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How can the density of a region of space go from finite density to infinite when there are no numbers larger than any Aleph0 number but smaller than any Aleph1 number (no decimal point in front of it, of course)? Aren't Planck volumes and strings designed to sidestep infinities?

My point there, stated differently, is how can the density go from finite to infinite when there is a 'no-number gap' between finite and infinite, with Cantor losing his mind contemplating that gap? And did he not prove that none was constructible/possible?

The entropy of the visible universe is $\sim (10^{122})^2$, if I'm not too far off the mark. This is a stupendous number but no closer to infinity than any other integer or real. And there is no room in it to tuck a singularity away.

In simpler terms. Our BH has a finite mass. If it has a region of infinite density, that region must be infinitesimal. But the Planck length is the lower limit on the size of regions of space. There are, therefore, no points in space, no infinitesimals, only punctoids, my term of convenience, the utility of which may become obvious in future posts.And no infinities.

stafusa
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george lastrapes
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    Can you clarify your thinking: are you asking where does the infinite density of a BH lie in the hierarchies of mathematical infinities? – Daddy Kropotkin Dec 29 '18 at 22:36
  • "Worauf man kann sprechen nicht, Darauf muss man schweigen." But I will, anyway--1)exact center of a nonrotating BH? 2) Distributed over the surface? 3) There is no singularity, nothing in the universe is infinite, per the Planck units, and as string theory seems to suggest. When a theory predicts an infinity, that is a cry for help. – george lastrapes Dec 29 '18 at 23:14
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    In addition to the answers provided, let me point out that you are conflating two different concepts of infinity. The aleph numbers, $\aleph_0, \aleph_1, ...$, are cardinal infinities, which apply to counting discrete objects. However the infinite density is a continuum infinity, involving measurement over a continuum of values. These are distinct mathematical concepts. – Paul Sinclair Dec 30 '18 at 05:06
  • Thank you. Indeed, divergences tell us quite clearly where a theory fails. Conceivably, from a cosmological Copernican principle, the universe itself might be infinite. @PaulSinclair $\aleph_{1}$ is the continuum infinity, if by continuum you mean "set of real numbers." – Daddy Kropotkin Dec 30 '18 at 06:04
  • @N. Steinle: It must be Aleph-null: the infinite density- which I doubt the possibility of- must occur at the future infinity which Roger Penrose (blessings be on him) calls Scri+-- the future timelike infinity. But the Hawking radiation would seem to make that impossible. Once again, Relativity vs. QM. – george lastrapes Dec 30 '18 at 07:27
  • @Paul Sinclair: not following you. Cardinal infinity as opposed to ordinal infinity-- but Aleph-one is the cardinal of the reals, is it not, and to call it a continuum infinity adds nothing to the description of it. What can measurement over a continuum of Aleph-one values mean, if the answer is always Aleph-one? No Aleph-one is of a different size than any other. – george lastrapes Dec 30 '18 at 07:35
  • Possible duplicates: https://physics.stackexchange.com/q/18981/2451 , https://physics.stackexchange.com/q/75619/2451 and links therein. – Qmechanic Dec 30 '18 at 09:21
  • @N.Steinle Common misconception. In actuality the continuum infinity is $2^{\aleph_0}$, which cannot be proven equal or otherwise to $\aleph_1$. – wizzwizz4 Dec 30 '18 at 09:50
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    Possible duplicate of Why singularity in a black hole, and not just "very dense"? – Kyle Kanos Dec 30 '18 at 12:46
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    "how can the density go from finite to infinite when there is a 'no-number gap' between finite and infinite, with Cantor losing his mind contemplating that gap?" Don't get confused just because several distinct things have similar names. The infinities Cantor was contemplating have nothing to do with the infinities here. – knzhou Dec 30 '18 at 13:05
  • @N.Steinle - no, $2^{\aleph_0}$ is called the "power of the continuum", which as wizzwizz4 points out may or may not be $\aleph_1$, depending on whether one accepts the continuum hypotheses. But it is another cardinal infinity, not the continuum infinity, by which I mean $\infty$, which is not a cardinal infinity at all. Nor is it an ordinal infinity such as $\omega$. $\infty$ is defined as a compactification point of the real numbers as a topological space. Its properties and uses are topological in nature, and it is the infinity appropriate for this physical description. – Paul Sinclair Dec 30 '18 at 18:28
  • My apologies to everyone for getting into a strictly mathematical discussion on a physics forum. I had only wanted to indicate that the discussion of aleph numbers in the question was not mathematically appropriate, but my use of the term "continuum infinity" unfortunately muddied the waters, because of its similarity to another terminology. If you want to find more about various infinities, and how they are differ from each other, Math Exchange is the place to ask. – Paul Sinclair Dec 30 '18 at 18:36
  • @PaulSinclair What is the difference between the "continuum" and "cardinal" infinity? This article seems to be just confusing me more and more https://en.wikipedia.org/wiki/Cardinality_of_the_continuum So the statement $\aleph_{1} = 2^{\aleph_{0}}$ is equivalent to the continuum hypothesis and has not yet been proven? – Daddy Kropotkin Dec 30 '18 at 22:07
  • I have read statements of this sort: "The set R, of the reals, i.e. the continuum, is infinitely more dense than the set of rationals." But 'continuum density' is a phrase which doesn't seem to occur on the internet. I think the continuum hypothesis has been proven to be unprovable, by Cohen (forget first name), and hence one is free to go either way with it, with valid math resulting in either case. But I won't hold my breath waiting for set S such that Aleph-0<S<Aleph1 to be discovered by the Platonists or devised by the Formalists! – george lastrapes Dec 31 '18 at 00:29
  • "continuum" describes an interval in the real numbers. It is one-dimensional, connected, and locally compact. "Cardinals" are numbers used to count, regardless of how big the set you are counting is. "Cardinality of the continuum" is the cardinal that represents the number of elements in a continuum (all continuums have the same count). It is not related to $\infty$, which is simply an additional point we add to the end of the real numbers when we need a point there. It can be identified with $\aleph_0$ or $\omega$, but is really a different concept. – Paul Sinclair Dec 31 '18 at 02:09
  • You say, "The cardinality of the continuum [Aleph-1]......It can be identified with Aleph-0......" Surely this is a typo. – george lastrapes Dec 31 '18 at 06:33
  • @knzhou I'll try not to get confused. But it's difficult to think of infinity without thinking of Cantor. My original question provoked a comment by a physicist (so I assume) who said that physicists will say 'infinite' when they will admit if pressed that 'intractably large' is what they really mean. That being said, no one has given me a clue to how a physical quantity can go from finite to infinite. Is there a finite number so large that its successor is infinite? Perhaps this is not a question for physicists or mathematicians, but for logicians. – george lastrapes Dec 31 '18 at 20:34
  • The Continuum is the Reals, all of them Any subset-- say between 0 and 1-- is a continuum, of course, of the same cardinality. – george lastrapes Jan 01 '19 at 08:07

