Markus Hanke

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.

Well, they are useful to represent the perspective of a stationary observer at infinity. The crucial point to realise is that that is the only situation where measurements of space and time made in Schwarzschild coordinates actually coincide with what physically happens, since such measurements are always purely local in curved spacetimes. Anywhere else other than for a stationary observer at infinity, Schwarzschild coordinates are only a bookkeeping device, but they do not necessarily reflect what actually happens there, locally speaking. They also don’t cover the entirety of the spacetime. Many students of GR either do not understand this, or refuse to acknowledge it, since it goes against Newtonian intuition. The unfortunate result is all manner of misconceptions and misunderstandings. So you have good reason to mistrust Schwarzschild coordinates - they can be useful in certain circumstances, but they are also dangerous if not understood correctly. I think the only reason why pretty much every GR text uses them is because they are algebraically simple.

Yes - this is trivially true, since a vanishing Riemann tensor means you are on a flat manifold. More generally, if a tensor vanishes then so do all of its contractions - this is true for any tensor. Not really. It establishes only that its contractions vanish if Riemann itself vanishes. The reverse is not true, however - the vanishing of the Ricci tensor and/or scalar do not necessarily imply that Riemann is zero. Ricci flatness is a necessary but not a sufficient condition for the absence of Riemann curvature; to make it a sufficient condition, you need to demand the vanishing of Weyl curvature as well. The Ricci tensor is the trace of Riemann, whereas the Weyl tensor encodes the trace-free part of Riemann (the decomposition isn’t exactly trivial, though).

Ok. As a little tip - conventionally, writing (t,x,y,z) will imply Cartesian coordinates to most readers; if you want to indicate a general coordinate basis, it is better to use the notation \({x^0,x^1,x^2,x^3}\), as it is less ambiguous. Ok, that’s an important difference. I’m not sure what you mean by “relative sizes”?

As I said previously, my response was based upon my own understanding of the OP. If it completely missed the point, then it is the OP’s job to clarify things. Unfortunately this poster seems to be in the habit of opening threads, and then abandoning them after a couple of comments are made. Yes that’s pretty much it. Accelerated motion involves a little more than simply allowing gamma to vary.

I’m sorry, but I can’t make heads nor tails of this at all. What kind of coordinates (t,x,y,z) are you using here? Are these Cartesian coordinates, or Schwarzschild coordinates, or something else entirely? And what is your reasoning behind the metric (1)? This form of metric is incompatible with both Schwarzschild coordinates and Cartesian coordinates in a Schwarzschild spacetime, so please explain how you arrived at this ansatz.

I am uncertain what it is really is that the OP is trying to show, so I can’t guarantee this. What I presented is my own understanding of what he has posted. Yes of course, but that isn’t what the OP has been doing. That is kind of my point. He’s essentially using an inertial coordinate system to show that there is no acceleration - which is trivially true. Ok, but then, what is his point? He starts with a metric in Cartesian coordinates, then manipulates it using relations that imply an inertial observer, and ends up with the conclusion that there is no acceleration...? I’m not sure I understand your question. Both charts cover the same spacetime, so the difference is merely one of coordinate basis. Rindler is what you naturally get when you apply the proper transformations that arise in a frame with non-zero constant acceleration (since such frames are not related by Lorentz transformations): \[ct\rightarrow \ \left(\frac{c^{2}}{a} +x’\right) sinh\left(\frac{at’}{c}\right) \] \[x\rightarrow \ \left(\frac{c^{2}}{a} +x’\right) cosh\left(\frac{at’}{c}\right) -\frac{c^{2}}{a} \] The advantage here is that the world line of such a uniformly accelerated particle is one of constant x (in this coordinate basis!) as t “ticks along”, so in some sense a particle in this coordinate frame is “at rest”, even though it is uniformly accelerating. It’s the most natural choice for uniformly accelerated motion. Very simply put, the line element \(ds^2=...\) does exactly what it says on the tin - it is an infinitesimal section (element) of a curve in spacetime. To obtain the entire geometric length of a curve C, you integrate: \[L=\int _{C} ds\] To describe a uniformly accelerated particle, you can do one of two things: 1. Keep the line element in Minkowski coordinates, but make C a hyperbola to describe the motion 2. Use the line element in Rindler coordinates, which makes C a “straight line” in these coordinates Of course, the result will be the same (since this integral is an invariant), so both are valid, but the computational effort differs - it turns out that option (2) is very much easier to do. It’s simply the more natural choice - in much the same way as (e.g.) spherical coordinates are the natural choice to describe the surface of a sphere, rather than Cartesian coordinates.

It’s how I understood it, based on the fact that the metric given is of a form that would generally be used by an inertial observer, so it is natural to assume that these are Minkowski coordinates, and not hyperbolic ones. The geodesics calculated from his metric ansatz are straight lines, not hyperbolas - unless the coordinate basis is not Cartesian, but the OP never indicated that. This is also consistent with the OP’s conclusion: “[...] we see that the particle cannot accelerate”, which is of course trivially true, based on that metric.

Oops, sorry