# Black hole riddle

1. Apr 27, 2005

### Zanket

General relativity tells us that an infinite number of houses can be built at fixed altitudes in the one meter (say) above the horizon, as measured in Euclidian geometry. Let this infinite number of houses exist. Let the street numbers of the houses increment with decreasing altitude. For someone crossing the horizon at the end of this street of houses, what is the street number of the last house passed?

2. Apr 27, 2005

### PeteSF

Insufficient data to provide a definite answer: "the last house passed" is not well defined.

How does this relate to black holes?

3. Apr 27, 2005

### JesseM

What do you mean by "as measured in Euclidean geometry"? In a continuous classical world it would also be possible to build an infinite number of ever-thinner, ever-more-closely-packed houses in any given meter of flat space--what role does the event horizon play in this problem?

4. Apr 27, 2005

### PeteSF

Oops... I skipped through too fast. For some reason, I missed the GR reference and was considering an abitrary Euclidian space...

Anyway, the core idea of my last post remains, I think... it is not possible to uniquely identify "the last house" in such a situation.

5. Apr 27, 2005

### Zanket

This means, “measured as if spacetime were flat,” and not curved by gravity. It’s easier to talk about circumferential measurements, which are not affected by gravity. Radially, between the circumference of the horizon and a circumference infinitesimally longer, or let’s say one meter longer, an infinite number of houses can fit, so says general relativity. The directly measured radial distance between these circumferences is infinite.

You can assume that these houses are all the same proper size; e.g. they can all fit people within. Just above the horizon is where the last house is. At and below the horizon, no house can remain at a fixed altitude.

6. Apr 27, 2005

### Zanket

On the way to the horizon, houses are passed. Upon reaching the horizon, there are no more houses. Then there must have been a last house, uniquely identifiable.

7. Apr 27, 2005

### JesseM

In order for all the houses to maintain a constant distance from the horizon, it will have to expend large amounts of energy--would the energy density of the ever-more-tightly packed houses (in some coordinate system at least, I don't know which system would be the relevant one) cause the houses to form a black hole themselves, or the horizon of the original black hole to expand to include them? If so, maybe this thought-experiment wouldn't be possible even in principle...

8. Apr 27, 2005

### JesseM

Do you think there is a uniquely identifiable smallest positive fraction of the form 1/n, where n is a positive integer? After all, as you approach zero when moving along the real number line in the negative direction, fractions of this form are passed, but upon reaching zero there are no more such fractions.

9. Apr 28, 2005

### Zanket

Such details can be ignored in a thought experiment like this. These are not ever-more-tightly packed houses, at least not in their own frames. They can all have the same proper sized lots.

True, but it doesn’t answer the riddle. It doesn’t explain, for example, how there is not a uniquely identifiable last house. Have you ever passed a row of houses, and upon coming to the end of the row, found that there was no last house in the row?

Last edited: Apr 28, 2005
10. Apr 28, 2005

### JesseM

But in terms of the formation of a new black hole, does GR say that the fact that the distance to their nearest neighbors in their own frames is the same for each house is enough to insure that no black hole will form? Isn't the fact that houses closer and closer to the horizon will have to thrust with greater and greater energy in their own frames, with the thrust energy approaching infinity as you approach the horizon, also relevant to answering the question of whether any of the houses form a black hole (or cause the horizon of the existing black hole to expand to swallow them)?
Well, even if this thought-experiment is possible in principle, won't a freely-falling observer see the houses more and more tightly packed as he approaches the horizon, rather than an equal distance apart? If so, he is experiencing the same thing you would experience if you walked along a number line towards zero with each number of the form 1/n marked off--at some point the markings would become too close to distinguish.

11. Apr 28, 2005

### PeteSF

That's certainly intuitive, but is it true? Unituitive things happen when you're dealing with infinities, which is why Zeno is remembered so fondly.

12. Apr 28, 2005

### Zanket

Such details put the cart before the horse, so they can be ignored.

That is indeed the way one of my books explains how an observer free-falling across a horizon can travel an infinite distance (an infinite number of houses) in a finite time. But it still does not answer the riddle. I pass an infinite amount of infinitesimally-sized stuff with every step I take. But that doesn’t explain how, when I come to the end of a row of houses, there is no last house.

