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nauthiz
2007-Jun-19, 10:34 PM
In reference to the BA story: Do black holes really exist? (http://www.badastronomy.com/bablog/2007/06/19/news-do-black-holes-really-exist/)

I took a look at the Wikipedia entry (http://en.wikipedia.org/wiki/Gravitational_time_dilation) on gravitational time dilation, and I think I can see why the event horizon can never form, or at least how it's impossible for an object falling in to ever reach it. That vertical asymptote at the Schwarzschild radius is easy enough to see in the equations.

I still have two questions, though:
1. The article refers to this as a paradox - what makes it paradoxical?
2. How is it that people thought quantum mechanics might circumvent this apparent problem?

John Mendenhall
2007-Jun-20, 04:22 PM
In reference to the BA story: Do black holes really exist? (http://www.badastronomy.com/bablog/2007/06/19/news-do-black-holes-really-exist/)

(snip)
2. How is it that people thought quantum mechanics might circumvent this apparent problem?

In quantum mechanics,there is a non-zero probablity that the particle may appear on the inside of the event horizon. See tunnel diodes and Hawking radiation.

nauthiz
2007-Jun-20, 05:15 PM
Ah, thanks. I think I got it. I took a look at the paper linked in the article last night, and I think I see where they say why this doesn't actually happen. From the bottom of page 4:


The width of the Gaussian wave packet remains fixed in the u coordinate while it shrinks in the R coordinate via the relation dR = Bdu which follows from Eq. 36. This fact is of great importance, since if the wave packet remained of constant size in R coordinates, it might cross the event horizon in finite time.

That said, I got a bit lost in the jargon so I don't really understand what the u and R coordinates are. I'm assuming R is the Schwarzschild radial coordinate. Is u the spatial coordinate? I gather that Schwarzschild coordinates don't give an accurate representation of distances in space.

(Then again, I also gather that I've found myself way out to sea without a map.)

kzb
2007-Jun-20, 05:33 PM
If the event horizon never forms, how can you have Hawking radiation? Isn't the existence of an event horizon a prerequisite for the production of Hawking radiation?

nauthiz
2007-Jun-20, 07:08 PM
I believe you are correct. However, the existence of Hawking radiation is supposedly still a contentious issue, and there are other models for black hole radiation.

Disinfo Agent
2007-Jun-21, 05:22 PM
I took a look at the Wikipedia entry (http://en.wikipedia.org/wiki/Gravitational_time_dilation) on gravitational time dilation, and I think I can see why the event horizon can never form, or at least how it's impossible for an object falling in to ever reach it. That vertical asymptote at the Schwarzschild radius is easy enough to see in the equations.The singularity at the Schwarzschild radius is removable. With a different choice of coordinates, you can get rid of it.

The time it takes for the horizon to form depends on the reference frame of the observer. For some observers (someone watching from afar), this time is infinite, but for others (e.g. an object plunging into the black-hole-to-be) it is finite. Anyway, this has been the traditional view; the article says otherwise.

nauthiz
2007-Jun-21, 06:54 PM
The singularity at the Schwarzschild radius is removable. With a different choice of coordinates, you can get rid of it.

I've heard that, but that just raises a bunch more questions for me:

What other coordinate systems are used, and what do the equations look like when we use them?

The singularity at the Schwarzschild radius doesn't look like a simple coordinate singularity where everything looks fine except that you've got this spot that's undefined. I'm certainly not mathematician, but it seems odd to me that a whole vertical asymptote could be a mere artifact of the notation.

What happens at the Schwarzschild radius when you're using other coordinate systems?

And wouldn't it be a worse paradox if a mere change in coordinate system gives results that contradict each other?

Disinfo Agent
2007-Jun-21, 08:34 PM
I've heard that, but that just raises a bunch more questions for me:

What other coordinate systems are used, and what do the equations look like when we use them?The Kruskal-Szekeres coordinates (http://en.wikipedia.org/wiki/Schwarzschild_metric#Singularities_and_black_holes ).


The singularity at the Schwarzschild radius doesn't look like a simple coordinate singularity where everything looks fine except that you've got this spot that's undefined. I'm certainly not mathematician, but it seems odd to me that a whole vertical asymptote could be a mere artifact of the notation.It can. Mathematically, the same function can be written with many different formulas. Sometimes, a particular formula creates problems at some points (like singularities), but those problems disappear when you find another formula that gives the same results everywhere except at that point. These are just mathematical technicalities.


And wouldn't it be a worse paradox if a mere change in coordinate system gives results that contradict each other?You don't really have two different results. It's just that, at the Schwarzschild radius, one formula (the Schwarzschild coordinates) happens to not be applicable; it does not produce any meaningful result. But other formulas do produce a meaningful, and unique result. This is a kind of singularity, but one that is just a mathematical artifact of the formulas we use. It doesn't mean anything physically, since it can be completely eliminated by switching to another formula (like the Kruskal-Szekeres coordinates).


What happens at the Schwarzschild radius when you're using other coordinate systems?
This region was isolated from the rest of the universe by a place where Schwarzschild's coordinates blew up, though nothing was wrong with the curvature of spacetime there. (This was called the Schwarzschild radius. Later, other coordinate systems were discovered in which the blow-up didn't happen; it was an artifact of the coordinates, a little like the problem of defining the longitude of the North Pole. The physically important thing about the Schwarzschild radius was not the coordinate problem, but the fact that within it the direction into the hole became a direction in time.)This quote is from the Usenet Physics FAQ -- What is a black hole, really? (http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_holes.html) (much more knowledgeable about black holes than I. ;)) See also the Black Holes FAQ (http://cosmology.berkeley.edu/Education/BHfaq.html).

nauthiz
2007-Jun-21, 08:57 PM
You don't really have two different results. It's just that, at the Schwarzschild radius, one formula (the Schwarzschild coordinates) happens to not be applicable; it does not produce any meaningful result. But other formulas do produce a meaningful, and unique result. This is a kind of singularity, but one that is just a mathematical artifact of the formulas we use. It doesn't mean anything physically, since it can be completely eliminated by switching to another formula (like the Kruskal-Szekeres coordinates).

Woah, I think that just blew my mind. In that case, how do the two formulas formulas differ at Rs + ε?

(Nobyd has to answer that, I realize these pedantic little questions are probably a PITA.

:think: BAUT needs a "teach me math" subforum. )

nauthiz
2007-Jun-21, 09:22 PM
:doh: Wait, I think I got it. It's that the numbers coming out of the equations using Schwarzschild coordinates won't even look like the numbers you get using a different coordinate system, right?

Disinfo Agent
2007-Jun-21, 09:26 PM
Woah, I think that just blew my mind. In that case, how do the two formulas formulas differ at Rs + e?For any nonzero epsilon, they give equivalent results. You can transform back and forth between coordinate systems, and get consistent answers. It's only when epsilon is precisely 0 that one of them has a nervous breakdown -- but the others hang on just fine. :)

The thing to keep in mind, as John Baez says, is that spacetime curvature always has perfectly decent, finite values, no matter how close you get to the Schwarzschild radius. It's only when you approach the point singularity at the centre of the black hole that curvature goes to infinity.

nauthiz
2007-Jun-21, 09:43 PM
Thanks for all the help, Disinfo Agent. I think I'm beginning see how it works, though I might have to figure out a way to draw myself a picture or something.

Gah, how come Penrose never used one of those lovely diagrams of his to explain this in one of his books?