Markus Hanke

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

Yes, proper acceleration (unlike coordinate acceleration) is a quantity that everyone agrees on, so the coordinate basis does not matter. I was referring to the fact that, if an accelerometer co-moving with an observer measures something other than zero, then that observer cannot be inertial by definition. I understood the OPs comment to say that a particle can be in a state of proper acceleration, yet still be inertial - hence my comment. If you want to use transformation laws, then those can be found on this page (scroll down to the relations shown in red and blue). As mentioned before, accelerated particles will undergo hyperbolic motion in Minkowski spacetime, so the coordinate transformations use hyperbolic functions and their inverses. Alternatively, and that’s what I have been suggesting above, you can just adopt an appropriate coordinate chart and metric from the beginning. The Rindler metric is the most commonly used, but it is valid only for constant accelerations; for arbitrary accelerations you can use Fermi coordinates (e.g.).

Not necessarily true. A region of spacetime being special relativistic means only that the Riemann tensor vanishes everywhere within that region, so that spacetime is flat. It does not mean that all frames necessarily need to be inertial, or that there is only uniform relative motion. You can have non-inertial frames in flat spacetime, so SR is perfectly capable of handling acceleration. Not true. If a frame is inertial, then by definition there is no proper acceleration. There can, however, be coordinate acceleration, but that is an artefact of how we choose to label events, and not something that a physical accelerometer would measure. This is incompatible with (1) and (2) in your last post. An accelerated particle undergoes hyperbolic motion in Minkowski spacetime, so you can’t just make the gamma factor variable, but retain a Minkowski coordinate chart (with proper time) and its corresponding metric. I suggest what you should do is start with the correct metric ansatz - i.e. a Rindler or Fermi metric -, and work through it again, and see what result you then get.

The combination of coordinate chart and metric you are using as an ansatz along with your constituent relation \[\gamma=\frac{dt}{d\tau}\] already implies inertial motion from the beginning, so the end result is hardly surprising. If you want to allow for the possibility of uniformly accelerated motion, you need to use a Rindler chart; and if you want to consider arbitrarily accelerated frames, you need to use a Fermi chart to cover your spacetime, along with the appropriate constituent relationships between the components of the velocity and acceleration vectors in each chart.

Of course...that’s why I was saying I was going off on tangents But you have to admit that oftentimes tangents are fun. There’s only one gauge condition in spacetime, because there’s only one potential field and one electromagnetic field. But once we decouple this into E and B fields, the gauge condition also decouples into a constraint on the temporal part of A (which is the scalar potential), and a constraint on the spatial part of A (which is the Newtonian vector potential). The U(1) symmetry applies to the electromagnetic field in spacetime, but not to the E and B fields in Euclidean 3-space. I think the conclusion is the same though, because of this: consider the Lorentz gauge (e.g.) \[\partial_{\mu}A^{\mu}=0\] This is a second order partial differential equation, so to obtain a unique solution, you need to supply precisely two boundary conditions. Since by definition in general field theory (omitting any constants) \[dA=F_{\mu \nu } dx^{\mu } \land dx^{\nu }\] the A’s should be fully determined, given Maxwell’s equations (which now essentially become relations between the components of A). So the boundary conditions above should be the only physical DOFs in the theory. I have not researched this, I’m just pulling this out of my hat as I go along, so it’s possible - likely even - that I’m missing something. This is strictly my own two cents. EDIT: Sorry, the above is of course nonsense, this isn’t a second order PDE at all. What I meant to say is that this conditions leads to the constraint equation for A, which is \[\square A^{\alpha } =\mu _{0} J^{\alpha }\] This requires exactly two boundary conditions for a unique solution.

Ok, point taken Actually, I feel a bit outside of my comfort zone here, since I was never particularly interested in EM theory. However, I am tempted to argue that in actual fact EM fields can have no more than 2 physical DOFs. Given conservation of sources, the EM field arises uniquely from a vector potential and a scalar potential - that’s 3+1=4 “free” quantities. But, both of these potentials are invariant under suitable gauge transformations, so in each instance we can arbitrarily pick a gauge, which eliminates one DOF each - leaving only two physical DOFs in the end. This seems to also be consistent with the fact that EM waves have exactly two possible polarisation states at a 90 degree angle. But perhaps I’m getting this all wrong, I’ve never really thought about this in any detail.

I have studied GR in some detail, so I am aware of all these possible scenarios; by personality I also tend to be a “natural worrier” who easily gets anxious even over minor things and life events. In addition, I am an Aspie too. Yet I feel no sense of depression, worry or anxiety over the possibility of a cyclical cosmology. Furthermore, you need to remember that the observational evidence we have at the moment is much more consistent with other global topologies, and not a cyclical universe. I believe you when you say that you yourself might find such an idea depressing, but remember that this does not imply that others necessarily relate to the concept in the same way. Most of us here understand the notion of a cyclical cosmology well enough, but don’t find it depressing. I find that people often tend to miss the salient point of Schopenhauer’s philosophy - he did not advocate despair, depression, or absolute nihilism. The main point he was trying to make was about acceptance. It is strictly necessary to fully understand and acknowledge the inherent limitations of the human condition - such as the impossibility to permanently satisfy desires and craving, and the futility of constant strife towards some ideal utopia -, but then it is also necessary to accept them for what they are, and thus arrive at a position of peaceful coexistence with those limitations. Philosophical pessimism does not imply despair and meaninglessness. And of course, philosophical pessimism is only one possible life philosophy, which is by no means shared by everyone.

I think it is better to look at it as a 4-dimensional geometric object in spacetime (the electromagnetic field), which is mathematically described by a rank-2 object (the EM field tensor and its dual). The electric and magnetic parts of the field aren’t independent, they are actually the same object seen from different vantage points, so the 6D description is a little problematic, as the basis vectors of that state space wouldn’t all be linearly independent. But that’s just me going off on tangents again 😄

This kind of depends on what exactly you mean by “seen to stop”; in particular it depends on which clock you use, as measurements of time are always purely local in curved spacetimes. As measured on a far-away, stationary clock (Schwarzschild time), the in-falling test particle never reaches the horizon, as that would require an infinite amount of Schwarzschild time. As you correctly say, it will just be “seen” from this vantage point to move more and more slowly as it approaches the EH, and become more and more red-shifted. On the other hand, for a clock co-moving along with the freely falling test particle (Gullstrand-Painleve time), nothing special happens at all - it will record a finite and well defined amount of proper time to reach the horizon, and also a finite and well defined amount of proper time to the “crunch” at the centre. The horizon itself won’t even be easily noticeable by such an in-falling observer; it’s locally as smooth and regular as any other region outside the BH. It’s only when they fire their thrusters in order to get back out to “far-away” that they will notice something being amiss, because no matter how much radial thrust is applied, they cannot stop themselves from falling further in; they can’t even remain stationary wrt to some external reference.