Markus Hanke

Riemann curvature is a rank-4 tensor, so it's little bit more complicated than this - but yes, its components in a particular coordinate basis can be calculated via derivatives of the metric. The formal definition of this object is usually given as the commutator of the covariant derivative, which is much more intuitive. Sure. But what does this have to do with my original comment you quoted me on? I fail to see the connection (pun intended). Yes of course. But you'd need extremely strong fields to get any kind of appreciable gravitational effect.

Just two things to note here: 1. Black holes don’t have a surface, they only have horizons of various kinds 2. The inverse square law doesn’t really apply in the immediate vicinity of a black hole; this is very much a relativistic scenario, so you’d have to use the full machinery of GR to correctly describe it

Any form of energy-momentum will be a source of curvature, not just mass. For example, if you have an electromagnetic field in an otherwise empty region of space, then this will have a gravitational effect too. Also, curvature has nothing to do with displacement, it’s purely a geometric phenomenon - it’s about geodesic deviation, i.e. the failure of initially parallel geodesics to remain parallel.

Yes, obviously a clock comoving with the test particle will show a non-zero reading, even though the world line connects back to the same event in spacetime. This is why I said before that the world line would be of non-zero geometric length.

The outcome is the same. The statistical distribution of many individual hits on the detector. The which-path information does not need to be physically observed, it only needs to be accessible in principle, for there not to be an interference pattern.

It returns to the same starting point at a later instant in time. In a CTC, it would return to the starting point at the same instant in time wrt to a clock stationary there - so the situation is different from an ordinary oscillator.

As I said, the outcome does not in any way depend on the experimenter, but only on whether or not which-path information is available within the apparatus.

The outcome of the experiment does not depend on the particulars of the detector apparatus, or even on the layout of the setup itself; it depends only on what information is accessible to the experimenter. So long as the only accessible information is position and time of each individual detection event at the receiver device, a large ensemble of such events will always form an interference pattern. On the other hand, if any form of which-path information is available from the setup, then there will be no interference pattern.

It's a world line of non-zero length connecting the same event. This means the world line will return to its original position in space at the same instant in time.

Due to the effects of gravity. Remember that in General Relativity, all forms of energy-momentum are sources of gravity, not just mass; so whether the universe contracts, expands, or stagnates just depends on the average energy-density, even if there is no conventional form of mass). Note also that in this scenario there would be no singularity; contraction just turns into expansion below a certain minimum scale.

A wormhole and a closed time-like curve are very different topological constructs - though you could conceivably use wormholes to ‘build’ something that behaves like a CTC, at least in principle. I don’t know how to answer this - I cannot think of any way to time travel without creating a paradox. Even the mere presence of a time traveler would already create all manner of problems, never even mind accidental or intentional interference in events.

Quantum entanglement is neither to do with non-locality, nor with action at a distance; it has to do with non-separability of states, and statistical correlation of measurement outcomes. This difference is crucially important. The outcome of each measurement is subject to the usual rules of quantum mechanics, so the entanglement is not apparent to either one of the observers until they compare measurement outcomes - which is of course only possible at or below the speed of light. So no exchange of information at superluminal speeds is possible; and of course it can’t be, since entanglement features in relativistic quantum mechanics.

The concept of 'electron' as we observe them today does not make a whole lot of sense prior to electroweak symmetry breaking, so they are necessarily younger than the universe. Also, not all electrons are remnants from the early universe, they also appear as the result of decay process - to name just one example, a muon (which is an elementary particle!) decays into an electron plus two neutrinos. So electrons can be as young as you want them to be. There is no such thing as 'outside the universe'. No. Quantum entanglement does not allow for the instantaneous transmission of information.

I have several issues with this statement, and other careless statements like this. First and foremost, GR does not predict the existence of CTCs; it's rather the other way around, in that CTCs are consistent with the laws of GR, in the sense that such spacetimes are valid solutions to the field equations. The problem here is that a) not every solution to the field equations is necessarily physically realisable, and b) spacetimes containing CTCs are known to be highly unstable under even miniscule perturbations of initial conditions - and those conditions are only approximately true in the real world to begin with. Personally I would be very surprised if CTCs existed in the real world, and they are certainly not "predicted" by GR. Secondly, in what sense would a CTC be a "time machine"? A topological construct like this would connect an event to itself via a world line of non-zero length; not only would you not be able to travel forward or backward in time with this in a global sense, you would in effect be doomed to just 'loop through' that same event over and over again, without any means of ever escaping (neither spatially nor temporally). To me, this is actually the opposite of a time machine - it's like a Groundhog machine, if that reference makes sense. In ordinary spacetime, we age forward in time, so at least we travel through time in that (very limited) sense; when trapped in a CTC, you can't even do that much, you can only repeat the same cycle again and again. Thirdly, they seem to forget to mention that CTCs can exist only in vacuum; so even if you could physically realise such a spacetime (a possibility at least in principle), it would be impossible to introduce any kind of test particle without collapsing the spacetime into some more conventional geometry.

