Markus Hanke

Yes, that’s right. Yes, correct. Do note though that this isn’t the same as geodesic completeness, which is a whole different issue.

Yes, the precise Schwarzschild geometry will never occur in the real universe, because the necessary conditions aren’t given. However, there are many circumstances when it is a very useful approximation, and fits things quite closely. The same answer here - while no region of real-world spacetime will ever be perfectly flat, there are many circumstances where this is very nearly the case, so it is again a very useful approximation to the real world. It all depends on the levels of accuracy you require for the problem at hand.

To give a very general answer - Schwarzschild spacetime relies on certain conditions that need to be in place for this particular geometry to arise. It is static, stationery, spherically symmetric, and asymptotically flat (ie there are no other distant sources of gravity). If any of these conditions is violated, we are no longer dealing with Schwarzschild spacetime, but something more complicated. In principle, yes. But remember, a Schwarzschild BH is stationery and relies on an otherwise empty universe, meaning it doesn’t permit any changes - so you can’t have anything falling into it. If you add even as much as a single particle falling in, it’s no longer truly a Schwarzschild BH, but some other geometry. Yes and yes. But again, this wouldn’t be a Schwarzschild BH any longer. That’s a really good question! I presume you mean a gravitational wave. You can certainly embed a BH into a background gravitational wave field. The result would be something pretty complicated. I don’t know for sure just exactly what would happen, because, since GR is a non-linear theory, metrics don’t just add - you’d have to actually derive an entirely new solution for this scenario, which is likely only possible with numerical methods. I can make an educated guess though - given the right wavelengths for your gravitational radiation, the event horizon of your BH would begin to oscillate and ‘vibrate’ (like a bell) and eventually achieve a state of resonance with the external wave field. But this also means that the BH itself becomes a source of gravitational radiation - so it would essentially reflect some of the radiation back out. I don’t know if it would re-radiate all of the energy, or absorb some of it and grow in mass; one would have to run the numbers to find out. What’s more, the re-radiated waves will interfere with the incoming background waves in complicated non-linear ways, changing the wave field in ways that I can’t predict here now. And to go even further - if you were to ‘turn off’ the external wave field somehow, the BH will slowly ‘ring down’ like a bell, and eventually become stationery; however, the surrounding spacetime will remain permanently altered by all these waves having gone through it. It’s called the gravitational memory effect. This is a really complicated scenario, but very interesting. Yes, the event horizon will deform and ‘bulge out’ - this happens, for example, when two BH approach one another and merge. No, because spacetime inside the horizon is empty (assuming no in-falling material), so there’s nothing there to experience stresses. Schwarzschild spacetime is always spherically symmetric. If it doesn’t have this symmetry, then it will be a different kind of geometry. Yes. No, it wouldn’t be spherical, and thus it wouldn’t be a Schwarzschild BH any longer. Schwarzschild geometry requires spherical symmetry.

Imagine you throw a pebble into a pond - what happens? Wave fronts ripple out at finite speed in all directions on the water. Now plot the position of the leading edge of the wave field against a time axis that runs orthogonally to the water surface - you get a cone. Light cones work in a similar way. They simply depict regions of causality centered on some event. Of course they don’t - that’s the meaning of “singularity”. It’s a region of geodesic incompleteness, past which geodesics cannot be extended. Well, it’s the simplest possible modification of GR, and it avoids the singularity issue even in the classical domain. I wouldn’t dismiss this so readily. No, it wouldn’t, because the region bounded by the horizon (the presence of which is required to have Hawking radiation in the first place) would still be geodesically incomplete.

Once the gravitational collapse is complete, none of these quantities - charge, mass, angular momentum - are localisable, in particular not anywhere within or on the horizon. Instead, these quantities are now a global property of the entire spacetime, in the sense that they are encoded in the overall spacetime geometry. There’s nothing on or within the horizon that could possess these properties - Schwarzschild spacetime is entirely empty vacuum everywhere. Some features of the event horizon itself depend on these properties, but that does not imply that anything is actually located there. Hence, nothing needs to propagate and escape, since there’s nothing there to escape from.

