Markus Hanke

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.

Essentially what you are getting at is the fact that in SR, all components of the metric tensor are constants, meaning measurements are the same regardless of where and when they are performed. In GR this is of course not the case, since spacetime is no longer flat - hence, the components of the metric are generally functions of coordinates, and not constants. Nonetheless, the spacetime interval is covariant in GR as well, because changes of coordinate basis leave the relationships between components of the metric tensor (if not their specific values) unchanged, so the overall tensor remains the same. So in other words, when you shift around clocks and rulers in spacetime, they will vary along with that shift (“covariance”) in just such a way as to leave any overall spacetime interval the same.

What they are measuring on their clocks is the length of their own world line between these events, which is of course different for each one of them, so yes, they will necessarily obtain different readings. When we say that the spacetime interval is invariant, then that means that all observers agree on it - for example, if X measures a certain length for his own world line, then Y and Z agree that he did. Of course that does not imply that Y and Z get the same amount for their own world lines, which might be very different, even if they connect the same two events. So, invariance of spacetime intervals does not mean that all world lines connecting two events are the same, it means only that all observers agree on the length of any given world line, even if they are not the ones tracing it out. This is why the path integral I gave earlier depends not just on the metric, but of course also on the path C itself. Specifically, it means that all observers will agree on the total accumulated time on a clock that is attached to and stationary with the light source. Which of the three will trace out the longest world line isn’t so easy to answer, due to the dependence of the line integral on the background metric. I would guess it’s the stationary observer, but I may be wrong, pending actual maths. They are all correct - it’s just that they represent different world lines connecting the same two events, so they are all of different lengths. The point is that each observer agrees on the length of the other two observers’ world lines.

Do you mean the observer who is at the same place as the blinking light? Yes indeed, for him the interval is simply whatever he reads on his clock. It would seem to me that Y will be spatially removed, i.e. up in orbit above the tower...? Perhaps I misunderstood. I can't give you a straight answer to this without having done the numbers, since this scenario mixes relative motion with a curved space-time background, so calculating the geometric lengths of their respective world lines isn't trivial. Z isn't a geodesic, because he is launched up, so he undergoes acceleration. Yes, indeed. The maths here aren't overly complicated, but they are definitely tedious. I am not sure I follow you. Are you essentially saying that, if you somehow introduce a gravitational source into a scenario that was hitherto flat spacetime, then the spacetime interval will be affected by this? If so, then you are correct. Both is correct The interval is usually written as a line element, which is an infinitesimally small section of a world line: \[ds^2=g_{\mu \nu}dx^{\mu}dx^{\nu}\] This is a local measure, and it is covariant under appropriate changes in coordinate system. The obtain the geometric length of some extended world line C in spacetime, you integrate this: \[\tau =\int _{C} ds=\int _{C}\sqrt{g_{\mu \nu } dx^{\mu } dx^{\nu }}\] This is a standard line integral, and it can be shown that it is also a covariant measure. Hence, all observers agree both on the line element, as well as on the total length of some given world line. There are always infinitely many possible world lines between any two given events, in any spacetime. Generally speaking though, only one of them will be a geodesic (unless the spacetime in question has a non-trivial topology). Yes, it is a function of the metric, see expression above. I think I lost you here, I am not sure what you are meaning to ask...? In SR, there will be one unique geodesic connecting two given events (that's an inertial observer travelling between the events), and then there are infinitely many world lines that are not geodesics (corresponding to observers who perform some form of accelerated motion between the events). Yes. When performing coordinate transformations, the components of a tensor can change, but the relationships between the components do not, meaning the overall tensor remains the same. I see what you are saying, and whether or not this is a mathematically rigorous deduction is a good question. This is probably better posed to a mathematician. I am hesitant to commit myself here, because I can think of other quantities where this is not true - for example, energy-momentum (in curved spacetime) is conserved everywhere locally, but not globally across larger regions. The geometry of spacetime near a binary system is not stationary, and the overall spacetime isn't asymptotically flat either, because of the presence of gravitational radiation. The geometry will be slightly different each time the binary stars complete a revolution. There really isn't any way to define a consistent (!) notion of 'gravitational potential' that all possible observers could agree on.

The spacetime interval is defined as \[ds^2=g_{\mu \nu}dx^{\mu}dx^{\nu}\] This quantity transforms as a rank-0 tensor (a scalar), so it is covariant under changes in reference frames. All observers will agree on it. If this weren’t so, then observers would disagree on l both the geometric length of world lines, as well as on the geometry of spacetime itself - which would be problematic. For example, a photon in one frame (ds=0) would appear as something different in another frame (ds<>0). One of the observers is located at the same place as the light, whereas the other observer is at the bottom of the tower, so he will be spatially removed by some distance (even if they are both at rest) - in the first case, there is only a temporal term in the line element, in the second case there is both a temporal as well as a spatial part. But the sum of the two is “balanced” in just the right way so that they both agree on the spacetime interval. The two observers are related by a simple coordinate transformation - which leaves the metric covariant. Any tensorial quantity will be unaffected by changes in reference frame, regardless of whether you are in a flat or a curved spacetime. The metric is an obvious example, as are the various curvature tensors, as well as the energy-momentum tensor, the electromagnetic field tensor etc etc. The notion of gravitational potential can only be meaningfully defined in spacetimes that admit a time-like Killing vector field, and which are asymptotically flat. Schwarzschild spacetime is one such example. A counter example would be spacetime in and around a binary star system.

