Eise Posted January 23, 2020 Share Posted January 23, 2020 3 hours ago, MigL said: But I've never seriously considered space-time as something that can 'bend' or curve. It is at least difficult to imagine... But this brings me to another point. If it is correct what Kip Thorne says (which I strongly assume, knowing what kind of expert he is), then we have 2 completely different 'pictures' based on which interpretation we choose: gravity curves spacetime, and all free moving objects follow geodesics in spacetime gravity is a force, which affects paths of free moving objects, durations and distances But especially my use of 'force' seems to completely contradict the way GR is usually presented: that gravity is not a force. How could one formulate the second interpretation in a better way? Maybe I should stick to Mordred's view: it seems to have the least metaphysical baggage of all descriptions of GR. 41 minutes ago, Markus Hanke said: diffeomorphisms Word of the day... So glad that Wikipedia exists... 48 minutes ago, Markus Hanke said: But that alone does not allow them to conclude anything about whether or not that surface is embedded in a higher dimensional space. OK, now I play for a moment devil's advocate (in this case as a Kantian philosopher): the concept of curvature only makes sense when there is a higher dimension. A line is said to be curved because there is one dimension more, namely the second (wow, that sounds stupid. And as I think to understand, for a 1-D organism there is no chance, not even with differential geometry, to discover that his universe is curved.); a surface is said to be curved, because there is one dimension more in which it is curved. But a 2-D organism can at least discover that, by measuring angles and distances. But how can he conclude that this 3rd dimension does not exist, and that the curvature of his world is intrinsic only? And of course the same questions for 3-D organisms, that discover that their world can best be described as if it is curved? Shouldn't they conclude that there is one dimension more? 42 minutes ago, uncool said: I think you mean the opposite: the spaces are not diffeomorphic, but they are locally isometric. Or locally diffeomorphic? Link to comment Share on other sites More sharing options...
MigL Posted January 23, 2020 Share Posted January 23, 2020 My ( limited ) understanding was that you could always tell using Pythagoras'... A triangle's vertices add to 180 deg in no curvature. To less than 180 deg in positive curvature. And to more than 180 in negative curvature. But that doesn't seem to apply to the FLAT Torus example I mentioned in a previous post. Maybe someone more familiar with diff geometry and topology can clarify. Link to comment Share on other sites More sharing options...
Strange Posted January 23, 2020 Share Posted January 23, 2020 43 minutes ago, MigL said: My ( limited ) understanding was that you could always tell using Pythagoras'... A triangle's vertices add to 180 deg in no curvature. To less than 180 deg in positive curvature. And to more than 180 in negative curvature. But that doesn't seem to apply to the FLAT Torus example I mentioned in a previous post. It does apply because your "flat torus" is, be definition, flat (which is why you can't make a physical one, just a virtual representation on a screen). But you have just made me realise that the "inner" (nearest the hole) surface of a "real" torus has negative curvature while the outer surface has positive curvature. So you could create triangles that meet all three of those conditions. Link to comment Share on other sites More sharing options...
MigL Posted January 24, 2020 Share Posted January 24, 2020 and on re-reading my post and your reply, I realized I had Pythagoras backwards for positive and negative curvatures. Positive ( like a sphere ) should be MORE than 180 deg. Negative ( like a saddle ) should be LESS than 180 deg. Link to comment Share on other sites More sharing options...
Markus Hanke Posted January 24, 2020 Share Posted January 24, 2020 22 hours ago, uncool said: I think you mean the opposite: the spaces are not diffeomorphic, but they are locally isometric. Why are they not diffeomorphic? I am not a mathematician, so it is very possible that I am misunderstanding something. However, it seems clear that all diffeomorphisms are also isomorphisms (but not vice versa); furthermore, the definition of a diffeomorphism I am familiar with is that of an invertible function that maps one manifold into another (1-to-1) such that the function is smooth and differentiable throughout its domain. Why does this not apply to the cylinder and flat sheet example? 21 hours ago, Eise said: the concept of curvature only makes sense when there is a higher dimension That is true for extrinsic curvature, but not for intrinsic curvature. 21 hours ago, Eise said: But how can he conclude that this 3rd dimension does not exist, and that the curvature of his world is intrinsic only? Concluding that the higher dimension does not exist is not trivial. There is nothing that rules out per se that such a dimension exists. However, one can look at the physics in such a universe - it is likely that at least some of the laws of physics in such a world are “sensitive” to the presence of the extra dimensions. For example, there are fundamental reasons why in a universe with 3 spatial dimensions we would expect to find inverse square laws of various kinds; if we were to empirically find (e.g.) inverse cube laws instead, then that could potentially be due to the presence of an extra spatial dimension. So there may be ways. However, it is also possible that all laws of physics are confined to within the embedded manifold; in such a world, detecting the extra dimensions might be very difficult. I don’t know if it would be impossible - that’s a good question. 21 hours ago, Eise said: gravity is a force, which affects paths of free moving objects, durations and distances A test particle in free fall (gravity only) experiences no forces - an accelerometer falling in the same frame as the test particle will read exactly zero at all times. So clearly, gravity is not a force. Link to comment Share on other sites More sharing options...
