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md65536

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md65536 last won the day on December 15 2020

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  1. I think it also has intrinsic curvature in the usual setup. Sure, but the gravitational time dilation should depend only on their relative gravitational potential (right?) and not how space between them curves to produce that difference in potential. For example if you have 2 locations at different potentials in a constant gravitational field, you have 2 locally Minkowskian regions and no curvature anywhere, but you still get gravitational time dilation. Yours is an interesting example because the regions are intuitively flat. I guess the Riemann tensor is zero at those locations? But for example a region just outside the hollow shell is also locally Minkowskian in the coordinates of a particle in free fall, and locally measurably "flat" to such a particle (but spacetime is not actually flat there and the Riemann tensor isn't zero). I think I'm too stuck on confusion about the meaning of "locally Minkowskian," and trying to visualize it as horizontal on a diagram showing local coordinates, but then don't know how distant curved space could be shown. Now I remember you mentioned the covariance of tensors to me before, and I looked up the components of the Riemann tensor and couldn't make sense of it (same as now!). I think I need to learn more basics before understanding curvature in different observers' coordinates.
  2. Tying this in with the rubber sheet analogy: The height on the sheet or on Earth corresponds (assuming constant g) with gravitational potential, proportional to how much energy it takes to lift an object to a specific height. The derivative of that, the slope of the sheet or ground, corresponds with gravitational force---how quickly a marble would accelerate as it rolls along that point. The derivative of that represents curvature. If we imagine the rubber sheet to be semi-rigid, and you actually have to bend it into a curved shape, the severity of the bending at a given point corresponds with curvature. You can tilt a sheet, and that corresponds with a constant gravitational force but no curvature. (Just to complete the analogy, curvature corresponds with tidal forces. If you have a toy car where the front and back are separate and connected by a spring, and roll it down a hill with constant slope, the spring doesn't stretch or compress. If you put it on a curve, eg. on top of a ball, the front and back are on different slopes and can pull away from each other.) The "local flatness of spacetime" implies that if you're looking only at a single point, you can't detect curvature, but you also can't detect gravitational force (is that right???). So for the Earth analogy, using it only to show what you can detect locally, and not how it affects the motion of objects, we could take the real Earth and modify it so that "up" is always normal to the surface you're standing on. Like in some cartoons where if you're standing on the side of a mountain, your body is tilted so your feet remain "flat" against the slope. Then if you can only look at the ground beneath your feet, you can't tell if you're on a slope or not. Is that a fair analogy to spacetime, or is it only curvature that is locally undetectable, but not gravitational force? Also I have a feeling I've asked a similar thing but still don't get it: Is the curvature relative, so it actually is locally zero but a different value from a distant? Or does local flatness merely mean, like you suggest, that the local value of the curvature only has measurable meaning across some distance? I think it's the latter??? Can curvature be called a scalar field, and is it invariant in a static universe?
  3. Can you please explain the statement that I bolded? You implied that I should back up assertions with detail, and when I looked closer at your analogy that you claim is a "better one", I find that it makes no sense at all. Can you explain what this is meant to show about GR, especially related to the topic of how curvature in space-time is shown? I don't see any questions asked by swansont. I agree with all the points he made.
  4. I already justified the opinion, your model doesn't show the paths of objects bending in a curved space, and the rubber sheet analogy does. But I don't see what your analogy is even trying to say. What I get from it is you're saying that objects can only travel along gridlines through space? Are the streets representative of dimensions? Are they both spatial, or is one meant to be time? (I guess spatial, since you said "ignoring time", but then I don't understand MigL's comment that it "considers the space-time interval".) You can only travel in one dimension at a time, one "first" and then another? And that should give someone an idea of how gravity works?
