1. Sep 7, 2015

### sillycow

I have some questions regarding things that boggle my mind about black holes. These things seem to me like paradoxes, and I was wondering if someone could explain them.

1. How do you measure the distance to a black hole? As far as I understand the distance to a black hole is defined as the distance from an observer to the event horizon. Distance, in physics is defined by the time it takes light from point A to reach point B. However, as time is dilated close to the event horizon, no observer will ever see anything fall in to the black hole. An observer would just see things red-shifted and slowing down. Meaning: "Light would take an infinite amount of time to travel *into* the black hole. Wouldn't that mean that the black hole is infinitely far away from any observer? (farther away than things which are "behind" the black hole)

2. This one involves several "paradoxes"
2a. As you approach the event horizon, it keeps moving away from you (space-time gets increasingly curved).
2b. If you look at someone falling towards an event horizon, they slow down until they almost freeze.
==>
2c. That leads me to believe that no observer can ever observe any object to cross an event horizon. How can a black hole come into existence, if it wasn't there beforehand. Or in other words, if nothing can be seen to cross an event horizon for any observer, how can an event horizon exist?

3. Not really a paradox, but a "cool question". If I stand in between 2 black holes (not inside their event horizons), in the exact point where their gravity nulls each-other out. Assuming I am a point particle... Do I get double the time dilation compared to someone standing far away? Or 0 time dialation?

2. Sep 7, 2015

### Staff: Mentor

As you mentioned the radar distance is infinite. So, instead, the Schwarzschild r coordinate is measured by measuring the circumference around the black hole and dividing by $2\pi$

3. Sep 7, 2015

### sillycow

I think there is some circular logic here. Since distance becomes meaningless around a black hole, how would you define circumference? For every circumference defined by "it would take light x seconds to orbit the black hole", I could theoretically find a slightly closer circumference that takes x+1 seconds because of time dilation. That is unless this function converges to some value (That is I cannot find an x+1). Does it? My ballpark hunch says that it doesn't, since at the event horizon itself, light would take forever to orbit the black hole.

4. Sep 7, 2015

### Staff: Mentor

I am not sure why you would think that distance is meaningless around a black hole. The geometry is curved, but not meaningless.

The circumference is measured locally. It could be measured using a ring of radars or a ring of rulers, either way is fine. Time dilation is not relevant because all of the measurements involved are at the same gravitational potential.

5. Sep 7, 2015

### Staff: Mentor

This one has been discussed in many other threads here - a search will find some of them.

It is true that if I accidentally drop you into a black hole, I will never SEE you cross the event horizon - the light from that event will never reach my eyes so that I can't SEE it. However:
a) You will experience falling through the event horizon and then reaching the central singularity rather quickly.
b) If I try sending you a radio message ("Dude! - I'm sorry! - I didn't mean to drop you! Please forgive me!") you will be able to reply ("You've killed me because you are a clumsy jerk! No, I don't forgive you!") only if I send it before a certain time. After that, you won't get my message until you've crossed the horizon and it's too late.
c) There is another point in time, only slightly after the one in #b, such that I have to send my message before then or you won't even receive it because you've died at the central singularity before it got to you.

Putting all of these together, it's hard to see how your conclusion in #2c above makes sense. It's much more natural to say that you do in fact fall through the horizon and hit the central singularity - I just don't get to see it.

Gravitational time dilation is not caused by the strength of the gravitational field, but rather by the difference in gravitational potential between the two clocks. Thus, the amount of time dilation between your clock and a clock at infinity will depend on how far away you are from the two masses (which don't need to be black holes - as long as you're outside the event horizon event horizon there's no difference between a black hole and any other massive object). But whatever it is, it will be twice the what you'd get if there was only one mass present and at that distance.

6. Sep 7, 2015

### Staff: Mentor

Distance doesn't become meaningless around a black hole. What does become meaningless is the distance along any line that passes through the central singularity - and that means the diameter and the radius. However, there's no problem with the tangential directions (as long as the black hole is not rotating - that's a whole new can of worms) so circumferences and surface areas work just fine and that's why we use them.

