geordief Posted September 24, 2020 Share Posted September 24, 2020 (edited) If we put an observer,E on Earth and another observer ,M on Mars and present both with 2 events , S(1) and S(2) on the Sun , and visible to both is it possible to show that E and M both measure the spacetime interval between S(1) and S(2) exactly the same? So ,if the expression is s^2 = (ct)^2 -r^2 and both E and M measure the same quantity from their respective frames of reference what would be the actual measurements needed to be performed by E and M,? I imagine that both would time the difference in signal capture between S(1) and S(2) and would also somehow measure the physical(spatial) distance between the sites of S(1) and S(2) Would those measurements be the ones that would show that the spacetime interval between the 2 events on the Sun as being invariant? (let's say they were 2 sunspots appearing one after the other) Or have I ,as I often have got the wrong end of the stick again? Edited September 24, 2020 by geordief Link to comment Share on other sites More sharing options...

md65536 Posted September 26, 2020 Share Posted September 26, 2020 (edited) While waiting for a better answer... Having the events on the sun unnecessarily complicates things because of spacetime curvature. You're measuring distances from afar, in a different gravitational potential, so there's not one "correct" way to measure those distances. I don't think curved-spacetime intervals are invariant. However, if you're treating the sun as just a location in flat spacetime, that's fine. You wouldn't directly compare the arrival time of light signals from the events, you'd want the time those signals were sent. Basically you'd subtract the travel time of light to find that. Edit: I think I'm misunderstanding "time the difference in signal capture between S(1) and S(2)". Signals aren't really a necessary part. You could use any clock(s) synchronized with the observer's, to measure the events' times. Typically a clock at S1 and one at S2 would be used, but in this example a single clock at E and then compensate for light travel time would be more practical. Then M in a different reference frame would measure its different set of times, either using a different set of clocks or by transforming the times from another reference frame. To measure distance, you can imagine all of space being covered in a lattice of rulers, at rest in an observer's inertial frame. Another observer would use a different lattice of rulers. Then the events are located at some place on those rulers, and you can measure the distance between them. Simply knowing the position of the sun in the observer's coordinates, or timing distances using light signals, or other ways, tells you distances and/or the locations of the events. An event has a time and a location in any reference frame's coordinates, ie. a place and a time on a lattice of rulers and synchronized clocks. Then r is the spatial distance between the two events, and t is the time between the two events. Edited September 26, 2020 by md65536 Link to comment Share on other sites More sharing options...

geordief Posted September 26, 2020 Author Share Posted September 26, 2020 26 minutes ago, md65536 said: While waiting for a better answer... Having the events on the sun unnecessarily complicates things because of spacetime curvature. You're measuring distances from afar, in a different gravitational potential, so there's not one "correct" way to measure those distances. I don't think curved-spacetime intervals are invariant. However, if you're treating the sun as just a location in flat spacetime, that's fine. You wouldn't directly compare the arrival time of light signals from the events, you'd want the time those signals were sent. Basically you'd subtract the travel time of light to find that. Edit: I think I'm misunderstanding "time the difference in signal capture between S(1) and S(2)". Signals aren't really a necessary part. You could use any clock(s) synchronized with the observer's, to measure the events' times. Typically a clock at S1 and one at S2 would be used, but in this example a single clock at E and then compensate for light travel time would be more practical. Then M in a different reference frame would measure its different set of times, either using a different set of clocks or by transforming the times from another reference frame. To measure distance, you can imagine all of space being covered in a lattice of rulers, at rest in an observer's inertial frame. Another observer would use a different lattice of rulers. Then the events are located at some place on those rulers, and you can measure the distance between them. Simply knowing the position of the sun in the observer's coordinates, or timing distances using light signals, or other ways, tells you distances and/or the locations of the events. An event has a time and a location in any reference frame's coordinates, ie. a place and a time on a lattice of rulers and synchronized clocks. Then r is the spatial distance between the two events, and t is the time between the two events. Sounds good to me.Yes ,I should have chosen a location removed from spacetime curvature,even though I still hope that the spacetime interval may be invariant under those conditions too. I wasn't aware it was easy to synchronize clocks like you said and that is why I assumed there would have to be signal capture at E and M. Link to comment Share on other sites More sharing options...

