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md65536 last won the day on December 15 2020

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  1. I bet reddit.com/r/Whatisthis/ or reddit.com/r/whatisthisthing/ would figure it out within an hour. I figured it might be part of a tool, like a plane depth adjust, or a proportional divider knob. The closest I've seen is on a horological tool, like in the top left: The knob slides and tightens to lock. A string might fit in the groove, and the part is used to set the tension on it. This one looks like it spins freely though.
  2. I wouldn't say that exactly. I'd also call it "inertia" instead of artificial gravity. But objects do tend to have some rotation, and inertia does contribute to forces acting on things (eg. the Earth bulges around the equator, along with a feeling of less pull of gravity than at the poles, at a fixed distance from the centre of the Earth), and for a static object that inertia does not require energy to maintain. My point is that if you choose a force that requires constant use of energy for one observer, and another force that doesn't for another observer, that's not a difference that concerns the equivalence principle. The principle doesn't say that any 2 different things you choose should be the same, and if there's a detectable difference (eg. in energy use) then the principle doesn't hold. Effectively it says if you make everything else the same, gravity is not detectably different from acceleration.
  3. It wouldn't take infinite energy and you wouldn't reach a speed of c relative to anything. If you accelerated away from Earth with a constant proper acceleration for seventy years, the coordinate acceleration of Earth away from you would decrease and approach zero as its speed approached c, because of the velocity addition or composition law of SR. I don't think the difference in energy use is related to localness of effects, or whatever. You could hover over the Earth on a stationary platform that uses energy to create 200 lbs of thrust for 70 years. The equivalence principle still applies. You can't tell whether you're accelerating away or overcoming gravity. You could also simulate gravity by being in a rotating ship in freefall for 70 years, and use no energy, and experience no gravitational acceleration. The equivalence principle might not apply since that would be a rotating frame of reference.
  4. Well, the curvature of the parabola decreases with increasing positive x. There's an analogy for what you're talking about. "Local" effectively means at small enough distances that the effects of distance don't matter, and that's not a fixed size. At small x, you have to zoom in to a smaller area to make the parabola appear flat, and at large x, it can appear flat over a larger range. Analogously, near to a gravitational mass "local" might be very small, but very big much farther away. If you're free-falling near a black hole and being spaghettified, spacetime is still locally Minkowskian but at distances much smaller than a human.
  5. 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.
  6. 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?
  7. 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.
  8. 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?
  9. It's still much better than your buildings and streets analogy.
  10. 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.
  11. 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?
  12. 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.
  13. 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.
  14. 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.
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