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In Newtonian mechanics, if we throw an object in against direction of gravity with speed $v$ and it achieve max height of $h$. Now if we allow object to fall from that height $h$, it will eventually attain speed $v$ when it reach position where we launch it.

Now applying same idea to a black hole in general relativity. Speed require to escape black hole gravity is greater than $c$, so if we throw something into black hole with almost the speed of light, the object speed will exceed speed of light $c$ before hitting black hole surface! How relativity explain this? Can space-time curvature reduce speed of this freely falling object from attaining speed of light?

Qmechanic
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someone_ smiley
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    Your analogy is based on Newtonian mechanics, which is not applicable to strong gravitation field around black holes. – Siyuan Ren Apr 24 '12 at 06:01

5 Answers5

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To answer your question you need to be clear what coordinates you're using. If you use coordinates that are co-moving with the rock falling into the black hole then the rock will see the event horizon pass at the speed of light.

External observers, using Schwarzchild coordinates, will see the rock slow down as it approaches the horizon, and if you wait an infinite time you'll see it stop.

External observers obviously can't comment on the speed of the rock after it has passed the event horizon because it takes longer than an infinite time to get there. If you use the rock co-moving coordinates then you can ask what speed you hit the singularity and ... actually I'm not sure what the answer is. I'll have to go away and think about it.

Incidentally http://jila.colorado.edu/~ajsh/insidebh/schw.html is a fun site describing what happens when you fall into a black hole.

Anixx
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John Rennie
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    Although I +1'd this at a quick glance, on second reading, you absolutely need to say that the "slowing down" is a coordinate effect, that you don't actually slow down relative to an outside observer, since the coordinate distance you travel per unit coordinate time (scaled by the metric) is going to c even for the external observer, it's just that the speed c in external coordinates is frozen at the horizon, since the external coordinates are symmetric between white hole and black hole, they do not distinguish the sense of the horizon, so there is no crossing of light through the horizon. – Ron Maimon May 01 '12 at 21:36
  • @Ron Maimon: So if they don't "actually" slow down, does that mean they even pass the horizon? If so, at what speed are they going then? – The_Sympathizer Nov 20 '14 at 00:26
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    @mike4ty4: The coordinates where the objects freeze are degenerate on the horizon, there is no paradox. The objects freeze because the coordinate time stops, not because their intrinsic velocity is slow. It requires knowing the metric form at a horizon, something you can work out for Rindler space easily, because it's just flat Minkowski space in disguise. – Ron Maimon Nov 21 '14 at 04:15
  • Has (almost) a decade been enough time to think about it? :) Do you think you'd hit the singularity at a speed that's higher than $c$ but the same speed at which light would hit it? Also, how would you respond to Ron's comment now? Is there any chance that he was right about the speed "going to $c$ even for the external observer"? – Gumby The Green Apr 03 '22 at 08:24
  • @GumbyTheGreen the speed any point that is distant from you is coordinate dependent. In your rest frame at your position you can approximate the spacetime as flat so speed has a nice clear definition. But as soon as you consider points that aren't at your position you need to consider the curvature and then the number you get for the speed depends on what coordinates you choose. I don't think Ron is correct, though this is a somewhat philosophical perspective because we probably just differ in what we mean by speed. – John Rennie Apr 03 '22 at 08:46
  • In relativity the equivalent of Newtonian speed is the four-velocity but this doesn't help because the norm of the four-velocity is always equal to $c$ for all observers everywhere. In this sense Ron is correct to say the speed is $c$ at the event horizon, but only because it's $c$ everywhere. – John Rennie Apr 03 '22 at 08:47
  • Right but I was just using the same definitions of those terms that you seemed to be using when you said "If you use the rock co-moving coordinates then you can ask what speed (presumably the 3-velocity in space) you hit the singularity..." for my first question. Regarding my second question, I think I misunderstood Ron's comment. I thought he was saying that things cross right over the horizon from an external frame but that doesn't seem to be the case. – Gumby The Green Apr 03 '22 at 09:35
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Nope, it will just fall in a reasonable amount of time (if you go with it, but watch out for tidal forces!), or take forever to fall in (if you are watching from outside).

Also, if I may be so bold as to suggest doing some quantum mechanics instead of kinematics while you are there, you could probably lock down some funding no problem.

tmac
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    +1 for you could probably lock down some funding :) – Manishearth Apr 24 '12 at 07:18
  • ohhh yes, that make sense, gravity make time slower, so it would take infinite time for object to hit the black hole. thanks. and well am novice to quantum mechanics. will explore it for more :) – someone_ smiley Apr 24 '12 at 08:31
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It will reach the speed of light exactly at the black hole surface.

Anixx
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  • What surface? I assume you mean the event horizon, but that's just a mathematical entity, not a physical thing. – PM 2Ring Jan 28 '18 at 16:51
  • I meant event horizon. Whether ir is physical or not is a matter of opinion/phylosophy. – Anixx Jan 28 '18 at 17:07
  • Sorry, but I can't see "surface" as ever being anything but an abstraction, which leaves this answer just like John Rennie's, except that he stresses the importance of coordinate systems, GR being local. – Edouard Jun 10 '19 at 16:40
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if we throw something into black hole with almost the speed of light, the object speed will exceed speed of light c before hitting black hole surface! How relativity explain this? Can space-time curvature reduce speed of this freely falling object from attaining speed of light?

If you throw (or drop someting for that matter) radially inwards towards a black hole the gravitational acceleration in coordinate time will go as:

$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\hat{r}(1-2\frac{v^2}{c^2(1-\frac{2GM}{rc^2})}-\frac{v^2}{c^2(1-\frac{2GM}{rc^2})^2})$$

The two extra terms will prevent any object moving radially inwards from reaching the speed of light. You can only reach the Schwarzschild radius of $r=2GM/c^2$ if you move infinitely slow. Usually you do not talk of the Schwarzshild radius as the "black hole surface", but I guess that is what you mean.

Agerhell
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  • Does this imply that the escape velocity at the event horizon is not actually the speed of light but is rather less than it? If so, how can the event horizon be the point beyond which light can't escape? – Gumby The Green Apr 03 '22 at 08:13
  • @GumbyTheGreen well you sort of get infinite redshift at the event horizon as viewed by a distant observer and also the speed of light as viewed by a distant observer would be infinitely slow but for a person at the event horizon time also goes infinitely slow compared to a distant observer leading him to believe the speed of light is normal. I do not know if you for sure can know what is going on beyond the event horizon. – Agerhell Apr 20 '22 at 10:08
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Imagine that the spaceship falling into the black hole broadcasts the frequency i.e. ticks of its atomic clock back to us. Due to the huge and increasing red shift, the cycles or ticks are received by us further and further apart in time. We will in fact never receive the last tick, and the total number of ticks we will count will be finite. That is how long the spaceship took to fall into the black hole, but it will take forever for us to find out.

Paul
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  • This is wrong! The last tick before crossing the horizon will always reach the distant observer in a finite time, regardless of how close together the ticks are. – D. Halsey Apr 22 '23 at 23:11