
#163
Oct2911, 06:12 AM

P: 1,115

I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flatlanders living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flatlanders now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here  there's a faint whiff of sanity to Dr Who's 'Tardis' if you like. 



#164
Oct2911, 08:24 AM

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I have explained this multiple times and you have continued to ignore it. 



#165
Oct2911, 09:17 AM

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"In other words, in nonEuclidean geometry, the relation between the sides of a triangle must necessarily take a nonPythagorean form." at http://en.wikipedia.org/wiki/Pythago...idean_geometry [Another grab, from: http://en.wikipedia.org/wiki/Curved_...flat.2C_closed "Triangles which lie on the surface of an open space will have a sum of angles which is less than 180°. Triangles which lie on the surface of a closed space will have a sum of angles which is greater than 180°. The volume, however, is not (4 / 3)πr^{3}" In context this appears to me to be a generalized statement applicable to higher than 2D curvature. As I say , haven't studied this subject at all.] 



#166
Oct2911, 10:02 AM

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One specific argument that flat planes are embeddable in a 4manifold is simply to note that for an arbitrary symmetric metric with 10 components, you can apply 4 coordinate conditions. This is sufficient to make e.g. the x,y components of the metric [[1,0],[0,1]], that is a Euclidean plane. As a result, any nonEuclidean behavior of a plane is just a function of coordinate choice, and is not telling you anything intrinsic about the manifold. 



#167
Oct2911, 10:40 AM

P: 1,115

"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a nonEuclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in nonEuclidean geometry, the relation between the sides of a triangle must necessarily take a nonPythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2." I can only take that one way  in *any* nonEuclidean geometry. Just how that applies to triangle in 3curvature is the question. 



#168
Oct2911, 10:55 AM

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#169
Oct2911, 11:09 AM

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Also recall my discussion with Peter: the SC geometry allows embedding of completely flat 3space regions. The K factor is actually a feature of a particular class of observers (static observers, which are noninertial observers), not something intrinsic to the geometry. It is analogous to distortions seen in an accelerating rocket in flat spacetime  a feature of the observer, not the intrinsic spacetime geometry. GP observers in the same SC geometry, experience absolutely flat 3space. 



#170
Oct2911, 11:52 AM

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This deserves further animations; I'd rather show you, if I can, rather than tell you. But suffice it to say, the animation above could be Lorentz transformed so as to place any particle in the center of the circle. However, because there are only a finite number of particles in the picture, there would be a gravitational asymmetry for every particle, except for one. There are a couple different ways I might specify the initial conditions; (1) starting with an equipartition of rapidity (2) starting with an equipartition of momentum. In this animation I assumed equal masses, and used equipartition of momentum. Let [tex]q\equiv \left \\frac{\vec p}{m c} \right \ = \left \\frac{(\vec v/c)}{\sqrt{1v^2/c^2}} \right \<3[/tex] [tex]\frac{v}{c}=\sqrt{\frac{q^2}{q^2+1}}[/tex] Once I calculated the velocities, I animated, finding position, by simply multiplying the velocities by t. The animation renders about 79,000 dots of equal mass with random q between 0 and 3 in random directions. The particle at the center of this distribution should experience no net acceleration; however, closer to the edges, there is more and more gravitational force. There's no limit to what q might be. If I set q=10^100, then all the particles that are in this animation would be so close to the center that they would experience essentially no net force. But then you'd have the problem, being inside a shell of infinite mass, that at any given point inside, you are at an infinite negative gravitational potential On the other hand, with a finite mass, the particles at the edge could experience, at least for a time, acceleration equivalent to a black hole. I'm not at all sure what theoretical ramifications that would have. Anyway, I'll work on the other animations, and hope that makes my meaning clearer. 



#171
Oct2911, 12:17 PM

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If, instead, you assembled the triangle carefully in a curved region, so that the "natural" configuration of its sides was as geodesics on the curved surface, and the "natural" angles at each vertex summed to more than 180 degrees, then the triangle would deform if you tried to make it conform to a flat Euclidean plane. Similarly, if you assembled a tetrahedron in a region of curved space, hovering over a gravitating body, so that it was unstressed in that configuration, and then took it far away from gravitating bodies where space was flat, it would undergo stress and deformation in the course of conforming to the flat space. http://en.wikipedia.org/wiki/File:Fl...paraboloid.svg ...and think of moving a triangle from the flat region to the curved region. 



#172
Oct2911, 12:21 PM

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#173
Oct2911, 01:28 PM

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But let's look at the other extreme now, from the very edge of the distribution. Here, obviously we don't have symmetry. The stationary particle should experience a net force to the right. Let me give you a couple of premises of how I would go about calculating the force, and see if you agree with these premises, or if I am hopelessly naive. #1 The net gravitational force on a particle is [tex]F = G m \sum_i \frac{M_i}{r_i^2} \vec u_i[/tex] #2 The force on a particle should be calculated based on the reference frame of the particle that is undergoing the force. #3 The mass of each particle affecting gravitation is the restmass of the particle. i.e. the Lorentz factor affects momentum, but not gravitational attraction. #4 The location of the particle is not the "simultaneous" location, but rather the speedoflight delayed location of the particle. i.e. the speed of gravity is the same as the speed of light, so we must find an intersection of the pastlightcone of our particle of interest with the worldlines of the particles involved. Another peculiarity of relativity, (whether using Galilean Transformation or Lorentz Transformation) is that when one accelerates toward a future event, it leans toward him, becoming directly in his future, but when one accelerates toward a past event, it leans away. We may find that even if the particles at the edge accelerate "toward" the center, that in fact, the end result is not at all what commonsense would suggest. By accelerating toward the center, the particle continually enters reference frames where the center is further and further away. 



