Does a contracting universe destroy its black holes?

  • #1
fhenryco
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I found related questions being debated on the web so i'm not sure wether the question is closed.
The following simple reasoning seems to imply that indeed the contracting universe is able to destroy its blackholes but what's wrong with it ?:
The black hole solution is usually computed outside the source where a static vacuum is assumed. But this assumption is not valid anymore if there is no actual vacuum outside the source but if the source is rather described as a fluctuation over an evolving homogeneous fluid. If this fluid density and pressure increases with time as in a contracting universe scenario , it might increase faster than the density of the black hole (total mass divided by volume inside Schwarzschild sphere) in which case the density contrast associated to the blackhole might decrease until destroying the blackhole before it is actually formed (for an outside observer a BH is formed in an infinite time so the universe crunch will occur before, for the one falling into BH and near to cross it's horizon, time of the whole outside universe accelerates and again the universe crunch occurs first).

If this is true it would mean that in a fast contracting universe, only linear fluctuations beyond the horizon can still grow (well known result), but linear fluctuations below the Horizon probably decay because at some point diffusion is expected to win against gravity when the universe reaches high pressures (even if DM is only weakly interacting), also non linear fluctuations could only grow until the cosmic pressure (for weakly interacting DM) and density (same argument as for BH) starts to win the battle and then decay , and at last even BH would be destroyed...

Eventually when we say that in a bouncing universe cosmology we in general expect a very inhomogeneous universe just following the bounce (except in epkyrotic scenario) , we are then only talking about beyond horizon fluctuations that have grown before they left the horizon at a time the univers was cold and had low density, and have subsequently also grown beyond the horizon.
 
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  • #2
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In a contracting universe that ends in a Big Crunch, it is impossible to have true black holes. A black hole is defined as a region of spacetime that cannot send light signals to future null infinity; but in a contracting universe that ends in a Big Crunch there is no future null infinity.
 
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  • #3
fhenryco
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your answer looks a bit like a more rigorous formulation of what i was trying to say... but i'm not sure
 
  • #4
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your answer looks a bit like a more rigorous formulation of what i was trying to say
No, I'm just stating a much simpler fact that answers the basic question. But the title question of this thread and the vast majority of your OP are based on the false premise that I rebutted in post #2; so what I said is not a more rigorous formulation of what you said, it's a quick explanation of why what you said is not even applicable because it's based on a false premise.
 
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  • #5
fhenryco
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The first time i hear about your definition: a black hole is most often merely described as a region of spacetime that cannot send light signals at all...But such region would take an infinite time to form from the point of view of an external observer, and your definition seems to take this into account.

but now what about my question in a bouncing cosmology (instead of big crunch) in which the universe reaches a greater density than that of the BH before bouncing : shouldnt the BH disappear in this case?
 
  • #6
Ibix
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The first time i hear about your definition: a black hole is most often merely described as a region of spacetime that cannot send light signals at all...
That isn't enough, unfortunately. Every event cannot send light signals to events outside its future light cone, but that doesn't mean a light cone is an event horizon. Wald provides a few more attempted definitions before noting that the only workable definition involves being able to send light signals to infinity, since that's the only way to make the destination invariant (more or less - there's more set theory in the way he says it). But we can only define "at infinity" in a rigorous sense in asymptotically flat spacetimes, which FLRW spacetimes are not. So black hole horizons in an FLRW spacetime are only apparent horizons.
but now what about my question in a bouncing cosmology (instead of big crunch) in which the universe reaches a greater density than that of the BH before bouncing : shouldnt the BH disappear in this case?
Peter's point is that there aren't really any black holes in this case. There are things that are (on human scales) functionally identical, but since they don't actually have event horizons there isn't an immediate conceptual problem with what happens when density outside the hole rises. I don't know what ghe consequences would be. I believe there's a metric called the Mcvitie metric that is a Schwarzschild black hole embedded in an FLRW universe. That might yield some insight.
 
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  • #7
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a black hole is most often merely described as a region of spacetime that cannot send light signals at all
No, it's not. It is always possible to send light signals from any event in spacetime.

uch region would take an infinite time to form from the point of view of an external observer
No, it wouldn't. This is a common misconception that has been discussed in multiple previous threads here at PF.

what about my question in a bouncing cosmology (instead of big crunch) in which the universe reaches a greater density than that of the BH before bouncing
There is no such thing as "a greater density than that of a BH." A black hole is vacuum; it has no density.

