Free falling into a black hole that evaporates by Hawking Radiation

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  • #26
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This is false, period. Please read the second paper I linked.
I doubt that. What did you find wrong in my argument? Did you read it or just the first few lines?
 
  • #27
PAllen
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I doubt that. What did you find wrong in my argument? Did you read it or just the first few lines?
The SC metric is invalid for the problem (distant metric is Vaidya). Since your starting premise is wrong, there is nothing further worth reading.

However, you make further errors as well.

Your argument is essentially the same as the one refuted in the paper I linked - misuse of SC coordinates, and no model of rate of infall versus evaporation in the region around the horizon.
 
  • #28
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The SC metric is invalid for the problem (distant metric is Vaidya). Since your starting premise is wrong, there is nothing further worth reading.

However, you make further errors as well.

Your argument is essentially the same as the one refuted in the paper I linked - misuse of SC coordinates, and no model of rate of infall versus evaporation in the region around the horizon.
The paper you linked doesn't appear relevant to answer the question raised. They argue that "We find that, in any realistic collapse scenario, the backreaction effects do \emph{not} prevent the formation of the event horizon."

They simply say that backreaction doesn't inhibit formation. My analysis was independent of whether back reaction is present or not.

You seem to think that going to one of the Vaidya metrics is going to change the transit time to less than infinity. Can you demonstrate this?
 
  • #29
PAllen
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The paper you linked doesn't appear relevant to answer the question raised. They argue that "We find that, in any realistic collapse scenario, the backreaction effects do \emph{not} prevent the formation of the event horizon."

They simply say that backreaction doesn't inhibit formation. My analysis was independent of whether back reaction is present or not.

You seem to think that going to one of the Vaidya metrics is going to change the transit time to less than infinity. Can you demonstrate this?
The transit time you compute is not a coordinate independent quantity. It is just a label assigned at a distance (that can become singular in a purely coordinate sense, thus the transfinite nonsense). Like it or not, you need near horizon model of BH evaportion (which will not be SC metric and will violate energy conditions) in order to ask whether the infaller crosses the horizon in finite proper time. If they do, this is a coordinate independent fact. If they don't, that is also coordinate independent. But your calculation is conceptually irrelevant and invalid.
 
  • #30
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It's obvious where this is going.

I'm not going to look at every metric that comes down the pike and demonstrate an infinite vs. infinite transit time. I'm sure there are an infinitude of them. Going over to the WF metric and it's relatives doesn't make the problem go away, and it's still just classical.

To compare two clocks we really need them adjacent, as you seem to imply. Try it in the metric of your choice.

Find the literature demonstrating something falling in. Good luck with that.

See you later.
 
  • #31
PeterDonis
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Nothing falls into a classical black hole.
Incorrect. See below.

It's perfectly fine to use Schwarzschild coordinates on the open set bound by the event horizon for an external observer and the local coordinates of the infalling astronaut.
But it's not fine to then assume that only the open set bounded by the event horizon is accessible to the infalling astronaut, because that's all that's visible to the external observer.

In the coordinate system of an external observer, we need to integrate the elapsed time of an infalling test particle from an initial radial displacement to [tex]R_s + \delta[/tex] In nonstandard analysis this is [tex]R_s + dR[/tex] This amount of time is transfinite.
Show your work, please. The integral that gives elapsed proper time on an infalling trajectory for a finite ##r > r_s## to ##r = r_s## is finite; this is a standard textbook problem in GR.

But the evaporation time is finite.
I thought you were leaving out Hawking radiation. You should; you need to get the purely classical case right first, before adding quantum complications.

For the infalling astronaut, the boundary would not be crossed but recede before him.
Incorrect.

If we assume time symmetry, modulo CP, and there are time-like paths for matter to enter or null curves for light to enter, then there are time-like paths for matter to escape and null curves for light to escape
Incorrect. Time symmetry does not require this. It only requires that if there is a "black hole" solution to the field equations, in which objects can fall in but not escape, then there must also be a time-reversed "white hole" solution, in which objects can escape but not fall in. In other words, time symmetry is a property of the solution space of the field equations, not of single solutions considered in isolation.
 
  • #32
PAllen
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It's obvious where this is going.

I'm not going to look at every metric that comes down the pike and demonstrate an infinite vs. infinite transit time. I'm sure there are an infinitude of them. Going over to the WF metric and it's relatives doesn't make the problem go away, and it's still just classical.

To compare two clocks we really need them adjacent, as you seem to imply. Try it in the metric of your choice.

Find the literature demonstrating something falling in. Good luck with that.

See you later.
You are trying to draw global conclusions. You have globally inconsistent assumptions: that there is an absolute horizon defined is inconsistent with the statement that the BH evaportates - thus the horizon vanishes. To answer questions of such a scenario you need a metric that incorporates vanishing of an apparent horizon.

Maybe a simple analogy will show your fallacy. Consider the family of Rindler observers in Minkowski space. Use Rindler coordinates exactly as you've done for SC coordinates to argue that if something makes this horizon disappear for me at any finite time, then no one could have crossed the Rindler horizon because that would be at transfinite Rindler time. However, this is utter nonsense. As soon as you include a mechanism for horizon vanishing (e.g. you stop accelerating at some point), you find that at one moment you get the news of all the infalls that really happened; then the you get the news of the history of what was beyond the Rindler horizon.

This is very similar to what various semi-classical or classical analogs predict for an evaporting BH. Once the fact of horizon disappearance is incorporated into the geometry, you find that distant observers get information about all who crossed the apparent horizon 'in time' at the moment they get news of horizon disappearance.

