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 JREF Forum Terra revolving round Sol question

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 2nd May 2012, 10:47 AM #241 Farsight Suspended   Join Date: Mar 2008 Location: Poole, UK Posts: 1,956 Originally Posted by H'ethetheth Sorry, but that still does not address anything about the question. Let me explain: NASA asked the question what Newton(-ian mechanics) would say about a stationary earth in a rotating universe, and assumed Newton(-ian mechanics) would say that such an earth would not bulge. That's right. If the Earth wasn't rotating, Newtonian mechanics says there's no bulge. If you put the Earth inside a shell of matter, Newtonian gravity says the Earth doesn't feel any resultant gravity. If you then spin the shell, neither say that the Earth bulges. Originally Posted by H'ethetheth However, I don't think he/it would say that, because (barring something like sol invictus' super-strong cables solution) Newton(-ian mechanics) would have to posit some way for the universe to freely rotate around the earth that is itself consistent with Newtonian mechanics. This would result in some set of fictitious forces that would necessarily behave exactly like the fictitious forces that arise in a rotating reference frame. This would then necessarily subject the mass of the earth to internal stresses that would cause it to bulge. You don't need a universe, just a shell of matter. Or even just a ring. In Newtonian mechanics all the forces are on the ring, and in Newtonian gravity the Earth is unaffected. You have to go to "GEM", which was Heaviside's extension to start predicting an effect. See gravitomagnetism on wiki for that. But take a bit of care, it says "a moving electric charge creates a magnetic field" which is wrong because it's the electromagnetic field and we can move at different speeds through it. Plus it gives a fluid analogy rather than the rubber analogy. And the important thing is that The main consequence of the gravitomagnetic force, or acceleration, is that a free-falling object near a massive rotating object will itself rotate. That's because it's "twisted space". If you flew through space twisted like this, you find yourself turning. And if you had a rubber Earth in that NASA image, the twist reduces the equatorial diameter rather than increasing it. At least I think so, but I'm happy to stand corrected on that.
 2nd May 2012, 11:13 AM #242 Kwalish Kid Critical Thinker   Join Date: Feb 2010 Posts: 388 Originally Posted by Farsight At least I think so, but I'm happy to stand corrected on that. OK, show us the equations that you are using to come to your conclusion and then we will correct you if necessary.
 2nd May 2012, 11:27 AM #243 H'ethetheth fishy rocket scientist     Join Date: Aug 2004 Location: among the machines Posts: 2,340 Originally Posted by Farsight That's right. If the Earth wasn't rotating, Newtonian mechanics says there's no bulge. If you put the Earth inside a shell of matter, Newtonian gravity says the Earth doesn't feel any resultant gravity. If you then spin the shell, neither say that the Earth bulges. You don't need a universe, just a shell of matter. Or even just a ring. In Newtonian mechanics all the forces are on the ring, and in Newtonian gravity the Earth is unaffected. Am I really making my point that badly? My point is that there is not such a shell or ring that can carry loads or provide a centripetal force across the orbit around the earth. It's just loose stars and planets and gravity, so if Newtonians have to say whether the universe as we observe it rotating around the earth would make it bulge at the equator, then I think the answer is yes, because there would need to be some universal force on the stars to keep them in orbit around the earth. To be consistent with Newtonian mechanics, this force must work exactly like the centrifugal force that arises in rotating formulations of Newtonian mechanics (only in the other direction), and it will not work on the mass of the earth, because it is not rotating in this example, and will therefore cause it to bulge. Also, I'd like to point out that we're arguing over what is basically a marginal objection to NASA's article by a stubborn "professional Newtonian". Last edited by H'ethetheth; 2nd May 2012 at 11:29 AM.
 2nd May 2012, 11:45 AM #244 ben m Illuminator   Join Date: Jul 2006 Posts: 4,657 Originally Posted by H'ethetheth Am I really making my point that badly? Before you blame yourself, I might suggest wandering over to any of the other recent Farsight threads (Black Holes, Relativity+ are long ones; Nuclear Strong Force Is A Fiction, though short, was the last straw for me) to see how many gallons of perfectly-clear physics explanations have vanished down Farsight's voluminous drains.
 2nd May 2012, 02:08 PM #245 H'ethetheth fishy rocket scientist     Join Date: Aug 2004 Location: among the machines Posts: 2,340 Ah, then I may have mistaken my ignorance of relativity for his expertise. I'll have a look at the threads you mentioned.