In any case, the statement I made in post #2, that there is no future null infinity, applies to bouncing cosmologies as well as Big Crunch cosmologies.
 
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  • #8
PeroK
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  • #9
fhenryco
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No, it's not. It is always possible to send light signals from any event in spacetime.
If you really mean that light can escape the horizon then almost all i read almost everywhere about the BH is wrong... probably because people want to present it in a pedagogical way and neglect rigor...
No, it wouldn't. This is a common misconception that has been discussed in multiple previous threads here at PF.
can you recommand one of these threads ?
There is no such thing as "a greater density than that of a BH." A black hole is vacuum; it has no density.
if a black hole has a mass M and a Volume enclosed within the Scharzschild spĥere, why can't i divide the two to get what i could call the mean density of the BH ... again i see such density numbers computed almost everywhere included in books about BH that emphasize that BH have very different densities depending on they being stellar BH or Giant BH or micro Primordial BH.
In any case, the statement I made in post #2, that there is no future null infinity, applies to bouncing cosmologies as well as Big Crunch cosmologies.
OK , as far as i understand even an FRW cosmology has no future null infinity if it is not asymptotically flat. What about my question in the case of a bouncing cosmology with k=0 (flat background)?
 
  • #10
Orodruin
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if a black hole has a mass M and a Volume enclosed within the Scharzschild spĥere
Because what you refer to as ”volume of the Schwarzschild sphere” is not what you think it is due to the curvature of spacetime. You cannot really think of a sphere in the Euclidean sense.

The Schwarzschild solution itself assumes that there is vacuum everywhere (the mass M is rather a parameter in the solution - it does not correspond to a density anywhere).
 
  • #11
fhenryco
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Because what you refer to as ”volume of the Schwarzschild sphere” is not what you think it is due to the curvature of spacetime. You cannot really think of a sphere in the Euclidean sense.

The Schwarzschild solution itself assumes that there is vacuum everywhere (the mass M is rather a parameter in the solution - it does not correspond to a density anywhere).
Right but people need to have an idea of a volume being cut out from the rest of the universe by the presence of the BH so extrapolating the euclidean notions that we have far enough from the BH to the BH itself is may be not so useless...may be all the more since there is negligible curvature near the horizon .
 
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  • #12
Orodruin
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Right but people need to have an idea of a volume being cut out from the rest of the universe by the presence of the BH
Why? Unless they want to fool themselves that they have understood anything about how a black hole works. I’d argue that it is precisely this kind of nonsense statement that will act to prohibit people from actually understanding.


so extrapolating the euclidean notions that we have far enough from the BH to the BH itself is may be not so useless...
It is utterly useless as it gives an impression of some sort of understanding where none exists.
 
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  • #13
fhenryco
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What i'm trying to understand is the difference between my sentence
"such region (BH) would take an infinite time to form from the point of view of an external observer"
and Weinberg sentence p347 in Gravitation and cosmology:
"The collapse to the Schwarzschild radius therefore appears to the outside observer to take an infinite time"
 
  • #14
PeroK
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and Weinberg sentence p347 in Gravitation and cosmology:
"The collapse to the Schwarzschild radius therefore appears to the outside observer to take an infinite time"
I'd like to see the context of that statement.
 
  • #15
jbriggs444
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I'd like to see the context of that statement.
Looks like he is talking about what an outside observer "sees", not about what coordinates such an observer would ascribe or what simultaneity convention would be used.

The outside observer "at infinity", looking through a telescope will never see photographic evidence of the completion of the formation of the black hole from the original collapsing body.