This issue is notorious, and you will find large numbers of papers written about it, many of them arriving at diametrically opposed answers. To make a long story short - yes, the electron in "free fall" does in fact radiate. The equivalence principle does not apply to electrically charged test particles, and when one does the actual maths for this (which turns out to be a surprisingly complex and subtle endeavour), one finds that the electron does not actually trace out a geodesic of spacetime (so it isn't truly in free fall), and is in fact surrounded by a radiation field. Perhaps even more surprisingly, an electron at rest within a curved spacetime background (e.g. an electron confined in a vacuum tube at rest relative to earth) does not radiate, even though a comoving accelerometer would show non-zero proper acceleration. In some sense this is expected, since anything different would violate local conservation of energy; nonetheless, it is somewhat counterintuitive result. This is one of those cases where common sense and intuition are at odds with GR, and one has no option but to work through the (extremely tedious, in this case) maths.

I do not really wish to get involved in this discussion, as I believe that understanding the human condition should not become a partisan issue. But I do wish to offer two observations: 1. It seems that almost everyone here equates religion with theism, or (even more narrowly) with Christianity. This is misguided - all theistic world views are to some degree religious, but not all religions are theistic, or even supernatural. There are religious systems that are expressly empirical, right here and now in this lifetime. I think it is important to clarify what you all actually mean by 'religion', in the context of this debate. 2. For those of you who know me from here and other forums, you will probably agree that I am all about science - it's a huge part of my life, and I spend a lot of time researching and teaching myself physics, and that's not likely to ever change. Nonetheless, there is also a religious dimension to me - in fact, I live full-time in a monastery, and have plans to ordain as a monk in a contemplative order next year. This dimension is equally as important to me as is science. For me personally, there has never been a conflict between the scientific and the religious/spiritual sides of me. I understand them as complementary domains of enquiry, that ask different questions about the same human condition. My scientific enquiries have helped me gain insights on my spiritual path, and the spiritual practice has helped me gain new angles on scientific issues. So, for both the religionists who reject science, and the scientists who reject anything religious, you need to ask yourselves the question - why does this need to become/continue to be a partisan issue? The "us vs you" mentality isn't helpful, and can - if taken to the extreme - frequently be dangerous. But when approached with wisdom, the two sides have the potential to coexist harmoniously, and inform each other constructively. Just my humble opinion and experience

Acceleration is not change in speed, but change in velocity. That's an important difference - velocity is a vector that, at a given instant, is tangential to the world line of the test particle at that instant in time. Speed is the magnitude of that vector, so it is just a real number. Consider two frames in spacetime, which are initially inertial. You can go from one frame to another by performing a Lorentz transformation; geometrically speaking, such a transformation is simply a hyperbolic rotation in spacetime. So in other words, the relative speed between two inertial frames can be geometrically interpreted as a rotation angle - you (hyperbolically) rotate space into time, and vice versa, as you go from one frame to another. This is where length contraction and kinematic time dilation come from. Applying acceleration to one of the frames means you are changing that rotation angle - and while the original angle itself depends on the relationship between the frames (so it is relative to the frames), the change in that angle is not relative to anything. It's like an oven - when you first turn it on and set a temperature (say e.g. 200 degrees), then that temperature is always relative to the room temperature it started off at. So for example, if you set it at 200 degrees starting at 20 degrees room temperature, you are actually setting it at 493K relative to absolute zero. So it's a relative value. But when you later on decide to dial up the temperature by another 20 degrees, then that change is not relative to anything - you simply add an absolute amount of 20 degrees, regardless of what the original temperature was. The change is by an absolute amount of 20 degrees, it isn't relative. Yes - once you 'get' it for the first time, the concept is very intuitive. Velocity in spacetime is a 4-vector, which has 1 component for time, and 3 components for space. When you apply acceleration, what you are effectively doing is changing the 'mix' of time and space components, so in general terms you are changing the relationship between 'motion in time' and 'motion in space', or (if the magnitude of the vector remains constant) you are changing the mix of the three spatial components only (=change in spatial direction). While this change is happening, the test particle is not in free fall, so the geometry of the world line it traces out in spacetime is different from that of an inertial frame, until acceleration is zero again.

There is no 'paradox' at all. It quite simply means that the two twins connect the same two events by two different world lines in spacetime - and since the accumulated reading on a clock is always identical to the geometric length of the world line traced out by that clock, their respective ages at the end of the experiment will differ. That is all there is to it - two world lines of differing length connecting the same two events. It's purely down to geometry. Yes, but acceleration is not. When you hold an accelerometer in your hand, the reading it shows is a proper quantity, it is not relative to anything else. Speed on the other hand is not a proper quantity, it is a relationship between frames, so it is relative to some reference point. What instruments measure locally in their own local frames without reference to anything else are called proper quantities; quantities calculated with respect to some other reference point are called coordinate quantities. For a thorough grasp on the theory of relativity, it is crucially important to understand the difference between these. Indeed - this is essentially like firing a rocket thruster. No, it's not relative to anything. An accelerometer carried along with you would have registered and recorded that acceleration locally in your own frame, irrespective of how it physically came about. Yes, you would have accumulated less proper time (aged less) between two fixed events, in comparison to some other reference clock which connects the same events, but without throwing a ball. In other words - you would have traced out a shorter world line between the same set of events.

