Markus Hanke Posted August 18, 2022 Share Posted August 18, 2022 3 hours ago, Mitcher said: Just a layman way of saying it is outside the manifold. It’s outside the particular coordinate chart that happened to be used; that doesn’t mean the manifold itself is not smooth and continuous there. 3 hours ago, Mitcher said: According to Weyl, when M grows bigger than 2.5 solar masses then the proper mass is inverted. I don’t know what this means? There’s no such thing as negative proper mass. 3 hours ago, Mitcher said: Oppenheimer and Wheeler had not read it when they invented the singularity-based BH model many years later. The first solutions found in and around 1917 were exterior metrics, meaning they described spacetime in a vacuum, ie outside the central body. Oppenheimer and others later found solutions that describe the interior spacetime of bodies, ie spacetime inside the central body, and by extension a full metric that consistently encompasses both regions. Since, once certain limits are exceeded, it is physically impossible to prevent gravitational collapse, these must contain singularities. 3 hours ago, Mitcher said: Tolman and Oppenheimer only forgot about the preriquisite condition r > Rs (that’s what I mentionned as "corrupted"). There is no such condition once you use a metric that covers both interior and exterior spacetime, which is what they did. 3 hours ago, Mitcher said: According to Schwarschild, if r becomes smaller than Rs then the ds squared becomes negative, hence we are outside the hypersurface. When the sign on the squared line element inverts, then that means that time and space trade places - which is to say that ageing into the future becomes equivalent to a diminishing radial coordinate. In other words, there are no longer any stationary frames, and one cannot avoid falling towards the center. 3 hours ago, Mitcher said: IMO only a mathematical singularity can make sense, believing in a physical one is like believing flat oceans could have no borders. The issue here is that Einstein’s GR is a purely classical model of gravity, so it does not and cannot account for quantum effects. Within the framework of the classical model, the appearance of physical singularities is inevitable (this can be mathematically proven). However, the real world isn’t classical below a certain scale, so the current assumption is that quantum gravity will remove such singularities. We don’t have such a model yet, but it’s an era of active research. To remove singularities from classical GR, you can make a small modification to it that leads to a model called Einstein-Cartan gravity. This is free of singularities. However, this modification has other consequences, which to date we haven’t seen in the real world. But you are right of course in that one does not expect singularities to be actual objects that occur in the real world. They are artefacts of the model, and generally mean that it breaks down under the given set of circumstances; they are “physical” only within the context of that model. 1 Link to comment Share on other sites More sharing options...
joigus Posted August 19, 2022 Share Posted August 19, 2022 16 hours ago, Markus Hanke said: It’s outside the particular coordinate chart that happened to be used; that doesn’t mean the manifold itself is not smooth and continuous there. This is so important, and so often forgotten... Link to comment Share on other sites More sharing options...
Eise Posted August 19, 2022 Share Posted August 19, 2022 (edited) 18 hours ago, Markus Hanke said: Since, once certain limits are exceeded, it is physically impossible to prevent gravitational collapse, these must contain singularities. I assume you mean 'these must contain singularities according GR'. Which is the reason to say that, as is often said, 'GR breaks down' in the centre of a black hole. What I am wondering is how Oppenheimer & co could be so sure about the 'total' gravitational collapse, without knowing all ins and outs of the strong nuclear force. How could they be sure, that gravitation would even overcome neutron degeneration? I mean even now, knowing that neutrons are made of quarks, the Pauli exclusion principle is also valid for quarks, in the end, these are also fermions. Maybe a '(top/bottom?) quark star' is hiding immediately behind the event horizon? How can we exclude such a scenario? So I assume I am just asking 'which limits, and how do we know these limits?'. Edited August 19, 2022 by Eise Link to comment Share on other sites More sharing options...