md65536 Posted September 26, 2020 Share Posted September 26, 2020 (edited) 4 hours ago, geordief said: I wasn't aware it was easy to synchronize clocks like you said and that is why I assumed there would have to be signal capture at E and M. It's more about the coordinates (an inertial frame is 3 Euclidean space dimensions and a time that is the same everywhere within the frame), the clocks just represent a measure of the frame's time at different points. If you never had to consider different frames, you could use a single clock to represent time everywhere. I'll just keep talking because I hope more people talk about the meaning of the spacetime interval being invariant! But with respect to that, do you know the 3 types of interval: space-like, light-like, and time-like? The type is invariant, and depends on if s^2 is respectively negative, zero, or positive (in the sign convention you used, s^2 = (ct)^2 - r^2). For any spacelike interval, there's a set of inertial frames where the two events are simultaneous. For a timelike interval, there's an inertial frame where the two events occur at the same place. I think the interval being invariant means that if S1 and S2 are more distant in M than E, then they must also be farther apart in time in M than E. An example is that a clock at rest ticks faster (smallest time between ticks) than if measured from any frame in which it is moving. If you have an interval where r=0, ct=1 (ie. a clock at rest ticking once, with appropriate choice of units) then s^2=1. In some other frame where it takes ct=2 for the moving clock to tick once, s^2 also must equal 1 = 4 - 3, so r must equal sqrt(3) units of distance. In that frame, there's more time between the two events, and more distance. And indeed, in such a frame, gamma=2, v=(sqrt(3)/2)c, and the moving clock moving for 2/c units of time moves a distance of sqrt(3). Does that make sense? I'm not sure I got it right because I have no experience dealing with intervals. Edited September 26, 2020 by md65536 Link to comment Share on other sites More sharing options...

geordief Posted September 26, 2020 Author Share Posted September 26, 2020 15 hours ago, md65536 said: It's more about the coordinates (an inertial frame is 3 Euclidean space dimensions and a time that is the same everywhere within the frame), the clocks just represent a measure of the frame's time at different points. If you never had to consider different frames, you could use a single clock to represent time everywhere. I'll just keep talking because I hope more people talk about the meaning of the spacetime interval being invariant! But with respect to that, do you know the 3 types of interval: space-like, light-like, and time-like? The type is invariant, and depends on if s^2 is respectively negative, zero, or positive (in the sign convention you used, s^2 = (ct)^2 - r^2). For any spacelike interval, there's a set of inertial frames where the two events are simultaneous. For a timelike interval, there's an in ertial frame where the two events occur at the same place. I think the interval being invariant means that if S1 and S2 are more distant in M than E, then they must also be farther apart in time in M than E. An example is that a clock at rest ticks faster (smallest time between ticks) than if measured from any frame in which it is moving. If you have an interval where r=0, ct=1 (ie. a clock at rest ticking once, with appropriate choice of units) then s^2=1. In some other frame where it takes ct=2 for the moving clock to tick once, s^2 also must equal 1 = 4 - 3, so r must equal sqrt(3) units of distance. In that frame, there's more time between the two events, and more distance. And indeed, in such a frame, gamma=2, v=(sqrt(3)/2)c, and the moving clock moving for 2/c units of time moves a distance of sqrt(3). Does that make sense? I'm not sure I got it right because I have no experience dealing with intervals. I struggle with the simple maths (and the implications thereof) of basic special relativity . So I can't with confidence say that it makes good sense or bad sense. (I will try to reread it a few times and it may become clearer) Edit: have dug up this very old old response from Markus to me where I asked him about the space time interval . http://www.thescienceforum.com/physics/41968-space-time-interval.html At the time I took it to mean that it was invariant in both inertial and non-inertial frames. As I said then ,I found that reassuring (because the physical world seemed to make a little(a lot) more "sense" to me as a result Hope I didn't misinterpret at the time. Link to comment Share on other sites More sharing options...