uncool Posted January 24, 2020 Share Posted January 24, 2020 (edited) 4 hours ago, Markus Hanke said: Why does this not apply to the cylinder and flat sheet example? Because the map is not invertible - only locally invertible. When you wrap the sheet around the cylinder, it will cover itself (infinitely many times, even). Also, "diffeomorphism" is a term properly from differential topology - the metric plays no role in it (e.g. a "smoothed" coffee cup is diffeomorphic to a donut). The term "isometry" is much stronger - literally "same metric". Edited January 24, 2020 by uncool Link to comment Share on other sites More sharing options...
studiot Posted January 24, 2020 Share Posted January 24, 2020 (edited) 1 hour ago, uncool said: (e.g. a "smoothed" coffee cup is diffeomorphic to a donut) with handle ? 1 hour ago, uncool said: Because the map is not invertible - only locally invertible. When you wrap the sheet around the cylinder, it will cover itself (infinitely many times, even). Also, "diffeomorphism" is a term properly from differential topology - the metric plays no role in it (e.g. a "smoothed" coffee cup is diffeomorphic to a donut). The term "isometry" is much stronger - literally "same metric". That's what my maths dictionary says. A general note to all on 'local' ; A great many (well respected) textbooks on differential geometry say somewhere in their introduction that "This book deals primarily or even exclusively with local differential geometry" Edited January 24, 2020 by studiot Link to comment Share on other sites More sharing options...
Strange Posted January 24, 2020 Share Posted January 24, 2020 24 minutes ago, studiot said: with handle ? A donut with a handle is diffeomorphic to a loving cup Link to comment Share on other sites More sharing options...
scuddyx Posted January 24, 2020 Author Share Posted January 24, 2020 On 1/18/2020 at 12:41 PM, scuddyx said: I find it hard to understand General Relativity when it is casually referred to as curvature in spacetime or as the sagging in a trampoline mat. Would a better explanation for the novice be to say things fall because they are seeking out the place where time runs the slowest? For instance, when explaining how light is deflected as it passes close to a star imagine it surrounded by voxels (3D blocks of space). Time runs slowest in the voxels close to the star. Because the speed of light is constant in all frames of reference and as speed = distance /time the voxels close to the star would appear to be smaller to the photons of light. Consequently the quickest route for the path of light would be to travel close the the star. This neatly explains the deflection. Do you agree? Does gravity dictate the flow of time or does time itself define gravity? Thanks I think the concept of spacetime makes the understanding of GR unnecessarily difficult. Closer you are to a massive object time slows down. Anything passing through space where time has slows down (to external observers) must also appear to experience travel shorter distances to ensure the speed of light is constant. If something wants to take the shortest path (why it would want do so is another issue) it would travel close to the massive object. Link to comment Share on other sites More sharing options...
swansont Posted January 24, 2020 Share Posted January 24, 2020 24 minutes ago, scuddyx said: I think the concept of spacetime makes the understanding of GR unnecessarily difficult. And that's an issue...for you. It does not make that statement true for other people. Quote If something wants to take the shortest path (why it would want do so is another issue) it would travel close to the massive object. That's incomplete, or wrong. If I want to go somewhere, detouring near a massive object will not, in general, shorten my path. Link to comment Share on other sites More sharing options...
scuddyx Posted January 24, 2020 Author Share Posted January 24, 2020 1 hour ago, swansont said: And that's an issue...for you. It does not make that statement true for other people. That's incomplete, or wrong. If I want to go somewhere, detouring near a massive object will not, in general, shorten my path. The first observation of light deflection was performed by Arthur Eddington, during the 1919 total solar eclipse, when stars near the Sun were observed to change position. Starlight took a shorter path thanks to gravitational time dilation caused by the Sun. Link to comment Share on other sites More sharing options...