 2nd May 2012, 02:23 PM #246 Perpetual Student Illuminator     Join Date: Jul 2008 Location: USA Posts: 3,707 Originally Posted by ben m To be honest, I'm getting tired of talking about it. I don't blame you. It must be quite tedious for you at his point. I guess my asking for confirmation of this aspect of GR so many times and in so many ways is because it is the most remarkable and counterintuitive thing I have ever encountered in science. In any case, I think it will remain difficult to accept until I can understand the underlying mathematical basis for myself -- if I am ever able to do so. Quote: There are two separate issues. (a) Are we allowed to label the coordinates of everything in the Universe in just such a way that your top's coordinates are constants, like an object sitting still, rather than rotating? Yes we are. If we do so, do the relabeled-laws-of-physics still predict that the top stays upright, gyroscope-style? Yes they do; the relabeled laws of physics include centripetal and Coriolis forces which indeed account for the stable angle between your top and the Earth. (b) "Well," you say, "I don't wanna deal with centripetal and Coriolis forces. I want to just look at the distant stars, pick a coordinate system in which those stars are at rest, and declare myself finished with wacky coordinates, can't I do that?" No you can't. In the presence of gravitomagnetic effects, centripetal and Coriolis forces are still there in the coordinate system in which distant stars are at rest. In the presence of gravitomagnetic effects, to find a centripetal-force-free coordinate system, you have to pick one where the distant stars are rotating. (c) Suppose you are locked in a room with your spinning top. You've waved your arms around and determined that the room is feeling no centripetal forces. But you can't look at the mass distribution around you. Can you predict, based on this information, whether the stars are sweeping past your window (as though you were rotating) or not (as though you were still)? No you can't---a nearby spinning mass, shell-like or otherwise, would make you make the wrong prediction. (d) You're not in a windowless room. Look into space. Is there actually a huge shell of matter out there, spinning? No. (Well, there could be one beyond the observable horizon.) Therefore there is no gravitomagnetic source-current that would make you spin. So don't worry about it. That's very helpful, thanks. __________________ It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman ξ Last edited by Perpetual Student; 2nd May 2012 at 02:33 PM.
 2nd May 2012, 05:44 PM #247 ynot Illuminator     Join Date: Jan 2006 Location: New Zealand Posts: 4,654 Originally Posted by Farsight You can tell it's got a mass because you fall down. Or does the mass fall up to you? When two things concurrently fall down from opposite sides of the mass towards it’s centre is the mass actually falling up to both of them? Does that mean the mass expands? When two things move toward a third thing at different speeds how does that third thing alternatively move toward the two things concurrently at different speeds? __________________ Rumours of a god’s existence have been greatly exaggerated. My post are all (IMO) unless stated otherwise. Last edited by ynot; 2nd May 2012 at 06:16 PM.
 2nd May 2012, 08:06 PM #248 Perpetual Student Illuminator     Join Date: Jul 2008 Location: USA Posts: 3,707 Quote: ben m Are we allowed to label the coordinates of everything in the Universe in just such a way that your top's coordinates are constants, like an object sitting still, rather than rotating? Yes we are. If we do so, do the relabeled-laws-of-physics still predict that the top stays upright, gyroscope-style? Yes they do; the relabeled laws of physics include centripetal and Coriolis forces which indeed account for the stable angle between your top and the Earth. In my recent review of the properties of tensors, my studies have revealed that a tensor equation holds true regardless of any transformation of coordinate system (I am currently trying to learn something about the mathematics of this stuff). So, when I look at $R_\mu_\nu - \dfrac{1}{2}g_\mu_\nu R + g_\mu_\nu \Lambda = \dfrac{8\pi G}{c^4}T_\mu_\nu$ and imagine a change in frame of reference on the left side of the equation (of the metric tensor), I see that things have to change in the stress energy tensor, for the equation to hold. Is it these changes to $T_\mu_\nu$ that result in these "centripetal and Coriolis forces"? __________________ It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman ξ
 2nd May 2012, 08:23 PM #249 ben m Illuminator   Join Date: Jul 2006 Posts: 4,657 Originally Posted by Perpetual Student Is it these changes to $T_\mu_\nu$ that result in these "centripetal and Coriolis forces"? I'll get out of my depth before long here, but no: rotating the coordinate system should show up primarily in the metric.
 3rd May 2012, 12:38 AM #250 Farsight Suspended   Join Date: Mar 2008 Location: Poole, UK Posts: 1,956 Originally Posted by H'ethetheth Am I really making my point that badly? There's obviously some kind of problem, sorry if it's me. Ben's unhappy bcause he and his friends peddle woo and I call them out. On the black hole thread they were trying to say space is infalling in a gravitational field, which is total trash. Originally Posted by H'ethetheth My point is that there is not such a shell or ring that can carry loads or provide a centripetal force across the orbit around the earth. It's just loose stars and planets and gravity... OK. And there's nothing that can keep some distant massive star moving in a circular orbit around the Earth. The Earth's gravitational field isn't up to the job. Originally Posted by H'ethetheth ...so if Newtonians have to say whether the universe as we observe it rotating around the earth would make it bulge at the equator, then I think the answer is yes, because there would need to be some universal force on the stars to keep them in orbit around the earth. The Newtonians wouldn't entertain this for a moment. They'd point to one distant massive star and say it's moving at a ridiculous speed, and the force required to keep it going round in a circle would have to be matched by an equal and opposite and force on the Earth. Then they'd point to all the other distant massive stars, and tell you this: the forces on the Earth wouldn't make the equator bulge, they'd pull it to bits in an instant. Originally Posted by H'ethetheth To be consistent with Newtonian mechanics, this force must work exactly like the centrifugal force that arises in rotating formulations of Newtonian mechanics (only in the other direction), and it will not work on the mass of the earth, because it is not rotating in this example, and will therefore cause it to bulge. Sure. I note Sol's cables, but he didn't spell out the upshot. It would be something like a bicycle wheel where you cut off the rim and solder a big lead weight on to each of the spokes. Then you cut out the hub and replace it with a ball of sand. Then you try spinning the thing.
