Markus Hanke

Such a concept does not make any physical or mathematical sense.

There is not really any such thing. There is a waterfall analogy that is sometimes used to illustrate certain concepts of GR that would otherwise be difficult to explain without going deeply into the maths; but that is just an illustrative analogy. Unlike is arguably the case in quantum mechanics, there is no space for differing “interpretations” in GR.

Precisely. Thanks for your efforts in putting together that excellent post.

To the best of my knowledge, the time dilation in the interior of a shell cavity wrt some clock at infinity is already a well established fact. I never mentioned anything about singularities, nor do I claim that GR breaks down anywhere in this type of spacetime. Obviously it doesn’t, since it is globally geodesically complete. Time dilation is directly related to differences in gravitational potential, so the Newtonian limit both in the cavity and at infinity are very relevant here. In fact, it is all we need to know - if there is a difference in gravitational potential between these regions, then there will necessarily be time dilation. And we know that there is a difference, even in the weak field Newtonian limit. This difference will increase, the more massive we make the shell.

No, it’s Minkowskian. That means that - within the metric - the time and space parts have opposite sign. This gives spacetime a type of hyperbolic geometry. In Euclidean geometry, all parts of the metric have the same sign. No, this cannot be done in a consistent manner. In Euclidean geometry, for example, speeds add linearly - if you ride on a very fast rocket, and shine a torch light into the direction of motion, this would give you a superluminal ray of light. Obviously that is not what happens in the real world.

The boundary conditions in this case are simply that the spacetime must be asymptotically flat at infinity (i.e. reduces to Newtonian gravity), and smooth and continuous everywhere else, including at the inner and outer boundaries of the shell. That is what it means to have a global spacetime manifold. This continuity condition is crucial - if you have points where spacetime is not continuous and differentiable, then the field equations do not apply there, and the whole thing becomes internally inconsistent. When you account for the continuity condition, the metric constants become fixed automatically. For an example of how boundary conditions are used in GR to match solutions and ensure continuity, see §23 of Misner/Thorne/Wheeler, which deals with the interior Schwarzschild metric. This way is how I learned to do it, and I don’t see how it could possibly be “wrong”, so long as the result matches the situation in the Newtonian limit, and reproduces the Birkhoff theorem outcome, which it does. My reference would be the well-understood Newtonian limit. We already know that in the Newtonian case, the gravitational potential in the interior cavity is not the same as the one at infinity. The GR solution must reproduce this boundary case, since the g{tt} component of the metric tensor is directly related to gravitational potential in the Newtonian limit. If time dilation was zero in the cavity, as compared to a reference clock at infinity, you would end up with a contradiction in the Newtonian limit. Therefore, globally speaking, g{tt} cannot be the same inside the cavity and at infinity. For a more direct confirmation of what the consensus on this is, physicsforums.com would be a good place to ask, since that is where the actual experts in the field go. The general argument though is along these lines: https://www.quora.com/Does-a-hollow-sphere-of-mass-still-cause-GR-time-dilation-inside-it-even-though-there-is-no-net-gravitational-field.

The difference is the geometric length of the observers’ world lines, which connect the two events - and that length physically corresponds to what a clock travelling along that world line records. The longest possible world line between two given events is always a geodesic in spacetime - which physically corresponds to an inertial frame. However, an accelerating rocket does not trace out a geodesic, because it isn’t an inertial frame, due to the presence of proper acceleration. Therefore, the world line of the rocket is shorter, meaning less time is recorded by a co-moving clock; that is your time dilation. No complications need to come into this, just simply compare the geometric length of the world lines traced out by the two observers. You’ll find they are not equal. It’s that simple.

As a matter of fact, yes, I have (though it was years ago) - and my result was precisely the one I had described above, which also corresponds to the scientific consensus on this matter. It certainly did not involve any step functions though, I don’t know how you got that. To me that’s a red flag right there, because in general, step functions are not smooth and continuous (and hence not differentiable) at the points where the steps occur. Then you must have done something wrong, because your result would mean that the energy-momentum of the massive shell has a gravitational effect in one radial direction, but not the other, as seen from the shell. That is obviously unphysical. I’m also pretty sure (but would need to check in more detail) that it would be mathematically inconsistent, since it can’t be asymptotically flat at both infinity and some other point that is not at infinity, while remaining smooth and continuous everywhere in between, in a spacetime that is not globally empty. Surely you wouldn’t seriously expect to have a mass-energy distribution without any gravitational effect in some region just outside it, would you? Again, this would mean you did something wrong, because if you have a region of flat spacetime that is surrounded by a thin massive shell, then you are guaranteed to have a Schwarzschild spacetime in the exterior, due to Birkhoff’s theorem. If you don’t get this, then there is an error in your maths somewhere. I would say check your boundary conditions at the shell - from what I remember (again, this was years ago, and I don’t know where my notes from that time are gone), these were quite tricky to get right, and I got stuck at that point too. Do remember that the overall global metric has to be both smooth and continuous everywhere, including at the two boundaries, as well as the non-vacuum interior of the thin shell itself. In your case it looks like you have no gravity whatsoever in the interior cavity, and then suddenly gravity in the non-vacuum shell itself, meaning there is a discontinuity in your solution at the interior boundary of the shell. Perhaps that is why you ended up with a step function...? Check your maths again.