Markus Hanke Posted August 19, 2022 Share Posted August 19, 2022 7 hours ago, Eise said: I assume you mean 'these must contain singularities according GR'. Which is the reason to say that, as is often said, 'GR breaks down' in the centre of a black hole. Yes, that’s what I meant. 7 hours ago, Eise said: What I am wondering is how Oppenheimer & co could be so sure about the 'total' gravitational collapse, without knowing all ins and outs of the strong nuclear force. Good question! What they did was to approximate the tightly packed neutrons in a neutron star as a special kind of gas, called a Fermi gas. The dynamics of this were fairly well worked out, which made it possible to derive a rough limit for when neutron degeneracy is able to resist gravitation. It’s called the Oppenheimer-Volkoff limit. If that limit is exceeded, gravity will be stronger than the degeneracy pressure. Oppenheimer/Volkoff did not use any results from QCD, which wasn’t fully developed until later. Nowadays we can speculate that there might also be a degeneracy state involving quarks, which then would also have a corresponding limit. This would lead to an astrophysical object called a quark star (purely hypothetical). However, it is safe to say that in this domain no classical approximations will suffice, this is where we need to use quantum gravity - which we don’t yet have. So we can’t say what happens when the quark degeneracy limit gets exceeded. 7 hours ago, Eise said: I mean even now, knowing that neutrons are made of quarks, the Pauli exclusion principle is also valid for quarks, in the end, these are also fermions. Yes, but if you keep collapsing the body, the pressures and energies will eventually get so high that the fundamental forces will re-unite into a GUT scenario - at which point the concept of ‘quark’ ceases to make sense. It is also not clear that the Pauli exclusion principle is meaningful in a scenario where quantum gravity plays a role. 7 hours ago, Eise said: Maybe a '(top/bottom?) quark star' is hiding immediately behind the event horizon? How can we exclude such a scenario? This is inconsistent with GR, because there can be no stationary frames beyond the horizon, due to the fundamental geometry of the spacetime there. So you can’t have stable objects of any kind. In other words, it doesn’t matter what specific mechanism you propose - so long as there is classical spacetime, a full collapse is inevitable (ref also singularity theorems). The only ways to avoid this is to either modify GR, or abandon classical (smooth, continuous) spacetime once certain limits are exceeded. 7 hours ago, Eise said: So I assume I am just asking 'which limits, and how do we know these limits?'. You can calculate them using the laws of QM and QFT: Chandrasekhar limit - electron degeneracy - white dwarfs Tolman-Oppenheimer-Volkoff limit - neutron degeneracy - neutron stars Quark degeneracy (don’t know if there’s a name for it) - quark stars (hypothetical) 1 Link to comment Share on other sites More sharing options...
Mitcher Posted August 19, 2022 Share Posted August 19, 2022 On 8/18/2022 at 9:18 PM, Markus Hanke said: It’s outside the particular coordinate chart that happened to be used; that doesn’t mean the manifold itself is not smooth and continuous there. The manifold can be smooth and continuous without necessarily passing through an event Horizon, it should be the throat circle of a Flamm's paraboloid according to Schwarschild's original metric. On 8/18/2022 at 9:18 PM, Markus Hanke said: I don’t know what this means? There’s no such thing as negative proper mass. There is a thing such as the theoretical apparent negative mass, at least theorized by Shakharov, Souriau, Kostant and others. It is supposed to appear when inverting time. On 8/18/2022 at 9:18 PM, Markus Hanke said: The first solutions found in and around 1917 were exterior metrics, meaning they described spacetime in a vacuum, ie outside the central body. Oppenheimer and others later found solutions that describe the interior spacetime of bodies, ie spacetime inside the central body In 1916 Schwarschild wrote a second paper which is less known, in which he wrote the metric inside the Sun. It was only traducted from german to english in 1999 if I remember right. On 8/18/2022 at 9:18 PM, Markus Hanke said: that means that time and space trade places - which is to say that ageing into the future becomes equivalent to a diminishing radial coordinate. In other words, there are no longer any stationary frames, and one cannot avoid falling towards the center. Spherical symmetry does not necessarily imply central symmetry. If it's impossible to penetrate into a BH then it's impossible to be crunched towards the center, be it by following time or space. Link to comment Share on other sites More sharing options...
MigL Posted August 19, 2022 Share Posted August 19, 2022 I remember reading how S Chandrasekhar first calculated the electron degeneracy limit, on his way from India by boat, to England, to study under Sir A Eddington. He considered confining an electron to a smaller and smaller box, such that its momentum becomes more and more indeterminate. An early use of the then new HUP, or QM applied to classical physics. At some point the box determines its position enough, that the velocity component of its momentum could be faster than light; a clearly unphysical situation. The electron can then absorb into a proton to become a neutron, which, for the same size confinement ( positional accuracy ), and momentum uncertainty, allows the velocity to be up to 2000 times smaller, as its mass is 2000 times larger. At least until neutron degeneracy is reached. Speculation about quark stars don't make much sense to me, in light of the previous arguments, as they involve going to a much lighter fermion, and unphysical velocities again. But I'm no expert on QCD, and have no idea whether these quarks would act as 'free' particles, or as a massive'combined' particle. Link to comment Share on other sites More sharing options...
