Robittybob1 Posted February 23, 2016 Posted February 23, 2016 (edited) Is it possible to puncture the fabric of spacetime? In the demonstrations the ball rolls around the depressed well caused by the more massive weight, and you could imagine the rubber sheet tearing on occasions. Could such a thing happen in nature? What would be the consequence of a mass not following the geodesic? What made me ask this question was in elliptical orbits the object does not go around the circular sided well of the fabric of spacetime. Is it taking a sloped path then or does it puncture the fabric of spacetime? Edited February 23, 2016 by Robittybob1
ajb Posted February 23, 2016 Posted February 23, 2016 Is it possible to puncture the fabric of spacetime? I am not sure what that question really means. In the demonstrations the ball rolls around the depressed well caused by the more massive weight, and you could imagine the rubber sheet tearing on occasions. We have to be careful with such analogies. They are analogies and always break down: they are no replacement for the actual mathematics. Could such a thing happen in nature? It is no clear what tearing space-time real means. Informally, one could be thinking about the formation of singularities. Singularity are points or regions where the structure of space-time breaks down: we no not have the structure of a smooth Lorentzian manifold at these points. What would be the consequence of a mass not following the geodesic? This would either imply that the geodesic equation is not the equation of motion for this mass under the influence of gravity only (which would be very strange), or more likely the mass bis being acted upon by some external force. What made me ask this question was in elliptical orbits the object does not go around the circular sided well of the fabric of spacetime. Is it taking a sloped path then or does it puncture the fabric of spacetime? I suspect this question comes from taking the analogy too far.
Robittybob1 Posted February 23, 2016 Author Posted February 23, 2016 I am not sure what that question really means. We have to be careful with such analogies. They are analogies and always break down: they are no replacement for the actual mathematics. It is no clear what tearing space-time real means. Informally, one could be thinking about the formation of singularities. Singularity are points or regions where the structure of space-time breaks down: we no not have the structure of a smooth Lorentzian manifold at these points. This would either imply that the geodesic equation is not the equation of motion for this mass under the influence of gravity only (which would be very strange), or more likely the mass bis being acted upon by some external force. I suspect this question comes from taking the analogy too far. Do the equations handle elliptical orbits ok? So the curvature of spacetime takes the orbiting body around on the ellipse, so it doesn't go through the sides gravitational well around the central mass which must always be concentric. How do you get the elliptical shape then? It would have to be climbing up out of the gravity well and later falling back in. Does that path describe an ellipse?
ajb Posted February 23, 2016 Posted February 23, 2016 Do the equations handle elliptical orbits ok? You can study geodesics in the context of the Schwarzschild metric quite explicitly. You recover the standard kinds of orbits in Newtonian gravity and some paths that are not. https://en.wikipedia.org/wiki/Schwarzschild_geodesics See if the above helps answer your questions.
Robittybob1 Posted February 23, 2016 Author Posted February 23, 2016 (edited) T You can study geodesics in the context of the Schwarzschild metric quite explicitly. You recover the standard kinds of orbits in Newtonian gravity and some paths that are not.https://en.wikipedia.org/wiki/Schwarzschild_geodesicsSee if the above helps answer your questions. This question follows from a thought in the thread on RB law http://www.scienceforums.net/topic/93568-robittybobs-law-orbital-issue/#entry907577. If in a situation where thrust was applied to a rocket while it was orbiting would the effects of thrust, increase in velocity rising in the gravitation well and slowing down all be simultaneous. Or else the rocket may come away or go through the spacetime curvature. Wouldn't it be possible to always stick to the curvature during a manoeuvre? The more I think it through I must be wrong. It is possible to be in a position that isn't along a geodesic. Edited February 23, 2016 by Robittybob1
ajb Posted February 23, 2016 Posted February 23, 2016 Or else the rocket may come away or go through the spacetime curvature. I do not understand this statement. If you have a thrust, so some external force, then the rocket will not follow a geodesic. Rather it will follow some other path which in principle you can calculate: the geodesic equation is then modified by an extra term on the RHS. Wouldn't it be possible to always stick to the curvature during a manoeuvre? I am not sure what this means.
