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Markus Hanke last won the day on September 12
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376 Beacon of HopeAbout Markus Hanke

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A mundane example of a spacetime interval being invariant
Markus Hanke replied to geordief's topic in Relativity
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 spacetime 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 nontrivial 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, energymomentum (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. 
Markus Hanke started following Antigravity  Is it Possible?, A mundane example of a spacetime interval being invariant, GRAVITY: Force or Farce? and and 3 others

A mundane example of a spacetime interval being invariant
Markus Hanke replied to geordief's topic in Relativity
The spacetime interval is defined as \[ds^2=g_{\mu \nu}dx^{\mu}dx^{\nu}\] This quantity transforms as a rank0 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 energymomentum tensor, the electromagnetic field tensor etc etc. The notion of gravitational potential can only be meaningfully defined in spacetimes that admit a timelike 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.

Split from Time dilation dependence on direction
Markus Hanke replied to Charles 3781's topic in Relativity
Indeed. 
Split from Time dilation dependence on direction
Markus Hanke replied to Charles 3781's topic in Relativity
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. 
Split from Time dilation dependence on direction
Markus Hanke replied to Charles 3781's topic in Relativity
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.

The equivalence principle and the curvature of spacetime
Markus Hanke replied to geordief's topic in Relativity
When you pass a unit timelike 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.

The equivalence principle and the curvature of spacetime
Markus Hanke replied to geordief's topic in Relativity
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. 
The equivalence principle and the curvature of spacetime
Markus Hanke replied to geordief's topic in Relativity
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).