Ok, so we are in agreement on this point 👍 You mean they possess a singularity? Yes, I agree, it’s unphysical. When a singularity appears in a model that cannot be removed mathematically, then that is generally taken to imply that the model breaks down at that point. For GR, that means that the final stages of a gravitational collapse are outside its domain of applicability, so it must break down there. This isn’t a surprise, since GR is a purely classical theory - but at the densities and energies encountered during gravitational collapse, quantum effects become important and cannot be ignored. That’s why the theory is doomed to fail there. This is inevitable and quite independent of coordinate choices, and thus can’t be interpreted away. Geodesic incompleteness is just a part of the topology in this spacetime. As I mentioned earlier, it is in fact possible to remove the singularity by making a small modification to GR, ie by choosing a different connection on the manifold so that you can have torsion in addition to curvature. It’s called Einstein-Cartan gravity. It’s still a purely classical model, but it doesn’t contain singularities in black holes or at the Big Bang. The trouble is that this necessarily implies a modification to the Dirac equation as well, which to date has not been observed in our world. But the effect would be very small under normal circumstances, so it can’t be definitively ruled out either; ECG is still a contender. Yes - we are using a classical model in a situation where quantum effects are not negligible. See above - have a look at Einstein-Cartan gravity. One other important point: because the Einstein equations only constitute a local constraint on the metric, the space of all mathematical solutions to these equations is much larger than the space of all physically possible spacetimes. In other words, you can always obtain solutions that formally are valid solutions to the Einstein equations, but which don’t correspond to any physically reasonable spacetime. This is why the issue of boundary conditions and initial values is so important. Furthermore, even if a solution is both mathematically valid and physically reasonable, it may still be topologically ambiguous in a global sense. GR really is a very subtle thing.

And I find it a bit strange that you are so fixated on historical papers that are 100+ years old. Would you go and solely rely on Maxwell’s original publications when wanting to learn electromagnetism? Would you read Newton when wanting to learn calculus? To be clear, there’s nothing wrong with doing this, but you shouldn’t just disregard all the progress that’s been made since. We now know a lot more about GR than Schwarzschild ever did. But regardless, I’ve done something better than just read about it, whatever the source - I’ve done the derivation of the Schwarzschild metric myself, using pen and paper, and some little extra help from MAPLE with some of the more complicated differential equations. So I understand the boundary conditions, how these give rise to the spacetime, and what the features of this geometry are. I thus have no need to rely on anyone’s words with regards to this subject, since I’ve acquired the tools and knowledge to do the maths myself. It’s certainly quite tedious and takes time, but there’s no mystery left in it. It’s quite straightforward, really. So no, Schwarzschild hasn’t been “misunderstood” - there’s literally nothing there to misunderstand or interpret, it’s all just standard differential geometry. You are essentially just working out the solution to a system of differential equations, that’s all. Well, we’ve been over this within the past three pages of discussion, so I think we may just have to disagree on this. There’s no grounds for any misunderstanding or ambiguity in the Schwarzschild solution, so far as I (and the physics community in general) am concerned. Saying the spacetime somehow terminates at the horizon (which seems to be what you’re imying? Please correct me if I’m wrong) is like covering Earth with a chart that terminates at the equator, and then claiming the Southern Hemisphere doesn’t exist. The properties are of a geometric nature. Even within the energy-momentum tensor, ‘mass’ does not directly appear. The Schwarzschild black hole spacetime (post-collapse) has T=0 and thus R=0 everywhere, so it’s a pure vacuum spacetime. There is no matter of any kind anywhere. The mass term M appearing in the metric is a global property of the entire spacetime, and is technically just a selection parameter for a 1-parameter family of metrics. Nonetheless, matter and radiation can exist just fine while they are in the process of falling in - but you need a different spacetime geometry to model this accurately, such as Vaidya and its generalisations.

It isn’t time that “rotates”, but the light cone associated with events. That’s not the same. When we say that time and space trade places at the event horizon, then what we really mean is that beyond the horizon, ageing into the future necessarily corresponds to a decay in radial position. This means, in practical terms, that there cannot be any stationary frames below the horizon - no matter how much force you exert to counter gravity, you will continue to fall downwards. This is due to the geometry of spacetime, so it is inevitable.

The number of photons (or any type of particle) within a given volume of space is also observer-dependent. This is called the Unruh effect. That’s also wrong. Gluons are massless and move at c as well, just as photons do. Wrong again. Gravity acts on all bodies equally, irrespective of their make-up - put a feather and a lead ball into a vacuum tube, and they will fall at the same rate, and hit the bottom at the same instant. This is basic high school physics.