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Most physicists believe that the prediction of an infinite-density singularity (though note that for a Schwarzschild spacetime, the singularity is a moment in time, NOT a point in space) is a flaw in general relativity rather than a real physical thing that happens, and that at some density roughly around $m_{p}/\ell_{p}^{3}$, where $m_p$ is the planck mass, and $\ell_{p}$ is the planck length, quantum gravitational effects will take over and prevent a true singularity from forming. Obviously, without a working quantum theory of gravity, no one can know exactly how this happens, but this is the expectation.

Zo the Relativist
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  • The Schw singularity is a physical singularity at $r = 0 $ where $r$ is the radial coordinate in Schw coordinates. How is that a singularity in the time coordinate? – Daddy Kropotkin Dec 29 '18 at 20:51
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    @N.Steinle : because the Schwarzschild $r$ coordinate is timelike in the interior of the horizon,. – Zo the Relativist Dec 29 '18 at 22:05
  • Ah, thank you for clarifying that you specifically refer to the interior metric of the Schw BH. – Daddy Kropotkin Dec 29 '18 at 22:28
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    Sure, but it's not about what I meant. The singularity lives inside of the horizon. For a schwarzschild spacetime, this makes it a spacelike surface. Note that this is NOT true of a Kerr-Nordstrom spacetime. – Zo the Relativist Dec 30 '18 at 14:22
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Physics uses mathematical models of the real world. The best models are simpler, more accurate, and/or have broader scope that other models. General relativity is currently the best (simplest, most accurate, broadest scope) well-tested model that we have for describing gravitational phenomena. However, any time any model predicts a "singularity" (or "infinite density", etc), that's a sign that we have exceeded the limits of that model's validity. That's just as true for general relativity as it is for any other model.