 3rd May 2012, 01:00 AM #251 H'ethetheth fishy rocket scientist     Join Date: Aug 2004 Location: among the machines Posts: 2,340 Originally Posted by Farsight OK. And there's nothing that can keep some distant massive star moving in a circular orbit around the Earth. The Earth's gravitational field isn't up to the job. So far, so good. Originally Posted by Farsight The Newtonians wouldn't entertain this for a moment. They'd point to one distant massive star and say it's moving at a ridiculous speed, and the force required to keep it going round in a circle would have to be matched by an equal and opposite and force on the Earth. No it wouldn't, because if they carefully observe the heavenly bodies, they will see that the mystery centripetal force on the bodies is proportional (among other things) to their distance from the rotational axis of the universe. Since this axis pierces the earth, the mass of the earth experiences relatively little of this mystery force. What they could also do is analyze this whole thing in a frame of reference rotating with the universe, and they will find that the mystery force precisely cancels out the centrifugal force, leaving only the elasticity of the earth to provide the centripetal force on the (now rotating) earth's mass, thereby making it bulge at the equator. Originally Posted by Farsight Sure. I note Sol's cables, but he didn't spell out the upshot. It would be something like a bicycle wheel where you cut off the rim and solder a big lead weight on to each of the spokes. Then you cut out the hub and replace it with a ball of sand. Then you try spinning the thing. Forget sol's cables. They are not part of the hypothetical observed universe I'm talking about.
 3rd May 2012, 01:02 AM #252 xtifr Muse     Join Date: Apr 2012 Location: Sol III Posts: 563 Originally Posted by Farsight Ben's unhappy bcause he and his friends peddle woo and I call them out. Just to be clear here, when you say "he and his friends", you would include people like Robert Wald, or Charles Misner, Kip Thorne, and John Wheeler, right? And when you say they "peddle woo", you are referring to the interpretation of General Relativity offered in widely used textbooks such as Wald's General Relativity or Misner, Thorne, and Wheeler's Gravitation. Is that correct? These are among the people who have failed to properly understand Einstein's theory, yes? __________________ "Those who learn from history are doomed to watch others repeat it." -- Anonymous Slashdot poster "The problem with defending the purity of the English language is that English is about as pure as a cribhouse whore." -- James Nicoll
 3rd May 2012, 03:02 AM #253 edd Graduate Poster     Join Date: Nov 2007 Posts: 1,560 Originally Posted by Farsight There's obviously some kind of problem, sorry if it's me. Ben's unhappy bcause he and his friends peddle woo and I call them out. On the black hole thread they were trying to say space is infalling in a gravitational field, which is total trash. The fact is you've been shown to be incorrect in your understanding of general relativity. You've even been shown to be wrong about fairly basic geometry of curved spaces, let alone GR. You don't seem to have recognised that and learnt from the errors however. Some analogies may not be perfect, but sol for example was perfectly happy to give more detail on the nature of the waterfall analogy and be quite explicit about what physics it was intended to elucidate and how it did so, if I recall correctly. Quote: They'd point to one distant massive star and say it's moving at a ridiculous speed, and the force required to keep it going round in a circle would have to be matched by an equal and opposite and force on the Earth. Then they'd point to all the other distant massive stars, and tell you this: the forces on the Earth wouldn't make the equator bulge, they'd pull it to bits in an instant. This is a strange thing to say. It's coming through perfectly loud and clear from those on that side of the argument that choosing a new coordinate system has no physical effects. How could you possibly come up with the idea that they would come up with a prediction that the Earth would be torn to bits? __________________ When I look up at the night sky and think about the billions of stars out there, I think to myself: I'm amazing. - Peter Serafinowicz
 3rd May 2012, 03:52 AM #254 Farsight Suspended   Join Date: Mar 2008 Location: Poole, UK Posts: 1,956 Originally Posted by edd The fact is you've been shown to be incorrect in your understanding of general relativity. You've even been shown to be wrong about fairly basic geometry of curved spaces, let alone GR. You don't seem to have recognised that and learnt from the errors however. In your dreams. I haven't been shown to be incorrect at all. Originally Posted by edd Some analogies may not be perfect, but sol for example was perfectly happy to give more detail on the nature of the waterfall analogy and be quite explicit about what physics it was intended to elucidate and how it did so, if I recall correctly. He was peddling woo. Space is not falling inwards in a gravitational field. It's garbage Originally Posted by edd This is a strange thing to say. It's coming through perfectly loud and clear from those on that side of the argument that choosing a new coordinate system has no physical effects. How could you possibly come up with the idea that they would come up with a prediction that the Earth would be torn to bits? It's not a strange thing to say at all. Go check up on Newton's laws of motion, and pick one massive star. Note the first law: The velocity of a body remains constant unless the body is acted upon by an external force. That star is supposed to be rotating around the Earth, and it is massive. Now note the third law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear. And then add another star to the mix, and another, and another.
