Markus Hanke

White holes are (in principle at least) a part of maximally extended Schwarzschild spacetime; this type of spacetime is static and stationary, so both black hole and white hole would have constant mass. Evaporating black holes belong to a different type of spacetime geometry, called Vaidya spacetime. I am not actually sure what the global topology of this kind of spacetime is like, but to the best of my knowledge (someone correct me if wrong) there is no white hole region here.

MTW never gave a formal definition for this operator, other than to say it acts only on contravariant vectors, but not on forms. I think they might just have invented it

No, we don’t make any assumptions for the collapse process beforehand. What you do is start by writing down the energy-momentum tensor for the interior of the collapsing body (a star, usually), and then impose the necessary condition that the metric is to remain continuous and differentiable at the boundary (the star’s surface). You then insert all of this into the field equations, and work out the solution. There are no assumptions about the collapse process itself. We already know that GR - which is a mathematical model - works really well in the classical domain, so there is no scientific reason to assume that it won’t work for a collapsing body, up to the point when quantum effects are no longer negligible. The appearance of the singularity means just that - that GR no longer applies in that region, since that stage of the collapse isn’t classical. But it is perfectly reasonable to take GR at its word in that at least for some distance beyond the horizon spacetime remains smooth and regular. The question is really only how far that region extends. On the other hand, suggesting that GR’s modelling of the beginning stages of the collapse is somehow wrong (bearing in mind the aforementioned caveat), and that regular matter is replaced with exotic matter in the process, is, to me, not especially reasonable. Then how do you know what the geometry of spacetime in that region will be? What is the nature of that exotic matter fluid, how does it arise from the (quite regular) matter of the original collapsing star, and what is its energy-momentum tensor? What holds it in place at the precise radius of the event horizon, since it would be gravitationally self-repulsive? How do you know any of these things, if you don’t have a metric or at least an energy-momentum tensor? We have never observed exotic matter, after all. Gravity is a highly non-linear thing, so it is dangerous to make any kind of assumption about how something will behave, without working through the maths first - remember that energy density is not the only component of the energy-momentum tensor, so it isn’t the only source of gravity here. For exotic matter, the influence of the stress components in the tensor would actually be larger than that of energy-density. Because you cannot join a region with vanishing cosmological constant to a region with non-vanishing constant, without violating the continuity boundary conditions. There also isn’t any known mechanism for that constant to change from one spatial region to another. How does it just “appear”? Where does it come from, and by which physical mechanism does it form? What type of particle is it made of, and how does that fit in with the Standard Model? If the interior region wasn’t empty space (as you appear to suggest), then the gravitational wave signature of these mergers would be quite different. In fact, the signature from merging event horizons as modelled by GR is pretty unique, so no, it couldn’t really be otherwise in any significant way. And if the interior was an exotic fluid with negative energy density, then the signature would be very different - in fact, there would be no merger, since two such black holes would be gravitationally repulsive towards one another. I’m somewhat confused now what exactly you really suggest - an empty interior with negative cosmological constant? Or an interior filled with a gravitationally repulsive exotic substance? For the former, you can’t join these regions without violating basic boundary conditions. For the latter, the exterior vacuum cannot be Schwarzschild, since it would have to have a very different geodesic structure - positive mass test particles would actually fall away from the event horizon.

You're right, I didn't read it carefully enough

No, you get magnetic jets in both black holes and pulsars - but what I am saying is that their dynamics are different, because the former arises from accretion disks, and the latter from the dipole field of the neutron star itself. They are observationally different. Yes, this is essentially what we have been saying for the past two pages.

Well, there are at least some relativistic effects that are directly observable in everyday life - two examples that immediately spring to me mind would be the colour of gold (it would be silvery without relativistic effects), and the effectiveness of lead-acid batteries used in cars etc.

It's an analogy of the principle that not everything is 'either A or B'. Sometimes there's a deeper reality, and what you observe depends on how you look at it, as is the case here. It looks precisely like quantum field theory. Photons are just that - excitations of the underlying quantum field. As such, they are neither classical particles nor classical waves, nor both together. They don't have to be either of those things - that's just a cognitive bias that we as humans have, since our direct experience is purely classical. This is completely inconsistent with quantum electrodynamics - an extremely successful and well-tested model.

Generally I'm the same, but I do enjoy some far-out speculation at times

Neutron stars generate strong magnetic fields around them, whereas black holes don't (only their accretion disks do, but those fields have different geometry). This is directly observable. Even for neutron stars themselves, gravity is strong enough to have an influence on the geometry of their magnetic fields, as well as their rotational frequency - both of which are strong field GR effects, and directly observable. And then of course you have gravitational wave signatures of the mergers of such objects, which are different for black holes and neutron stars. All of this has been mentioned before, so I don't know why you keep going on about this.

