I've been studying the basics of General Relativity, and my question is: does it make sense to say that near a black hole (or any massive body), distances increase, the way on a topographical map the contours get closer and closer near a steep hill? So that an object escaping from near the event horizon is actually having to travel thousands of times more than the 'real' distance to reach a destination far away.  This seems to jive with the idea that an infalling object (from an outsiders perspective) seems to travel an infinite distance, fading away but never crossing the horizon, and also with the Shapiro time delay phenomenon.  But I want to ask in case this is a false analogy. And if it is true, how would that "stretched distance" (i.e. all the local distances on the trajectory added together) relate to the distance seen by an outsider?  How could the outsider ever say "I am 100 million miles from a black hole," if any test particle would actually travel many times greater distance to traverse it?  And how could a distant observer measure distance to a black hole anyway? How would a distant observer measure a valid distance to any object, if not by sending and receiving a light ray?

  • $\begingroup$ Related: physics.stackexchange.com/q/21319/123208 & physics.stackexchange.com/a/170506/123208 Also see my description of Schwarzschild coordinates here: physics.stackexchange.com/a/552874/123208 $\endgroup$
    – PM 2Ring
    Sep 14, 2021 at 5:23
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    $\begingroup$ While your topological map analogy is correct to describe the gravitational length contraction, the total distance to the horizon measured by a fee falling observer is finite and only somewhat larger than the coordinate distance. The “infinite distance” premise of your question is incorrect. $\endgroup$
    – safesphere
    Sep 14, 2021 at 14:51
  • $\begingroup$ @safesphere So in the case of AdS spacetime, light emitted outward from r=0 returns to r=0 in some finite time as measured by the observer at r=0. Does that mean that AdS space is finite? $\endgroup$
    – Razor
    Sep 14, 2021 at 14:55
  • $\begingroup$ @Razor This question is about the Schwarzschild spacetime. If you are interested in the AdS spacetime, please feel free to ask a separate question. $\endgroup$
    – safesphere
    Sep 14, 2021 at 15:16
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    $\begingroup$ @RC_23 You can however get a notion of the diverging coordinate distance in the coordinate system of a hypothetical observer hovering arbitrarily close to the horizon. $\endgroup$
    – safesphere
    Sep 14, 2021 at 15:36

1 Answer 1


You can form $\int ds$ along a radial line at some given $t$ and thus create a notion of ruler distance. The ruler distance from the horizon is finite. You can see this also by looking at Flamm's paraboloid: distances along the paraboloid are finite.

However the time required for a particle to move outwards from the horizon is infinite.


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