 3rd May 2012, 04:13 AM #255 edd Graduate Poster     Join Date: Nov 2007 Posts: 1,560 Originally Posted by Farsight In your dreams. I haven't been shown to be incorrect at all. The thread is there for all to read. I'll just let people 'do their own research' on it, as you're fond of saying. Quote: It's not a strange thing to say at all. Go check up on Newton's laws of motion, and pick one massive star. Note the first law: The velocity of a body remains constant unless the body is acted upon by an external force. That star is supposed to be rotating around the Earth, and it is massive. Now note the third law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear. And then add another star to the mix, and another, and another. Actually I misunderstood what you were trying to say there. Apologies. I'm not sure I do understand what you are trying to say there but never mind. It certainly wasn't as I originally misread it. __________________ When I look up at the night sky and think about the billions of stars out there, I think to myself: I'm amazing. - Peter Serafinowicz
 3rd May 2012, 04:19 AM #256 DazzaD Critical Thinker   Join Date: Jul 2006 Location: Romford Posts: 303 Originally Posted by Farsight It's not a strange thing to say at all. Go check up on Newton's laws of motion, and pick one massive star. Note the first law: The velocity of a body remains constant unless the body is acted upon by an external force. That star is supposed to be rotating around the Earth, and it is massive. Now note the third law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear. And then add another star to the mix, and another, and another. I presume you mean looking at the effect that the gravitational "pull" of surrounding stars has on the Earth itself? So as you add more stars in a homogenous (or isotropic) universe the net effect would be? Additionally what is the numerical value of these forces? Large or small?
 3rd May 2012, 05:42 AM #257 Farsight Suspended   Join Date: Mar 2008 Location: Poole, UK Posts: 1,956 Originally Posted by H'ethetheth No it wouldn't, because if they carefully observe the heavenly bodies, they will see that the mystery centripetal force on the bodies is proportional (among other things) to their distance from the rotational axis of the universe. Since this axis pierces the earth, the mass of the earth experiences relatively little of this mystery force. They wouldn't think of it as a mystery force, they'd call it the force of gravity. They'd be familiar with heavenly bodies rotating about their common centre of mass, and if they carefully observe the heavenly bodies they know about the rotation of Jupiter's moon's. Maybe they'd be familiar with galactic rotation. A galaxy rotates in about 15 million years, so they'd small a rat with rotation every 24 hours. The force of gravity would have to be millions of times bigger to maintain the rotation. Originally Posted by H'ethetheth What they could also do is analyze this whole thing in a frame of reference rotating with the universe, and they will find that the mystery force precisely cancels out the centrifugal force, leaving only the elasticity of the earth to provide the centripetal force on the (now rotating) earth's mass, thereby making it bulge at the equator. I'm not clear what you mean there. If the Earth is now rotating they're back to standard Newtonian mechanics, and saying the Earth has an equatorial bulge because of the laws of motion. There's a "fictitious" force in the Coriolis effect, but that's just plain-a-day roundabout inertia, and they aren't using Mach's principle because it isn't part of Newtonian mechanics. It isn't part of relativity either. E doesn't equal mc² because of all the other mass in the universe, it's because of the energy-momentum there in front of you. Yes in the wikipedia article there's a section on Modern General Relativity along with a reference to the Lense-Thirring effect, but note the quote that says "If one rotates [a heavy shell of matter] relative to the fixed stars about an axis going through its center, a Coriolis force arises in the interior of the shell; that is, the plane of a Foucault pendulum is dragged around". This is what Gravity Probe B detected and it's incredibly weak. It isn't the thing that makes the Earth oblate. Originally Posted by H'ethetheth Forget sol's cables. They are not part of the hypothetical observed universe I'm talking about. OK.
