# How massive objects affect distance measurements, or "But how far away is a black hole singularity, really?"

My question basically boils down to: Is the singularity of a black hole infinitely far away?

I think questions like mine have been asked before, but they encounter responses like "no meaningful answer," so I thought I'd try to ask a modified version that will help laypeople like me understand.

It's my understanding that, in empty spacetime, an observer may measure the circumference $$C$$ of some sphere, and then measure that sphere's diameter $$d$$, and happily find that $$C = \pi d$$.

Let's say then that they fill that sphere uniformly with something that has mass (something like dark matter, so as to not disturb our measuring tape). The mass distorts spacetime, and when they repeat their procedure, the observer is surprised to find that the diameter is slightly greater than they expect: $$C < \pi d$$.

(if this isn't correct, you can stop reading here)

So, perhaps it is incoherent to ask what the distance is to a black hole's singularity, even in principle. But, I'm still hoping to understand how the ratio of the sphere's measured diameter to its [constant] measured circumference changes as the the massive stuff our observer filled the sphere with begins to collapse into a black hole. Does it remain fixed? Does it change up to some point and then become undefined?

• How are they supposed to measure the diameter? Commented Aug 29, 2022 at 23:42
• With a measuring tape. Sure, it won't work for a black hole, but I'm curious how the ratio changes during the process. Commented Aug 29, 2022 at 23:45
• The mathematics of black hole singularities is tricky. See physics.stackexchange.com/a/144458/123208 Maybe you should ask a new question about how mass (& other forms of energy) affect volume in scenarios that don't involve black holes. Commented Aug 30, 2022 at 5:57

The trouble with using a measuring tape is that it only shows the correct length if it's neither stretched nor compressed. You can't just hold it taut at both ends, because the collapse time of a black hole from a small multiple of the Schwarzschild radius to the singularity is not much larger than the time for light to cross the measuring tape once. Even if the speed of sound in the tape is the speed of light, you can't conclude much about its true length during the collapse from how it feels in your hands.

The problem is unfixable because it's rooted in the intrinsic oddness of the black hole geometry. If you try to define the diameter abstractly as a metric distance at a fixed $$t$$, you have to specify what $$t$$ is, and there is no natural choice. Dividing spacetime into space and time only makes sense in some sort of Newtonian limit, and black holes are far from Newtonian.

What you can do is consider a uniform sphere of matter that is stabilized by internal pressure so it doesn't collapse, and look at the limit of circumference over diameter as the reduced circumference approaches the Schwarzschild radius. This is a static solution, so you have as much time as you need to let the measuring tape equilibrate.

You can model the matter with the Schwarzschild interior solution (not to be confused with the black-hole-interior part of the Schwarzschild vacuum solution). It has the nice property that a constant-$$t$$ slice (where $$t$$ is a physically meaningful static time coordinate arising from the symmetries of the geometry) is just a portion of a hypersphere. A constant-$$t$$ slice of the vacuum exterior is a paraboloid of revolution, and you can graft them together as shown in this diagram from the Wikipedia article:

In the limit $$r_g\to r_s$$, the diameter of the matter approaches a half-circle (in this embedding; intrinsically it's straight), and so $$C/d$$ approaches $$2$$. At smaller radii, there is no static solution and you can't define a $$C/d$$ this way. I don't know that this answers your question.

• This is really helpful, thanks! If I understand you correctly, in my thought experiment, as we shrink our sphere of still-internally-stabilized matter, $C/d$ shrinks from $\pi$ to $2$, after which point the system lacks the properties that allow for a meaningful solution. At this point, I believe that the distance through the sphere between two antipodal points is equal to the distance traveling along its surface between the two points. It's intuitively nice that this is where things stop making sense--since beyond that means, counterintuitively, it's farther to go 'through' than 'over' Commented Aug 30, 2022 at 5:10
• @eriophora Right, $C/d\to π$ monotonically as $r_g\to\infty$, and the distance through the center approaches the distance along the surface as $r_g\to r_s$, but the latter is probably a special property of the Schwarzschild solution, not a general principle (though I'm not certain that it doesn't follow somehow from some energy condition). Commented Aug 30, 2022 at 6:58

This question already has a really nice answer on the difficulties of measuring lengths, so I'll focus on something else. The question

Is the singularity of a black hole infinitely far away?

is way more similar to "How far away is tomorrow?" than to "How far away is Egypt?", because the singularity in a black hole is not a place, but an instant. It is not a where, but a when.