Halc Posted September 26, 2020 Share Posted September 26, 2020 20 hours ago, geordief said: Sounds good to me.Yes ,I should have chosen a location removed from spacetime curvature,even though I still hope that the spacetime interval may be invariant under those conditions too. The interval cannot be invariant between observers at different gravitational potentials. Consider a light blinking at the top of a radio tower (on a non-spinning planet if we want to be more precise). We have an observer at the top and one on the ground, each measuring the interval between the blinks. They're both stationary relative to the tower, so the interval is trivially just the time between the blinks, but the two observers will measure different elapsed times on their local clocks because they're at different potentials and don't run in sync. So the spacetime interval being invariant is a property only of flat Minkowskian spacetime. Link to comment Share on other sites More sharing options...

geordief Posted September 26, 2020 Author Share Posted September 26, 2020 2 minutes ago, Halc said: The interval cannot be invariant between observers at different gravitational potentials. Consider a light blinking at the top of a radio tower (on a non-spinning planet if we want to be more precise). We have an observer at the top and one on the ground, each measuring the interval between the blinks. They're both stationary relative to the tower, so the interval is trivially just the time between the blinks, but the two observers will measure different elapsed times on their local clocks because they're at different potentials and don't run in sync. So the spacetime interval being invariant is a property only of flat Minkowskian spacetime. Thanks for putting me right. Are there other things that those observers do agree on ? Can they agree i your scenario when they compensate for the difference in their gravitational potential? Is that kind of information always available to them in principle? Link to comment Share on other sites More sharing options...

md65536 Posted September 26, 2020 Share Posted September 26, 2020 (edited) 1 hour ago, geordief said: I struggle with the simple maths (and the implications thereof) of basic special relativity . So I can't with confidence say that it makes good sense or bad sense. (I will try to reread it a few times and it may become clearer) The Pythagorean theorem seems to crop up a lot, and if you move one of the terms to the other side it looks like an equation for a hyperbola. Often you can see that in diagrams, where you have eg. a length in the x dimension, and one in ct, and the hypotenuse is meaningful in some way. Anyway I'm still figuring out things about that. The numbers I used were just an interval s^2=1, with (ct)^2=1,r^2=0 in one frame, and (ct)^2=4, r^2=3 in another. A common speed used in examples is approx 0.866 c ie. sqrt(3)/2, because the Lorentz factor in that case is a simple factor of 2. You might try setting gamma to 2 and solve gamma=1/sqrt(1-(v/c)^2) for v, if you can, to see why those numbers come about. Or get Wolfram Alpha to solve it for you if not. Otherwise, you can get away with following a lot of examples just using v=.866c, gamma=2. (v=.6c, gamma=1.25, Doppler factor=2 is also common.) Edited September 26, 2020 by md65536 Link to comment Share on other sites More sharing options...

Halc Posted September 27, 2020 Share Posted September 27, 2020 3 hours ago, geordief said: Are there other things that those observers do agree on ? Can they agree i your scenario when they compensate for the difference in their gravitational potential? I can think of few frame-invariant things. In fact, the interval seems to be a rare such thing. Yes, anybody can compute what is being measured relative to a different frame. We're all doing it in this forum all the time. That doesn't make the thing being measured frame invariant unless there is somehow a way to determine the objective frame that all can determine independently instead of an arbitrary choice that all decide to use. There is no such frame. In particular, there is no objective measure of gravitation potential. It is always relative. Ideally one would assign 0 gravitational potential to a place infinitely distant from all matter, but of course there is nowhere you can go to get away from it all. I've never seen the gravitational potential of Earth expressed in any absolute way. If you could do that, you could compute the objective dilation of a clock here on Earth and then get a figure for the actual rate of time passage in the universe. It would probably change over time. Again, I've never seen any valid attempt at this, just religious presumptions that we're the standard and the universe is all about us, just like we're also by chance the only stationary thing in the universe, and thus the center of it. Link to comment Share on other sites More sharing options...

Markus Hanke Posted September 27, 2020 Share Posted September 27, 2020 12 hours ago, Halc said: The interval cannot be invariant between observers at different gravitational potentials. 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). 12 hours ago, Halc said: We have an observer at the top and one on the ground, each measuring the interval between the blinks. They're both stationary relative to the tower, so the interval is trivially just the time between the blinks, 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. 8 hours ago, Halc said: I can think of few frame-invariant things. In fact, the interval seems to be a rare such thing. 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. 8 hours ago, Halc said: In particular, there is no objective measure of gravitation potential. 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. 1 Link to comment Share on other sites More sharing options...