Markus Hanke Posted August 19, 2022 Share Posted August 19, 2022 (edited) 28 minutes ago, Mitcher said: it should be the throat circle of a Flamm's paraboloid according to Schwarschild's original metric. If the embedding diagram terminates in a throat, then that means the coordinate chart isn’t continuous at that point. This doesn’t imply anything about spacetime itself. 28 minutes ago, Mitcher said: In 1916 Schwarschild wrote a second paper which is less known, in which he wrote the metric inside the Sun. It was only traducted from german to english in 1999 if I remember right. Ok, but then he would have realised that that spacetime is everywhere continuous and doesn’t terminate at any horizon surface. I don’t really understand the significance of these historical references, to be honest. We nowadays know of a large number of exact solutions to the field equations, including maximally extended metrics that cover the entirety of this particular spacetime, so we know in depth its complete geometry and topology - which is a lot more than was known back in early 1900s. Why do you keep referring back to the state of affairs a hundred years ago? Our state of knowledge and maths has moved on greatly since then. Remember also that metrics aren’t invented - they are derived solutions to the field equations. Edited August 19, 2022 by Markus Hanke Link to comment Share on other sites More sharing options...
Mitcher Posted August 19, 2022 Share Posted August 19, 2022 10 minutes ago, Markus Hanke said: If the embedding diagram terminates in a throat, then that means the coordinate chart isn’t continuous at that point. But that's the point, a throat does not terminate, after passing the vertex of a parabola it just continues. The inside of the throat is outside the manifold. 15 minutes ago, Markus Hanke said: 17 minutes ago, Markus Hanke said: Ok, but then he would have realised that that spacetime is everywhere continuous and doesn’t terminate at any horizon surface. I don’t really understand the significance of these historical references, to be honest. We nowadays know of a large number of exact solutions to the field equations, including maximally extended metrics that cover the entirety of this particular spacetime, so we know in depth its complete geometry and topology - which is a lot more than was known back in early 1900s. Why do you keep referring back to the state of affairs a hundred years ago? Our state of knowledge and maths has moved on greatly since then. He had realized that, I keep suggesting we completely misunderstood Schwarschild's equation I'am afraid. Link to comment Share on other sites More sharing options...
Markus Hanke Posted August 21, 2022 Share Posted August 21, 2022 On 8/20/2022 at 7:52 AM, Mitcher said: But that's the point, a throat does not terminate, after passing the vertex of a parabola it just continues. The inside of the throat is outside the manifold. I’m really confused now - could you try to rephrase for me what your main point is? I can’t really make sense of the progression of the last few posts. I should remind you again that an embedding diagram concerns a coordinate chart, which is a separate thing from the manifold itself. To say that any part of an embedding diagram - whether missing or not - is ‘outside the manifold’ is meaningless, since you can have regions that aren’t covered by that particular chart. On 8/20/2022 at 7:52 AM, Mitcher said: He had realized that, I keep suggesting we completely misunderstood Schwarschild's equation I'am afraid. Schwarzschild spacetime is the simplest and most straightforward solution to the Einstein equations - both its geometry and topology are well understood and have been studied ad nauseam by generations of physicists and mathematicians. Precisely which aspect of it do you think we are misunderstanding? Link to comment Share on other sites More sharing options...