[Edit: even more simply, SC coordinates and metric describe an eternal Bh, and static geometry. An evaporating BH sits in a geometry that is not stationary (let along static), and has no absolute horiozn (this follows from the formal definition of absolute horizon; the horizon is formally an apparent horizon). How can you think it is valid draw conclusions from the distant future using the first geometry about the second situation??!!]
 
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  • #33
PAllen
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I thought it would be worth summarizing the key differences between SC geometry and any classical or semi-classical model of an evaporating BH (that is assumed to evaporate completely)

0) SC geometry (or collapse glued to SC along a hypersurface)
--------------
- static (after collapse, at least)
- vacuum (thus energy conditions don't even apply in the late exterior)
- eternal absolute horizon
- (at least future) eternal BH

1) Any classical or semi-classical model of a BH that evaporates completely
--------------------------------
- neither static nor stationary
- not vacuum
- non-eternal horizon; light eventually escapes from the horizon itself
- energy conditions apply and are violated (negative energy must be taken to exist to shrink the BH)
-BH neither past eternal nor future eternal

Thus, whenever I read an argument that attempts to analyze an evaporating black hole using SC coordinates (or geometry in general), it reads to me like a mathematical argument that begins with:

Let us assume we can take 0=1

[Edit, a few corrections due to input from Peter and George, below]
 
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  • #34
PeterDonis
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An evaporating BH sits in a geometry that ... has no absolute horizon
This isn't quite correct; there are classical geometries describing evaporating black holes that have absolute horizons--the original geometry that Hawking suggested has one, for example; see the first Penrose diagram on the Wikipedia entry for the black hole information paradox:

http://en.wikipedia.org/wiki/Black_hole_information_paradox

The 45-degree line going up and to the right from the vertical r = 0 line to the right end of the singularity (the jagged line) is the absolute horizon.

Of course the current controversy over the information paradox is about whether this, or something like it, is the correct geometry to describe an evaporating BH; but since that controversy is not resolved, I don't think we can say definitively that an evaporating BH geometry has no absolute horizon. We don't know for sure at this point.
 
  • #35
PAllen
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This isn't quite correct; there are classical geometries describing evaporating black holes that have absolute horizons--the original geometry that Hawking suggested has one, for example; see the first Penrose diagram on the Wikipedia entry for the black hole information paradox:

http://en.wikipedia.org/wiki/Black_hole_information_paradox

The 45-degree line going up and to the right from the vertical r = 0 line to the right end of the singularity (the jagged line) is the absolute horizon.

Of course the current controversy over the information paradox is about whether this, or something like it, is the correct geometry to describe an evaporating BH; but since that controversy is not resolved, I don't think we can say definitively that an evaporating BH geometry has no absolute horizon. We don't know for sure at this point.
An absolute horizon must have no light from it escaping to future null infinity. For an evaporating BH, the light from the horizon does reach future null infinity. Therefore it is not, strictly speaking, an absolute horizon.
 
  • #36
PeterDonis
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An absolute horizon must have no light from it escaping to future null infinity.
No; an absolute horizon is the *boundary* of the causal past of future null infinity. That means it is the null surface composed of outgoing light rays that just barely reach future null infinity.

Another way to see this is to work from the other end, so to speak: if the causal past of future null infinity is not the entire spacetime, then a "black hole" is present: the black hole is the region of spacetime that is not in the causal past of future null infinity. The absolute horizon is then the boundary of the black hole region. In other words, if there is a black hole present (which there clearly is in the Penrose diagram I referred to), there must be an absolute horizon. The only way there can be no absolute horizon is if there is no black hole at all--i.e., if the entire spacetime is in the causal past of future null infinity.
 
  • #37
George Jones
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Expanding on what Peter wrote:

Let B be a subset of topological space M. Define the interior of B, Int(B), as the union of all open subsets of M that are subsets of B. Define the closure of B, Cl(B), as the intersection of all closed subsets of M that contain B. The boundary of B is Cl(B) - Int(B). B might contain all (iff B is closed), some (iff B is neither open nore closed), or none (iff B is open) of its boundary.

Here B is the black hole region of spacetime M, i.e., B is set of all events that are not connected to future null infinity by a future-directed causal (lighlike or timelike) path. The absolute (event) horizon is the boundary of B.

The only way for an absolute horizon not to exist is if there isn't a black hole. This is what Hawking now argues, but doesn't prove.
 
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  • #38
PAllen
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Expanding on what Peter wrote:

Let B be a subset of topological space M. Define the interior of B, Int(B), as the union of all open subsets of M that are subsets of B. Define the closure of B, Cl(B), as the intersection of all closed subsets of M that contain B. The boundary of B is Cl(B) - Int(B). B might contain all (iff B is closed), some (iff B is neither open nore closed), or none (iff B is open) of its boundary.

Here B is the black hole region of spacetime M, i.e., B is set of all events that are not connected to future null infinity by a future-directed causal (lighlike or timelike) path. The absolute (event) horizon is the boundary of B.

The only way for an absolute horizon not to exist is if there isn't a black hole. This is what Hawking now argues, but doesn't prove.
Ok, that's clear and sharp for me. The key is the interior world lines never escape (because they all reach a spacelike singularity; and the details of what evaporation means for the singularity is typically left out semi-classical treatment). However, then we must distinguish an evaporating horizon from an eternal horizon in that it no longer has the property that light from it never escapes. It is an absolute horizon, but no longer an eternally trapped surface! This would substantively change the behavior of an SC style time coordinate in the near horizon distant future.
 

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