Halc Posted September 27, 2020 Share Posted September 27, 2020 (edited) 7 hours ago, Markus Hanke said: This quantity transforms as a rank-0 tensor (a scalar), so it is covariant under changes in reference frames. Of course. I don't know my tensor calculus, but I know that tensors are frame invariant, and any scalar quantity like that must thus also be frame invariant, so that answers that. Thanks. Some specific examples then: Quote 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. OK, it makes more sense for the guy present at both events to have an easier job of determining the interval than the remote guy. Is the interval then just the time on his clock, or is it more complicated than that? I ask because I can have three observers X,Y,Z. The light blinks every ~90 minutes. X is our guy located at the top of a tower on a spherical non-rotating Earth (effectively Schwarzschild spacetime). Y is in low orbit and comes around exactly when it blinks. Z is lauched straight up a somewhat under escape velocity and comes back exactly in time for the second blink. All three observers are present at both events. Y I think will measure the shortest time and Z the greatest. Those two are inertial in that they follow geodesic worldlines. X does not and is under 1g of proper acceleration between the events. My guess is that all three are going to have to do some complicated mathematics to compute the invariant interval between those two events. None of them has the luxury of simply looking at his watch. I also admit not being able to explain concepts like time-killing vector fields. Heard of them, but I'm just a novice at this. Edited September 27, 2020 by Halc Link to comment Share on other sites More sharing options...

md65536 Posted September 27, 2020 Share Posted September 27, 2020 (edited) 8 hours ago, Markus Hanke said: The spacetime interval is defined as [...] 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. So instead of saying the spacetime interval isn't invariant in curved spacetime, I should have said the interval defined for Minkowski spacetime, ie. the quantity (ct)^2 - r^2 isn't invariant in GR. I was going to ask if the spacetime interval in GR is a local thing only, that applies only to intervals between nearby events, but if it implies world line lengths are invariant, it might apply to any arbitrarily separated events? Oh but then, there can be multiple world lines between two events in GR, so the spacetime interval is local only??? and a world line's length depends on local variances in spacetime curvature. Conversely, in Minkowski spacetime there is only one straight line between any 2 points, so the spacetime interval equation for SR is invariant no matter how far apart the events are. Am I on the right track? The interval in GR is based on a set of values of g for each pair of the 4 space and time dimensions??? Different observers would have different components for the 4 dimensions, but the interval itself would be invariant. Can that be paraphrased as, "The spacetime interval in GR is invariant because curved spacetime is locally Minkowskian", so that any variations in the interval's components are like those seen in SR's interval, or is that wrong or incomplete? 8 hours ago, Markus Hanke said: 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 don't understand. If you have two events in spacetime in a binary star system, where you might carry a mass from one event to the other, isn't there a fixed gravitational potential energy difference between the masses at the two events? Doesn't that imply a meaningful notion of gravitational potential? Edited September 27, 2020 by md65536 Link to comment Share on other sites More sharing options...

Markus Hanke Posted September 28, 2020 Share Posted September 28, 2020 11 hours ago, Halc said: Is the interval then just the time on his clock, or is it more complicated than that? 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. 11 hours ago, Halc said: All three observers are present at both events It would seem to me that Y will be spatially removed, i.e. up in orbit above the tower...? Perhaps I misunderstood. 11 hours ago, Halc said: Y I think will measure the shortest time and Z the greatest. 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. 11 hours ago, Halc said: Those two are inertial in that they follow geodesic worldlines. Z isn't a geodesic, because he is launched up, so he undergoes acceleration. 11 hours ago, Halc said: My guess is that all three are going to have to do some complicated mathematics to compute the invariant interval between those two events. None of them has the luxury of simply looking at his watch. Yes, indeed. The maths here aren't overly complicated, but they are definitely tedious. 10 hours ago, md65536 said: So instead of saying the spacetime interval isn't invariant in curved spacetime, I should have said the interval defined for Minkowski spacetime, ie. the quantity (ct)^2 - r^2 isn't invariant in GR 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. 10 hours ago, md65536 said: I was going to ask if the spacetime interval in GR is a local thing only, that applies only to intervals between nearby events, but if it implies world line lengths are invariant, it might apply to any arbitrarily separated events? 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. 10 hours ago, md65536 said: Oh but then, there can be multiple world lines between two events in GR 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). 10 hours ago, md65536 said: a world line's length depends on local variances in spacetime curvature Yes, it is a function of the metric, see expression above. 10 hours ago, md65536 said: Conversely, in Minkowski spacetime there is only one straight line between any 2 points, so the spacetime interval equation for SR is invariant no matter how far apart the events are. 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). 10 hours ago, md65536 said: The interval in GR is based on a set of values of g for each pair of the 4 space and time dimensions??? Different observers would have different components for the 4 dimensions, but the interval itself would be invariant. 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. 11 hours ago, md65536 said: The spacetime interval in GR is invariant because curved spacetime is locally Minkowskian 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. 11 hours ago, md65536 said: I don't understand. If you have two events in spacetime in a binary star system, where you might carry a mass from one event to the other, isn't there a fixed gravitational potential energy difference between the masses at the two events? Doesn't that imply a meaningful notion of gravitational potential? 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. 1 Link to comment Share on other sites More sharing options...