  5. It's still much better than your buildings and streets analogy.
  6. 1) It is meant to represent the curvature of a space (just not real space), and the paths of objects through that space. What I meant was that I don't think the sheet's curvature is intended to actually model real spacetime curvature. I don't think they curve in the same way. But that's fine, it is an analogy of real spacetime curvature, not a model of it. My problem is that it is not clear how closely it is analogous, or even what properties exactly it is representing. Usually with a physical analogy you can clearly see the differences between the analogy and the real thing, and you don't confuse them. Here, the "fabric of spacetime" is such an elusive abstract thing that people see the sheet as a model of that "fabric". 2) I don't understand. It doesn't represent the real path of an object through spacetime (see (1)), yet a marble initially at rest on the trampoline does follow the line to the COG of the mass. 4) Why can't a 1D line curve? Can't you map one 1D space onto another different 1D space? 5) Again it's an analogy. You can't practically show the curvature of a sheet due to mass alone, the thing that causes the curvature in the sheet is only an analogy to the thing that causes curvature in spacetime.
  7. The mass on a trampoline shows how paths of objects bend in a curved space. studiot, I don't see how your model shows that. The curved sheet doesn't technically need gravity to show this; distort the sheet some other way, and run a "straight" line of tape over the curve and the path will bend (analogous to a null geodesic). Yes the analogy has problems. The mass represents mass, but the curvature is not representative of spacetime curvature, which I think is 0 at the center of the mass?* Showing the Earth resting on the sheet incorrectly suggests that it's the volume of matter displaces spacetime. I think this fails Einstein's "as simple as possible, but no simpler" criterion. Instead of saying "fabric", it could be called a manifold made up of events---would that stop people from asking what it's made of? Maybe the rest of the analogy could be fixed by labeling things similarly abstractly, instead of using concrete things like an Earth. But I don't know how you'd label it because I don't know what the curvature of the trampoline is actually meant to represent. Is it gravitational potential? Or is it just a toy example of an abstract curved space? I think the trampoline model could be set up and described differently, "no simpler than possible", so that it would both be clearer what it's meant to show, and not suggest other things. At the very least, I feel it should make people think something like "curved spacetime bends the paths of objects" and not "gravity pulls on the fabric of space" or whatever. Speaking of space vs. spacetime, the inclusion of time in the curvature is what makes masses at rest gravitate toward each other (is that right? along with constancy of 4-velocity magnitude?) but I don't see how that could be represented on a curved sheet. * Edit: now I'm confused because the curvature of the trampoline is also zero under the center of the mass, so maybe it does fairly represent curvature?
  8. Maybe this was already explained better but I didn't see it. An inertial object is only moving relative to something else (ie. an observer). If they're inertial and their distance isn't changing, then they're relatively at rest. If you then say that they're moving at the same speed but still at rest relative to each other, then that's relative to another frame, which you haven't mentioned. In non-relativistic physics, you might assume you're talking about some universal frame, but probably no one else here would do that; if you only mention frames A and B and then a speed, I think everyone would assume you mean A's speed relative to B and vice versa. You would have to specify a third frame ("Earth frame" for example) for people to get what you mean. Also, an observer is a frame of reference in SR. https://en.wikipedia.org/wiki/Observer_(special_relativity) This is because you observe ie. measure the same distances and times no matter where you are in a given inertial frame, so you don't have to distinguish different viewpoints in the same frame as different observers. The first part sounds right........ if the star is moving towards you at 80% of the speed of light and is 6 LY away, then you accelerate instantly so it is at rest, it should now be 10 LY away (not "would have". It still is 6 LY as measured in your first frame, and is 10 LY in your second frame). The second sentence is kinda wrong and this is where it gets fun! You could ignore this until you get the rest of the replies in the thread. Because of relativity of simultaneity, you and the star don't measure the distance between you as 6 LY at the same time. For example if the star is 6 LY away, approaching at 0.8c, it will take 6LY/.8c = 7.5 years for it to arrive. But as observed by the star, your clock is ticking at a rate of 0.6x its own. So while you measure 7.5 years to reach the star (from the event where you measure the star being 6 LY away), the star measures 12.5 years in its own frame, during which you travel 12.5 y * 0.8c = 10 LY. The star measures you as 10 LY away when (according to you!) you measure the star being 6 LY away. You also measure the star's clock ticking at .6x your own. You and the star are symmetric: each of you measure the other as 6 LY away at the moment (in your own frame) that the other measures being 10 LY away. Or another way to see it is: at the moment (according to you) that the star passes the 6 LY mark on your rulers (which are at rest relative to you), the star's rulers are length-contracted by a factor of 0.6, and you are passing the 10 LY mark on its ruler. At the event where you say "we're 6 LY away", the star measures you at the mark that's 10 LY away. ... But then, you might also see, if you're at the 10 LY mark on the star's ruler and it's your ruler that's length-contracted according to the star, then when you're at the star's 10 LY mark, the star is at a 16.666 (repeating of course) LY mark on your ruler! According to you, the event of you passing the star's 10 LY mark is simultaneous with the event of the star passing your ruler's 0.6 LY mark. According to the star, the event of you passing its 10 LY mark, is simultaneous with it passing your 16.666 LY mark. This is no problem because the relative simultaneity of distant events is different for different frames of reference. There are a lot of ways to describe this, I edited it to try to simplify, others probably have clearer and simpler ways to say it.
  9. I'm assuming flat spacetime (no mass, SR only), and inertial motion unless specified. Generally yes. Also "when are you talking about?" matters and is more complicated than a Newtonian description. The positions of things on a map are coordinates within a coordinate system, and those are different for different observers. You could have a map where the Milky Way is at a fixed location and Andromeda is moving, or one where Andromeda is fixed and the Milky Way is moving. Those correspond to the coordinate systems of 2 observers at rest in the respective galaxies. Yes, the distance to the star is different in the different frames of reference. There are invariant measures of distance, eg. the "proper length" of a 1 m stick at rest is always 1 m and everyone will agree on that, even if the stick is moving relative to some observer and is length-contracted ie. has a coordinate length less than a meter in that observer's coordinates. We can say that. We can measure the distances to both galaxies using one frame of reference (eg. the one in which we're at rest), and you can measure the motion of objects using the coordinates of that frame. Consider the map analogy. The spatial coordinates can be represented by a grid drawn on the map. The same grid coordinates can be shown by putting a lattice of rulers throughout space. In our own frame of reference, our rulers are not moving and so they don't length-contract. An object light years away can wobble at speeds near c, but yet stay near one place in the grid of rulers. Meanwhile, that distant wobbling object is moving relative to our lattice of rulers, and our rulers do length-contract, in its frame of reference. For simplicity consider two different inertial frames of reference F1 and F2 that the wobbling object switches between. Each of those frames has its own set of rulers making up a lattice throughout space, each at rest and not length-contracted in its own rest frame. Say I'm at 1 LY from Earth, as measured by Earth, and I'm wobbling relative to Earth. I stay near the 1 LY mark on Earth's set of rulers, but those rulers are contracted by different amounts in F1 vs F2. For example, in F1 Earth might be 0.8 LY away from me and the 1 LY mark, and only 0.6 LY away in F2's frame. The reason that the distance as measured by Earth isn't changing much, and the distance measured by me is changing drastically, is that I'm switching between different frames of reference. The distance between Earth and the 1 LY mark, which has a proper distance of 1 LY, is length-contracted by different amounts to different observers, depending on their relative speed.
  10. You're making up definitions but it sounds like you're using "duration" to describe proper time. You're not talking about velocity here. Velocity is a measure of distance/time as measured by a single observer (aka inertial frame). There is a measure of rate of motion called proper velocity or celerity that instead of measuring time using the observer's clock, it uses proper time as measured by the moving clock. Celerity approaches infinity as velocity approaches c. It's not a measure of velocity because you're measuring distance in one frame and time in another.