 3rd May 2012, 05:45 AM #258 Farsight Suspended   Join Date: Mar 2008 Location: Poole, UK Posts: 1,956 Originally Posted by edd Actually I misunderstood what you were trying to say there. Apologies. Apology accepted. I'm sorry that I didn't twig that H'ethetheth was talking about heavenly bodies rotating like a galaxy with the Earth in the middle.
 3rd May 2012, 07:02 AM #259 H'ethetheth fishy rocket scientist     Join Date: Aug 2004 Location: among the machines Posts: 2,340 Originally Posted by Farsight They wouldn't think of it as a mystery force, they'd call it the force of gravity. No they wouldn't. When they look carefully at the planets, they see that the planets influence each other at a distance through a force that decreases with the square of their respective distances and their masses in addition to a force that increases proportionally to the distance from the universal axis. Originally Posted by Farsight They'd be familiar with heavenly bodies rotating about their common centre of mass, and if they carefully observe the heavenly bodies they know about the rotation of Jupiter's moon's. Maybe they'd be familiar with galactic rotation. A galaxy rotates in about 15 million years, so they'd small a rat with rotation every 24 hours. The force of gravity would have to be millions of times bigger to maintain the rotation. So that's why gravity can't be the mystery force. Originally Posted by Farsight I'm not clear what you mean there. If the Earth is now rotating they're back to standard Newtonian mechanics,... No, not quite. We're analyzing our rotating universe in a rotating frame of reference. That's not an inertial frame of reference.
 3rd May 2012, 07:06 AM #260 Farsight Suspended   Join Date: Mar 2008 Location: Poole, UK Posts: 1,956 Originally Posted by DazzaD I presume you mean looking at the effect that the gravitational "pull" of surrounding stars has on the Earth itself? So as you add more stars in a homogenous (or isotropic) universe the net effect would be? Nil. There would be no net pull. Originally Posted by DazzaD Additionally what is the numerical value of these forces? Large or small? Zero. Sorry, H'ethetheth's equatorial bulge and sol's cables had me thinking tetherball rather than rotating galaxy.
 3rd May 2012, 08:11 AM #261 Perpetual Student Illuminator     Join Date: Jul 2008 Location: USA Posts: 3,707 Originally Posted by ben m I'll get out of my depth before long here, but no: rotating the coordinate system should show up primarily in the metric. Would not the metric also be imbedded in the $T_\mu_\nu$ term? The left side of the equation seems to deal with space curvature while the right side seems to define the forces on any x. Perhaps this is an oversimplification of what is going on. __________________ It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman ξ
 3rd May 2012, 08:42 AM #262 Farsight Suspended   Join Date: Mar 2008 Location: Poole, UK Posts: 1,956 Originally Posted by H'ethetheth No they wouldn't. Yes they would. Originally Posted by H'ethetheth When they look carefully at the planets, they see that the planets influence each other at a distance through a force that decreases with the square of their respective distances and their masses addition to[/i] a force that increases proportionally to the distance from the universal axis. When they look carefully at the planets our Newtonian mechanics would notice they were spinning, and would work out that the Earth was spinning too. They'd entertain the rotating universe for about five minutes flat. If they decided to humour the idea, they'd draw a comparison with a flat galactic rotation curve, and some of them would propose dark matter to explain it. Others would say why invent a mystery substance? and point to what Newton said about density varying by degrees, and how it matched what the upstart Einstein said about inhomogeneous space. Then they'd look at the plot of gravitational potential and say that on a large scale the force of gravity increases like it increases from the centre of the Earth. Then somebody would slap them and remind them that the Earth was spinning. Originally Posted by H'ethetheth So that's why gravity can't be the mystery force. There is no mystery force, the Earth is spinning that's all, like just about everything else. Originally Posted by H'ethetheth No, not quite. We're analyzing our rotating universe in a rotating frame of reference. That's not an inertial frame of reference. Fine. But see the http://en.wikipedia.org/wiki/Coriolis_effect]Coriolis effect[/url] and look at the picture. Somebody referred to it somewhere, maybe you. We're slap in the middle of a fictitious rotation which ought to cause things to move through the universe in curved lines, only they don't, presumably because of some mystery force. It's all getting out of hand.
 3rd May 2012, 08:51 AM #263 Farsight Suspended   Join Date: Mar 2008 Location: Poole, UK Posts: 1,956 Originally Posted by ynot Or does the mass fall up to you? If you were the same mass as "the mass", you'd fall towards each other. In practice you are so small compared to the Earth that's the Earth's motion towards you is not detectable. Originally Posted by ynot When two things concurrently fall down from opposite sides of the mass towards its centre is the mass actually falling up to both of them? Does that mean the mass expands? If you had three big masses like pearls on a string the outer two fall inwards and the middle one gets stretched. Originally Posted by ynot When two things move toward a third thing at different speeds how does that third thing alternatively move toward the two things concurrently at different speeds? It doesn't. You take some reference point between all three masses, then when you let them fall towards each other you express their speeds with reference to your point.