It is true that even the radial distances that do not cross the central singularity behave strangely, but that strangeness doesn't mean that they're undefined, just that they don't behave the way they would in flat space. Suppose we have two spherical shells around the black hole: the inner one has surface area $4\pi{R}_1^2$ and the outer has surface area $4\pi{R}_2^2$. In an uncurved spacetime you would expect that the gap between the two spheres would be $R_2-R_1$, but in the curved spacetime that gap is somewhat larger. That doesn;t make the distance meaningless, it just makes it larger than you'd expect if there was no gravity.

7. Sep 8, 2015

### Staff: Mentor

This is the most obvious way to define it, yes. However, there is an important caveat to this definition; see further comments below.

No, it isn't. You can measure distances without using light; use a ruler, for instance.

The current SI unit of distance, the meter, is defined so that the speed of light is exactly 299,792, 458 meters per second, which effectively means it's defined in terms of light travel time. But that's a definition of a unit, not the definition of the concept of distance. Before that SI definition was adopted, the meter was defined in terms of a standard number of wavelengths of a particular atomic emission line; and before that, it was defined in terms of a standard meter stick kept in France.

No; but it does mean that we need to be careful in defining exactly what is meant by "distance to the horizon".

The key point is that the event horizon is not a "place". It's a null surface--a surface formed of radially outgoing light rays. That means nothing can "stand still" at the horizon, so you can't put one end of a ruler at the horizon and the other end at a chosen point outside and measure the distance between them--the end of the ruler at the horizon would have to move outward at the speed of light, like the light rays that form the horizon. (The reason these light rays don't actually increase their radial coordinate, even though they're moving outward at the speed of light, is that spacetime at the horizon is curved just enough to keep them at the same radial coordinate.)

To rigorously define the distance along a radial line to the horizon, what you need to do is envision a limiting process, where we imagine a series of rulers with their upper ends at our chosen point, and whose lower ends approach the horizon more and more closely. We then define the distance to the horizon as the limit of the lengths of those rulers as the radial coordinate of their lower ends approaches the radial coordinate of the horizon. This limit can be computed mathematically, and it is finite, so it can serve as a definition of a finite "distance to the horizon" from any point outside the horizon.

8. Sep 8, 2015

### Staff: Mentor

There are complications lurking here that need to at least be mentioned to avoid giving a wrong impression. As my previous post shows, distance along any radial line can be defined outside the horizon; and by using an appropriate limiting process, radial distance to the horizon from any point outside can also be defined.

What can't be defined is distance of this kind inside the horizon. That is because inside the horizon, spacetime is not static; you can't think of a particular radial coordinate $r$ inside the horizon as marking a "place in space" so that you can measure distances between two different $r$ coordinates. So even the idea of a radial line that passes through the singularity is meaningless (let alone distance along such a line); there is no static "space" for such a line to pass through. So a black hole doesn't even have a diameter or a radius in the usual sense. Another way of putting this is to say that the singularity is not a place in space; it's a moment of time (and this moment is to the future of any event inside the horizon).

(Even the above doesn't completely cover all the complications lurking here, but I'll stop now since this is a "B" thread.)

9. Sep 8, 2015

### sillycow

Thanks for taking the time guys, I really appreciate it.
@PeterDonis, Here is what I don't get.: You can define distance in several ways. For example, as you have suggested, let's use a ruler instead of light.
So let's say that we are using 1 meter rulers to make things simple.
If we defined "distance" relative to the length of said rulers, how can we say that the "distance" converges? (distance is in quotes, because I am starting to feel that the word is getting "mushy" on me :-). ) . If there are infinite rulers, and a unit of length is a ruler, then doesn't that mean that the distance is diverges to infinity instead of converging to some value? Would the 100th ruller dissapear (besides growing fainter)?

I realize that as they approach the event horizon, rulers will seem "shorter" and fainter. (I also realize that at some quantum point "fainter"=="does not exist anymore", but let's ignore quantum mechanics, as that just makes it more complicated). However, notice that "shorter" is in quotes: If I defined rulers as my unit of distance, how can I say that they are "shorter"?