Mitcher Posted August 21, 2022 Share Posted August 21, 2022 11 hours ago, Markus Hanke said: I’m really confused now - could you try to rephrase for me what your main point is? I can’t really make sense of the progression of the last few posts. I should remind you again that an embedding diagram concerns a coordinate chart, which is a separate thing from the manifold itself. To say that any part of an embedding diagram - whether missing or not - is ‘outside the manifold’ is meaningless, since you can have regions that aren’t covered by that particular chart. Schwarzschild spacetime is the simplest and most straightforward solution to the Einstein equations - both its geometry and topology are well understood and have been studied ad nauseam by generations of physicists and mathematicians. Precisely which aspect of it do you think we are misunderstanding? The parameter r Schwarschild used has been wrongly interpretated as a radial parameter from the start. If the surface forms a paraboloid of revolution it createss a hole that is not reachable excepted by jumping in a higher dimension, this is easy to understand, like if there is a hole in a sheet of paper it is possible to go on the other side of the surface but impossible to go inside the hole. I mentionned the work of Shakharov, Souriau, Kostant or Petit on this specific. Link to comment Share on other sites More sharing options...
joigus Posted August 22, 2022 Share Posted August 22, 2022 (edited) 15 hours ago, Mitcher said: The parameter r Schwarschild used has been wrongly interpretated as a radial parameter from the start. If the surface forms a paraboloid of revolution it createss a hole that is not reachable excepted by jumping in a higher dimension, this is easy to understand, like if there is a hole in a sheet of paper it is possible to go on the other side of the surface but impossible to go inside the hole. I mentionned the work of Shakharov, Souriau, Kostant or Petit on this specific. The coordinate r is not the radial coordinate. That much I, for one, will concede. For starters, Schwarzschild's r in the interior maps time, not space. But even outside it doesn't have to be, even though the temptation to call it so is very strong. If you want an argument that shows this very clearly, consider this: Scharzschild coordinates separate the angular factor so that the total solid angle gives 4*pi. Well, that cannot be. And the reason is that in a space with (positive/negative) spatial curvature, there must be an angular (deficit/excess). Of that, Scharzschild's coordinates tell us nothing. Here's the thing: Coordinates in GR are meaningless. Scharzschild's coordinates are only meaningful at spatial infinity. What you do in GR is pick out a set of coordinates that suits you to solve the equations, then do an analytic extension that smooths out all the singular points of the coordinate chart, then you identify the real singularities (infinite curvature). Finally, you discuss causality and the like in terms of the best set of coordinates(Kruskal-Skezeres, conformal, with Penrose diagrams, etc). Mind you, if your favourite coordinate chart that you used to solve the equations was "contaminated" with spurious singularities, you will have to introduce a singular change of charts that undoes the damage. It's not you, it's not me, it' generations of physicists that have been confused by the mirage of their coordinates charts. GR is a lanscape to tread very carefully. Edited August 22, 2022 by joigus correction 1 Link to comment Share on other sites More sharing options...
Eise Posted August 23, 2022 Share Posted August 23, 2022 Thanks, Markus, for your answer. On 8/19/2022 at 11:03 PM, Markus Hanke said: What they did was to approximate the tightly packed neutrons in a neutron star as a special kind of gas, called a Fermi gas. The dynamics of this were fairly well worked out, which made it possible to derive a rough limit for when neutron degeneracy is able to resist gravitation. It’s called the Oppenheimer-Volkoff limit OK, looked it up. Seems that they had the order of magnitude correctly, but they seem to have come at a value that was even lower of that of a white dwarf. So still pretty rough, yes. On 8/19/2022 at 11:03 PM, Markus Hanke said: Yes, but if you keep collapsing the body, the pressures and energies will eventually get so high that the fundamental forces will re-unite into a GUT scenario - at which point the concept of ‘quark’ ceases to make sense. But that is rather hypothetical too, isn't it? GUT is more a GUH (hypothesis) than a theory. AFAIK GUTs predict that the proton is not stable, but experiments already pushed the lifetime of the proton higher than predicted. On 8/19/2022 at 11:33 PM, MigL said: Speculation about quark stars don't make much sense to me, in light of the previous arguments, as they involve going to a much lighter fermion, and unphysical velocities again. That's why I referred to a bottom/top quark star... Both are heavier than the proton. But I know this is 'extreme speculation'. And of course there would be a degeneration limit for them too. Link to comment Share on other sites More sharing options...