An accelerated observer is one who traces out a world line that is not a geodesic of the underlying spacetime. It is thus fundamentally geometric in nature, and has nothing to do with forces, virtual particles etc. Thinking of it in terms of varying velocity - though formally correct - isn't especially helpful in understanding the actual physics at play, IMHO. Acceleration is simply a measure of a test particle's failure to be in free fall. An accelerating observer will measure a non-zero reading on a comoving accelerometer, someone in an inertial frame will show zero on his instrument. So these two situations are physically distinct, and easily distinguishable - there is no symmetry. Yes.

I think just demanding the interval to be invariant is not enough to uniquely determine the geometry of the underlying manifold. For example, you can write out this interval in ordinary 3D Euclidean space, the kind you learn about in school - which just gives you the Pythagorean theorem. The theorem is obviously true no matter what coordinate system you use, so there is also a notion of invariance there - but it won’t give you SR (e.g. vectors always add linearly, which they don’t in SR). Or you could have a curved spacetime - again, the interval is covariant, but it’s not SR. I think in addition to invariance, you also need to give the full metric, which has to be of a specific form in order to give you SR.

No, because this doesn’t allow you to derive what that interval actually is. It will give you most of SR within small, local patches, but it won’t give you the field equations which are necessary to derive the metric from given distributions of energy-momentum. It depends on what you consider ‘distinct’. In addition to the ones already mentioned, what comes to mind is the geometric reasoning put forward by Misner/Thorne/Wheeler in their book, the reasoning of String Theory where GR emerges as a necessary consistency condition on the background spacetime, and possibly a thermodynamic approach from quantum information theory (entropic gravity).

I think I may have been a bit sloppy with terminology here, in that I have used “spacetime interval” interchangeably with “world line between events”. Technically speaking however, the spacetime interval is an infinitesimal measure, so it is the interval between neighbouring events on the manifold - thus, it is essentially the metric, the components of which are functions of coordinates. The thing is, if you consider two events that are not neighbouring, i.e. distant in spacetime (as is the case in this scenario), then you need to integrate the spacetime interval along some path connecting these events - which just gives you the length of a world line. This of course depends on which path you choose. Both of these measures are invariant, but they are technically different concepts. I should have made this clearer. The distinction is less crucial in SR, because the Minkowski metric is constant and not a function of coordinates, so you can define a spacetime interval even between distant events. In GR this is generally not the case though, hence it is better to simply consider line elements and world lines instead, which avoids any confusion. Yes, if you want to be really awkward about it, you can. This would essentially be the point of view of some distant observer who is in relative motion to the light and the clock that is attached to it. You’d end up with the same result, but the maths would be vastly more complicated, and the essential principle would be obscured. The central point in this is the principle that the geometric length of a world line is identical to what a clock that follows that world line physically reads. So it is far more intuitive to look at the situation in a coordinate system that puts the rest frame with the light (even if this is a non-inertial rest frame). But of course, you are free to set up your coordinate chart in any way you like. Yes. I am not entirely sure if I get what you are trying to say, but essentially this looks ok to me.

Apologies, it should have been \[\Delta \tau =1=\int ^{t_{1}}_{0}\sqrt{g_{00}} \ dt\\ \]

You need to read this more carefully - it allows to determine position and momentum of the LIGO mirrors, which is a macroscopic and classical system. This does not work with quantum systems, since non-commuting observables are inherent in the very nature of such systems, and not due to measurement limitations. In fact, it is that very non-classicality which allows entanglement relationships in the first place.

A spacetime interval is always a line integral, and hence depends on a world line. I am beginning to think there might be some confusion about terms here - are you specifically referring to geodesics, by any chance? All geodesics are world lines, but not all world lines are geodesics. Furthermore, world lines (whether geodesics or not) depend on initial and boundary conditions, so in general there may be more than one way to connect two events via a geodesic, depending on initial conditions of the test particle tracing out that world line, as well as the geometry of the underlying spacetime. No, it being invariant means that it is unaffected by changes in coordinate basis, i.e. all observers agree on it. I think what you might have in mind here is the way you obtain a geodesic - this doesn’t reference any world lines, you just set proper acceleration to zero and replace ordinary derivatives with covariant derivatives. The result is a differential equation, the solution to which is a geodesic. Note that the solution depends on initial and boundary conditions, though. They all measure a spacetime interval between these events - they are three different world lines connecting the same two events. The two events happen at the same spatial location, so they are separated only in time, but not in space. Hence only the time-part of the metric can be relevant to the spacetime interval, hence: \[\Delta \tau =\int _{C} ds=\int _{C}\sqrt{g_{\mu \nu } dx^{\mu } dx^{\nu }} =\int _{C}\sqrt{g_{00} dx^{0} dx^{0}} =\int ^{1}_{0}\sqrt{g_{00}}dt \] which is the reading on the clock that is stationary with respect to the blinking light. This is of course consistent with the physical meaning of the geometric length of world lines - it’s the proper time accumulated on a clock that traces out this world line. Since the light and the clock are not in free fall, this particular world line is not a geodesic of this spacetime.