I found a text format archive here:
https://archive.org/stream/WeinbergS.GravitationAndCosmology..PrinciplesAndApplicationsOfTheGeneralTheoryOf/Weinberg+S.+Gravitation+and+cosmology..+principles+and+applications+of+the+general+theory+of+relativity+%28Wiley%2C+1972%29%28ISBN+0471925675%29%28685s%29_djvu.txt said:
Now w^e return to the problem of calculating the behavior of light signals
emitted from the surface of the collapsing sphere. A light signal emitted in a
radial direction at a standard time l will have drjdt given by Eq. (11.9.27) and the
condition dr = 0, so it will arrive at a distant point r at a time


r c. -


V = i +


2 MG\~


dr


(11.9.39)




9 Gravitational Collapse


347


The most striking consequence of Eqs. (11.9.39) and (11.9.34) is that both i and
V approach infinity when the radius (11.9.33) of the sphere approaches the
Schwarzschild radius 2 GM, that is, when

R(t) -► = lea 2 (11.9.40)

a

The collapse to the Schwarzschild radius therefore appears to an outside observer to
take an infinite time
, and the collapse to R = 0 is utterly unobservable from outside.

Although the collapsing sphere does not suddenly disappear, it does fade out
of sight, because light from its surface is subject to an increasing red shift. The
proper time for a light source on the sphere’s surface is just the comoving time t ,
so the comoving time interval between emission of wave crests at the surface
equals the natural wavelength X 0 that would be emitted by the source in the
absence of gravitation. The standard time interval dt' between arrivals of wave
crests at 7 is the observed wavelength X' ; thus the fractional change of wavelength
is [...]
 
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  • #16
Ibix
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If you really mean that light can escape the horizon then almost all i read almost everywhere about the BH is wrong...
What you said was that a black hole was a region of spacetime that cannot send light signals at all. Nearby observers inside a black hole can happily communicate by exchanging light signals, although they have limited time to do so. You may have meant to add "...to its exterior" to your definition, but you didn't. And, as noted, the formal definition requires an asymptotically flat spacetime.
What i'm trying to understand is the difference between my sentence
"such region (BH) would take an infinite time to form from the point of view of an external observer"
and Weinberg sentence p347 in Gravitation and cosmology:
"The collapse to the Schwarzschild radius therefore appears to the outside observer to take an infinite time"
I believe Weinberg is using "appears" in the literal sense of "receives light signals from". That does take infinite time, but that isn't really the point here. There is a finite time before which I can send a radar pulse and will get a return off the infalling matter, athough I may have to wait millenia or longer. After that, though, the event "infalling matter crosses the horizon" is out of my causal future and I am free to say it is in my past. Whether I do or not depends on my choice of simultaneity criterion.

So your statement is stronger than Weinberg's, or at least can be read as such, particularly in the context of a discussion on the formation and fate of black holes.
 
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  • #17
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If you really mean that light can escape the horizon
I said no such thing. Light can be emitted inside the horizon without escaping the horizon. So a black hole is not "a region of spacetime that cannot send light signals at all", which was the incorrect claim you made that I was responding to.

if a black hole has a mass M and a Volume enclosed within the Scharzschild spĥere
Which it doesn't. The region inside the event horizon does not have a well-defined volume V and does not enclose a mass M. The black hole spacetime is vacuum; the "mass" M is a geometric property of the spacetime and does not mean there is matter inside the hole.

people need to have an idea of a volume being cut out from the rest of the universe by the presence of the BH
No, they don't, since this idea is incorrect. The region of spacetime inside the horizon is not "cut out from the rest of the universe". It is simply unable to send light signals outside the horizon. Spacetime is still continuous across the horizon, and things can continue to fall into the hole.

extrapolating the euclidean notions that we have far enough from the BH to the BH itself is may be not so useless
Space is not Euclidean even outside the event horizon. Also, the spacetime inside the horizon has some very different properties from the spacetime outside the horizon, so the "extrapolation" you refer to, even if done using the correct properties of spacetime outside the horizon, does not work.
 
  • #18
fhenryco
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To summarize: In weinberg chapter 9 entitled gravitationnal collapse and badly reproduced in jbrigs444 post, he makes the simplifying assumption of negligible pressure to study the spherical collapse of dust thus acted on by purely gravitational force so that it falls freely and can be considered a physical basis for a comoving coordinate system in which unsurprisinggly the metric solution turns out to be the same as an FRW metric (so not only isotropic but also homogeneous) except that R(t) is not the scale factor but something like the radius of the star which vanishes in a finite time T: so in this comoving coordinate system "we see a fluid sphere collapsing from initial rest to a state of infinite density in a finite time T"

But outside the sphere in vacuum the metric according the Birkhoff theorem must be static and can be described by a Schwarzschild metric. So the next step is to match internal and external solutions which can be done by first converting the internal solution into standard Schwarzschild coordinates.