Markus Hanke Posted August 23, 2022 Share Posted August 23, 2022 On 8/22/2022 at 5:30 AM, Mitcher said: The parameter r Schwarschild used has been wrongly interpretated as a radial parameter from the start. If the surface forms a paraboloid of revolution it createss a hole that is not reachable excepted by jumping in a higher dimension, this is easy to understand, like if there is a hole in a sheet of paper it is possible to go on the other side of the surface but impossible to go inside the hole. I mentionned the work of Shakharov, Souriau, Kostant or Petit on this specific. joigus has beaten me to it with his excellent answer (+1). As I said earlier, the manifold and a particular coordinate chart chosen on it are not the same things at all - to put it succinctly, having a ‘hole’ in an embedding diagram does not necessarily imply that there is a corresponding ‘hole’ in the manifold, in a topological sense. These are different things. You can in fact have patches (or entire manifolds) without coordinates defined on them. For Schwarzschild spacetime, you need only transform the metric to a different, more complete coordinate basis to see this. But if you want to be absolutely sure and precise, it is always best to use tools that are coordinate-independent. 10 hours ago, Eise said: OK, looked it up. Seems that they had the order of magnitude correctly, but they seem to have come at a value that was even lower of that of a white dwarf. So still pretty rough, yes. Yes, indeed. Arriving at a precise value is actually not easy, also because external conditions play a role during the collapse. But I think the salient point is that there is such a limit, for any given level of degeneracy. 10 hours ago, Eise said: But that is rather hypothetical too, isn't it? It’s hypothetical to some degree, yes. But just as in the case of quantum gravity, there are good reasons to believe that the GUT domain is quite real, even if we don’t know for sure which of the numerous GUT candidate models will apply. That being the case, quarks and gluons are by-products of a broken GUT symmetry, so once energy levels are high enough, the strong interaction will cease to exist in its ordinary form. In more general terms, I very much agree that singularities are not real-world objects, but artefacts of our models being pushed beyond their domains of applicability. 1 Link to comment Share on other sites More sharing options...
MigL Posted August 24, 2022 Share Posted August 24, 2022 On 8/23/2022 at 4:49 AM, Eise said: That's why I referred to a bottom/top quark star... Both are heavier than the proton. Sorry, I had not considered that. Not quite sure of the mechanism for going to a 'top' uark, as it has a mass comparable to a gold atom, but even the W and Z bosons of the weak interaction ( 80 and 90 GeV/c2 ), equivalent in mass to an iron atom, don't have enough energy to decay to a 'top' quark. Unfortunately, the centers of neutron stars are just as difficult to examine, as the interior of an event horizon. Link to comment Share on other sites More sharing options...
Mitcher Posted August 25, 2022 Share Posted August 25, 2022 On 8/22/2022 at 12:58 PM, joigus said: The coordinate r is not the radial coordinate. That much I, for one, will concede. For starters, Schwarzschild's r in the interior maps time, not space. But even outside it doesn't have to be, even though the temptation to call it so is very strong. If you want an argument that shows this very clearly, consider this: Scharzschild coordinates separate the angular factor so that the total solid angle gives 4*pi. Well, that cannot be. And the reason is that in a space with (positive/negative) spatial curvature, there must be an angular (deficit/excess). Of that, Scharzschild's coordinates tell us nothing. Here's the thing: Coordinates in GR are meaningless. Scharzschild's coordinates are only meaningful at spatial infinity. What you do in GR is pick out a set of coordinates that suits you to solve the equations, then do an analytic extension that smooths out all the singular points of the coordinate chart, then you identify the real singularities (infinite curvature). Finally, you discuss causality and the like in terms of the best set of coordinates(Kruskal-Skezeres, conformal, with Penrose diagrams, etc). Mind you, if your favourite coordinate chart that you used to solve the equations was "contaminated" with spurious singularities, you will have to introduce a singular change of charts that undoes the damage. It's not you, it's not me, it' generations of physicists that have been confused by the mirage of their coordinates charts. GR is a lanscape to tread very carefully. As much as Kruskal has been pushing the equations beyond reason (0 power 0 = 1, for one), he still could not get rid of his central singularity. A singularity cannot be real, it is singularly wrong IMO, so I also disagree with the space-time switch. The BH basic model is supposed to be static but it magically becomes dynamic. On 8/23/2022 at 9:37 PM, Markus Hanke said: As I said earlier, the manifold and a particular coordinate chart chosen on it are not the same things at all - to put it succinctly, having a ‘hole’ in an embedding diagram does not necessarily imply that there is a corresponding ‘hole’ in the manifold, in a topological sense. These are different things. You can in fact have patches (or entire manifolds) without coordinates defined on them. As far as I know it is impossible to put patches in the space-time itself, maybe an extremely advanced civilisation could it it but this site is not about SF. Link to comment Share on other sites More sharing options...
