curvature of spacetime
in [Relativity]
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From: Tom Roberts on 29 Mar 2006 09:21 GSS wrote: > Let us agree to use only that spherical polar coordinate system which > has been adopted in the Schw. solution for the static spherically > symmetric gravitation problem. Kindly point out the *irrelevancies* of > this coordinate system which hinder the identification of 2-d surfaces > in the Schw space the curvature of which is induced by the components > of the curvature tensor. Consider the case M=0, in which the Riemann curvature tensor of the manifold is identically zero. The surfaces r=constant have nonzero Gaussian curvature. Tom Roberts tjroberts(a)lucent.com
From: GSS on 30 Mar 2006 11:23 Tom Roberts wrote: > GSS wrote: > > Let us agree to use only that spherical polar coordinate system which > > has been adopted in the Schw. solution for the static spherically > > symmetric gravitation problem. Kindly point out the *irrelevancies* of > > this coordinate system which hinder the identification of 2-d surfaces > > in the Schw space the curvature of which is induced by the components > > of the curvature tensor. > > Consider the case M=0, in which the Riemann curvature tensor of the > manifold is identically zero. The surfaces r=constant have nonzero > Gaussian curvature. How do you still consider the notion of *curvature of space* to be really valid? The term *curvature* basically applies to the bending of curves and 2-d surfaces. It is the forced extension of this term used to describe the new notion of *curvature of space* in GR. Inherently the new notion applies to the deformation of space induced by the change in metric (for M>0). Effectively we have been discussing whether this notion of *curvature of space* does adequately describe the induced deformation of space or not. The discussions so far only show that it does not appear to be feasible to properly describe the induced deformation of space through the notion of *curvature of space* . Hence it is not wrong to say that this new notion of *curvature of space* has been the most misleading notion of GR. I hope you will agree. GSS
From: GSS on 30 Mar 2006 11:43 JanPB wrote: > GSS wrote: > > JanPB wrote: > > > GSS wrote: > > > > > > > > As shown earlier, there are three non-zero distinct components of > > > > Riemann curvature tensor for this space as given below: > > > > > > > > R_1212 = -(GM/c^2 r).(1/(1 - 2GM/c^2 r)) > > > > R_1313 = -(GM.Sin^2 (q)/c^2 r).(1/(1 - 2GM/c^2 r)) > > > > R_2323 = 2GM r.Sin^2 (q)/c^2 > > > > > > > > Kindly identify three 2-d surfaces in this space, the Gaussian > > > > curvature of which is represented by the above mentioned three > > > > components of the curvature tensor. > > > > > > These values depend strongly on certain irrelevancies of the coordinate > > > system. To obtain values reflecting the geometry (as opposed to > > > geometry *together with* coordinate choice irrelevancies) it's much > > > better to use either the corresponding sectional curvatures or > > > components with respect to a Cartan moving frame since they are equal > > > to Gaussian curvatures of something that's actually in there. > > > > Let us agree to use only that spherical polar coordinate system which > > has been adopted in the Schw. solution for the static spherically > > symmetric gravitation problem. Kindly point out the *irrelevancies* of > > this coordinate system which hinder the identification of 2-d surfaces > > in the Schw space the curvature of which is induced by the components > > of the curvature tensor. I hope you can mentally visualize the surfaces > > under discussion. If so the same can then be simulated on computer if > > required. > > The difference between R_1212 and R_1313 is one example of > coordinate-induced irrelevancy. The situation is spherically symmetric > so any coordinate-independent quantification of something actually > *there* should give the same result both in the (d/dr, d/dtheta) and > (d/dr, d/dphi) directions. > > Another example of a coordinate phenomenon is that the length of > d/dtheta is not equal to the length of d/dphi (r and r sin(theta), > resp.) - yet by symmetry there is no difference between these two > directions. > > > > Either way, you get: > > > > > S_12 = -M/r^3 > > > S_13 = -M/r^3 > > > S_23 = 2M/r^3 > > > > > > The first two are curvatures of what looks like regular 2D planes > > > through the origin (minus the origin). Because of the odd metric they > > > are not really flat. > > > > Do you mean the planes passing through the polar axis theta=0? > > Any "plane" through the origin. The S_12 and S_13 at any given point > (r0, theta0, phi0) refer to Gaussian curvatures at that point of the > planes spanned by (d/dr, d/dtheta) and (d/dr, d/dphi), resp. > > > > The third one is the curvature of the surface made > > > (at a fixed point) of geodesics starting (from that point) in > > > directions which are linear combinations of d/dtheta and d/dphi (at > > > that point). Because of the symmetry, it is sufficient to calculate > > > just one of such geodesics and then make the surface of revolution out > > > of it (spinning it around the radius). > > > > Kindly confirm that you can visualize this surface and can it be > > simulated on computer? > > You have to solve some differential equations to get the exact > geodesics but you can imagine at every point (r0, theta0, phi0) a > snippet of surface of positive curvature tangent to the (d/dtheta, > d/dphi)-plane at that point and curving toward the origin. The exact > equation of the geodesic is quite messy and even when you reduce the > order from 2 to 1 by using the constants of the motion, the resulting > equations are still nasty. ....... > > Can you visualize how the shape of this surface will change when we > > vary the value of M in the curvature components from zero to M1 say? > > Well, it'll be uncurling and approaching the plane spanned by > (d/dtheta, d/dphi) as the surface moves away from the origin. Assuming that after solving some differential equations, you are in a position to visualize the surface of some finite curvature passing through a particular point (r0, theta0, phi0). Now if you repeat this process at all neighborhood points (r1, theta1, phi1) etc. then don't you think that the curved 2-d surfaces passing through such points will appear to be intersecting or crossing one another? GSS
