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I've read that as one approaches the event horizon of a black hole, time is dilated relative to time measured farther away from the event horizon (clocks tick slower near the event horizon).

I've also read that within the event horizon, the curvature of space becomes very large and approaches infinity near the singularity.

So I'm trying to understand with the time dilation and high curvature of space (or perhaps rather space-time) what an observer within the event horizon would measure as distance.

For example - from the outside of the black hole one might estimate the black hole's diameter to be 500 km. Would an observer inside the event horizon also estimate this distance or would the diameter appear to be more extensive?

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  • $\begingroup$ How are they to estimate the distance? $\endgroup$
    – Ryan Unger
    Jan 15, 2015 at 21:36
  • $\begingroup$ @Ocelo7 This of course has to take the advantage of a thought experiment. Each observer carries with them a huge yardstick. $\endgroup$
    – docscience
    Jan 15, 2015 at 21:41

5 Answers 5

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When people talk about the "diameter" they're just talking about twice the radius in some coordinate system like Schwarzschild coordinates--I talked about the physical meaning of the radial coordinate in these coordinates in this answer. The only coordinate-independent measure of "distance" is relativity is proper distance along a spacelike path (see my answer here), so you could choose some spacelike path from a point on the observer's worldline to some point in spacetime that lies on the event horizon. But you'd have to make some decision about what point you wanted to choose, the proper distance could be made arbitrarily short by picking a point that was arbitrarily close to where your past light cone intersected the horizon (since the spacetime "length" along a lightlike path, as defined by the metric, is always zero--see the spacetime wiki article if you aren't familiar with spacetime intervals, and John Rennie's answers here and here about how the metric is used to calculate proper time along timelike paths might also be helpful).

One approach to picking a spacelike path would be pick some simultaneity convention, like slices of constant time coordinate in a Kruskal Szekeres coordinate diagram, and then look at the proper distance along a spacelike curve that was confined to a single "moment", and this would correspond to a type of ruler distance, namely the sum of measurements on a bunch of short rulers whose ends line up at points along the spacelike path (I talked about the physical meaning of proper distance in that second answer of mine I linked to above) but since simultaneity is relative to your choice of coordinate system, you'd get different distances depending on what convention you chose, corresponding to different possible series of points in spacetime that the short rulers' ends line up.

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  • $\begingroup$ Upvoted. But I add that the natural spacelike path to choose is one that is purely spacelike as determined by the observer. That is, orthogonal to the observer's 4-velocity. Still, the resulting measurements are only meaningful locally. $\endgroup$
    – Colin MacLaurin
    Feb 9, 2018 at 23:42
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When you take a Schwarzschild (i.e. static, noncharged and non-rotating) black hole, it does not make much sense to talk about an observer trying to map the interior of the black hole because the singularity is fundamentally in the near future of the observer and the event horizon is fundamentally in the past of the observer.

When you are even above the event horizon the black hole seems much bigger than you would predict from naive Euclidean considerations. In Euclidean geometry you would predict the surface of a ball to fill a half of your view only when it is very large compared to you and only when you are on the surface itself. In the case of a black hole, the blackness covers a half of your view much earlier and when you are on "the surface" - on the event horizon, the blackness has covered your whole view and the whole sky is squeezed into one very bright point.

Once you plunge into the black hole, the featureless blackness has covered everything. You do not see anything, no instrument can reach the horizon and the surrounding seems isotropic, just the slap forces seem to be dynamically increasing. You send out light signals but they never return. In this sense, the answer most close to truth would be that the event horizon seems infinite from inside, because you cannot reach it or find any trace of it no matter how hard you try.

Just btw., the apparent shadow of a black hole from far away is actually about ten times larger than that of a planet with a radius equal to the event horizon radius. This is because the radius $2GM/c^2$ is fatal to photons going in the purely radial direction but photons which are just passing by are devoured much earlier.

I believe it is not very instructive to talk about the size of the black hole because it often lures us into thinking about it in terms of a black planet with a definite surface. When you study relativity further, you actually find out it even isn't a surface in any good sense, rather a very peculiar light-cone.