When writing down the Schwarzschild metric for a black hole, the singularity sits at radial position $$r=0$$, so often people look at $$r=0$$, understand radial distance zero, and hence the singularity should be at the spatial origin, right? Not really. Inside the black hole, time and space "switch" in a quite literal manner (you may want to check PBS Spacetime's video How Time Becomes Space Inside a Black Hole for more on this), and in fact $$r=0$$ turns out to be not a position, but an instant. So when you ask how far away the singularity is, the answer to your question can't be measured with a ruler, but rather with a clock!

I can't resist mentioning this is the reason you can't escape a black hole: going outside a black hole would literally mean traveling to the past. When you enter a black hole, the future always points inwards.

Something particularly nice about this is that it is sort of easier to do measurements with clocks. You can simply have your watch on your wrist and check how many times it ticked. Of course, your measurement might not match other people's measurements (I discussed this in other posts, such as this one and this one, but there are posts on the relativity of time throughout the whole site). Still, your measurement is still valid in your reference frame.

The question "Is the singularity of a black hole infinitely far away?" then boils down to "When I'm falling down a black hole, will it take infinitely long for me to hit the singularity as measured by my wristwatch?". The answer is no: you hit the singularity in finite time. This is actually the very definition of a spacetime with a singularity: there are observers (actually, geodesics, if you want to be precise) that disappear of the spacetime within a finite time (affine parameter, technically). By "disappear" I mean quite that: in classical General Relativity, the singularity is not a point in spacetime, it is a "hole". This bothers a lot of scientists, who hope this weirdness will be cured by a theory of Quantum Gravity.

• Thanks for this! A similar point was described to me as (something like) "you can no more escape from within the event horizon than you can drive to yesterday" and that always stuck in my head. Commented Aug 30, 2022 at 5:16
• @eriophora GR gives us maps of spacetime. But the singularity isn't simply a blank region of the map that GR can't describe, it's more like a hole in the map. Commented Aug 30, 2022 at 6:03
• This does not answer the question, unless you are arguing that the experiment cannot be done as stated in the question. The collapse as described isn't governed by the Schwarzschild metric. Commented Aug 30, 2022 at 6:09
• @ProfRob I aimed an answering the question that OP's question "boils down to", instead of how to measure distances in the collapsing spacetime. Nevertheless, I think it is useful to point out that any spacetime which as a singularity has the property I mentioned of geodesics ending in a finite affine parameter. In particular, a star that collapses to a black hole does have observers that fall into the singularity in a finite (proper) time, even though the spacetime isn't Schwarzschild spacetime Commented Aug 30, 2022 at 10:28
• Also, I do argue that the experiment of measuring the distance to the singularity cannot be done. You can't measure the distance until tomorrow. But you can measure time till then Commented Aug 30, 2022 at 10:30

In relativity, length is relative to the observer's velocity. In general, there are ambiguities about extending the reference frame of a localised observer to a large region. So, consider instead an entire field of observers (velocities). Each measures a tiny region of space near themselves, and we combine their experiments into a macroscopic result. So consider a whole bunch of astronauts falling straight down into a non-rotating black hole. It turns out they measure radial proper distance $$dr/|e|$$, where $$e$$ is their energy (Killing energy per mass). This is easiest to see using a generalisation of Gullstrand-Painleve coordinates due to Gautreau & Hoffmann (1978). If they "fell from rest at infinity", they have $$e = 1$$, and hence the radial coordinate $$r$$ is precisely the radial proper distance, according to them! Hence if you and all your friends are falling this way, and at one instant your radial coordinate is $$r_0$$, then the distance to the singularity is simply $$r_0$$ also. See my paper arXiv:1911.07500.

Now suppose they are not falling straight down, but are twisting around one another. Now, their individual space measurements do not connect together. (In technical language: with vorticity, the orthogonal hyperplanes within each tangent space are not integrable to a hypersurface, by the Frobenius theorem.) Hence you have to be careful to combine the local measurements into a macroscopic one. This is related to the infamous problem of the relativistic rotating disc, whose proper circumference is greater than $$2\pi r$$! See my forthcoming work "Length-contraction in general relativity".

The reason the above seems unfamiliar is that there is usually far too much attention paid to the Schwarzschild $$t$$-coordinate, and not enough to the different choices of observer velocities.