Markus Hanke

The Schwarzschild solution is static and stationary, and it requires some boundary conditions that one wouldn’t find in the real world for this type of event - so it is not a suitable model to describe the actual collapse process itself. It can only be used to approximately model either the end result or the original gravitating object, i.e. the original star or the resulting black hole, which are then assumed to be static and unchanging. Describing the collapse process itself is a very much more complex task, which can really only be done using numerical methods. The reason why the event horizon radius in Schwarzschild spacetime happens to agree with a suitable Newtonian calculation is that this type of spacetime is spherically symmetric and asymptotically flat. But this is not necessarily true in general - if you relax any of the boundary conditions of Schwarzschild (e.g. allow the central mass to carry angular momentum, or electric charge), the location of the event horizon will no longer agree with Newtonian theory. In fact, you will quickly find that there is in fact more than one horizon surface, giving quite a complex structure. The existence, nature, and location of horizon surfaces is intrinsically a relativistic phenomenon which has to do with the causal structure of the spacetime in question. It can, and (given certain conditions) inevitably must - but you have to use the correct model.

If you look carefully, you’ll notice that he hasn’t really addressed most comments made so far. He either ignores them, or just repeats earlier posts.

Yes, that’s not what I am trying to do - though I do think it is possible to a large extent to talk about gravitational physics without making reference to coordinates. But of course, you need a suitable choice of coordinates to extract numerical predictions from the model, I wouldn’t deny that at all. What I am attempting to emphasise is merely that the coordinate system doesn’t encode the actual physics on your spacetime, anymore than street names on a map encode the layout of a city; so while it might be necessary in practical terms, it isn’t fundamental as such. The physics are found in the metric and the connection. This is ultimately (in a physical sense) why one is free to change coordinate basis without affecting any physical prediction the model makes.

I’ve also pointed this out to him already on his “Norm of 4-Acceleration” thread, and gave the explicit expressions for the necessary transforms of x and ct, but he ignored it completely. In fact, so far he has ignored pretty much everything that has been pointed out to him - it doesn’t seem like there is any genuine openness to feedback. It feels more like a personal blog to me. Just for the casual reader’s reference - in the special case of motion starting from rest at the origin, with uniform acceleration, the integral becomes straightforward, and the expressions evaluate to: \[v( \tau ) =c \tanh\left(\frac{a\tau }{c}\right)\] \[v(t) =\frac{at}{\sqrt{1+\left(\frac{at}{c}\right)^{2}}}\] Which is, of course, hyperbolic motion, as we would expect in a Minkowski spacetime.

Yes, and it also doesn’t imply that Riemann is always identically zero for all manifolds. Which it evidently isn’t.

Well, in a pure mathematics context, the metric has a pretty abstract definition in terms of something called the “first fundamental form” - which essentially boils down to defining the notion of an “inner product” on your manifold. Physically, this means that a metric enables you to quantify lengths, areas, volumes, angles etc - so it makes it possible to relate physical laws to real-world measurement outcomes. More specifically in GR, it is used to quantify the separation between events in spacetime. Given a connection, this determines all the various curvature tensors etc. I must reiterate that it is not my intention to deny that some coordinate system needs to be chosen in order to actually perform the calculation - however, the point is that the physics aren’t in the coordinate system itself, they are in structure of the metric, i.e. in how the components of the metric tensor (in a given coordinate basis) are related to one another. This is an invariant property, unlike the value of the components themselves. I meant it in this sense. Indeed, that’s pretty much what I am attempting to point out, with the addition that in the context of GR, the metric is the more fundamental object. Indeed.

Ghideon has essentially said it already (+1) - gravity isn’t actually a force (except as an approximation in the Newtonian domain), so you can’t model it using a vector field approach, such as the one you have presented. It also exhibits different dynamics than electromagnetism, some superficial similarities notwithstanding. It is, in fact, possible to mathematically show that no vector field theory can accurately model gravity, since such models are intrinsically incapable of encoding the necessary degrees of freedom (for fundamental, albeit technical, reasons). You need at the very least a rank-2 tensor theory, such as GR for example.