  11. Classically, a point particle with a precisely located energy makes sense. With QM wouldn't such a thing require a completely undefined momentum, maybe other consequences? Wouldn't you need to start with a particle with infinite energy in QM as well? This example sounds like less of an incompatibility of the two, and more a problem of applying certain aspects of modern physics while ignoring others, and as expected reaching a conclusion that doesn't agree with measurements of reality.
  12. Yes, that's what I should have said, thanks for the correction. I was mixing up having a space and time coordinates drawn as a grid on the paper, and then mistakenly thinking of the paper as a space that exists in time. It should be the two worldlines of the respective ends of the wormhole, joined and then effectively sewed together. As you hint at, and I argued in another thread, it wouldn't be enough to have two wormholes, with one going backward in time, to allow causal loops. The mouths would also have to be close enough together. There are no closed time-like curves if the curve is space-like. For example, if you go through a wormhole that brings you 100 years back in time, but also puts you 200 light years away, there is nothing at the entrance that you can affect before you leave. Or, if you travel far away "instantly" using one wormhole, and then back 100 year and ending up near Earth with another, but you have to travel 200 light years to get from one wormhole to the other, you also can't affect anything from before you leave. I didn't mean they wouldn't be. The clock goes through at vanishing speeds. I was calling it "the Earth frame" but they share that frame. Yes, that sounds like what I was thinking of. When I wrote "long time" I had the Doppler effect on my mind. But really, only relativity of simultaneity (3 hours) and time dilation (gamma=2) contribute to the long time, so let's say "it's a stretch" to call it a long time. The delay of light due to the (length-contracted) distance to Earth as well as the (not length-contracted) increasing distance of the Earth between the rocket's now and 6 hours later when it enters the wormhole at the other end, contribute to the much longer time until the rocket sees the Doppler-shifted image of itself entering the wormhole. I wasn't thinking that far, I was only trying to describe the simplest case I could. But I guess you'd have to decide if the wormhole remains set up for the Earth frame, or if it behaves as an object that changes to a new inertial frame. In the paper analogy, eg. if you have 2 joined worldlines and you move them separately across the paper, do they stay sewed together like they originally were, possibly stretching and distorting? Or do they slide across each other as you skew the paper, and end up like they were set up for their new reference frame as much as they were for the old one? I have no thoughts on that, but I think that since this wormhole is not based on anything real or theoretical, you'd have to make it up. Yes, I agree completely. I was only talking about "according to the rocket", meaning the coordinate time of the distant Earth in the rocket's standard (Minkowski?) coordinates. The whole point is the acceleration doesn't have a causal effect, because the change in coordinate time on Earth when the rocket accelerates also corresponds with a change in relative simultaneity between the mouths of this wormhole. So for example, if two twins leave Earth and travel together "the long way" to the far end of the wormhole, and only one turns around (thereby having the distant Earth's time in their coordinates "jump ahead"), and they enter the wormhole roughly at the same time, they exit at roughly the same time too (travelling in opposite directions here). As in your example, they both age 2 hours while Earth aged 4, despite only one experiencing a change in relative simultaneity in its coordinates.