You mean the original version? You can of course do that, but personally I don't really see the advantage. But in any case, for the exterior region it makes the same physical predictions. As I said, we already know that a point singularity is in all likelihood not what happens in the real world. And even for the exterior region, the Schwarzschild solution is only an approximation. We are not 'forcing' the singularity, it is simply what happens during the collapse process in a purely classical model. Even if we take into account quantum effects as far as we can model them, once you go beyond a certain total mass, there is nothing there that can stop a complete collapse. 'Independent of what math' suggests is not a very scientific approach. We already know that the event horizon cannot be a curvature singularity (both in terms of theory, and in terms of observational evidence), so it is quite reasonable to work off the assumption that GR gives correct predictions for at least part of the interior region. Because it's meaningless. No we aren't. The boundary conditions used to derive the solution to the field equations make no reference to singularities, horizons, or interior regions at all. Yes, you can 'glue' different regions together, but with two caveats: 1. The overall metric must remain continuous and differentiable everywhere at the boundary 2. The new metric must itself be a valid solution to the field equations In practice, because the field equations are non-linear, this places very stringent constraints on what geometries the regions can have. You have very little freedom in choosing the metrics. In particular, if the exterior region is Schwarzschild, and the interior region beyond the horizon is a vacuum, then the interior geometry must be Schwarzschild as well. There is no other mathematical possibility. An Einstein-Rosen bridge is already a feature of the maximally extended Schwarzschild metric, it isn't a different interior metric. You just need to pick a suitable coordinate chart that covers the entire spacetime before you can see it in an embedding diagram. Can you write down a suggested metric? This essentially would mean that the interior region has a non-zero cosmological constant. You can do this of course, but then the exterior region will not have Schwarzschild geometry. We have direct observational evidence for how light and massive test particles behave close to the horizon; we also have direct observational evidence of the gravitational wave signature of black hole mergers. Both of these place very stringent constraints on the structure of spacetime near the horizon, and both of these are in perfect accord with what GR predicts in those circumstances. My thoughts are that you cannot glue an exterior Schwarzschild vacuum to an interior vacuum with negative cosmological constant (which would have the effect of a negative energy density) in any self-consistent way.

Consider this: In the same way, a quantum object is neither particle nor wave, nor both - but it may appear as either one of these, depending on how you look at it. Note that the 'either-or' categories do not apply here, because that duality belongs into the classical world; but a quantum object is not classical. So there is not actually any problem here that needs to be solved.

It's 1916 - Einstein didn't find his field equations until 1915. They describe the same physical spacetime, but Schwarzschild's original coordinate chart covers only the exterior region of that spacetime, so of course it does not contain an event horizon. That's not because such a horizon doesn't exist, but because the coordinates used simply don't cover it. No, it's the exact opposite - the boundary condition used to derive a maximally extended Schwarzschild solution (which covers the interior region as well) is asymptotic flatness, i.e. it is assumed that far enough away from the black hole gravity is described by Newton's laws. When you solve the field equations, you are then left with a single free parameter in the metric - which, via the asymptotic flatness condition, can be interpreted as mass. It is very important to understand here that unlike in Newtonian physics, the mass in Schwarzschild spacetime is not localisable, meaning it is a property of the entire spacetime. Einstein's GR is a purely classical theory, it does not account for any quantum effects, which during a real-world gravitational collapse cannot be neglected. Hence, in real world black holes there almost certainly are no singularities, we know this already. The question is then, what happens in the central region, if not a singularity? To answer this we need a model of quantum gravity, which we currently don't have yet. To make a long story short, GR is best understood as a classical approximation to a more complete theory of quantum gravity. Actually, there are many ways to avoid the formation of the central singularity, even in classical physics. For example, one can allow spacetime to have intrinsic torsion - that gives a model called Einstein-Cartan gravity, and no singularities occur there. But such models have other issues that are potentially problematic. Mathematically speaking it is quite clear that nothing special actually happens at the event horizon - you can see this easily by simply choosing a different coordinate chart for your Schwarzschild spacetime, such as Kruskal-Szekeres coordinates for example. The event horizon isn't really the problem (things would be far more problematic if it wasn't there). The problem is that GR as a theory is purely classical, so its description of the actual collapse process that gives rise to the black hole in the first place is evidently wrong or incomplete, since it can't account for quantum effects. It's simply outside its domain of applicability.