#174
Oct2911, 02:51 PM

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As I noted above, using #1 requires the problem to be nonrelativistic, and you don't seem to be imposing that limitation. For the relativistic case, you can't really use a "force" equation for this problem, or at least it does not seem to be the easiest way to approach it. A better model would be an expanding matterdominated FRWtype model, as I mentioned in a previous post, especially since it doesn't seem like your particles are a "shell", since there's no interior vacuum region, as far as I can see. This type of model does not view gravity as a "force"; it just solves for the dynamics of a curved spacetime using the EFE and an expression for the stressenergy tensor of the matter. You can still view individual particles as being subject to a "force", but that force can be more easily calculated *after* you have constructed the model from the EFE. If the spatial extent of the expanding "shell" is limited, then at the surface of the shell, the FRWtype solution would be matched to an exterior Schwarzschild vacuum solution; this would basically be the time reverse of the OppenheimerSnyder solution for the gravitational collapse of a star. It looks like this discussion might be better moved to a new thread, since it appears to be getting further from the topic of this one. 



#175
Oct2911, 06:13 PM

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If indeed, the gravitational field from a receding object has some predictable variation based on its relative velocity, then we should, of course, use that modification. But if you want to claim that there is no predictable variation; no predictable "meaningful" position we can use for distant objects, I beg to differ. On another topic, if possible, can you support your argument that "ordinary Lorentz frames...are valid only locally." I've heard this time and time again, but no one has ever explained what it means. Are you saying that when I point at a distant galaxy, that that direction that I am pointing only exists locally? Are you saying that the very concept of direction is only a local phenomenon? 



#176
Oct2911, 06:59 PM

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Here's a simple example: take an object that is freely falling towards the Earth from far away (say halfway between the Earth and the Moon), and set up a Lorentz frame using its worldline as the t axis. Since the object is in free fall, i.e., inertial motion, its worldline can be used this way according to the laws of SR. Now take a second object which is also in free fall towards the Earth, but which is slightly lower than the first object. Suppose there is an instant of time at which the second object is at rest relative to the first; we take this instant of time to define t = 0 in our Lorentz frame, and the position of the first object at this instant to define the spatial origin, so the frame is centered on that event on the first object's worldline. It should be evident that, for a given accuracy in measuring the relative velocity of the two objects, there will be some time t > 0 at which it becomes apparent that the second object is no longer at rest relative to the first. (This is because it is slightly closer to the Earth and therefore sees a slightly higher acceleration due to gravity.) But SR says that' can't happen: both objects are moving inertially, both were at rest relative to each other at one instant, and SR says that they therefore should remain at rest relative to each other forever. They don't. So we can only set up a Lorentz frame around the first object's worldline for a short enough period of time that the effects of tidal gravity (which is what causes the difference in acceleration of the two objects) can't be measured. Similarly, if the second object were further away from the first, its relative acceleration would become evident sooner; so there is a limit to how far we can extend a Lorentz frame around the first object in space as well, before the effects of spacetime curvature become evident. This also applies to what you were saying about assigning a "distance" to distant objects. There is no unique definition of "distance" in a curved spacetime; there are at least three different ones that are used in cosmology. You can uniquely define distance in a local Lorentz frame, but that only works locally; once you are out of the local region where the laws of SR can be applied to the desired accuracy, your unique definition of distance no longer works. 



#177
Oct2911, 07:54 PM

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There's a well known solution for an expanding (or collapsing) sphere of pressureless dust. Which isn't quite what was asked for, but you might be able to graft onto it to get the solution. The expanding pressureles dust sphere solution is just the FRW solution of cosmology, by the way.
As far as forces go, they're pretty much not used in GR. You can calculate them as an afterthought when you have the metric by evaluating [itex]u^a \nabla_a u^a[/itex], where u^a is your velocity vector. [add]Conceptually, what the above expression does is calculate what an accelerometer following the worldline would read. The reason forces aren't used much in GR that they transform in a complex manner  i.e. they don't transform as a tensor. In SR, you can still deal with 4forces as a tensor under Lorentz boosts. In GR, with accelerated coordinate systems, forces obviously cannot transform as a tensor. For instance, if you have an unaccelerated system with zero force, an accelerated system will have a nonzero force, but a tensor that is zero in one coordinate system must be zero in all. 



#178
Oct3011, 09:00 AM

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#179
Oct3011, 09:03 AM

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#180
Oct3011, 04:02 PM

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Although, depending on what you mean by "set up a Lorentz frame using its worldline as the t axis," I may have an even greater objection: Are you suggesting that geodesic paths in a gravitational field can represent straight lines in global inertial Lorentz Frames (in which case, I cannot agree; curved paths are not straight lines) or are you saying that a momentarily comoving reference frame with time axis tangent to the geodesic worldline at one particular event represents an inertial Lorentz Frame? In any case, this no longer has anything to do with shells, so I can invite you to read my comments here: http://www.physicsforums.com/showthread.php?t=545002 or if you prefer, you or I could start a new thread regarding whether Lorentz Transformations are purely a local phenomenon. 


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