 3rd May 2012, 09:22 AM #264 W.D.Clinger Master Poster     Join Date: Oct 2009 Posts: 2,442 Originally Posted by Perpetual Student Would not the metric also be imbedded in the $T_\mu_\nu$ term? The left side of the equation seems to deal with space curvature while the right side seems to define the forces on any x. Perhaps this is an oversimplification of what is going on. All three of the tensors that appear in Einstein's field equations (Ricci, metric, and stress-energy) are covariant tensors of rank 2, so the components of all three tensors transform covariantly. Hence $ $T^\prime_{\alpha\beta} = \frac{\partial x^{\mu}}{\partial x^{\prime \alpha}} \frac{\partial x^{\nu}}{\partial x^{\prime \beta}} T_{\mu\nu}$$ For the Ricci tensor, change every T in that equation to R. For the metric tensor, change every T to g. That equation applies to every pair of coordinate systems that's allowed by the definition of a differentiable manifold. That covariant transformation law is equation (11) in Einstein's Die Grundlage der allgemeinen Relativitätstheorie. By denying the applicability of that cooordinate transformation, Farsight is arguing with Einstein.
 3rd May 2012, 09:37 AM #265 ben m Illuminator   Join Date: Jul 2006 Posts: 4,657 Originally Posted by Perpetual Student Would not the metric also be imbedded in the $T_\mu_\nu$ term? The left side of the equation seems to deal with space curvature while the right side seems to define the forces on any x. Perhaps this is an oversimplification of what is going on. Clinger's response is correct, but you specifically asked where the centripetal and Coriolis forces come in. My intuition (which is not super-confident) is tat they have to come in from the left-hand side of the equation. Consider a flat, empty Universe, with lambda=0, no mass, no gravity waves. Describe it in ordinary spherical coordinates, it should be that T=0 everywhere. Now change to a rotating coordinate system; Clinger's transformation says that T=0 in the new coordinates. But there are nonzero pseudoforces in these coordinates; they must have come in from something on the right hand side.
 3rd May 2012, 10:02 AM #266 H'ethetheth fishy rocket scientist     Join Date: Aug 2004 Location: among the machines Posts: 2,340 Originally Posted by Farsight Yes they would. When they look carefully at the planets our Newtonian mechanics would notice they were spinning, and would work out that the Earth was spinning too. They'd entertain the rotating universe for about five minutes flat. [...] ...Then somebody would slap them and remind them that the Earth was spinning. [...] There is no mystery force, the Earth is spinning that's all, like just about everything else. Er, do you even remember what we are discussing, or why? Let me remind you: Originally Posted by NASA ...imagine that the earth were standing still and that the rest of the universe were rotating around it: would its equator still bulge? Newton would have said "No". So in this thought experiment the earth is not rotating, but universe is rotating around it. That is given. All I wanted to point out is that Newton would not have said "No", in the case that this rotating universe is anything like the one we have.
 3rd May 2012, 10:03 AM #267 sol invictus Philosopher     Join Date: Oct 2007 Location: Nova Roma Posts: 8,419 Originally Posted by ben m Clinger's response is correct, but you specifically asked where the centripetal and Coriolis forces come in. My intuition (which is not super-confident) is tat they have to come in from the left-hand side of the equation. Consider a flat, empty Universe, with lambda=0, no mass, no gravity waves. Describe it in ordinary spherical coordinates, it should be that T=0 everywhere. Now change to a rotating coordinate system; Clinger's transformation says that T=0 in the new coordinates. But there are nonzero pseudoforces in these coordinates; they must have come in from something on the right hand side. In the case you describe (flat spacetime), both the left and right-hand side of the equation are zero in all coordinate systems, as are both R_{\mu \nu} and R separately. The only thing that isn't zero is the metric. To see where the "fictitious forces" come in, you need a different equation - the geodesic equation, the equation that describes the free motion of point particles. That contains a Christoffel connection, which is a non-tensorial object with three indices. It's zero in flat space in Minkowski coordinates, but it's not zero in rotating or spherical coordinates (which is possible because it isn't a tensor). Last edited by sol invictus; 3rd May 2012 at 10:05 AM.
