 # Markus Hanke

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## Everything posted by Markus Hanke

1. 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.
2. 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.
3. 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.
4. 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.
5. 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.
6. 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.
7. 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.
8. 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.
9. 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.
10. 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.
11. 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.
12. 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.
13. ! 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.
14. 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.
15. 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).
16. No, this is a common misunderstanding. Spacetime in a uniformly accelerated reference frame in otherwise empty space is perfectly flat - acceleration is not a source of spacetime curvature. Only distributions of energy-momentum (like planets, stars,...) are. Physically speaking, the defining characteristic of a curved spacetime is the presence of tidal gravity - however, if the reference frame in question is small enough, these tidal effects become negligibly small, which is why in a small local area only uniform acceleration looks like a uniform gravitational field. This is just the equivalence principle, and this is somewhat akin to the surface of Earth looking flat so long as you only look at a small enough section of it. Anyway, the upshot is that a uniformly accelerated frame in otherwise empty space has no spacetime curvature (the Riemann tensor vanishes).
17. It has been pointed out to you multiple times already that there is no such thing as "multiple realities" in a classical model such as SR/GR. It seems to me that in fact you are the only one here attempting to use this as an argument.
18. Lol, I do that every day That’s one of the big positives about being just an amateur - there’s no obligation, pressure or expectation on us to deliver anything, so we are free to pursue what we like and find beautiful. I know for myself that I would not have performed very well in the pressurised environment of professional academia. I used to have a lot of regrets about not choosing that as a career path, but now I’m almost glad I didn’t.
19. You can’t possibly be serious...? Lorentz invariance is one of the most thoroughly tested phenomena in the history of science, it is difficult to even estimate how many experiments have been done to test (and confirm) it over the past 100 years, but it’s certainly many thousands. You might want to take a look at this, just for starters, and remember that it continues to be tested and confirm directly in every particle accelerator run we do.
20. Yes, of course there are differences, which is why I used the term “on equal footing” instead They are treated the same, which is different from being the same.
21. Exactly Relativity treats time and space on equal footing - which not only underpins relativistic effects in the macroscopic realm, but also gives rise to the Dirac equation in QM, as well as the concept of spin for elementary particles. Relativity is a purely classical model based on the notion of local realism, so of course there is only one reality. It couldn’t be any different. The one major difference compared to Newtonian mechanics is that measurements of time and space are now purely local concepts, so in order to describe a global reality, you need to consider a different set of quantities - namely those that are invariant. I think this is a gross overgeneralisation, because it depends on what the question is. The original question of the thread (What is time?) is a very good one, and many aspects and facets of it are indeed unanswered, or at the very least subject to ongoing debate and investigation. Only a fool would dispute this. I personally found the discussion we have had here on this subject to be interesting and constructive. Who would have though that a tea cup can spawn 14+ pages of constructive exchanges? On the other hand though, the question as to whether the world is actually Lorentz invariant or not within the domain we can observe and probe (i.e. is SR a valid model?), and what that means in physical and mathematical terms, is not in any way subject to debate - this question has been definitively answered long ago, and the model itself is one of the most studied and well-understood in all of physics. It can even be formally proven that it is internally self-consistent, and that no physical paradoxes can be constructed using the axioms of SR. So you really are flogging a dead horse on this particular issue. This is honestly not meant to be personal but...the fact that relativity doesn’t make sense to you really is down to yourself. But you know you could fix that, if you wanted. What is subject to debate however is how far exactly the domain of applicability of relativity extends, and what exactly happens once you go outside of it. This is an area of very intensive research right now. Symmetry means that a transformation that brings you from one frame to another is invertible - so if you apply the inverse of the same transformation, you arrive back in the original frame. In the case of the twin travellers, there will be at least one section of the journey when this does not hold.
22. I am not entirely sure what you mean. Basically, you’d calculate the interval in one frame first; you then apply Lorentz transformations to the time and space parts in order to “go” into the other frame, and check what this does to the overall interval. What you’ll find is that the interval remains the same - Lorentz transformations are just hyperbolic rotations in spacetime, so when you go from one frame to another, you essentially rotate some portion of the space part into the time part, or vice versa. This leaves the overall quantity unchanged. In the classical Twin Paradox setup, one of the frames is inertial, whereas the other one isn’t - so these frames are not symmetric, and they can be physically distinguished (by accelerometer readings). This asymmetry is what leads to the difference in total accumulated times. If both frames were inertial, then they would be physically indistinguishable, and there would be no residual difference in total accumulated times.