joigus Posted August 25, 2022 Share Posted August 25, 2022 1 hour ago, Mitcher said: As much as Kruskal has been pushing the equations beyond reason (0 power 0 = 1, for one), he still could not get rid of his central singularity. A singularity cannot be real, it is singularly wrong IMO, so I also disagree with the space-time switch. The BH basic model is supposed to be static but it magically becomes dynamic. As far as I know it is impossible to put patches in the space-time itself, maybe an extremely advanced civilisation could it it but this site is not about SF. Why do you say Kruskal has been "pushing the equations"? The equations give a catalogue of exact solutions, and Kruskal didn't touch Einstein's equations to do what he did. He just re-arranged the solution. He re-expressed a well-known exact solution by analytic continuation (maximal analytic extension). He introduced a change of charts, which is singular at the point where the initial chart was singular. Sure enough, if you plug in the solution in Einstein's equations again, but in terms of Kruskal coordinates, it still satisfies them. Are you saying it doesn't? I agree with you that the actual singularities probably point to a region were GR alone probably is not the whole story, so they're not "real", if you will. I totally agree with, On 8/23/2022 at 9:37 PM, Markus Hanke said: [...] In more general terms, I very much agree that singularities are not real-world objects, but artefacts of our models being pushed beyond their domains of applicability. 2 hours ago, Mitcher said: The BH basic model is supposed to be static but it magically becomes dynamic. No, it doesn't. Schwarzschild's solution looks the same for all times. In fact, the solution is unrealistic, among other things, due to this. It was always there and it never grows. The fact that there is a time doesn't imply the metric is "dynamic". It doesn't display collapse, it doesn't display accretion, it doesn't display evaporation. But Markus will explain this better, I'm sure. Link to comment Share on other sites More sharing options...
Markus Hanke Posted August 26, 2022 Share Posted August 26, 2022 (edited) 13 hours ago, Mitcher said: As much as Kruskal has been pushing the equations beyond reason (0 power 0 = 1, for one) This is the generally agreed upon definition in algebra. 13 hours ago, Mitcher said: he still could not get rid of his central singularity. Of course not. The singularity is inevitable in purely classical Schwarzschild spacetime, as can be formally proven using the singularity theorems. As I mentioned earlier, the appearance of a singularity in any theory of physics (not exclusive to GR) generally means that the model has been extended past its domain of applicability. In this case, GR, as being purely classical, fails to account for quantum effects during the collapse. It does most emphatically not mean that we expect a singularity to be a real-world object. If you want to eliminate the singularity, you can choose a connection other than Levi-Civita on your manifold, which allows for the presence of torsion in addition to curvature. This model is called Einstein-Cartan gravity, and is singularity-free. But this is not the same theory as General Relativity. 13 hours ago, Mitcher said: The BH basic model is supposed to be static but it magically becomes dynamic. Schwarzschild spacetime is static and stationary by definition. There are no dynamics whatsoever - you actually use this fact as boundary condition to derive the solution in the first place! You can translate the entire manifold in time without changing anything in its geometry. In technical terms, the manifold admits a time-like Killing vector field. When we say that time and space trade places below the event horizon, what we really mean is that ageing into the future inevitably corresponds to a radial decay - meaning there cannot be any stationary frames, no matter how much force you exert in trying to counter gravity. It’s inherent in the causal structure of spacetime itself. This is of course independent of the choice of coordinates. 13 hours ago, Mitcher said: As far as I know it is impossible to put patches in the space-time itself, maybe an extremely advanced civilisation could it it but this site is not about SF. ‘Patch’ is simply the technical term for a particular region on a manifold, it has nothing to do with any manipulations of same. Edited August 26, 2022 by Markus Hanke Link to comment Share on other sites More sharing options...