Yes, I did indeed say that the universe is fundamentally quantum. Nonetheless, in this thread I am only talking about classical gravity, being the kind we know about and can mathematically describe. Whether aforementioned topological principle plays a role in quantum gravity is not a question I can answer, since we don’t have a model for quantum gravity just yet. It remains to be seen.

In some sense, gravity is indeed the result of a conservation law of sorts, but it has nothing to do with energy. The law in question is the topological principle that the boundary of a boundary is zero. Misner/Thorne/Wheeler have described this very nicely in “Gravitation”. So far as energy is concerned though, it should be remembered that the energy associated with a region of curved spacetime is a difficult to define concept. It is also not localisable.

I can fluently switch without having to consciously think about it. I sometimes also mix languages, so I could have more than one language in a single thought. I’m nonetheless aware of myself using different languages, mostly because I am also a synaesthetic, and the German/English languages have very different patterns, colours, textures, and general “feel” to them (for me). There’s no mistaking their differences.

I disagree with this (and yes, I’m aware you are quoting somebody else here) - firstly, GR is about spacetime, not space. Secondly, it is most certainly not conceptualised as any kind of medium; the nature of spacetime is a collection of events, which are causally related in certain ways. This is very different from any kind of mechanical medium.

Yes, this is exactly what I said in my answer, and it is just a result of Birkhoff’s theorem. The local patch inside the shell is Minkowski, as is the asymptotic behaviour of the exterior metric at infinity. However, the global metric that spans the entire spacetime - shell cavity, shell, and exterior vacuum - shows that the locally flat patch inside the cavity still sits at a different potential than the asymptotically flat limit at infinity. Mathematically, if you look at the global situation, all components of the metric tensor are constants at asymptotic infinity, and they are also all constants in the interior cavity of the shell, but the two constants aren’t the same ones. Therefore, in this spacetime you have two regions that are flat (i.e. the metric tensor is isomorphic to Minkowski in those regions), but they nonetheless sit a different gravitational potential. Again, a table mountain - the top of which is as flat as the surrounding countryside, but nonetheless at a different level - is a good analogy for this. To fully understand how a clock at infinity compares to a clock in the cavity, one must look at the global metric that spans the entire spacetime, not just one local patch, because there is a local gauge freedom in what value the constants in the metric tensor can take. The constants are fixed by the boundary conditions at the shell, as well as by the asymptotic behaviour at infinity. If, at infinity, we choose the metric to be diag{-1,1,1,1), then inside the hollow cavity it will be diag{-a,1,1,1}, with some constant a that is determined by boundary conditions. That is where the global time dilation comes from. Alternatively - as you did - you can reverse this and say that in the interior you want to have diag{-1,1,1,1}, which is fine, but then you will get a different metric constant at infinity. Both metrics are still isomorphic to Minkowski (so they are flat patches of spacetime) - but they are nonetheless different. My advice to you - steer well clear of that site. It’s a repository for cranks and crackpots, with no scientific credibility whatsoever. If you want confirmation of your results, you are better off putting them to the experts on www.physicsforums.com, they will give you instant feedback.

If it cannot be observed or detected, then it has no impact whatsoever on the laws of physics - meaning it is irrelevant to physics as a discipline. Its existence or non-existence thus becomes mere speculation. This does not make any sense as a concept, because gravity is a geometric property of spacetime.

I honestly do not understand what you are actually trying to do here - the purpose of physics is to make models about physical systems, from which we can then produce new predictions that weren’t known before. If we first have to “watch” the system before we can describe it, then this completely defeats the purpose. Furthermore, the universe is inherently quantum and not classic, and in the quantum world you cannot “watch” the system before you describe it, because any act of observation will irrevocably change the system. Your overall idea does not make any sense to me. In the specific case of classical gravity, the motion of a test particle does not depend on its internal composition or energy configuration - it depends only on the geometry of spacetime. Furthermore, all relevant quantities in this context are covariant in nature (they are tensors), so they specifically do not depend on the observer at all. Lastly, gravity as we observe it in the real world would not exist if the universe had only three dimensions, so time most definitely is quite physically real.

I was born in Germany, but I think and dream mostly in English, having spent all of my adult life in other countries. It is only for specific purposes that I slip back into German - reason being that German has a much larger vocabulary, so sometimes it is easier to express subtle nuances in a language that has separate words for them. On very rare occasions I would also use other languages such as Chinese or Pāli, on account of them having words for concepts that do not exist in the European languages.

Again, I am unsure what you mean by “stretchy frames”. Do you mean patches of spacetime with a non-zero Riemann tensor?

How do you get g00=1 between the two masses? There’s a massive rod connecting the two black holes, so on account of this alone one cannot have zero time dilation in or near this region.