Yes of course, that much should be obvious to all of us. It was not my intention to imply anything different. I meant my comment in the sense that the structural elements I had listed are what I think (!) is minimally required to construct the specific model of General Relativity, as opposed to other models of gravity. For example, change the Levi-Civita connection to a Weizenböck connection, and you get “Teleparallel Gravity”. So I meant it in that sense. To be honest, the list I gave was what spontaneously popped into my head when I read your previous post, I haven’t actually thought about it too deeply. You are right that diffeomorphism invariance is important, but rather than putting this as a separate requirement, I would amend (3) to say that every small enough patch is to be locally Minkowskian. Along with the automatic conservation of the Einstein tensor, this would imply both the correct metric signature, as well as diffeomorphism invariance, and the smoothly connects to SR as a limiting case. As for the coordinate system, I think this is already implied by the requirement to have a metric. The important point is that you can choose any suitable coordinate basis that fulfils basic analytic conditions of continuity and differentiability, so I think a choice of coordinates is secondary to the metric. One could easily formulate all of GR in abstract geometric language, without any reference to specific coordinates at all - as Misner/Thorne/Wheeler have done in their text “Gravitation” for example. Is it really? I would tend to think that it is only the metric that is essential, but not any specific coordinate system. What makes spacetime is its structure, not the labels we assign to each event. The conservation of the stress-energy tensor follows from Noether’s theorem, so it will locally hold whether gravity is described by GR or not. The question whether the automatic (!) conservation of the Einstein tensor is a necessary condition, is far more interesting. I think that it is, because by demanding this we essentially fix the general form of the local constraint on the metric (up to constants), i.e. the Einstein equations. If we don’t demand this, then we no longer have a clear mechanism to couple stress-energy to any specific notion of curvature, and the field equations could take other forms - for example, they could contain higher powers of Riemann and its derivatives (such as e.g. Lovelock gravity), or functionals of Ricci (such as f(R) gravity e.g.). So I think the requirement is necessary to uniquely recover GR, as opposed to other possible metric models. I think it is worthwhile also to note that this conservation requirement has an underlying deeper structure, being the topological principle of "the boundary of a boundary is zero". Again, Misner/Thorne/Wheeler have a very good presentation on this in "Gravitation". The point here is that this is not a completely arbitrary requirement. Spacetime is a mathematical model, it is not any kind of physical substance or object - so I don’t think “what is it physically made of” is necessarily a meaningful question. Nonetheless, if I was to attempt a classical answer, then I would say spacetime is a network of relationships between events, and those relationships are constrained in specific ways. In a deeper quantum sense, one could go with one of the newer ideas, for example that spacetime arises from entanglement relationships, or that it is a spinfoam network. Or perhaps one can look at it in terms of information. Upshot is, I think it is reasonable to (preliminarily) say that the ontology of spacetime is a mathematical model, whatever specific form it might take. This of course raises the question - is spacetime an actual feature of the physical world at all, or is it just the mind’s way to structure information and construct its PRM (phenomenological reality model - ref Thomas Metzinger)? Because that’s all our directly experienced “reality” is - a model constructed by the mind. How this maps to an external reality (if that is even a meaningful concept) is anyone’s guess. Indeed! Subjective phenomenology (=direct experience) being another example.

1. A topological manifold 2. The Levi-Civita connection 3. A metric with the correct signature 4. A local constraint on the metric which guarantees the automatic conservation of the Einstein tensor (=the Einstein field equations) This is pretty much the minimum structure required to get GR, as opposed to other models of gravity.

This doesn’t make any sense. The apparent motion of the sun through the sky in the course of a day is due to Earth’s rotation, not due to relative motion between Sun and Earth. So the Sun-Earth distance does not come into this at all. The issue of course is that Flat Earth rejects the notion of a rotating planet, so pointing this out will just result in hand-waving dismissal. My advice: don’t bother. I’ve been there too, and all it ever resulted in was unnecessary grief. That particular community rejects not only basic scientific observations (such as gravity e.g.), but even the very scientific method itself; there simply isn’t any common ground to base a meaningful debate on.

Let’s look very briefly at what the symmetries of Riemann actually constrain. We have two sets of symmetries - first, those that are present even in the absence of a metric: \[R{^\alpha}{_{\beta \gamma \delta}}=R{^\alpha}{_{\beta [\gamma \delta]}}\] \[R{^\alpha}{_{[\beta \gamma \delta]}=0}\] \[R{^\alpha}{_{\beta [\gamma \delta ||\mu]}}=0\] Second, we have metric-induced symmetries: \[R_{\alpha \beta \gamma \delta}=R_{[\alpha \beta]\gamma \delta}\] \[R_{\alpha \beta \gamma \delta}=R_{\gamma \delta \alpha \beta}\] \[R_{[\alpha \beta \gamma \delta]}=0\] What this means: in dimension n, every index pair can take on n(n-1)/2 independent values (due to their anti-symmetry), which initially leaves us with a symmetric matrix with \[\frac{1}{2}\left(\frac{n( n-1)}{2}\right)\left(\frac{n( n-1)}{2} +1\right) =\frac{n( n-1)\left( n^{2} -n+2\right)}{8}\] independent components. The Bianchi identities (last relation in non-metric symmetries above) constrain a further n!/(n-4)!/4! components. This finally leaves us with \[c_{n} =\frac{n^{2}\left( n^{2} -1\right)}{12}\] functionally independent components of the Riemann tensor. In n=4 dimensions, this evaluates to 20. So on a spacetime manifold with 4 dimensions, the symmetries of Riemann leave 20 tensor components unconstrained and functionally independent, meaning those components are not identically zero in the general case. Hopefully this clears things up, since it is trivially obvious that geodesic deviation does in fact exist in the real world (contrary to the OP’s claim), just like the theory says it does.

Indeed not. The cosmological model we are using (the Lambda-CDM model) is, at large scales, based on homogeneity and isotropy.

All you have done is repeat what you had already stated, you did not actually address any of the points raised.

I would tend to agree.