joigus Posted August 26, 2022 Share Posted August 26, 2022 (edited) Here's the Kruskal-Szekeres change of charts: It's singular at r=RSchwarzschild=2GM because the Jacobian is zero there. I don't know where you got the 00 problem from. 00 is no problem if you can calculate the limit. The problem is the Jacobian is zero. This is not allowed for good reasons I'm not going to delve into. But:What is a chart? It's an assignment of coordinates: \[ \phi:\mathcal{U}\subseteq\mathcal{M}\rightarrow\mathbb{R}^{4} \] This could be the Schwarzschild chart. And in comes Kruskal: \[ \psi:\mathcal{U}\subseteq\mathcal{M}\rightarrow\mathbb{R}^{4} \] What's a change of charts? In this case from Schwarzschild to Kruskal: \[ \psi\circ\phi^{-1}:\mathbb{R}^{4}\rightarrow\mathbb{R}^{4} \] The initial chart is singular at the 2-surface r=2GM. Now Kruskal introduces a singular change of charts that restores smoothness on that region. It stands to reason that you must do something drastic on r=2GM if you want to restore smoothness. This is a particular feature of analytic continuation in this case. Now "stasis". You do not look at the metric in a particular coordinate set and infer anything. That's a sure way to make mistakes. What you do is what Markus told you: On 8/23/2022 at 9:37 PM, Markus Hanke said: For Schwarzschild spacetime, you need only transform the metric to a different, more complete coordinate basis to see this. But if you want to be absolutely sure and precise, it is always best to use tools that are coordinate-independent. In this case, if one wants to prove that the metric is static, what one does is define a Lie derivative, and from there introduce time-like Killing fields. Then prove that your metric is invariant under those infinitesimal transformations. Edited August 26, 2022 by joigus LateX rendering not working 1 Link to comment Share on other sites More sharing options...
Markus Hanke Posted August 26, 2022 Share Posted August 26, 2022 1 hour ago, joigus said: It stands to reason that you must do something drastic on r=2GM if you want to restore smoothness. ...or you can just look at the invariants of the Riemann tensor in that region, in particular the Kretschmann scalar. Since it exists and is regular and well defined on the horizon, spacetime must necessarily be smooth and continuous there. 1 Link to comment Share on other sites More sharing options...
Mitcher Posted August 26, 2022 Share Posted August 26, 2022 8 hours ago, Markus Hanke said: This is the generally agreed upon definition in algebra. If zero power zero = 1 then zero = 1 power(1/0), with undefined exponent. Hence zero power zero cannot equal 1 and is also undefined, excepted in programming usually. Link to comment Share on other sites More sharing options...
Mitcher Posted August 26, 2022 Share Posted August 26, 2022 22 hours ago, joigus said: I agree with you that the actual singularities probably point to a region were GR alone probably is not the whole story, so they're not "real", if you will. I totally agree with, That's a rather profound admittance since GR is a theory expected to describe reality perfectly, and now it would point at an incomplete result ? 23 hours ago, joigus said: Sure enough, if you plug in the solution in Einstein's equations again, but in terms of Kruskal coordinates, it still satisfies them. Are you saying it doesn't? I do. Kruskal's transcendant f function is not at all smooth on the Schwarzchild's sphere. When the (r - 1) exponent tends to zero from the upper or the lower limit it takes different forms when passing this limit and both sides don't connect with each others. And that's normal since the sphere is not reducible. Link to comment Share on other sites More sharing options...
joigus Posted August 26, 2022 Share Posted August 26, 2022 1 hour ago, Mitcher said: That's a rather profound admittance since GR is a theory expected to describe reality perfectly, and now it would point at an incomplete result ? No. GR's job is to describe gravity. Nobody has ever said it's supposed to describe reality. That's a pretty tall order. That's why Einstein himself immediately started working on unification. This is probably Maxwell's most valuable legacy. 1 hour ago, Mitcher said: I do. Kruskal's transcendant f function is not at all smooth on the Schwarzchild's sphere. When the (r - 1) exponent tends to zero from the upper or the lower limit it takes different forms when passing this limit and both sides don't connect with each others. And that's normal since the sphere is not reducible. So you don't believe in calculus then. I see no other way to interpret what you're saying. What does that mean, "the sphere is not reducible"? Link to comment Share on other sites More sharing options...