Robittybob1 Posted February 23, 2016 Author Posted February 23, 2016 I do not understand this statement. If you have a thrust, so some external force, then the rocket will not follow a geodesic. Rather it will follow some other path which in principle you can calculate: the geodesic equation is then modified by an extra term on the RHS. ..... So when you add that extra term would the orbiting mass still be on a geodesic? I imagine it as a surface that determines the path through space. Can you have an object above or below that surface as opposed to just somewhere else on that surface? I know if I was to look at the formulas I would also be none the wiser. It is hard to describe sorry.
ajb Posted February 23, 2016 Posted February 23, 2016 So when you add that extra term would the orbiting mass still be on a geodesic? Test particles (masses) only travel along geodesics when acted upon by gravity only. As you have some external force (ie. the thrust of your rocket), the path taken will no longer be a geodesic. I imagine it as a surface that determines the path through space. What surface? The path is a one dimensional submanifold of space-time. Can you have an object above or below that surface as opposed to just somewhere else on that surface? I am not sure what you mean by this surface. You maybe taking the analogy you started with too far.
geordief Posted February 23, 2016 Posted February 23, 2016 Going ,with permission slightly off topic I heard recently (in relation to the recent Black Hole detection) that these BH s were not filled with matter but filled with spacetime (or a similar term). Does that sound like a scientific description of what is happening or was it just pop speak for the viewers?
ajb Posted February 23, 2016 Posted February 23, 2016 Does that sound like a scientific description of what is happening or was it just pop speak for the viewers? Black holes are not solid objects as such, they are defined in terms of their event horizon, which any object can pass from the outside in, but not vice versa. Of course inside the horizon we would expect lots of particles, but again the black hole is not really a solid lump of stuff. So the pop-sci description you give is okay.
MigL Posted February 23, 2016 Posted February 23, 2016 The 'rubber sheet' analogy is a reduced dimensionality representation of space-time, Robittybob. The sheet itself , in fact, represents space-time. All motion is constrained to the 'sheet'. How could it not be ? There is no outside or inside it. And there is no tearing through it. ( disregarding singularities for the moment, as the model cannot handle them )
Robittybob1 Posted February 23, 2016 Author Posted February 23, 2016 (edited) The 'rubber sheet' analogy is a reduced dimensionality representation of space-time, Robittybob. The sheet itself , in fact, represents space-time. All motion is constrained to the 'sheet'. How could it not be ? There is no outside or inside it. And there is no tearing through it. ( disregarding singularities for the moment, as the model cannot handle them ) Understood the rubber sheet is an analogy and so I'll never mention it again, but going back to the physics of the situation I am trying to comprehend how mass "knows" how to make spacetime the right shape immediately after someone intentionally accelerates a rocket? Test particles (masses) only travel along geodesics when acted upon by gravity only. As you have some external force (ie. the thrust of your rocket), the path taken will no longer be a geodesic. .... That might be it, thanks Ajb. "It follows a geodesic when acted upon by gravity only" but when the acceleration occurs "the path taken is no longer a geodesic", so that is further out than the original geodesic for it will orbit at an increased radius, so how does matter rediscover where it is once the acceleration is stopped and it goes back to a circular orbit or more likely an elliptical orbit? MigL and ajb seem to be slightly at odds here for one says "all motion is constrained to the 'sheet'. How could it not be ?" and the other "It follows a geodesic when acted upon by gravity only". That is the heart of the topic: which solution is it? How does matter determine where it and the other is with or without forces acting on it? Edited February 23, 2016 by Robittybob1
geordief Posted February 23, 2016 Posted February 23, 2016 Understood the rubber sheet is an analogy and so I'll never mention it again, but going back to the physics of the situation I am trying to comprehend how mass "knows" how to make spacetime the right shape immediately after someone intentionally accelerates a rocket? Showing off some new knowledge I was given recently , it seems that acceleration does not curve space-time -.Acceleration (as distinct from gravity) ,apparently can be modeled just using Special Relativity as only flat space-time is involved. If I have understood my lesson correctly of course.
swansont Posted February 23, 2016 Posted February 23, 2016 Showing off some new knowledge I was given recently , it seems that acceleration does not curve space-time -.Acceleration (as distinct from gravity) ,apparently can be modeled just using Special Relativity as only flat space-time is involved. If I have understood my lesson correctly of course. Acceleration is not distinct from gravity, though. They are indistinguishable. I think you have over-simplified the lesson; the usual claim is that SR can't handle acceleration, which is false. But that's not to say that SR can handle any acceleration (or at least do so gracefully) or that it will predict all effects from it.