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  • $\begingroup$ Not all points on the event horizon lie in the observer's past light cone, there are points on the event horizon that can only be connected to the observer by spacelike curves. This is easiest to see if you use Kruskal-Szekeres coordinates, where light cones are easy to draw because all radial light rays are lines at a 45 degree angle from vertical, and all spacelike curves have a slope that's everywhere > than 45 degrees from vertical, while all timelike curves have a slope that's everywhere < 45 degrees from vertical. $\endgroup$
    – Hypnosifl
    Jan 15, 2015 at 23:17
  • $\begingroup$ Yes, but a space-like separation means that it is fundamentally inaccessible to the observer. Even when poking with a yardstick, you first send a pulse at the speed of sound in the stick and then wait for the response coming back at the same speed. I am trying to avoid the formal discussions and to jump straight to physics. A Penrose diagram might seem intuitive to a relativist but it usually just confuses lay people. $\endgroup$
    – Void
    Jan 15, 2015 at 23:37
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    $\begingroup$ But yes, there are ways to somehow form a conspiracy of observers which allow you to compute certain quantities which have the dimension of distance. You could program a large number of probes to plunge into the black hole afterwards you jump into it and send a light signal towards you once they stop seeing the sky. You would then compare the time they were predicted to reach the horizon in terms of your proper time with the time you received the signal and divide the difference by $c$ to get a certain luminal distance to the horizon. But it seems rather contrived at least. $\endgroup$
    – Void
    Jan 16, 2015 at 9:24
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    $\begingroup$ Another comment on a different issue: you say "on the event horizon, the blackness has covered your whole view and the whole sky is squeezed into one very bright point" -- but this is only true if you are considering a series of observers hovering at constant Schwarzschild radius, in the limit as that radius approaches that of the event horizon. For a freefalling observer the outside world does not shrink to a point, nor is it true that "the featureless blackness has covered everything" for the freefalling observer inside the horizon. $\endgroup$
    – Hypnosifl
    Jan 17, 2015 at 20:52
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    $\begingroup$ (continued) to see how the outside world looks to a freefalling observer both before and after they cross the horizon, see the animations on this page, as well as the more schematic ones on its sister site here. Note that the outside world always fills more than half of the visual sphere surrounding you, only approaching filling half in the limit as the falling observer approaches the singularity. $\endgroup$
    – Hypnosifl
    Jan 17, 2015 at 20:55
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The simple answer is that you can't measure anything from within the event horizon.

Ignoring the question of whether you'd survive crossing the event horizon or not, you'd have the issue that you can't move your yardstick and you also can't observe it (e.g. with light bouncing off it).

To observe a yardstick light needs to fire at it and bounce off. E.g. light travels up and away from the black hole, bounces off the yardstick then back to your eyes. Obviously this can't happen as by definition light can't escape the gravity of the black hole within the event horizon so can't travel up and away from it.

Also an observer outside the event horizon will notice time slowing down for someone falling in and essentially stopping at the point of the event horizon. Although we don't know what happens past this point we can imagine you essentially have a frozen person travelling straight towards the black hole with no ability to move in any direction (so they can't take or observe measurements).

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  • $\begingroup$ You can have a spacelike curve (whose proper distance can be measured by a series of freefalling rulers whose ends touch at points along the curve) between you and the horizon, such that light from each point on the curve can reach you before you hit the singularity. So in that sense you can in principle measure the distance from yourself to the horizon along the chosen spacelike curve. As for the fact that an observer outside the horizon sees you slow down, this is just a visual effect, your own clock and motions could proceed normally from your point of view even after crossing. $\endgroup$
    – Hypnosifl
    Jan 15, 2015 at 22:20
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    $\begingroup$ The answer has some mistakes. You can move your yardstick / ruler inside the horizon. To the falling observer, there are 3 dimensions of space and 1 dimension of time just like normal. You can observe it; let's say the ruler points upwards, then you can shine light onto the end of the ruler, yes the light is moving to decreasing $r$-coordinate but relative to the observer and yardstick it moves outward; it can certainly reflect back and catch up to you also. Theory does predict what happens at the horizon. Also the "time slowing down" claim depends on the simultaneity convention. $\endgroup$
    – Colin MacLaurin
    Feb 9, 2018 at 23:49
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The distance measured depends on the observer. To articulate distance, we pick a coordinate and describe how it relates to physical measurements, such as a ruler or radar ranging. Now the coordinate does not, in general, have any direct physical meaning, it is simply a choice for labelling or mapping out spacetime events.

The Schwarzschild $r$-coordinate is the obvious choice. Outside the event horizon, any observer at fixed $r$ measures a radial proper distance $(1-2M/r)^{-1/2}dr$. This relates the coordinate interval $dr$ to the ruler distance. Inside the horizon, I suggest the most natural observer is one who initially fell from rest far away from the black hole. And these measure $1\cdot dr$.

Let's interpret your 500km as meaning the event horizon is at Schwarzschild coordinate $r=250$km (another option would be optical appearance). Generally, measurements are only meaningful when made locally. So take a whole bunch of the above falling observers; they will measure the internal diameter of the black hole to be 500km! The only flaw is they can never communicate this to the outside. :(

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I just think that light never "escapes" the event horizon because from outside observer, time has slowed to a point that Speed of Light is no longer Speed of Light. Speed is indicative of time/distance however time is now so slow, that from outside, the light seems to break down. I think that the event horizon is relative to the observer. the closer you get, the more time returns to the same plane as the light, and we can then see farther in. Farther in we go, the more time keeps us in line with light law.... and we can see farther in. No longer out, because time has now completely changed from observer perspective, but still see around in this time dilation bubble. That said, if we sent a train of sensors in,as the event horizon moves onward in toward the black hole, one transmitting back to the former(with Doppler shift of course), then it repeats back to be next in line until finally work from one time dilation on to the next. I think eventually we could extract inner BH telemetry back out layer by layer by layer(of time dilation) till we could actually get something from within the event horizon back out to a "normalized" timeframe outside the horizon.

A relative horizon based on how slow time has become around you to break the speed of the universe(LIGHT). If you move in slow enough, the horizon just gets smaller and smaller till you see something within..... and get squashed to atomic screening.

Just some thought to try and see something from a different perspective!

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