When the internal metric is expressed in standard coordinates it turns out that in standard time it takes an infinite time for the sphere to reach the Schwarzschild radius. To this infinite time we can add another infinite time for a light ray emitted radially from near the Schwarzschild sphere to reach an external observer when the emission point converges to the Schwarzschild radius. So there are 2 infinities (Eq 11.9.39) : it's not only the duration of the light trip which is infinite in standard coordinate but also the collapse to the Schwarzschild radius itself.
 
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  • #19
Ibix
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When the internal metric is expressed in standard coordinates it turns out that in standard time it takes an infinite time for the sphere to reach the Schwarzschild radius.
That's because the time coordinate being used is a radar-based one (or equivalent to one), and you can't establish a radar time for something you can't (even in principle) get an echo off. But you are not obligated to use these coordinates! So this is not "the" point of view of a distant observer except in the literal sense that it reflects what he sees, corrected by radar time.
 
  • #20
PeroK
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When the internal metric is expressed in standard coordinates it turns out that in standard time it takes an infinite time ...
Standard time is not the only option.
 
  • #21
fhenryco
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So a black hole is not "a region of spacetime that cannot send light signals at all"
A region of spacetime that cannot send light signals outside itself (outside the horizon) is of course what i meant from the begining !
 
  • #22
fhenryco
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Standard time is not the only option.
but curiously it's the time coordinate Weinberg uses to estimate at which time the external observer will receive the signal so he is apparantly assuming that this is also the time of distant observer clocks...
 
  • #23
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outside the sphere in vacuum the metric according the Birkhoff theorem must be static and can be described by a Schwarzschild metric
This is not quite correct. The correct statement is that the Birkhoff theorem shows that there must be a fourth Killing vector field in a vacuum spherically symmetric spacetime, in addition to the three that are present because of spherical symmetry. This fourth KVF is hypersurface orthogonal, but it is only timelike outside the horizon; so the Birkhoff theorem only implies that the spacetime is static (i.e., has a timelike Killing vector field) outside the horizon.
 
  • #24
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A region of spacetime that cannot send light signals outside itself (outside the horizon) is of course what i meant from the begining !
That might be what you meant, but it isn't what you said. "Cannot send light signals at all" is not the same as "cannot send light signals outside itself".
 
  • #25
Ibix
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but curiously it's the time coordinate Weinberg uses to estimate at which time the external observer will receive the signal so he is apparantly assuming that this is also the time of distant observer clocks...
You are missing the point. The coordinate system is one way of defining what you mean by "at the same time as" something that happens at a distance. The method Weinberg uses doesn't assign a time coordinate at the event horizon, but that doesn't mean other such systems cannot.

To put it another way, Weinberg says (correctly) that his choice of coordinate system cannot assign a time to a horizon crossing event, but not that it is impossible to assign a time to a horizon crossing event.
 
  • #26
fhenryco
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That might be what you meant, but it isn't what you said. "Cannot send light signals at all" is not the same as "cannot send light signals outside itself".
it's what i said because it was hard to believe that given the context someone would "understand" it literally
 
  • #27
fhenryco
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This is not quite correct. The correct statement is that the Birkhoff theorem shows that there must be a fourth Killing vector field in a vacuum spherically symmetric spacetime, in addition to the three that are present because of spherical symmetry. This fourth KVF is hypersurface orthogonal, but it is only timelike outside the horizon; so the Birkhoff theorem only implies that the spacetime is static (i.e., has a timelike Killing vector field) outside the horizon.
I avoided the blabla about killing vectors as did Weinberg. I notice that Weinberg again was not quite correct...you should apply for the Nobel prize !
 
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  • #28
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it's what i said because it was hard to believe that given the context someone would "understand" it literally
You might find it hard to believe, but my response should have led you to reconsider.

I avoided the blabla about killing vectors as did Weinberg
Killing vector fields are not "blabla". Either you want to understand the correct meanings of the terms you use, or you don't.
 
  • #30
DrClaude
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The original question has been answered, so the thread will stay closed.

Thanks to all that participated.
 

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