Even ignoring singularities, we have other good reasons to think that general relativity is only approximately valid and that it breaks down in some extreme circumstances that are currently beyond our ability to explore experimentally. The black hole information paradox is a famous example. The basic problem is that general relativity doesn't account for quantum physics, and one of the lessons of the Black Hole Information Paradox (when analyzed carefully) is that any way of reconciling general relativity with quantum physics will necessarily require some radical change(s) in our current understanding of nature.

General relativity is not expected to be a good approximation under the extreme conditions where it would predict a singularity, and the black hole information paradox gives us reasons to suspect that it might break down under even less extreme conditions. According to [1],

The black hole information paradox forces us into a strange situation: we must find a way to break the semiclassical approximation in a domain where no quantum gravity effects would normally be expected.

(In this excerpt, the "semiclassical approximation" is an approximation in which gravity is treated as a non-quantum thing and everything else is treated as quantum. That's the approximation we use today to describe everyday situations involving both gravity and quantum effects, like when individual atoms fall under the influence of earth's gravity. We don't need a theory of "quantum gravity" for that kind of thing, because the gravitational field itself is not exhibiting significant quantum behavior in that case.)

Page 2 in same paper summarizes the black hole information paradox like this:

...in any theory of gravity, it is hard to prevent the formation of black holes. Once we have a black hole, an explicit computation shows that the hole slowly radiates energy by a quantum mechanical process. But the details of this process are such that when the hole disappears, the radiation it leaves behind cannot be attributed any quantum state at all. This is a violation of quantum mechanics. Many years of effort could provide no clear resolution of this problem. The robustness of the paradox stems from the fact that it uses no details of the actual theory of quantum gravity. Thus one of our assumptions about low energy physics must be in error. This, in turn, implies that resolving the paradox should teach us something fundamentally new about the way that physics works.

Reference:

[1] Mathur (2012), "Black Holes and Beyond," http://arxiv.org/abs/1205.0776

Chiral Anomaly
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    From a distant observer's point of view, objects in free fall to a black hole, seem to brake asymptotically near the event horizon, even considering the irresistible action of a possible singularity at this place. This observer does not see these objects crossing the event horizon. For him there is no event horizon, no Hawking radiation, no paradox of information, no singularity, and no black hole, as Hawking admitted in 2014. It would be simpler if we accepted that the black hole has only one side, the outside. I think that every black hole can be only a gravitacional sphericall shell. – João Bosco Dec 30 '18 at 02:21
  • @JoãoBosco, the observer in free fall would have a very different opinion than the outside aysmptotic (static) observer. The falling guy will encounter the singularity. – Cham Dec 31 '18 at 22:08
  • @JoãoBosco, what you wrote isn't true. The falling observer will cross the event horizon without noticing it. In his falling frame, gravity appears to be non-existent (locally), because of the equivalence principle, until he reaches the central singularity where his timelike geodesic ends abruptly. This is a well known behavior that can be shown using the Schwarzschild metric. It's a basic thing in General Relativity. – Cham Jan 01 '19 at 03:16
  • @JoãoBosco, the horizon crossing and not-crossing events are observer dependant. In the static exterior frame, the falling dude is never crossing the event horizon, or it will take an infinite time to do so. In the falling reference frame, the time to cross the horizon is finite. It's an extreme case of relativistic time dilation. – Cham Jan 01 '19 at 23:22
  • Of course. Delete what you want to be deleted. – Cham Jan 02 '19 at 01:16
  • An observer outside a BH, approaching it, will observe the EH receding, and will continue to do so, even after she has crossed the EH as calculated by a distant- and 'safe'- observer who cannot actually observe that event. – george lastrapes Jan 02 '19 at 20:49
  • @georgelastrapes - Nobody, in free fall to a black hole observes or sees the approach of the event horizon. Someone who falls in the terrestrian (for example) gravitational field does not know when he crosses a certain surface corresponding to an especific orbit. The same happens when approaching a black hole, even more if we consider that it doesn't emit signs of its presence. Who falls into a gravitational field doesn't perceive the fall, because he thinks he is at rest. – João Bosco Jan 04 '19 at 02:36
  • to the infalling observer, the EH recedes, seems to be unreachable. For a while, The unfortunate observer. however, will approach the singularity, infinitely dense or not, and begin to be dismantled by the tidal forces. – george lastrapes Jan 10 '19 at 06:52
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    @JoãoBosco "It would be simpler if we accepted that the black hole has only one side, the outside. I think that every black hole can be only a gravitacional sphericall shell." - This is correct despite a widespread misconception. There is nothing inside a black hole. – safesphere Jan 11 '19 at 09:16
  • @Cham "In the falling reference frame, the time to cross the horizon is finite" - This is incorrect and a wide spread misconception. An infalling observer does not have a frame at the horizon. His worldline is not continuous, but is interrupted at the horizon. There is no valid frame of reference, in which anything ever crosses the horizon. – safesphere Jan 11 '19 at 09:20
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    @safesphere, your last comment is completely wrong. The infalling observer wordline extends down to the singularity, where it ends abruptly. There are several coordinates systems which shows this clearly. All this is shown in any textbook on General Relativity and is very standard. I don't know where you got the idea that the observer doesn't have a frame at the horizon, but this is clearly false and even against the equivalence principle. – Cham Jan 11 '19 at 14:21
  • @DanYand Just the Schwarzschild solution. Of course the geodesic is continuous mathematically, but the world line of the infalling observer is interrupted in the sense that he does not have a frame at the horizon. Thus there is no frame, in which anything ever crosses the horizon. "$\mathcal{N}$ [a family of observers] cannot be complemented in a continuous way by an observer meeting $\gamma$ [the observer] in $h$ [the horizon] (such an observer would have to move with the speed of light)" - See, for example, section 3 here: https://arxiv.org/pdf/0804.3619.pdf – safesphere Jan 12 '19 at 07:38
  • @Cham "this is shown in any textbook" - Yes, I did say the misconception was wide spread. Time stops at the horizon in any frame. You may argue, "not in the proper frame, because the speed of local time (a.k.a. the local speed of light) is always the same". It is... except when it is zero. Example. You have wristwatches on both hands. The speed of one according to the other is one second per second no matter how close you are to a BH (ignoring the size). At the horizon though you cannot make this measurement, because either watch no longer moves. Your time stops, in all frames of reference. – safesphere Jan 12 '19 at 08:06
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    @safesphere I looked the paper you cited. Page 7 says, "The second observer, $\gamma$, falls freely, i. e., his world line is a radial geodesic... The most important fact about $\gamma$ is that at some moment $\tau_h<\infty$ of its proper time it unavoidably meets the horizon. ... Once $\gamma$ enters $M_+$ it cannot cross the horizon back and inevitably terminates at the singularity..." Page 9 in section 3 says: "Thus in any orthonormal basis (i. e., in a proper reference system of any observer located in $h$) $\gamma$ crosses the horizon moving slower than light." – Chiral Anomaly Jan 12 '19 at 15:08
  • @safesphere Although the parts that I highlighted in boldface make the meaning clear, the author's choice of words is unfortunate. The author's wording is "it cannot cross the horizon back..." The context (boldface, etc) indicates that what the author means is that $\gamma$ cannot turn around and cross back out after it has crossed the horizon going inward. – Chiral Anomaly Jan 12 '19 at 15:13
  • @safesphere In the abstract, the author says "he will not cross the horizon at the speed of light". Unfortunate wording again! The author is only trying to dispel a misconception that $\gamma$ crosses the horizon at the speed of light. The author is not saying that $\gamma$ doesn't cross the horizon, and in fact section 3 says explicitly the opposite: that $\gamma$ does cross the horizon, but it does so moving slower than light. – Chiral Anomaly Jan 12 '19 at 15:18
  • @DanYand I did not refer to this paper or its author's opinion for anything other than one specific point that I have quoted - the infalling observer does not have a frame at the horizon. He proves this point clearly, but doesn't see its full consequences. For any physical observable, there must be an observer in a valid frame. Answer a simple question: in which frame goes an infalling object crosses the horizon? The logical answer is, in none. Crossing the horizon is not a physical observable. It does not happen. It's a persistent misconception based on a logical flaw of the "proper frame". – safesphere Jan 12 '19 at 16:39
  • @DanYand Please see my comment with the example of two wristwatches. The proper time is not always one second per second, but only when it moves. This condition is commonly overlooked. Once both your wristwatches stop, you no longer have a proper time. It doesn't have a speed or rate, it doesn't move in any frame. Crossing the horizon is physically equivalent reaching the speed of light by a massive object, a big no-no, yet no one seems to mind. My goal is not argue or prove my point. You are a pro and a great asset on this site. I'm just asking to give this a thought. Have a great weekend! – safesphere Jan 12 '19 at 17:00
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There is no singularity associated with a Black-hole. At the final nanosecond in the formation of a black hole, essentially all of the matter confined within the envelope of a developing Schwarzschild sphere (whatever its size) can no longer be transformed into particle kinetic-energy, i.e., particle momentum.