Again, this essentially comes down to the difference between topology and geometry. When we say the universe is spatially infinite, what we actually mean by this are three things: 1. Spacetime has no boundary 2. For any arbitrary pair of (spatial) points {A,B}, there exists another pair of points {C,D} the spatial separation of which is greater than that of {A,B}. 3. Spacetime is singly connected Herein, (2) actually implies (1), but I’m listing them separately for added clarity. These three conditions are true at all times t>0, including immediately after the BB, and at the present time; so this does not change, and it - roughly - represents an aspect of the global topology of the universe. On the other hand, when we say that the universe was singular at the BB, what we mean is that as t -> 0, the separation between any pair of arbitrarily chosen spatial points will tend towards zero; and it means that no geodesics can be extended beyond the hyperslice t=0, without them extending into the future again (so this is a bit like a “pole” in spacetime). It does not really mean - at the danger of straying into the disciplines of metaphysics and philosophy here - that only a single point existed; the spacetime manifold was already there in some sense, but there was no notion of “separation between events” yet. So it’s the geometry that was singular, but not necessarily the topology. Of course, this is the purely classical picture, it does not account for any quantum effects (which will likely change the story quite radically).

Good point!

This is in “Speculations”, so it should be related to science in some way. There are other forum section for general conversation.

And the point of this is...?

It is important to remember that this is only an analogy - it’s a pedagogical device that can be useful to illustrate certain aspects of certain spacetime geometries (it’s called the “waterfall analogy”), but it isn’t to be taken literally. You will find this analogy often used for black holes that carry angular momentum and/or electric charge. In actual fact though spacetime isn’t some kind of mechanical fluid, and it is not embedded into anything higher dimensional, so it doesn’t physically “flow” anywhere. The proper mathematical description of spacetime geometry doesn’t use such concepts. It’s just a visualisation aid.

I don’t fully understand the question - what do you mean by energy density being “spatially infinite”? In general terms, we need to distinguish between local geometry, and global topology of the universe. When we say that the universe expands, what that actually means, in general terms, is that measurements of spatial distances are dependent on when they are made; so if you pick two arbitrary (not gravitationally bound, and at relative rest) points in space, and measure their separation (e.g.) 10 million years after the BB, and then again 10 billion years later, you will get a different result, even though you are using the same pair of points, and the same ruler. That’s geometry. The global topology roughly speaking tells you something about the overall “shape” of the universe - it is constrained, but not necessarily uniquely determined, by local geometry. It also doesn’t change - if the universe had a certain topology at the beginning, it will still have that same topology later on.

To me it is an enigma why you keep trying to propose things that are just trivially wrong, and easily contradicted by simple examples. Even in ordinary Euclidean space (the kind you deal with in high school), the metric tensor is evidently not null. Have you considered what it would mean for all tensors to be null? It would mean (among other things) that there would be no notion of distance, area, volume, or angles, and hence that the length of any arbitrary curve in that space is identically null, irrespective of your choice of parametrisation. It would be a world without any concept of separation in space, or duration in time. This is evidently nothing like the world we actually live in, so why would you make such a claim? This is only true across two indices though: \[A_{\alpha \beta \gamma ...} =A_{( \alpha \beta ) \gamma ...} +A_{[ \alpha \beta ] \gamma ...}\] See above - the vanishing of symmetric and antisymmetric parts does not necessarily yield a null tensor. Not true. For example, the Minkowski metric is locally the same in all reference frames, yet it is most certainly not a null tensor. It doesn’t need to. For example, you can cover a patch of Schwarzschild spacetime with a Gullstrand-Painleve chart, or with a Schwarzschild coordinate chart; they are related by a simple coordinate transformation, but the former contains off-diagonal elements in the metric, whereas the latter doesn’t. They both describe the same spacetime.

Unfortunately not, at least not that I know of. It’s a little bug bear of mine too, but not a massive issue.

I think the problem is one of physical meaning. It is not enough to just blindly do index gymnastics, if one does not understand the physical significance of the objects that one is manipulating, as well as of the manipulations themselves. All physical situations can be described mathematically, but not everything that is mathematically doable is automatically physically valid or meaningful. This has been a recurring issue across all of the threads he has opened here. It also isn’t clear what the actual claim really is, which has also been an issue on the other threads. Based on his last sentence, I understood the OP’s claim to be that if the Riemann tensor is zero, then all contractions derived from it are zero as well, which is trivially true. Only when reading Kino’s excellent (+1) post did I realise that what he actually seems to claim is that Riemann is always zero - which is of course both trivially wrong and physically meaningless. It is, in fact, so meaningless that it didn’t even occur to me that this might be what he trying to show!

Quite possibly so! To be honest, I am still struggling to understand what it is the OP is actually trying to do. It also seems he has abandoned this thread, just like all the other ones he opened before this. Indeed - very well summarised! +1

Yes, you can indeed - the result is the hyperbolic transformation I gave earlier. This just doesn’t seem to be what the OP was doing or getting at - and if it is, then the conclusion he reaches is of course meaningless (“in this article we see that the particle cannot accelerate”). It is, however, correct and fully consistent (albeit trivial) with the absence of proper acceleration in an inertial frame. Hence my reading.