The reference frame (i.e. coordinate choice) is arbitrary here; the entire formalism used in the post I made is covariant under coordinate transformations. So the 4D volume (which is thought of as sufficiently small) can be anywhere and anytime. Yes, you can define such a 4D volume. Yes, the boundary of any n-dimensional volume is a (n-1)-dimensional volume. It isn't the boundary that is zero, but the boundary of the boundary. It is easier to visualise this in fewer dimensions. Suppose you have a ball in normal 3D space - its boundary is just its surface, which is a 2D sphere. And what is the boundary of the 2D sphere (which is the boundary of the boundary of the 3D ball)? It doesn't have any - there's no edge or discontinuity to it, you could draw continuous lines on it in any direction you choose. The boundary of a boundary is zero. The same is true for the 4D spacetime hypercube, but it's much less obvious, because the sides of the 3D cubes that make up its boundary are oriented surfaces, so each side is counted twice, but with opposite sign, giving a net total of zero.

He didn't do any coordinate transformations, he simply chose a coordinate system in which the determinant of the metric tensor was unity, as an ansatz to solve the field equations. It's essentially just a modified version of polar coordinates, with r=0 being the event horizon. It's a legitimate - but unnecessarily awkward - coordinate choice. I don't know what you mean by this. The geodesic structure of the spacetime is an invariant, and quite independent of the choice of coordinate systems. They are the same metrics in the sense that they describe the same physical spacetime. The original is from 1916 though, not 1912. You haven't actually made any assertions, so I am not saying you are wrong on anything. I only wish to point out that a coordinate singularity does not necessarily imply a physical curvature singularity, simply because the choice of coordinate chart is arbitrary. Coordinate values do not carry any physical meaning, only the events they refer to do. I assume this is what you are getting at, as it is a very common question. The easiest (but not the only) way to tell whether there's a curvature singularity or not is to calculate the curvature invariants of the Riemann tensor. The thing is that Schwarzschild did not know this when he found his solution - GR was a brand new theory back then, and not yet well understood. Hence, there was a lot of debate about the status of the event horizon then. Nowadays though, after studying the model for well over a century, we understand the situation much better, both in terms of physics, and in terms of the maths. I don't know why you would think that? Spacetime at the horizon is smooth and regular, so the equivalence principle holds there just as it holds everywhere else. Well, the Schwarzschild solution works well as a rough approximation, and it is a useful teaching tool. However, the boundary conditions used to find this solution aren't physical. In order to have a more accurate model of real-world black holes, you need to look at other (more complicated) solutions to the field equations - such as Vaidya spacetime for example -, or use numerical GR.

The singularity at the horizon is purely a coordinate singularity, but not a physical one. Spacetime is smooth and regular at the horizon, which you can easily see either by calculating the curvature invariants of the Riemann tensor (they all stay finite and well defined there), or by performing a simple coordinate transformation. This stands in contrast to r=0, where all such invariants diverge, so that one is a physical curvature singularity. Also, Schwarzschild spacetime is a vacuum spacetime, so there is no mass anywhere. The 'M' value in the metric is just a parameter for a 1-parameter family of metrics.

No, it also depends on the metric of the underlying manifold, and obviously permeability and permittivity of the vacuum. But provided the above things are given, an interesting question arises - are the relative positions and potentials on the surface of the sphere sufficient boundary conditions in order to fix a unique solution to Maxwell's equations? I don't know the answer, but my guess would be no, because there is a gauge freedom in where the zero point of the electric potential is, so I think we need precisely one additional boundary condition that specifies whether or not there is a background potential/field permeating the space. The relative positions of the charges alone don't uniquely fix this. But that's just a guess now, without having done any actual maths.

No problem. It is far more understandable though if one has access to the textbook, where all of this is described in much more detail.

Actually, what I am hoping to find is precisely a connection between ER=EPR and the issue of locality/separability/realism. What if the violation of Bell's inequalities precisely implies that spacetime is in fact multiply connected? Shouldn't it be possible, at least in principle, to retain both locality and realism, while still violating Bell's inequalities, if the underlying spacetime is multiply connected in just the right ways?