geordief Posted February 23, 2016 Posted February 23, 2016 Acceleration is not distinct from gravity, though. They are indistinguishable. I think you have over-simplified the lesson; the usual claim is that SR can't handle acceleration, which is false. But that's not to say that SR can handle any acceleration (or at least do so gracefully) or that it will predict all effects from it. Was the part of my post where I said "acceleration does not curve space-time" also incorrect?
swansont Posted February 23, 2016 Posted February 23, 2016 Was the part of my post where I said "acceleration does not curve space-time" also incorrect? Since acceleration and gravity are indistinguishable, I would guess that one can model any arbitrary acceleration as curved spacetime. There's the notion that the curvature tells mass how to move. It would be nice if one of our members who is well-versed in GR can confirm that, though. Spacetime is not a physical object that physically deforms, like pliers+a vice+effort will bend some sheet metal. It's a coordinate system we use to describe what's going on. So it's not so much that gravity distorts spacetime as that a distorted spacetime is how we describe gravity. The former description is often used, but it tends to reify the geometry, which IMO is a mistake.
Mordred Posted February 23, 2016 Posted February 23, 2016 (edited) Its the stress energy/momentum tensor that tells space how to curve. In the Minkowskii form the usual equation to describe this is [latex]T^{\mu\nu}=(\rho+p)U^{\mu}U^{\nu}+p\eta^{\mu\nu}[/latex] Edited February 23, 2016 by Mordred
swansont Posted February 23, 2016 Posted February 23, 2016 Its the stress energy/momentum tensor that tells space how to curve. In the Minkowskii form the usual equation to describe this is [latex]T^{\mu\nu}=(\rho+p)U^{\mu}U^{\nu}+p\eta^{\mu\nu}[/latex] So the question is whether one could have a tensor that models an arbitrary acceleration (i.e. a configuration that gives the same acceleration as an energy/momentum distribution would). e.g. there's a tensor that would give you the result of constant gravitational acceleration. But since we can't distinguish this from some other acceleration, the curvature would be exactly the same, and would describe the same motion. There's nothing I can see that would invalidate this.
Mordred Posted February 23, 2016 Posted February 23, 2016 This equation specifically described how energy density/pressure influence the stress energy momentum equation that tells space how to curve. I'll dig up some of the momentum transforms later on after my flight. So the question is whether one could have a tensor that models an arbitrary acceleration (i.e. a configuration that gives the same acceleration as an energy/momentum distribution would). e.g. there's a tensor that would give you the result of constant gravitational acceleration. But since we can't distinguish this from some other acceleration, the curvature would be exactly the same, and would describe the same motion. There's nothing I can see that would invalidate this. Correct
Phi for All Posted February 23, 2016 Posted February 23, 2016 ! Moderator Note Are we done with the original OP topic? I'd like to close this when we are, otherwise we'll get all kinds of stretching-the-analogy-too-far replies, based on the title.
Mordred Posted February 23, 2016 Posted February 23, 2016 the short answer is no you can't puncture spacetime as it isnt a fabric. The OP can refer to the 'What is space" pinned topic on top. So other than showing the stress/momentum elements to the stress momentum tenser I'd say were done
Robittybob1 Posted February 23, 2016 Author Posted February 23, 2016 So the question is whether one could have a tensor that models an arbitrary acceleration (i.e. a configuration that gives the same acceleration as an energy/momentum distribution would). e.g. there's a tensor that would give you the result of constant gravitational acceleration. But since we can't distinguish this from some other acceleration, the curvature would be exactly the same, and would describe the same motion. There's nothing I can see that would invalidate this. @Swansont - could you say this again in any other way that I can relate it to what happens when a rocket is accelerated in orbit please.
swansont Posted February 23, 2016 Posted February 23, 2016 @Swansont - could you say this again in any other way that I can relate it to what happens when a rocket is accelerated in orbit please. You can describe any acceleration with curved spacetime.
Robittybob1 Posted February 23, 2016 Author Posted February 23, 2016 You can describe any acceleration with curved spacetime. So at all times it stays on a geodesic, a different one accepted. Thanks
Mordred Posted February 23, 2016 Posted February 23, 2016 (edited) Not quite, geodesics are for free fall. The subject for more detail is the equivalence principle https://en.m.wikipedia.org/wiki/Equivalence_principle If you want I recommend a new thread and I'll help you out with it if you need. Though it will have to wait till I land lol. (Not supposed to have cell phones going on flights hehe) Edited February 23, 2016 by Mordred
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