Is there a definition of "distance" for which this doesn't happen?

Last edited: Sep 8, 2015
10. Sep 8, 2015

### sillycow

R1 and R2 are such distances (they pass through the black hole). I can however deduce them by measuring circumference: R=C/2Pi. I would measure this circumference using the amount of time it takes an object to orbit it. That way I can deduce R2. However, as I approach the black hole wouldn't the object start orbiting more and more slowly because of time dilation, thereby making the circumference paradoxialy larger? Or would it only accelerate more slowly thereby giving me a limit?

I have read, and perhaps misunderstood, that objects traveling towards the black hole seem to freeze in time on it's surface, because their time slows down. If their time slows down, wouldn't their orbit slow down as well? Or would the acceleration caused by gravity "even out" with time dilation cause the orbit speed to converge to a fixed speed? If so, then I see how I would deduce a limit on the radius. Is this the case?

To reiterate: If R1 is behind R2 (I have to pass R2 to reach R1) is R1 allways shorter than R2 ? Geomtrically speaking, can a black hole be curved like an "hourglass"?

Last edited: Sep 8, 2015
11. Sep 8, 2015

### Staff: Mentor

They are not distances, they are just numbers that we've calculated. The circumference is a distance because I can measure it the old-fashioned way using a meter stick; but the number I calculate from $R=C/2{\pi}$ isn't the distance between anything and it doesn't "pass through" anything. It's just a number that pops out of the calculation.

You can, but you could also (in principle) actually build the spherical shell around the black hole, land your spaceship on the surface of the shell, and use an ordinary meter stick and similar tools to directly measure the distance around the circumference pf the shell. That way you can be sure that you really have measured the circumference, with no complications from time dilation, orbital dynamics, and the like.

If the spherical shell whose area is $4\pi{R}_1^2$ is inside the spherical shell whose area is $4\pi{R}_2^2$, then $R_1\lt{R}_2$. That's "less than", not "shorter" - they're numbers, not the lengths of anything, so "longer" and "shorter" aren't words that apply here.

12. Sep 8, 2015

### Staff: Mentor

I am not sure how you would change that into a measure of distance. You would have to assume some relationship between the orbital period and the circumference of the orbit. You could certainly do so, but it would be as much a measurement of the mass as a measurement of the circumference.

Why not directly measure the circumference using rulers or radar as mentioned above?

13. Sep 8, 2015

### stevebd1

From http://www.mathpages.com/rr/s7-03/7-03.htm (Falling Into and Hovering Near A Black Hole) -

'..relative to the frame of a particle falling in from infinity, a hovering observer must be moving outward at near light velocity. Consequently his axial distances are tremendously contracted, to the extent that, if the value of $\Delta r$ is normalized to his frame of reference, he is actually a great distance (perhaps even light-years) from the r=2M boundary, even though he is just 1 inch above r=2M in terms of the Schwarzschild coordinate r. Also, the closer he tries to hover, the more radial boost he needs to hold that value of r, and the more contracted his radial distances become. Thus he is living in a thinner and thinner shell of $\Delta r$, but from his own perspective there's a world of room...

Quantitatively, for an observer hovering at a small Schwarzschild distance $\Delta r$ above the horizon of a black hole, the radial distance $\Delta r'$ to the event horizon with respect to the observer's local coordinates would be-

$$\Delta r'=\frac{\Delta r}{\sqrt{1-\frac{2M}{2M+\Delta r}}}$$

where $\Delta r'$ is proper distance, $\Delta r$ is coordinate distance and $M=Gm/c^2$

14. Sep 8, 2015

### sillycow

Didn't want to change the observer's position by coming closer to the black hole. Having something orbit the black hole at the speed of light seemed "cleaner", as I could measure how many times that object revolves around the black hole without comming any closer myself.

This depends on how space-time is curved. In a torus (mmm... doughnuts.. :-) ) curvature ,this can be untrue. A bigger circular circumference could be placed "inside" a smaller circumference. Where "A is inside B" means that you have to pass through a to get to B. So "inside" and "contains" are not directly related to sizes,lengths,areas,and volumes.

