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the information paradox


joshuam168

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ah yes...the tiresome attempt at trying to make one look foolish.

 

I think I did a pretty good job. You should hope that you're never one of my students :)

 

You really don't want to see the full display, and I don't want to compose it. It ought to be clear why I said that we logarithmically approach the horizon... but also that "we get" infinitesimally close in a reasonable coordinate time.

 

Yes, so the time scales as [math]r_0^{3/2}[/math]. So the time to crossing the horizon goes to infinity only as [math]r_0[/math] goes to infinity.

 

No?

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NO! The entire first term becomes insignificant compared to the log, so basically you have to figure the limit with the two terms of the log argument involving [math]r_o[/math] forming a constant other than 8. I tried to warn you, this'll make you stupid. I won't tell you how long I trashed yesterday, and we'll give you credit for lunchtime, also. Basically the RHS becomes: [math] -2m \log K \frac{\sqrt r -\sqrt{2m}}{\sqrt r + \sqrt{2m}}[/math]. Last year I enjoyed the following exchange with Hal Puthoff:..........(NA) You mean someone else dares talk sense? I'm still asking about the geometric stretching, basically the distinction between radial and transverse.

----- Original Message -----

From: Puthoff@aol.com

To: singularities@clab.net

Sent: Wednesday, October 18, 2006 6:21 PM

Subject: Re: isotropics

 

 

In a message dated 10/18/2006 6:32:38 P.M. Central Daylight Time, singularities@clab.net writes:

(NA) As time crunches to a crawl even in your dark gray singularity we can say things are not happening at our clocking.

(HP) Absolutely.

(NA)It seems to me immense mistakes have been made here, by people talking about blithely going through a horizon, etc, not that you say this in particular.

(HP)I agree. Another way of saying it. As the velocity of light seen from our frame drops to zero (approaching an event horizon) we (in our frame) we'll never see a particle go thru the event horizon, since its velocity cannot exceed the local velocity of light "down there," and that has dropped to zero. (For the exponential case, we will never see a particle reach a singularity at the origin.) So even though a rider on a particle would not note anything strange in his frame, he might notice that external clocks seem to be going so fast that eons will go by "out there" while he approaches the event horizon (or singularity for the exponential case) and therefore rightly assumes that the outsider would never see him make the progress he himself considers himself making.

 

Pretty straightforward, in a way!

 

Hal

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Norm---

 

I claim to be no GR expert, but just taking a limit [math]r_o\rightarrow\infty[/math] it is clear that the second term is constant. If r is constant, the r_o^3/2 dominates every time.

 

What is it I'm missing? x^3/2 grows faster than log x---I know because I just graphed it...

http://www.coolmath.com/graphit/index.html

 

Maybe we are talking past each other---the idea I have is that a particle starts at rest, at some distance r_o, and crosses the horizon (at r) at a time t. As r_o tends to infinity, the particle takes infinite time to fall into the horizon. You've probably fixed your frame at r_o, so this is the time it takes an observer to watch the particle falling in.

 

Also, you're failing to take into account gravitational backreaction---that is, because the infalling particle has a mass, then the mass of the black hole increases, which makes the radius bigger. So, for a Schwarzchild black hole, we have r = M. For a particle mass m, the black hole's new radius is r + r' = M + m. So when the particle is a distance OUTSIDE the horizon (r'-r), the horizon size INCREASES, absorbing the particle.

 

This is a subtle point, and I don't know if it's generally addressed. My guess is that most people assume that the infalling particle is massless, or something. It's all pretty confusing to me, actually.

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The position of the infalling object is described by coordinate r. What happens as this approaches the event horizon? Correct, here I am not dealing with any back reaction; this is what is meant by describing an infalling "small test particle or object". When we allow ourselves to feel confused we are in danger of learning something.

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Ahh yes... I was confused about what r and r_0 meant.

 

I believe you know, pursuant to you having derived the above equation correctly, of course :)

 

So when you ignore the gravitaitonal backreaction, I will agree that it takes an infinite amount of time for a test particle to fall in to a static, unchargedm non-expanding black hole.

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Cool, now we can get to duh fizziks. With each passing time constant unit measured by a somewhat distant observer, the coordinate r by which they measure the infalling "test" is reduced by another order of magnitude, so in not too long it seems they are very close but it is an exponentially stretched-out deal. Radius is a funky marker if you don't realize what's going on here. One can relate the differential change in radius by the ratio of the metric terms evaluated at the two locales, the outer observer and the infaller. You must also acknowledge the stretching of time. All this can only be stated differentially, and there is singularity at the horizon. You must be careful, these things really do totally suck. The first solution I wrote is completely general, I think.

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Light also seems to slow approaching an event horizon. Remember it is observed from outside going to infinite redshift, meaning it is becoming not observable. You can only describe crossing a horizon if you choose the reference frame there. Yes, truly compared to the far observer, this never happens in a finite time, and you have to deal with cosmologic questions of what ever gets to happen. The point about duh fizziks is that we try to do quantum mechanics on a horizon, and this is beyond me now. Bogoliubov is one name I associate here.

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