I didn’t really follow the entirety of this thread too closely, so if there was a ‘Planet X’ mentioned, then I missed that. So then the answer depends on exactly what form this world line of the travelling twin takes. It is of course possible that the travelling twin is almost fully inertial, meaning that the vast majority of his world line is inertial - if your Planet X coincides with those inertial portions, then you are right, the two frames will be symmetric. However, there must be at least one portion of the journey - however short and small - that is not inertial; during that portion the symmetry is broken, which leads to the difference in total proper time recorded on the clocks. This is all closely related to the difference between coordinate quantities, and proper quantities - understanding the difference is crucial to understanding relativity.

I’m actually having difficulty understanding what exactly it is you are asking. Can you rephrase or reformulate the question? A test particle in free fall cannot not follow (if that makes sense) the geometry of spacetime, just as you cannot not follow the curvature of Earth’s surface as you move about on it. You can of course compensate for these effects by equipping yourself with suitable thrusters - but then you are no longer in free fall. Furthermore, there are scenarios where you cannot counteract gravity at all, regardless of how much you fire your thrusters; so it is evidently quite a real thing, in the sense that it has real consequences for the motion of bodies.

The travelling twin, upon his return, will find that this brother who stayed behind is older than himself. This is as expected, since the travelling twin is not fully inertial, so these frames are not globally related by a Lorentz transformation - there is no symmetry between them. Note that both of them agree on who is older and who is younger.

Because both twins agree that the world line of the travelling twin is shorter than that of the stationary twin (one reality!). A purely inertial frame always represents the longest possible world line between two given events - since the travelling twin is not purely inertial, his world line will be shorter, so he ages less in comparison.

Indeed.

Fair enough - though I couldn't imagine which bona fide physicist would possibly disagree with SR, given the overwhelming evidence, and on which grounds. That notwithstanding, the consensus on the subject matter is clear and unambiguous.

This is an amateur Internet forum, it is not representative of the state of the scientific community (even though some members here are professional scientists). Within the scientific community itself, there are no doubts or disagreements about the validity of Special Relativity - it is one of the most well understood and thoroughly tested models in the history of science. Or perhaps I should put this differently - since relativity is fundamental to all of particle physics, you wouldn't exist and be here to ask this question if relativity was wrong. It really is that simple.

Just to add to what has been said already - velocity (or better: it's magnitude, being speed) isn't something that any one observer "has", it's a relationship between two frames in spacetime. What's more, this relationship is of a geometric nature - using the concept of what is called rapidity, speed is actually equivalent to a rotation angle. So in that sense, @michel123456 does have a point - SR is all about perspectives. But these perspectives aren't optical/visual ones, and they aren't purely spatial ones either. When I talk about rotations above, then those aren't rotations in Euclidean space, but in a 4D hyperbolic spacetime. So we rotate from space to time, and vice versa. That's exactly what a Lorentz transformation is - a hyperbolic rotation in spacetime, so inertial observers in relative motion are related by a rotation in spacetime. This is why their measurements of space and time, taken separately, do not necessarily agree. But it is important to understand that this is physically real, it isn't just a matter of visual appearance; it has real physical consequences, which can be directly observed and measured.

When you pass a unit time-like future oriented vector to both slots of the Einstein tensor, you get a scalar that is just precisely the average (!) Gaussian curvature in the spatial directions within a small neighbourhood. So the Einstein tensor is a measure of average curvature around an event. In vacuum, geodesics will diverge in some directions and converge in others, in such a way that the average balances to exactly zero - hence the vanishing of the Einstein tensor in vacuum. Indeed - it’s average Gaussian curvature, and thus it captures only a particular subset of overall Riemann curvature.

I’m pretty sure you were told not to bring this up again...

! Moderator Note It is not acceptable to just post external links - as per the rules of this forum, you are required to clearly present a comment or discussion point. Also, necrothreading (resurrecting old threads, in this case 12+ years old!) is very much frowned upon, as many of the original contributors are likely no longer active.

It’s like longitudinal lines on the surface of the Earth - the closer you get to the pole, the less the distance between these lines becomes (and vice versa). It’s rather the other way around - if you only look at a small enough region of the field, then it will appear almost flat.

That is just the point - an accelerated frame does not trace out a geodesic in spacetime. In the original example, you have an enclosed box under uniform acceleration, so the box itself traces out a world line that is not a geodesic. If you now place a test particle into the interior of the box, and release it, then that test particle will be in free fall, and will thus trace out a geodesic (until it makes contact with one of the walls). The fact that the test particle’s spatial trajectory is curved with respect to the box is precisely due to the difference in the nature of their world lines - that’s one way to look at it, anyway. They will remain parallel, if there is only acceleration, but no sources of gravity - this is a flat spacetime. If there are sources of gravity, then spacetime is curved, and they will not remain parallel. However, iff the room is small enough, then their geodesic deviation will be so small as to be negligible - in which case these two scenarios become equivalent. But this is true only if the room is small enough. It wasn’t meant as an insult, please don’t take it personally. It really is a very common misunderstanding - and one which I myself have fallen afoul of in the past (and I made an utter fool of myself on some forum trying to argue the point).