 3rd May 2012, 10:06 AM #268 Perpetual Student Illuminator     Join Date: Jul 2008 Location: USA Posts: 3,707 Originally Posted by ben m Clinger's response is correct, but you specifically asked where the centripetal and Coriolis forces come in. My intuition (which is not super-confident) is tat they have to come in from the left-hand side of the equation. Consider a flat, empty Universe, with lambda=0, no mass, no gravity waves. Describe it in ordinary spherical coordinates, it should be that T=0 everywhere. Now change to a rotating coordinate system; Clinger's transformation says that T=0 in the new coordinates. But there are nonzero pseudoforces in these coordinates; they must have come in from something on the right hand side. T could have non-zero components even if T = 0. It was my naive thinking that these non-zero components of the T tensor would represent these so called ficticious forces. I have been reviewing the Stanford University (by Leonard Susskind, consisting of twelve 1.5 hour segments) lectures on GR. I can tell that we will get to this question soon enough. I kind of jumped the gun with my question here. By the way, this Internet access is a real joy for someone like me. Imagine lectures by someone like Susskind at no cost in my home. When there is a point I don't get (like recently, some questions about covariant differentiation), I can pause, do a little digging and go back to the lecture. Fantastic! __________________ It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman ξ
 3rd May 2012, 10:06 AM #269 W.D.Clinger Master Poster     Join Date: Oct 2009 Posts: 2,442 Originally Posted by ben m Clinger's response is correct, but you specifically asked where the centripetal and Coriolis forces come in. My intuition (which is not super-confident) is tat they have to come in from the left-hand side of the equation. Einstein's field equation can be read in two different ways, both legitimate. If you regard the equation as a coordinate-independent relationship between four tensors (counting the Ricci scalar as a tensor), then its meaning is independent of coordinate systems, and it has nothing to do with the fictional forces. If you regard the equation as a relationship between the coordinate-dependent numerical components of the three non-scalar tensors, then the values of those components incorporate the fictional forces. Originally Posted by ben m Consider a flat, empty Universe, with lambda=0, no mass, no gravity waves. Describe it in ordinary spherical coordinates, it should be that T=0 everywhere. Now change to a rotating coordinate system; Clinger's transformation says that T=0 in the new coordinates. But there are nonzero pseudoforces in these coordinates; they must have come in from something on the right hand side. If spacetime is flat and the cosmological constant is zero, then the left hand side of Einstein's field equation is zero, so the right hand side must be zero also. The coordinate-dependent pseudo-forces show up only in the Christoffel symbols and in the coordinate-dependent components of the metric. For flat spacetime, the coordinate-dependent components of the metric don't affect the equation because they're multiplied by the Ricci scalar, which is zero for flat spacetime. Note that flat spacetime and T=0 is a special case. If T is nonzero, then the numerical components of all non-scalar tensors will may depend on your coordinate system. ETA: sol invictus gave a simpler explanation, three minutes before I posted. Last edited by W.D.Clinger; 3rd May 2012 at 10:17 AM. Reason: strikeout, text in gray, ETA
 3rd May 2012, 10:36 AM #270 Perpetual Student Illuminator     Join Date: Jul 2008 Location: USA Posts: 3,707 Well, it seems that I have a lot more road to travel. Thanks for all the responses. __________________ It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman ξ
 3rd May 2012, 10:40 AM #271 Kwalish Kid Critical Thinker   Join Date: Feb 2010 Posts: 388 Originally Posted by H'ethetheth No they wouldn't. When they look carefully at the planets, they see that the planets influence each other at a distance through a force that decreases with the square of their respective distances and their masses in addition to a force that increases proportionally to the distance from the universal axis. You are trying to talk to someone completely ignorant of the actual way that Newton established his theory of universal gravity. Farsight does not understand the importance of the inverse square relationship between the force of gravity and the distance because it plays a role in the theory and the evidence. He only knows a few buzz words.
 3rd May 2012, 10:52 AM #272 H'ethetheth fishy rocket scientist     Join Date: Aug 2004 Location: among the machines Posts: 2,340 You may well be correct. I think I should just join Perpetual Student in studying up on modern physics. I've always thought it's fascinating, but I never got around to learning it.
 3rd May 2012, 12:10 PM #273 ynot Illuminator     Join Date: Jan 2006 Location: New Zealand Posts: 4,654 Originally Posted by Farsight If you were the same mass as "the mass", you'd fall towards each other. In practice you are so small compared to the Earth that's the Earth's motion towards you is not detectable. If you had three big masses like pearls on a string the outer two fall inwards and the middle one gets stretched. It doesn't. You take some reference point between all three masses, then when you let them fall towards each other you express their speeds with reference to your point. I was referring to “All motion is relative” not “Every action has an equal and opposite.” __________________ Rumours of a god’s existence have been greatly exaggerated. My post are all (IMO) unless stated otherwise.
 3rd May 2012, 12:54 PM #274 sol invictus Philosopher     Join Date: Oct 2007 Location: Nova Roma Posts: 8,419 Originally Posted by Perpetual Student T could have non-zero components even if T = 0. Hmm. An equation like {some tensor with indices}=0 means that every component individually is zero. Since tensors transform by getting multiplied by something, if that equation is true in one coordinate system it's true in all. So unless by "T" you mean the trace of the stress-energy tensor - which can indeed be zero even if the components aren't - I can't make much sense of your comment.
