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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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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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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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It will reach the speed of light exactly at the black hole surface.

Anixx
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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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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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