studiot Posted January 24, 2020 Share Posted January 24, 2020 2 minutes ago, scuddyx said: Starlight took a shorter path thanks to gravitational time dilation caused by the Sun. A 'deflected' path is surely longer ? Link to comment Share on other sites More sharing options...
scuddyx Posted January 24, 2020 Author Share Posted January 24, 2020 1 hour ago, studiot said: A 'deflected' path is surely longer ? The 'deflected' path is actually a geodesic as described by general relativity and is by definition a 'straight line' or shortest path in 'curved spacetime'. Physicists skilled in general relativity may be able to visualise curved spacetime, stress-energy tensors and Riemannian space. But if you accept that gravity causes time dilation - it follows that objects 'simply' take the shortest path (for whatever reason) by travelling close to massive objects. Gravity may eventually be shown (consistent with Einstein's GR) to be an emergent phenomenon from the nature of time itself. Admittedly this is a working hypothesis and, as yet, I don't have supporting maths. 😉 Link to comment Share on other sites More sharing options...
swansont Posted January 24, 2020 Share Posted January 24, 2020 2 hours ago, scuddyx said: The first observation of light deflection was performed by Arthur Eddington, during the 1919 total solar eclipse, when stars near the Sun were observed to change position. Starlight took a shorter path thanks to gravitational time dilation caused by the Sun. Starlight on a particular path — that which ended up at the telescope — followed that path. Light heading to Jupiter would not have traveled a shorter path by passing near the sun. But the question here is : shorter than what? There's only one path to get from point A to point B that follows the laws of nature. It sounds like you are trying to apply the principle of least time (Fermat's principle). That the path taken by light will be the one that has the smallest travel time, but that's not physically the shortest path in situations like materials with different indices of refraction. Link to comment Share on other sites More sharing options...
Mordred Posted January 24, 2020 Share Posted January 24, 2020 38 minutes ago, scuddyx said: The 'deflected' path is actually a geodesic as described by general relativity and is by definition a 'straight line' or shortest path in 'curved spacetime'. Physicists skilled in general relativity may be able to visualise curved spacetime, stress-energy tensors and Riemannian space. But if you accept that gravity causes time dilation - it follows that objects 'simply' take the shortest path (for whatever reason) by travelling close to massive objects. Gravity may eventually be shown (consistent with Einstein's GR) to be an emergent phenomenon from the nature of time itself. Admittedly this is a working hypothesis and, as yet, I don't have supporting maths. 😉 I would suggest you look into the principle of least action and how it applies to the geodesics equation. I would also suggest you look at the Principle of equivalence. [math]m_i=m_g[/math] Link to comment Share on other sites More sharing options...
studiot Posted January 24, 2020 Share Posted January 24, 2020 2 hours ago, scuddyx said: The 'deflected' path is actually a geodesic as described by general relativity and is by definition a 'straight line' or shortest path in 'curved spacetime'. So if it is travelling its natural course, in what way is it deflected? I don't mean by what agent deflects it. We agree that is gravity. But 'deflected' implies passing through a different sequence of coordinates in some manner. Do you really mean this? I think the 'deflection' occurs in 3D space only and in Eddington's case can be reduced to a 2D analysis. Link to comment Share on other sites More sharing options...
Markus Hanke Posted January 25, 2020 Share Posted January 25, 2020 20 hours ago, uncool said: Because the map is not invertible - only locally invertible. When you wrap the sheet around the cylinder, it will cover itself (infinitely many times, even). Indeed, you are correct. I did not consider that. Thanks for pointing it out! But one could still say that they are locally diffeomorphic, right? 18 hours ago, scuddyx said: I think the concept of spacetime makes the understanding of GR unnecessarily difficult. GR is a model of spacetime. You can’t really have one without the other, in my opinion. Reducing GR down to time dilation alone is an extreme simplification, which works only in some highly specific cases, and even then omits some important features of gravity. I honestly don’t think one does a beginner any favours by casting the model in this highly inaccurate light. 14 hours ago, scuddyx said: The 'deflected' path is actually a geodesic as described by general relativity and is by definition a 'straight line' or shortest path in 'curved spacetime'. Actually, it is the longest path in spacetime. Remember that the geometry we are dealing with here is not Euclidean. Link to comment Share on other sites More sharing options...
uncool Posted January 25, 2020 Share Posted January 25, 2020 12 minutes ago, Markus Hanke said: But one could still say that they are locally diffeomorphic, right? Yes (even better: that the map is a local diffeomorphism; the difference being whether there is *some* map or whether it is *this* map), but that is a very weak statement; any two smooth manifolds of the same dimension are locally diffeomorphic - because locally, the differential topology is that of Euclidean n-dimensional space. The term I think you want is that they are locally isometric (or more specifically, that the "wrapping map" is a local isometry), that is, that the metric, or the geometry, matches. 1 Link to comment Share on other sites More sharing options...