I don’t know what you mean by “elastic” - a reference frame has no such property as elasticity, it’s essentially just a way to set up a coordinate system. Yes, I do indeed mean a small enough patch of spacetime - my terminology was a little sloppy here. A patch is usually considered “small” if its dimensions are on the order of 1/a, where a is the proper acceleration. However, this is just a rule of thumb, not an exact definition, and it does very much depend on the specific scenario at hand. This is quite a broad area of study - is there a specific question you have? Gravity in close vicinity of a black hole substantially deviates from the Newtonian theory, so in most cases you will not be able to use the inverse square law. Tidal distortions in particular vary with an inverse cube law, in Schwarzschild spacetime.

That is not actually the right way to look at it. Time passes at exactly the same rate (1 second per second) in all reference frames, including non-inertial ones. The only thing that changes is the relationship between frames in spacetime - and that is what time dilation and length contraction are: a relationship between frames. They are not local phenomena, but global ones. The longest possible amount of proper time between two events is always traced out by a geodesic in that spacetime. In other words - a locally inertial frame in free fall.

You are correct, it would depend on the radius also. Sorry for not explicitly mentioning that!

Actually, you can, at least in principle. The equivalence principle tells us that uniform acceleration is locally equivalent to a uniform gravitational field - this is just the elevator example you brought. However, gravity due to sources of energy-momentum - such as planets, stars, etc etc - is not uniform, but tidal. Hence, if you are under the apparent influence of gravity, you can look for tidal effects to see whether this is gravity due to a source of energy-momentum, or just acceleration. Of course, if your reference frame is small enough, it will be difficult to detect the difference, so the equivalence principle always holds locally in a small enough patch of spacetime.

Acceleration and gravity are not analogous. You can have accelerated frames in flat spacetime, just as you can have approximately flat local frames in curved spacetimes. In fact, the geodesic equation is exactly the statement that acceleration vanishes for free-fall test particles. As for spin, it is a form of intrinsic angular momentum, but it is not a rotation in space (which would be meaningless in this context) we are talking about here, but rather a rotation in spacetime of the wavefunction itself. To be more explicit, spin arises from how the wavefunction that describes the particle must “look like”, in order to behave correctly under Lorentz transformations. For example, an electron is quantum mechanically described by the Dirac equation; the solutions of this equation are objects that transform like the (0,½)x(½,0) representation of the Lorentz group - in other words, they are bispinors, and hence describe particles with spin-½. A scalar field (=boson) on the other hand might arise from the (0,0) representation, giving it a spin-0. And so on. So spin essentially tells you what happens when you rotate a wavefunction in spacetime, which is determined by the type of object the wavefunction is, which is in turn given by Lorentz representation theory.

And if you look at it more closely, you will find that this is exactly what happens in the field of physics; new ideas are constantly proposed, but the vast majority is rejected again quickly because they either don’t work, are based on false premises, or offer no improvement on existing models. You see, the trouble in your particular case is that you see a flaw where in actual fact there is none. The wave-particle duality - which, by the way, applies equally to all quantum systems, not just elementary particles or a specific kind - is really just that: a duality. That means there are two equally valid ways to regard the same thing, and what you see depends on how you look at it. The actual nature of a quantum system is neither wave nor particle, and it’s also not “both” or “neither”. It’s simply a class of object that has no equivalent in the classical world of human perception. That’s all there is to it. The trouble here is that you think “well, they claim it’s both a wave and particle...but that’s impossible, because in my experience nothing can be both at the same time!”. But that’s a fallacy - you are attempting to apply classical thinking and logic to a system that is not classical; you then (rightly) notice that there’s a contradiction, and hence (wrongly) conclude that there must be some problem in quantum mechanics. But in reality there is no problem, and so there is no flaw to be corrected here. It’s merely your own conclusion that is invalid, because it is based on the wrong premise that classical logic applies to the quantum world. But we have known for nigh on a century now that it doesn’t. The wave-particle duality is much like shining a light on a cylinder from different angles, and looking at the shadow. From one angle, the shadow looks like a rectangle; projected from another angle, the shadow looks like a circle. You could come along now and say: “Hold on, this is not possible!”. But in actual fact it is, because the object itself is neither a rectangle nor a circle, nor both nor neither - it’s a cylinder. There is no contradiction. The same in QM - there is no contradiction or problem in the wave-particle duality, it’s just two different projections of the same thing, which is in itself neither one of the two. So without wanting to appear rude or anything, I can still tell you straight out that what you suggest will not be taken seriously by anyone in the physics community, because you are attempting to solve a problem that does not actually exist - it’s based on a misconception, a wrong premise. The other thing then of course is that it is in direct contradiction to many things which we already know - as it happens, the various particles do not stand in isolation, they are part of an overarching hierarchy and scheme, called the Standard Model. This model is certainly not perfect or complete, but it works exceptionally well within a specific energy domain, and has been extensively tested for the past 50 years. And what you suggest is simply not compatible with that model.

The fact that some observables are incompatible is not just an “effect” - it’s a fundamental feature of our world, and the defining characteristic that distinguishes a quantum system from a classical one. Pretty much everything else in quantum theory emerges from this fundamental feature, and the HUP merely serves to quantify it. A white hole (as the term is normally used) is a static construct, and cannot expand. Also, it can only exist in a spacetime that is completely empty, and asymptotically flat. This is not at all what we observe the universe to be like.