Markus Hanke Posted August 27, 2022 Share Posted August 27, 2022 14 hours ago, Mitcher said: If zero power zero = 1 then zero = 1 power(1/0), with undefined exponent. Hence zero power zero cannot equal 1 and is also undefined, excepted in programming usually. I think you should debate this with a mathematician, such as @studiot. 0^0 is an indeterminate form, so whichever value you assign to it will inevitably lead to contradictions in some contexts, while it works just fine in others. So far as I am concerned, in this particular context - algebra and calculus - there are good reasons to treat it as =1, and that works. That’s the consensus. Nonetheless, if you don’t want to use K-S coordinates, then don’t - you are free to choose any coordinate system you like, since this is an additional structure separate from the manifold itself. And that’s the salient point here - as already pointed out multiple times. A coordinate singularity or discontinuity does not necessarily imply that the manifold is singular or discontinuous. Besides, there are coordinate choices that are perfectly smooth and regular at the horizon, such as Gullstrand-Painlevé, or Eddington-Finkelstein. But why do we even need to be talking about specific coordinate choices? That’s precisely why one should use coordinate-independent methods here. The fact of the matter is that the Riemann tensor and its invariants exist and are well-defined on the horizon, so spacetime is necessarily smooth and regular there. That’s all there needs to be said on this. 11 hours ago, Mitcher said: That's a rather profound admittance since GR is a theory expected to describe reality perfectly No. Firstly, GR is purely classical, so it is incapable of accounting for quantum effects. Secondly, the field equations are only a local constraint on the metric, but don’t determine it uniquely in and of themselves - you separately need to supply boundary conditions (most notably information about distant sources) in order to obtain specific local solutions. So you get out exactly what you put it - if you supply unphysical boundary conditions, you get metrics that are at best approximations to reality. That’s how it is for Schwarzschild - it assumes, among other things, asymptotic flatness, which isn’t something we find in the real world. So expecting exact Schwarzschild geometry to occur in the universe is foolish. But it is a useful approximation, so long as you understand the limitations. 11 hours ago, Mitcher said: I do. You are in conflict with well-established results then, as can be easily found in most major GR texts. Can you provide a coordinate-independent proof that the event horizon is geodesically incomplete - which is what you seem to be claiming? Link to comment Share on other sites More sharing options...
Mitcher Posted August 27, 2022 Share Posted August 27, 2022 16 hours ago, joigus said: No. GR's job is to describe gravity. Nobody has ever said it's supposed to describe reality. That's a pretty tall order. That's why Einstein himself immediately started working on unification. This is probably Maxwell's most valuable legacy. So you don't believe in calculus then. I see no other way to interpret what you're saying. What does that mean, "the sphere is not reducible"? Sorry, please read non-contractible hypersurface. 16 hours ago, joigus said: No. GR's job is to describe gravity. Yes, that's what I meant, it is the tool to describes physical gravity. 16 hours ago, joigus said: So you don't believe in calculus then. I mean the Kruskal ksi function = r + Ln(r - 1) and r + Ln(1 - r) is not regular for r = 2m since the log of something tending towards 0 tends towards minus infinity. 7 hours ago, Markus Hanke said: That’s how it is for Schwarzschild - it assumes, among other things, asymptotic flatness, which isn’t something we find in the real world. I do not understand this. ST would be flat very far away from the Sun if it was not for the planets. 7 hours ago, Markus Hanke said: Can you provide a coordinate-independent proof that the event horizon is geodesically incomplete - which is what you seem to be claiming? If a mere change of variables allows for central or spherical singularities to disappear then what is real ? On 8/18/2022 at 9:18 PM, Markus Hanke said: I don’t know what this means? There’s no such thing as negative proper mass. The proper mass and proper time remain as they are indeed, they change due to a change of coordinates, I agree. Link to comment Share on other sites More sharing options...
joigus Posted August 27, 2022 Share Posted August 27, 2022 1 hour ago, Mitcher said: Sorry, please read non-contractible hypersurface. I suppose what you mean is that a 2-sphere cannot be mapped with just one chart. I fail to see how this is relevant here. 1 hour ago, Mitcher said: I mean the Kruskal ksi function = r + Ln(r - 1) and r + Ln(1 - r) is not regular for r = 2m since the log of something tending towards 0 tends towards minus infinity. You're still trapped in the mirage of singularities of the coordinate chart. By the same token, you'd probably think there's a problem at the north pole of a sphere... I rest my case. Link to comment Share on other sites More sharing options...
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