Special relativity does NOT allow an inertial system to exceed the speed of light; and in the interior of a collapsing star, particles colliding at speeds approaching that of light cannot absorb a further increase in the momentum produced by gravitational forces. At this point, gravitational energy is transformed directly into radiation (entropy) rather than particle momentum.

This critical event, representing a change-in-state from matter to radiation, is preceded by an exponential increase in the momentum of particles confined within the volume of a contracting star (or any object). As particle velocities approach the speed of light, and as distance and time between particle collisions approach zero, the energy-density of a collapsing object will reach a limit where the transformation of gravitational force can no longer be defined in terms of particle collisions.

As the distance between colliding particles gets smaller, quantum mechanical factors require that the uncertainty in particle momentum correspondingly gets larger. At a critical point in this combination of events, the collision-distance between particles has decreased to a nanometer range that corresponds to the frequency of particle collisions; and particle interactions can be expressed in terms of a series of discrete quantum-mechanical wave-functions (distance and time between particle collisions) that approach the restricting value of 'h' (the Planck constant), producing a potential, quantum-mechanical catastrophe.

In order to preserve thermodynamic continuity, the thermodynamics of the system must change; consequently, particle matter is transformed into radiant energy by means of quantum mechanical processes. Gravitational energy is now expressed as a function of the total radiation-energy distributed over the surface of the ensuing Schwarzschild sphere, and any additional energy impacting the Black-hole produces an expansion of the Schwarzschild radius, while maintaining a constant energy-density, and a constant boundary acceleration, corresponding to a constant (Unruh) temperature.

A clue to the transformation of kinetic energy to radiation is seen in the function: e^hf/KT (from the Planck "key" to the ultraviolet paradox). In this function, particle kinetic-energy, "KT", increases due to an increase in particle velocity and effective particle temperature; the frequency of particle collisions, represented by "hf", increases with particle density due to increased gravitational confinement within a developing Schwarzschild object (Black Hole). But temperature and frequency do not rise to infinity, as one might expect from the Planck relationship.

The formation of a Schwarzschild boundary is coincidental with a maximum particle acceleration and a maximum temperature. This critical event represents the maximum energy-density (rather than the maximum energy) permitted by nature. Temperatures cannot rise beyond this critical point. Instead, these variables now become Black-hole constants and are conserved in all Black-Holes, regardless of their size and total energy. Subsequent to the formation of a Black-hole, temperature, acceleration, gravity and energy-density remain constant at their maximum values...even as more energy is added and the Schwarzschild envelope grows correspondingly larger.

RobertO
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