geordief Posted September 28, 2020 Author Share Posted September 28, 2020 I think one of the axioms of GR is that physical laws are the same everywhere,no matter what frame of reference. Is there any connection between that requirement and this invariance of the spacetime interval ,again under whatever frame of reference? Might it actually be possible to derive GR from using this latter requirement as an axiom in its own right? And what would it say if the opposite was the case? Would we be looking at a universe of chaos as opposed to a universe with "integrity"? ("Integrity" just intended to signify "wholeness" as opposed to disparity - no moral implications as applied.) Link to comment Share on other sites More sharing options...

Halc Posted September 28, 2020 Share Posted September 28, 2020 (edited) 13 hours ago, Markus Hanke said: 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. All three observers are present at both events, and they all measure a different time on their clocks, so it can't be as simple as just reading the clock. Quote It would seem to me that Y will be spatially removed, i.e. up in orbit above the tower...? Perhaps I misunderstood. You picture an orbit too high. One meter altitude is plenty to orbit an ideal sphere. The tower is just there to keep Y from getting belly burns and to give Z some acceleration space. Quote 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. Let's just say the blinks are far enough apart that the Z observer is going to measure the most time. If 90 minutes isn't enough for that, we can make it a month. X just sits at the light, experiencing proper acceleration. Remember, the planet is not spinning and has no air. Y orbits multiple times, but is inertial, and follows a geodesic at constant potential, so I assume he's dilated due to velocity, per H-K. Z goes up, fast at first, but quickly slowing to a near halt for most of the month at an altitude of considerable higher potential, so the +dilation due to potential will be greater than the -dilation due to the motion at either end of the trajectory. Quote Z isn't a geodesic, because he is launched up, so he undergoes acceleration. Z does his acceleration before the first blink, so the entire duration between the two events is inertial and follows a geodesic, albeit a different one than Y's worldline. This seems a nit-picky objection. Work with me here. I'm not after exact figures. I'm just pointing out that these three observers are going to measure different times between the events, so they can't all have measured the frame-invariant interval between the two events. If one of them is by definition correct, then which? Your equation doesn't immediately shed light on that. You hint that it is X, the one experiencing proper acceleration. This seems to contradict the equivalence principle which I think would have chosen Z. I can have observer X and Z in flat SR space, but in equivalent accelerating Rindler space. In that scenario, the two blinks take place at the same location in some inertial frame for Z and in the Rindler frame for X. In the inertial frame, X goes out and back, and Z is stationary at that location the entire time. In the Rindler frame, X is stationary and Z goes up and down, changing potential along the way, kind of like with Earth above, but different velocity relative to X. I cannot work out a path equivalent to Y in that scenario. Edited September 28, 2020 by Halc Link to comment Share on other sites More sharing options...

md65536 Posted September 29, 2020 Share Posted September 29, 2020 5 hours ago, Halc said: I'm just pointing out that these three observers are going to measure different times between the events, so they can't all have measured the frame-invariant interval between the two events. If one of them is by definition correct, then which? Sure, they can measure different coordinate times. Those times can be a component of the invariant spacetime interval, without being invariant themselves. Different observers have different components that combine to the same interval. It's the proper time that is the invariant length of a spacetime interval. Everyone agrees on the time that a clock measures on its world line between two events. But to different observers, the clock's path with have different coordinate times, and different coordinate lengths. Consider an infinitesimal line element of such a world line. For some observers, the clock can be stationary over that line element, with spacial components of 0 and an infinitesimal time component. For other observers the clock is moving, and the line element contains infinitesimal spacial component. They disagree on the components, but agree on its invariant length that results each set of those components. 1 Link to comment Share on other sites More sharing options...