  13. Yes, that sounds like what I'm describing. Another way to describe it is with paper. If you draw out gridlines representing time and one spatial dimension on a flat sheet of paper, that can represent flat spacetime, and the rules of SR can be applied. If you bend the paper without stretching it, you're not distorting the relations between events along paths drawn on the paper. The rules of SR still apply. The wormhole represents bending the paper and making 2 points touch, and adding a zero-length path or connection between those 2 points. I specified that the wormhole mouths are at rest in the Earth frame. It sounds like if you added the ability to accelerate them, you'd have a choice of what happens, and would have to further specify that? As specified, a clock from Earth that goes through the wormhole at vanishing speeds at time t, comes out at time t. So the clock would remain synchronized with Earth's clock, in Earth's frame. To be consistent in other frames, the clocks would have to be out of sync in other frames. Let's say there are 2 clocks at either end of the wormhole, sync'd in Earth's frame. For a rocket traveling away from Earth in the direction of the far mouth, the clock on Earth is behind the one at the far mouth. In this frame, a rocket that enters at Earth and exits "right now" at the far mouth, will not enter until a long time passes! For a rocket travelling in the opposite direction, the rocket that exits "now" has entered a long time ago. None of this is a problem at all. With the wormhole mouths not accelerating or changing their configuration, the rockets can communicate with each other. If a message can be passed in the Earth frame, it can be passed in any frame (light cones are invariant). Also, if a rocket exits and can say either "I have not yet entered the other side" or "I just entered the other side" or "I entered long ago", it can also switch between those statements simply by accelerating to a different inertial frame where another statement becomes true. It sounds like, by underspecifying the wormhole details, I've made it possible to add details that violate SR, but it's also possible to leave it so that no added details violate SR. So for example, say a rocket leaves Earth at high speed at Earth-time t and goes through the wormhole, exiting at local wormhole time t. But the time on Earth, according to the rocket, might now be t-0.6 years. The rocket turns around in negligible time. Local wormhole time is t+epsilon, but the time on Earth is now say t+0.6 years. It enters the wormhole at t+epsilon and returns to Earth at t+epsilon. Neither twin has spent significant time moving relative to the other, and both have aged negligibly. (This isn't any of the 4 scenarios above, all of which involve the rocket making a long-duration one-way trip.) Or you could say, the rocket that exits could see Earth at two different times, one across flat spacetime, say .6 years in the past or future but the image delayed by ... a little or a lot by the travel time of light, and one at time t seen nearby through the wormhole.
  14. The calculations I get are that The wormhole I described is apparently not realistic. But, the same calculations can be done by removing the worm hole and replacing it with 2 planets on each end, with synchronized clocks, and then calculating a one-way rocket trip. You're then comparing the rocket's proper time and the Earth's coordinate time so it's not the same as the twin paradox, but that's fine, SR lets you calculate that. The wormhole is set up to connect events that are simultaneous in the Earth's frame, and an object traversing it in an Earth-frame's instant is sort of like saying that two clocks at either end and synchronized in the Earth frame, represent the same clock. It doesn't matter who goes through the wormhole or what their inertial frame is, the wormhole represents an instant jump in the Earth frame, generally not in other frames. I wrote out a bit of time dilation and Doppler analysis for each of the cases, but it ended up being more tedious than I expected. I could post that if anyone's interested. As usual, if you take out any magic or mystery from SR "paradoxes" and look at it geometrically, it's straightforward.
  15. That makes sense, to treat objects in the same inertial frame the same, regardless of what's in their black box. I think it would be easier to deal with proper time, because if you have the proper time of B between 2 points, then you really don't care what it does between those 2 points, or even if it remains inertial or has constant speed. I think it doesn't work with coordinate time in this case. According to A, two different objects say B and B' that travel between the same two events, will age the same if they travel any path between the events at the same speed. By necessity, if they do that (same speed ending up at same place, regardless of what path they took to get there through a simply connected flat spacetime ie. no wormhole) then they'll have travelled the same odometer distance relative to A's inertial frame, for the same amount of time. But if B travels through the wormhole and B' travels through space, they can end up at the same location, with the same speed, and yet have travelled different distances. They couldn't travel different distances at the same speed and end up at the same event (same time). The way I set up the wormhole, I'm assuming that B ages negligible proper time when traversing the wormhole. If B' makes the same trip through space at a speed faster than light relative to A, it should age a negative proper time I think. I think it wouldn't matter, if it's possible to say "the proper time to traverse the wormhole is negligible", but I don't know if that's valid, eg. if what I described has a singularity. The relative time that passes outside the wormhole during the trip, is already specified. Effectively, if an object goes through the worm hole at Earth time t, proper time 0, they emerge at time t (in Earth's frame, aka. according to a local clock on the far end of the wormhole which is Einstein synchronized with Earth's clock) and proper time 0. Basically, the shortest shortcut possible. That might be geometrically impossible for some reason I'm unaware of.
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