Very interesting +1 This brings to mind the Sierpinski cube - infinite surface area enclosing zero volume

I haven’t read any of Schwarzschild’s originals papers, since to me these are only of historical interest, and I’m not much into the history of science. My main focus is on the contemporary foundations of GR - its underlying symmetries, why the field equations have the specific form they do, what kinds of solutions they admit and how to find and classify them, what kind of covariant objects can “live” on spacetime, possible candidate models for quantum gravity, generalisations of GR etc etc. Things of that nature. I acquired all my knowledge from a wide range of contemporary texts and sources on physics and maths, with Milner/Thorne/Wheeler being the oldest of them. Both the internal and external Schwarzschild solutions can be found in depth in nearly all these texts, presented in different ways, and maximally extended metrics that cover the entirety of this particular spacetime are also found in many of these texts. I find it unwise to rely too heavily on a single source or author for one’s understanding of GR, so I always make sure I consult multiple sources.

It’s the Dirac delta function, and you have \[\int_{-\infty}^{\infty}\delta(x)dx=1\]

Yes. I should clarify here though that my comments were specifically about the Schwarzschild solution - this wasn’t meant to imply that BHs don’t exist at all in the real world. It’s just that they would be of a different kind than Schwarzschild. At a minimum you have to consider angular momentum and the absence of asymptotic flatness - which leads to Kerr-Vaidya spacetime as a starting point. That’s a considerably more complex geometry than Schwarzschild.

The relativistic formulation of quantum mechanics respects all the usual laws of SR, so there can be no physical interaction (as in: exchange of information) if the events are not within or on each other’s light cones. There can, however, be a correlation between measurements that are space-like separated, as is the case with quantum entanglement. What swansont did yesterday can very well affect your state of affairs today - but not vice versa. So this isn’t a mutual “interaction” as such, but rather a one-way causal influence. Just be careful with the term “event” - in relativity, this term has a very specific meaning, being a point in space at a single instant in time. It doesn’t mean an occurrence with temporal extension, such as a car accident.

But that’s the thing - you can remove the singularity at the horizon by a simple change in coordinates (while all curvature tensors are regular there), so it isn’t a physical singularity, merely a coordinate one. At the same time you can not remove the central singularity in the same way, because the region is geodesically incomplete (all curvature tensors diverge or become undefined there). ‘Real’ in this sense is that which does not depend on choice of coordinates, ie covariant quantities such as tensors, but not coordinate charts and coordinate singularities. While it may conceivably be possible to construct such a solution (eg as a special case of Ellis-Bronnikov spacetime), this would not be Schwarzschild spacetime, but a different type of geometry. Using the boundary conditions for the Schwarzschild solution, it can be shown that the central singularity is in fact inevitable. Schwarzschild black hole, in its collapsed state, is a vacuum solution, it assumes an entirely empty spacetime. Yes, that was my point - in the real world there are all kinds of distant sources, so asymptotic flatness cannot actually occur, meaning Schwarzschild ST is just an approximation that cannot be found in the real world. It’s still very useful though.

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. 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. 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?

...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.

They are looking at it - if you search for this on arXiv, you’ll find quite a number of papers on this subject. I would imagine it hasn’t become part of the general consensus, because there are also problems and issues with these models. You are right, yes. I think the problem here is that the error bars for the precise value of the Hubble constant are far bigger than the effects of the acceleration, so at the moment we aren’t yet in a position to draw a meaningful graph for this. This is a work in progress.

This is the generally agreed upon definition in algebra. 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. 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. ‘Patch’ is simply the technical term for a particular region on a manifold, it has nothing to do with any manipulations of same.

I don’t know, to be honest - I’ve never looked into these models. Over the last few years my focus has been elsewhere, so I haven’t been keeping up with latest developments as much as I would have liked to. Another one for the future to-do list 👍

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. 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. 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.

That’s a really good question! It is indeed possible to vary lambda with time, location, or both - the resulting models are called “agegraphic dark energy models”. There are both advantages and problems associated with these, but I must admit that this isn’t something I’ve been following, so I don’t know where things stand on this. It hasn’t caught on in the mainstream though.

Einstein, despite having come up with the equations initially, didn’t know about the full set of principles underlying their form (the crucial topological concepts underpinning this were worked out by Ellie Cartan at a later date) - so he wouldn’t initially have been aware that the presence of the constant was the ‘normal’ state of affairs, hence it didn’t appear in his original formulation. So unfortunately any fine-tuning to precisely zero still lacks a physical mechanism or reason. Again, I’m not saying it can’t be zero, just that this would be an example of unexplained fine-tuning.

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 trouble with this is that the Einstein equations aren’t just invented out of thin air. There are some fundamental principles of consistency and topology that greatly constrain the form these equations can take (See Meisner/Thorne/Wheeler for details). As it turns out, the equations including the constant are the most general form that fulfils all these conditions - so there needs to be a reason why the constant should be exactly zero. Im not saying it can’t be zero, just that there would have to be a reason for it.