Markus Hanke Posted September 29, 2020 Share Posted September 29, 2020 11 hours ago, Halc said: All three observers are present at both events, and they all measure a different time on their clocks, so it can't be as simple as just reading the clock. 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. 11 hours ago, Halc said: If one of them is by definition correct, then which? 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. Link to comment Share on other sites More sharing options...

geordief Posted September 29, 2020 Author Share Posted September 29, 2020 5 hours ago, Markus Hanke said: 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. Extrapolating now,if we take an event such as the death of Julius Caesar and we take the set of all ("subsequent") events (eg my typing this) can we say that ,provided these events occur after the initial event (as judged from that event's frame of reference ).... if all that is true then there is no event that is not connected by a spacetime interval? Must there be a chain(s?) of causation no matter how tenuous There are no "blind spots"....no "causation shields"(except black holes or gravitational waves)? The "causation cone" which is the set of events in the light cone is thus interconnected like a moving net. And ,as you have said the distance between every pair (or chain ends ) of events will be judged as identical no matter from what vantage point in the set of events one is judging? Link to comment Share on other sites More sharing options...

md65536 Posted September 30, 2020 Share Posted September 30, 2020 (edited) 21 hours ago, Markus Hanke said: 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. There are two things that could mean, without context. Two different objects passing through the same pair of events can have different world lines (eg. twin paradox, with 2 paths of different proper time (geometric length)). Or, a single object passing through two events. In the latter case, there's only one world line. The object passes through a specific set of events, and everyone agrees that it passes through those events. The shape of that world line is different in different coordinates. For example, in a coordinate system that moves along with the object, the object is stationary all along the single world line. In other coordinates, the same world line might span a great distance. However, the 4d length of each infinitesimal part of that world line (representing local measurements) is the same no matter whose coordinates you use, and the geometric length of the whole world line is invariant, even though it can be made up of different spacial lengths and coordinate time spans. On 9/27/2020 at 10:19 PM, Markus Hanke said: 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. I didn't mean that, but I think what I said didn't make sense. In Minkowski spacetime, the interval s^2 = (ct)^2 - r^2 is invariant. The value r is the spacial length between say two distant events in flat spacetime, say in the coordinates of some distant observer. I have this notion that there's no such measure r in curved spacetime, because the local measurements of length along a world line between the two events will be different than the local measure of lengths near the observer. In SR there's not that problem, because a ruler has the same length throughout all of a given inertial frame's spacetime. In GR, the observer could make up a meaning of a distance between two remote points in its coordinates, but that wouldn't necessarily correspond with anything meaningful in local measurements of distance along the world line. So (ct)^2 - r^2 doesn't have enough information, and the notion of distance between two points in curved spacetime is not meaningful on its own, because it depends on the path between those points. For events that are "nearby" each other, with a single geodesic between the points, there is enough info, because the length would be along that geodesic, so an interval like (ct)^2 - r^2 only applies locally (to events that are near each other) in GR. (Even if the observer is distant from the events and has a different local measure of length, she can still meaningfully describe the distance between the two events that are near each other.) Does that make sense? Edited September 30, 2020 by md65536 Link to comment Share on other sites More sharing options...

Markus Hanke Posted September 30, 2020 Share Posted September 30, 2020 2 hours ago, md65536 said: Does that make sense? 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. Link to comment Share on other sites More sharing options...

Halc Posted October 1, 2020 Share Posted October 1, 2020 On 9/30/2020 at 1:19 AM, Markus Hanke said: 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. This is what I am getting at, and what nobody is answering. I am after the invariant spacetime interval between a pair of events separated in a time-like manner, under GR rules. In flat spacetime, the interval between two events is not a function of any worldline connecting the two events. Nevertheless, the one inertial worldline connecting the two events in question happens to have a proper temporal length equal to that interval (or to the square root of it). It is the worldline that maximizes this proper time between the events. Any other worldline will give a smaller time. On 9/29/2020 at 1:44 AM, Markus Hanke said: 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. This is by definition, and all observer will agree that Fred's clock must measure X time between this event and that. I don't care about this. I'm asking if any one of their time measurements happens to equal the spacetime interval between those two events. The measured times (worldline lengths) are obviously different, and the interval is not different for each observer, so at most only one of them can directly measure it. It's either X, Y, Z, some 4th worldline, or none of them. Z seems the only viable candidate (the one on a ballistic trajectory straight up, falling back just in time for the 2nd event. Z measures the maximal value, just like the one 'correct' observer in the flat scenario. I cannot think of a worldline that can log more time in my gravity scenario. 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. 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. How is a bunch of observers agreeing on the lengths of each other's worldlines in any way relevant to the spacetime interval being invariant? Said interval being invariant means that it isn't dependent on a choice of worldline connecting the two events. Again, this isn't my question. My question is this: If the interval between the two events in my example is 1 (we choose our units so the interval is 1), does any of the three clocks read 1? 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.I can tell you that Z will always trace the longest worldline, using the equivalence principle logic in my prior post. I'm not asking which traces the longest worldline. I'm asking if any of the three is taking a direct measurement of the one worldline-independent interval between those two events. I don't know how to do the tensor computation, else I'd not be asking. Link to comment Share on other sites More sharing options...