 3rd May 2012, 01:51 PM #275 Perpetual Student Illuminator     Join Date: Jul 2008 Location: USA Posts: 3,707 Originally Posted by sol invictus Hmm. An equation like {some tensor with indices}=0 means that every component individually is zero. Since tensors transform by getting multiplied by something, if that equation is true in one coordinate system it's true in all. So unless by "T" you mean the trace of the stress-energy tensor - which can indeed be zero even if the components aren't - I can't make much sense of your comment. This all too new for me to be making these kinds of leaps, sorry. I have been looking at tensors as a kind of function of vectors and scalars and like any function, assumed the function could take on a zero value with the argument not zero. For example, the dot product of two vectors can be zero with the components non-zero -- isn't the dot product a kind of simple tensor? __________________ It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman ξ
 3rd May 2012, 03:09 PM #276 W.D.Clinger Master Poster     Join Date: Oct 2009 Posts: 2,442 Originally Posted by Perpetual Student This all too new for me to be making these kinds of leaps, sorry. I have been looking at tensors as a kind of function of vectors and scalars and like any function, assumed the function could take on a zero value with the argument not zero. For example, the dot product of two vectors can be zero with the components non-zero -- isn't the dot product a kind of simple tensor? Yes, but it isn't zero. In three-dimensional Cartesian coordinates, the components Aμν of the dot product tensor areAμμ = 1 Aμν = 0 for μ ≠ ν So you need to distinguish between the tensor being zero (which means the result of applying it to the appropriate number of vectors or differential forms will always be zero) from the situation in which zero can result from applying the tensor to some particular vectors/differential forms. Last edited by W.D.Clinger; 3rd May 2012 at 03:11 PM. Reason: added word in gray
 3rd May 2012, 04:04 PM #277 Perpetual Student Illuminator     Join Date: Jul 2008 Location: USA Posts: 3,707 Originally Posted by W.D.Clinger Yes, but it isn't zero. In three-dimensional Cartesian coordinates, the components Aμν of the dot product tensor areAμμ = 1 Aμν = 0 for μ ≠ ν So you need to distinguish between the tensor being zero (which means the result of applying it to the appropriate number of vectors or differential forms will always be zero) from the situation in which zero can result from applying the tensor to some particular vectors/differential forms. If two vectors are orthogonal in Cartesian space, their dot product is zero even if both vectors may not have non-zero components. Would that not make the tensor zero with the components not zero? __________________ It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman ξ
 3rd May 2012, 04:14 PM #278 W.D.Clinger Master Poster     Join Date: Oct 2009 Posts: 2,442 Originally Posted by Perpetual Student This all too new for me to be making these kinds of leaps, sorry. I have been looking at tensors as a kind of function of vectors and scalars and like any function, assumed the function could take on a zero value with the argument not zero. For example, the dot product of two vectors can be zero with the components non-zero -- isn't the dot product a kind of simple tensor? Originally Posted by Perpetual Student If two vectors are orthogonal in Cartesian space, their dot product is zero even if both vectors may not have non-zero components. Would that not make the tensor zero with the components not zero? That does not make the tensor zero. Let A be a covariant tensor of rank 2, and let u and v range over (contravariant) vectors. ((∃ u) (∃ v) A (u, v) = 0) does not imply A = 0. ((∀ u) (∀ v) A (u, v) = 0) does imply A = 0. Do you see the difference?
 3rd May 2012, 04:18 PM #279 Perpetual Student Illuminator     Join Date: Jul 2008 Location: USA Posts: 3,707 Originally Posted by W.D.Clinger That does not make the tensor zero. Let A be a covariant tensor of rank 2, and let u and v range over (contravariant) vectors. ((∃ u) (∃ v) A (u, v) = 0) does not imply A = 0. ((∀ u) (∀ v) A (u, v) = 0) does imply A = 0. Do you see the difference? Got it. Thanks. Addendum I On second thought not quite: ((∀ u) (∀ v) A (u, v) = 0) does imply A = 0, but it does not imply both u and v have all zero components or that u and v are both zero. Addendum II No that's wrong. I am confusing the vectors with the tensor A, which must have zero components for ((∀ u) (∀ v) A (u, v) = 0) does imply A = 0. Finally got it. __________________ It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman ξ Last edited by Perpetual Student; 3rd May 2012 at 04:30 PM.
 3rd May 2012, 09:49 PM #280 ynot Illuminator     Join Date: Jan 2006 Location: New Zealand Posts: 4,654 Does the distant shell of rotating mass theory hold up when there's a stack of tops spinning in different directions at different speeds? By "stack" I mean tops on on top of each other so they all share the same axis. I was just in a toy store and saw a product called "Totem Tops" that does this. If two "shells" share the same axis and rotate in different directions why don't their opposing forces cancel each other out? __________________ Rumours of a god’s existence have been greatly exaggerated. My post are all (IMO) unless stated otherwise.

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