Markus Hanke Posted January 25, 2020 Share Posted January 25, 2020 31 minutes ago, uncool said: Yes (even better: that the map is a local diffeomorphism; the difference being whether there is *some* map or whether it is *this* map), but that is a very weak statement; any two smooth manifolds of the same dimension are locally diffeomorphic - because locally, the differential topology is that of Euclidean n-dimensional space. The term I think you want is that they are locally isometric (or more specifically, that the "wrapping map" is a local isometry), that is, that the metric, or the geometry, matches. Ok got it. Thank you! Link to comment Share on other sites More sharing options...
Mordred Posted January 25, 2020 Share Posted January 25, 2020 (edited) Informative side note on local diffeomorphism in GR which is diffeomorphic invariant through its use of one forms and a vector. (This also results in the invariance under coordinate change) vectors are not invariant under coordinate change however one forms are. A one form being a covector. The principle of covariance is preserved through the local diffeomorphism of the [math]g_{\mu\nu}[/math] metric by the push forward of the Minkowskii tensor [math]\eta[/math] with the Poincare group SO(3.1) being the ISO group for the Minkowskii tensor. ISO(n). The above will also apply to the killing vectors of causality connected fields. (Localization) Edited January 25, 2020 by Mordred Link to comment Share on other sites More sharing options...
Markus Hanke Posted January 26, 2020 Share Posted January 26, 2020 7 hours ago, Mordred said: This also results in the invariance under coordinate change I think you mean (general) covariance, not invariance...? The components of the metric tensor do change under coordinate transformations, but they change in such a way that the relationships between said components are preserved, so the overall tensor remains the same one. 7 hours ago, Mordred said: vectors are not invariant under coordinate change however one forms are 3-vectors aren’t generally covariant, but 4-vectors are, since they are rank-1 tensors. Am I being a pain now by pointing out these things? Link to comment Share on other sites More sharing options...
Mordred Posted January 26, 2020 Share Posted January 26, 2020 (edited) Nope your not being a pain in pointing those out. The four vectors differ from Euclidean vectors. The magnitude of the four vector invariance is preserved through the Lorentz transformations. I should have been a bit more clear on that. I've been trying to figure out how to introduce one forms and why they are used in tensors. However how to introduce them in a simple manner is a bit daunting. Particularly since you need the components of a vector and vector basis. Edited January 26, 2020 by Mordred Link to comment Share on other sites More sharing options...
Markus Hanke Posted January 27, 2020 Share Posted January 27, 2020 11 hours ago, Mordred said: The magnitude of the four vector invariance is preserved through the Lorentz transformations. Yes, that is indeed true. Link to comment Share on other sites More sharing options...
studiot Posted January 28, 2020 Share Posted January 28, 2020 On 1/23/2020 at 7:37 AM, Eise said: OK, now I play for a moment devil's advocate (in this case as a Kantian philosopher): the concept of curvature only makes sense when there is a higher dimension. There is more than one sort of curvature in geometry. Curvature is also a measure of 'twist'. (called torsion). Think of a line of cotton reels strung together tightly on a string. The string as a whole can snake about on 2 or 3 dimensions but is still a string of reels. But now imaging we start with the string stretched out in a straight line and every reel has a line mark (just like on the dial of a cooker,) and that every one of these lines is pointing the same way. Now twist the string so that the lines point in different directions, yet the string is still straight. This is torsion and the curvature is measured as the angle between two adjacent cotton reels. There is thus no sideways displacement associated with this form of curvature, as there is when the string curves around a bend. Finally I should mention that one dimensional beings on the string will be unable to measure anything to detect this torsion curvature, that is left to a two or higher dimensional being. Torsion is still extrinsic. Link to comment Share on other sites More sharing options...
Markus Hanke Posted January 29, 2020 Share Posted January 29, 2020 9 hours ago, studiot said: Curvature is also a measure of 'twist'. (called torsion). Well pointed out. It bears mentioning though that in standard GR the Levi-Civita connection is used, which is torsion free. Link to comment Share on other sites More sharing options...
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