Markus Hanke Posted October 2, 2020 Share Posted October 2, 2020 14 hours ago, Halc said: In flat spacetime, the interval between two events is not a function of any worldline connecting the two events. 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. 14 hours ago, Halc said: Said interval being invariant means that it isn't dependent on a choice of worldline connecting the two events. 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. 14 hours ago, Halc said: I'm asking if any one of their time measurements happens to equal the spacetime interval between those two events. They all measure a spacetime interval between these events - they are three different world lines connecting the same two events. 15 hours ago, Halc said: If the interval between the two events in my example is 1 (we choose our units so the interval is 1), does any of the three clocks read 1? 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. Link to comment Share on other sites More sharing options...

Markus Hanke Posted October 2, 2020 Share Posted October 2, 2020 Apologies, it should have been \[\Delta \tau =1=\int ^{t_{1}}_{0}\sqrt{g_{00}} \ dt\\ \] Link to comment Share on other sites More sharing options...

md65536 Posted October 2, 2020 Share Posted October 2, 2020 15 hours ago, Markus Hanke said: They all measure a spacetime interval between these events - they are three different world lines connecting the same two events. So that's 3 clocks, each measuring different spacetime intervals between two events? Each interval corresponds to the world line of the respective clocks? In GR the interval isn't just a separation of time and of space, but along a path? If you use one clock to measure the spacetime interval of another clock's world line, the first clock is measuring coordinate time, and the second is measuring proper time. I think what Halc was asking is, if you express the second clock's world line in terms of the first's coordinate measures of time and distance, do you still get the same invariant spacetime interval? I suspect that the answer is generally 'no', because the coordinate time measured by the first clock, which is generally distant from the second clock, is measuring something different than what the second clock is measuring locally. The clocks, traveling on different world lines, do not each measure the same local effects of curvature as the other. 15 hours ago, Markus Hanke said: The two events happen at the same spatial location, so they are separated only in time, but not in space. But isn't it also possible to express that same world line, in different coordinates where they are also separated in space? The difference here is that instead of talking about 2 clocks on 2 different world lines, measuring different things, I'm talking about one clock, but described using different coordinates. I suppose you would say, (as you explained before) the components of the spacetime interval for the 4 dimensions could be different, but the resulting spacetime interval is invariant, which simply means that all observers agree on the proper time that the clock measures between the two events, even if they describe the clock differently in their own coordinates. I think both Halc and I are struggling with the meaning of that, maybe in different ways? I think that the meaning of different observers having different time coordinates for the single clock's world line, yet calculating the same proper time is this: Even if different observers locally measure the space and time coordinates of the clock, over an infinitesimal line element, the different observers can still describe that invariant line element using different mixes of space and time. Same local measurements, different coordinates = same invariant spacetime interval. Remote measurements = different world lines, with different spacetime intervals. Sorry if I'm repeating things, it's not yet clear to me. Link to comment Share on other sites More sharing options...

Markus Hanke Posted October 3, 2020 Share Posted October 3, 2020 7 hours ago, md65536 said: So that's 3 clocks, each measuring different spacetime intervals between two events? Each interval corresponds to the world line of the respective clocks? In GR the interval isn't just a separation of time and of space, but along a path? 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. 8 hours ago, md65536 said: But isn't it also possible to express that same world line, in different coordinates where they are also separated in space? 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. 8 hours ago, md65536 said: which simply means that all observers agree on the proper time that the clock measures between the two events, even if they describe the clock differently in their own coordinates Yes. 8 hours ago, md65536 said: Same local measurements, different coordinates = same invariant spacetime interval. Remote measurements = different world lines, with different spacetime intervals. I am not entirely sure if I get what you are trying to say, but essentially this looks ok to me. Link to comment Share on other sites More sharing options...

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