13. Apr 28, 2005

### PeteSF

Sure it does. If there are infinity houses, there is no uniquely identifiable last house, just like there is no uniquely identifiable last number. What's left to explain?

14. Apr 28, 2005

### Zanket

Infinities can be used to mask absurdity. The solution to Zeno’s Paradox is logical.

15. Apr 28, 2005

### Zanket

What is left to explain is how the last house can be passed without it being uniquely identifiable. Houses are passed, and then there are no more houses. There must have been a last house. (But now we’re going in circles.)

16. Apr 28, 2005

### JesseM

What do you mean "put the cart before the horse"? Do you agree that houses closer and closer to the horizon will have to be thrusting with greater and greater G-forces to maintain a constant height? And wouldn't it be true that if a given mass thrusted with greater and greater G-force in flat spacetime, eventually the energy density would be high enough to form a black hole? If that's correct, then it seems the situation you describe is physically impossible, even in principle.
OK, but the black hole doesn't seem to be relevant to this particular "paradox". Do you agree that the exact same problem occurs in my thought experiment where you are walking towards 0 along a number line where every number of the form 1/n has been marked? I would suggest looking into some real analysis, particularly the concept of open sets vs. closed sets.

17. Apr 28, 2005

### pervect

Staff Emeritus
The main resolution of this "paradox", which I think needs a little work as well, is that it is not possible to "walk" across the event horizon of a black hole in any physically meaningful sense. An observer falling into a black hole will always be travelling at 'c', the speed of light, when he crosses the event horizon.

The reason I think the paradox needs work is that the coordinate system where one could hypothetically build an infinite number of houses is not even approximately Euclidean or flat. It's a highly non-physical geometry. One of the properties of this coordiante system is that the proper acceleration of gravity will tend towards infinity. Infinite proper accelerations can hardly be called a "flat" or "euclidean" geometry.

This is actually not a coordinate system I've seen discussed much - I'm more or less taking your word for it that it's possible to build an infinite number of houses in it. Where did this riddle come from originally? It seems like it might be related to Milne cosmology.

Anyway, these infinite proper accelerations are why it's impossible for an observer to cross the event horizon at any speed other than 'c', BTW. If you pick an observer with a finite proper acceleration, you will find that his velocity when he crosses the event horzion is equal to 'c'.

Last edited: Apr 28, 2005
18. May 1, 2005

### Zanket

Sorry for the delay in replying…

The theory allows for an infinity of houses along a segment of a radius above the horizon. We can discuss the logical ramifications of that before accounting for the mass of the houses. If we say that discussion is precluded by the mass issue, then IMO that is putting the cart before the horse; i.e. putting things in the wrong logical order.

A similar example in relativity discussions is the infinitely rigid rod. Such a rod cannot exist in principle, but you’ll see thought experiments using it.

Agreed.

I agree that it’s a similar situation. And I agree that there simply won’t be an infinite number of houses. Not because of the mass issue per se, but that’s one way to enforce the limitation. Even though there is room for an infinity of houses, there will be a finite number of house always, hence a last house with a uniquely identifiable number.

19. May 1, 2005

### Zanket

I just thought of it off the cuff. I think it is impossible to have an infinite number of houses, but might be worth imagining.

That raises another riddle: If everyone crosses at c, then if you and I are crossing independently but beside each other, what happens when I accelerate to pass you? It seems that we must stay in lockstep since we both must cross at exactly c. Our ships could be in a light-year-across region of almost perfectly flat spacetime for a large enough black hole. Our ships could each be a light-hour long, so it would take an hour on our clocks to completely cross. It would be strange if I accelerated and didn't make any headway relative to you. And if I did make headway then I'd be crossing at a different velocity than you and we could not both be crossing at c.

Last edited: May 1, 2005
20. May 1, 2005

### Mortimer

From the perspective of which observer??
An observer at infinity does not measure the falling object cross at c. At the event horizon he measures the velocity $$V(r)$$ to be zero (assuming zero starting velocity at infinity):

$$V(r)=\left(1-\frac{2MG}{rc^2}\right)\sqrt{\frac{2MG}{r}}$$

If you put that in a graph it looks like this:
http://www.rfjvanlinden171.freeler.nl/blackhole2.jpg

Last edited by a moderator: Apr 21, 2017