developed.) $\endgroup$
    – safesphere
    Jun 23, 2022 at 15:10
  • $\begingroup$ The issue is in relation to the photons reflected from the object, not so much the object itself. If it slows down or has stopped, it still has to reflect light in order to observe it. $\endgroup$
    – Bisk
    Jun 23, 2022 at 15:31
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    $\begingroup$ Also, note that more detailed calculations show that all but an infinitesimal amount ends up inside the black hole in finite time. (Calculations that show infinite time, to an over-approximation, neglect that the object falling into the black hole affects the black hole, as it has nonzero mass. See e.g. arxiv.org/abs/0907.2574 ) $\endgroup$
    – TLW
    Jun 24, 2022 at 7:02
  • $\begingroup$ @TLW This claim in the paper of the.Chinese researchers is incorrect. They neglect the linear frame dragging that pushes the falling matter out as the horizon expands. Nothing ever crosses the horizon in any external coordinates. $\endgroup$
    – safesphere
    Jun 24, 2022 at 14:48
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    $\begingroup$ Nevermind just found arxiv.org/abs/1112.3638 $\endgroup$
    – TLW
    Jun 24, 2022 at 23:21

3 Answers 3


At a correct moment enough light to last for an eternity was send after the object. A one second pulse for example, and infinite brightness of course.

The front part of the pulse arrives back after 10 years, the middle part arrives back after 1000 years, the rear end arrives back after eternity.

The middle part will be seen being reflected from an object that was slightly above the horizon.

And let's formulate the thing about light not escaping from the horizon like this: It takes an eternity for the light to escape the horizon.


Instead of dealing with light being reflected off the falling body, I'd like to give the falling body a flashlight, so we won't deal with ingoing and outgoing light rays, only outgoing.

I'll give you both intuitition and math on this question.

Firstly, you have a misconception about black holes. A far observer never sees the falling body touch the event horizon. As the falling body falls, its time dilation increases. It takes an infinite amount of time for the object to reach the event horizon according to the far observer. And as long as the flashlight did not reach the event horizon, you can see its light.

If you want a mathematical part, the Schwarzschild metric is given by (neglecting angles):

$$ds^2 = (1-\frac{r_s}{r})dt^2 - (1-\frac{r_s}{r})^{-1}dr^2$$

Light rays move such that their proper distance $ds=0$.

substituting, you find out that:

$$dt=\pm \frac{rdr}{r-r_s}$$

Taking the positive solution for the outgoing light ray:

$$dt= \frac{rdr}{r-r_s} \implies t=r+r_s\ln|r-r_s|+c$$

where I got the second result by integrating and $c$ is an integration constant.

Graphically, it looks like this:Where the vertical axis is time, the horizontal axis is the distance, the black region is inside of the Event Horizon, and the red lines are the light rays.

Where the vertical axis is time, the horizontal axis is the distance, the black region is inside of the Event Horizon, and the red lines are the light rays.

You can see that near the black hole, light rays are not moving quickly away from the event horizon, but as long as they did not start out from the event horizon, they leave the black hole.

So in other words, you still see the falling body forever, because it taken infinite time for him to reach the event horizon.

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    $\begingroup$ The flashlight helps with the problem as does the equation. $\endgroup$
    – Bisk
    Jun 23, 2022 at 17:27
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    $\begingroup$ This is a fine illustration of the confusion that taking the parameter r to be radius causes. The constancy of the speed of light is a postulate of the theory, so the theory can't possibly predict that light slows down. The apparent slowdown is due to the fact that r is defined as the circumference divided by 2π. But the geometry describes a hole: consider that the circumference of a well doesn't tell you how deep you are in it. $\endgroup$
    – John Doty
    Jun 23, 2022 at 20:03
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    $\begingroup$ True, r is just the radius one far observer would see if he took a photo and drew a line on it, but in reality, close to the horizon, the actual distance (no change it time) $ds^2 = (1-r_s/r)^{-1} dr^2$ where there is a factor to account for curvature. $\endgroup$
    – Habouz
    Jun 23, 2022 at 20:09
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    $\begingroup$ Be careful with the statement 'it takes an infinite amount of time for the object to reach the event horizon according to the far observer.'. This is an overapproximation. See e.g. arxiv.org/abs/0907.2574 ("[...] we conclude that [...] A dust shell of finite thickness can indeed cross the black hole’s event horizon, as observed by an external observer, except for its outer surface. This means that asymptotically only an infinitesimal amount of matter remains just outside the event horizon."). $\endgroup$
    – TLW
    Jun 24, 2022 at 7:05

The fundamental reason for this - i.e. independent of coordinates and observation methods - is the "one-way" nature of the horizon, in that information cannot flow from the interior region to the exterior.

Consider the counterfactual: suppose that you could see the object cross the horizon after a finite time. Then by combining your information from before and the new information from after, you could make a logical inference that within the horizon volume there is now an object. That means you now have a piece of information about what is going on inside part of the Universe you are not "supposed" to be able to have information about the goings-on in.

Hence, something must happen so as to make your reasoning fail. And the way the real world does this is by dragging out the light signals coming from that object with more and more "latency", becoming an infinite amount of such latency for a signal emitted right at the horizon. You thus never get to establish the truth of one of your premises, and your argument fails to go through. And so you gain no valid information about what is inside the horizon. This is not the only way the information could be blocked, but it is the way that our world does it.


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