That would probably help things - it's in chapter 15 of the book. The essential train of thought is this - suppose you have an elementary 4-cube of spacetime \(\Omega\). We know that energy-momentum within that cube is conserved, so (in differential forms language): \[\int _{\partial \Omega } \star T=0\] If we want to obtain a metric theory of gravity, the question becomes - what kind of object can we couple to energy-momentum, that obeys the same conservation laws, in order to obtain the field equations? For this, consider that the boundary of our 4-cube consists of 8 identical 3-cubes, each of which is in turn bounded by 6 faces. We can now associate a moment of rotation with each of the 3-cubes; to provide the link to energy-momentum, we then associate that moment of rotation with the source density current in the interior of each 3-cube. Let \(\bigstar \) define a duality operation that acts only on contravariant vectors, but not on differential forms (the Cartan dual). The moment of rotation is then \[\bigstar ( dP\ \land R) \] wherein R denotes the curvature operator. We also find that this expression is just the dual of the Einstein form: \[\bigstar ( dP\ \land R) = \star G\] So let's put this all together. First, we create a moment of rotation in our 4-cube of spacetime: \[\int _{\Omega } d\star G\] Apply Stoke's theorem: \[\int _{\Omega } d\star G=\int _{\partial \Omega } \star G \] Rewrite in terms of the curvature operator: \[\int _{\partial \Omega } \star G=\bigstar \int _{\partial \Omega } ( dP\ \land R) \] To associate this with total net energy-momentum (at the moment it's associated with source current density), we must sum not just over the boundary of the 4-cube (which are 3-cubes), but also over the faces of each 3-cube; so we must apply Stoke's theorem again, to get \[ \bigstar \int _{\partial \Omega }( dP\ \land R) =\bigstar \int _{\partial \partial \Omega }( P\ \land R) \] But because \(\partial \partial =0 \), this automatically yields \[\bigstar \int _{\partial \partial \Omega }( P\ \land R) =0 \] But because the bracketed expression is just the dual of the Einstein form, this implies \[\int _{\partial \Omega } \star G=0 \] This is the exact same as the conservation law for energy-momentum given above. So we can associate the two: \[G=T\] So, using the concept of a moment of rotation, some elementary geometric considerations, and the "boundary of a boundary is zero" principle, the form of the Einstein equations is uniquely fixed up to a proportionality constant. MTW even obtain this constant somehow, though I don't quite follow their thoughts on this minor detail. Anyway, that's the general idea. If you can get access to the text, it is all described in much more detail there.

That is essentially what I was hinting at, actually...I think our current concept of what is fundamental (quantum fields and their interactions and excitations) just doesn't really cut it, from a philosophical point of view (not saying those things don't work!!). I think there is a whole lot more going on than we currently realise, and it will need a major paradigm shift to reveal it. Something else just occurred to me earlier today - your mathematical definition of locality makes a tacit assumption: that the underlying manifold on which the field 'lives' has a trivial topology. But what happens if that is not the case? For example, what happens if the manifold is multiply connected, or has closed loops, or whatever else may be the case? Properly defining 'locality' becomes more difficult, then. P.S. I'm aware of course of ER=EPR, but haven't really arrived at a conclusion about what this really implies.

It would certainly be non-trivial. However, what I was trying to point out is that, if someone develops QFT first, then they would very quickly realise that a standard QFT for spin-2 bosons that couple to energy-momentum (a reasonable ansatz if one wants to find a model for gravity) yields something that is physically meaningless. So they would begin to wonder if perhaps gravity can't be described via a QFT at all. This would eventually bring them to consider metric theories instead - and GR is the simplest example of that. Yes, this is the strong field regime I mentioned in the last post.

The only textbook I know of that explicitly mentions this is Misner/Thorne/Wheeler "Gravitation". It's a very beautiful connection between GR (and also electromagnetism!) and that topological principle. If you don't have access to this text, I can try and summarise the derivation here, if I have time over the next few days. Interesting observation, but...isn't the AdS/CFT duality the exact opposite of this? Only spacetime within the bulk would have a time dimension, whereas the boundary on which the CFT lives is purely spatial. However, I may have this wrong, so please correct me if necessary.

Have we not answered this already earlier on this thread? Any scenario where the non-linearity of GR cannot be neglected - so essentially the strong field regime - will display effects that differ from the weak field approximation. Even the article on GEM which you shared mentioned some of these effects. I just see it as the simplest possible metric theory that fulfils all relevant self-consistency criteria. This is an issue only so long as one tacitly assumes that GR must be the classical limit of some quantum field theory - which implies the assumption that gravity works in the same way as the weak, strong, and EM interactions. But actually, there is nothing in physics that indicates that this must necessarily be the case. Also, it is actually trivially easy to write down a QFT for a spin-2 boson interaction, but it is just as easy to see that such an attempt yields something that is physically meaningless. So clearly, gravity doesn't work in the same way as the other forces, hence it is not a surprise that it isn't renormalizable. Personally I think gravity isn't a fundamental interaction at all, so attempting to apply the usual quantisation schemes to it is simply an error on our part; a category mistake, if you so will.

Quite possible, I am not sure what the convention for the notation is, on this one. I thought I have seen people use circles on both integrals...? Indeed. Did you know that the form of the GR field equations follows from this seemingly simple topological principle? Both this principle, and the generalised Stoke's theorem above, are IMHO among the most beautiful results in all of mathematics The name kind of rings a bell somewhere, but I wouldn't be intimately familiar with what he did (even though I live in Ireland). This is probably a good time to reiterate that all my maths are entirely self-taught, so there are large holes in my mathematical knowledge. I really only ever looked at those areas that are directly relevant to the areas of physics I am interested in. Gladly I'm somewhat out of my depths on this one, since I've never really studied QFT in any detail. That's a shortcoming I am intending to rectify when I have the time and inclination.

Lol...I given no guarantees, this is second hand information. Coming from what I would consider a reliable source, I simply assumed it's true