15. Sep 8, 2015

### Staff: Mentor

That's true, but you started you this thread asking about one particular way that space-time is curved, namely what you find around a black hole (actually, any spherically symmetrical mass, whether a black hole or not - the effects we're discussing here happen around the earth as well, but are so weak we seldom notice). That curvature is given by the Schwarzschild solution to the equations of general relativity, and it does behave as I describe.

The best way of visualizing the spatial curvature around a black hole is 'Flamm's paraboloid' - google will find many good references. However, it comes with a few cautions:
1) Flamm's paraboloid illustrates curved space, not curved spacetime. It's accurate but incomplete, just as a two-dimensional cross section of the earth through the equator will show that the earth's surface curves as you travel east or west, but doesn't show the north-south curvature.
2) Do not just look at a picture of the paraboloid and think "OK now I get it" - that almost guarantees that you will misunderstand what you're looking at. You have to read and understand the math that comes with it to understand what you're looking at.

(and a digression: although the surface of a physical doughnut is curved, it turns out that the surface of an ideal mathematical torus is actually flat. This is a digression so no need to follow up, but if you want to, googling for "intrinsic vs extrinsic curvature" will get you started).

Last edited: Sep 8, 2015
16. Sep 8, 2015

### sillycow

17. Sep 8, 2015

### stevebd1

This is correct (though it's not a 'faller', it's a static observer).

18. Sep 8, 2015

### Staff: Mentor

"Converges" is a technical term; it just means we are taking a limit as a particular parameter (in this case, the radial coordinate of the lower end of our measuring device) approaches a particular value (in this case, the radial coordinate of the horizon). If my use of the word "ruler" is confusing you (because the length of the "rulers" in question varies), think of it this way: we have a tape measure with one end fixed at a particular point above the horizon. We slowly extend the tape measure so that its other end approaches the horizon. The distance registered by the tape measure, which is defined in terms of how many standard rulers it takes to extend from one end of the tape measure to the other, will increase slowly, and as its other end gets closer and closer to the horizon, the distance registered will approach some particular finite value--it won't increase to infinity. That finite value is the distance to the horizon. We can't actually put the other end of the tape measure at the horizon, for the reasons I gave in my last post; that's why, to be careful and precise, we have to phrase things in terms of the limiting process I just described.

19. Sep 8, 2015

### sillycow

Hi, I am still reading stevebd1's link. However this thought experiment has me vexed:
To paraphrase: Say I am standing outside the black hole, and I drop a chain of 1m rulers towards it in freefall. (Thereby eliminating any rigidity of said tape measure and dividing it into discrete units, ). I let the black-hole accelerate them towards it by pulling them with it's gravity. I have an infinite amount of rulers to drop. If the distance is to converge to some fixed number, then one of the following needs to happen:
1. At some point I will see the first ruler I dropped disappear into the black hole.
or
2. At some point I will see two rulers overlapping: One's front end will pass the other's back end.

I gather that both of these things would happen at the event horizon: The rulers overlap because they are in singularity & They disappear because light cannot escape. However, I have read that no matter how much time I wait, I will never observe the rulers crossing the event horizon. So from my point of view, I can keep on dropping an infinite amount of rulers. and hence my problem with convergence. That is unless I actually see them overlap or disappear beforehand.

20. Sep 8, 2015

### PAllen

There is no singularity at the event horizon. As for 'seeing' it depends who is 'looking'. Each of the falling rulers (granted eyes), will see the ruler in front remaining a 'normal' distance in front of it, same for ruler in back, until nearing the singularity well inside the horizon. A observer hovering a distance outside the horizon will see the distance between rulers getting ever smaller, and they will be red shifted such that they become invisible on approach to the horizon. Given any sensitivity of detector for the hovering observer, no matter how good, the rulers will each 'disappear' fairly quickly - but this will happen just outside the horizon.

The upshot: both of your choices are wrong. No one outside will see a ruler cross the horizon, nor will they see any overlap.