From: JanPB on 30 Mar 2006 14:38 GSS wrote: > JanPB wrote: > > GSS wrote: > > > JanPB wrote: > > > > You have to solve some differential equations to get the exact > > geodesics but you can imagine at every point (r0, theta0, phi0) a > > snippet of surface of positive curvature tangent to the (d/dtheta, > > d/dphi)-plane at that point and curving toward the origin. The exact > > equation of the geodesic is quite messy and even when you reduce the > > order from 2 to 1 by using the constants of the motion, the resulting > > equations are still nasty. > ...... > > > Can you visualize how the shape of this surface will change when we > > > vary the value of M in the curvature components from zero to M1 say? > > > > Well, it'll be uncurling and approaching the plane spanned by > > (d/dtheta, d/dphi) as the surface moves away from the origin. > > Assuming that after solving some differential equations, you are in a > position to visualize the surface of some finite curvature passing > through a particular point (r0, theta0, phi0). Now if you repeat this > process at all neighborhood points (r1, theta1, phi1) etc. then don't > you think that the curved 2-d surfaces passing through such points will > appear to be intersecting or crossing one another? Yes, in this case they will. BTW, these surfaces are to be thought of as "surface snippets" - potentially very small surfaces passing through each point. Their exact size is unimportant from the point of view of curvatures (since curvature of a surface at a point is determined by any open neughbourhood in the surface around the point). The size is determined by geodesics emanating from the point and "weaving" the surface: as soon as any two geodesics intersect again, you cannot continue the surface past that point. (This is where the exponential map ceases to be 1-1.) -- Jan Bielawski
From: Tom Roberts on 31 Mar 2006 14:31
GSS wrote: > Tom Roberts wrote: >>Consider the case M=0, in which the Riemann curvature tensor of the >>manifold is identically zero. The surfaces r=constant have nonzero >>Gaussian curvature. > > How do you still consider the notion of *curvature of space* to be > really valid? Because geometrically it _is_ valid. That is, there are indeed geometrical manifolds with nonzero curvature. <shrug> And you never responded to how a 2-d surface in a flat 4-d spacetime can have nonzero curvature, and why that shows that the curvature of such 2-d surfaces is useless in "describing" the geometry of the 4-d manifold.... > The term *curvature* basically applies to the bending of curves and 2-d > surfaces. Not in differential geometry or GR. The term "curvature" was borrowed by analogy with 2-d surfaces, and has come to mean the Riemann curvature tensor. That is, a manifold of _any_ dimension with nonzero Riemann tensor is said to be curved. <shrug> Ironically, no 1-d manifold (a curve) has nonzero Riemann curvature tensor. For 2-d surfaces, Riemann curvature is directly related to Gaussian curvature. > It is the forced extension of this term used to describe the > new notion of *curvature of space* in GR. It may be new to you, but it most definately was not new in GR. Riemann, Ricci, Levi-Civita, Christoffel, et al investigated this in the 1800s. Einstein used _existing_ mathematics in his development of GR. <shrug> And it's the curvature of spaceTIME that is important in GR. > Inherently the new notion > applies to the deformation of space induced by the change in metric > (for M>0). I repeat: there is no "change" in the metric. These are _different_manifolds_. The Schw. manifold with M=0 is both geometrically and topologically different from those with M>0. <shrug> You keep trying to discuss "change of a manifold" in ways that simply do not make sense. Analogy: how would one "change" the Euclidean plane E^2 to "become" the surface of a sphere S^2? -- the question _does_not_make_sense_, and illustrates a very basic lack of understanding of geometry. <shrug> > Effectively we have been discussing whether this notion of > *curvature of space* does adequately describe the induced deformation > of space or not. That is inadequate in GR. One must discuss the curvature of spaceTIME. Indeed, for many situations the curvature of space is negligible (e.g. for objects falling or orbiting near earth). > The discussions so far only show that it does not > appear to be feasible to properly describe the induced deformation of > space through the notion of *curvature of space* . I have no idea what you are trying to say. The Einstein field equation plus the usual notions of differential geometry do indeed adequately describe the geometry of GR. <shrug> > Hence it is not wrong to say that this new notion of *curvature of > space* has been the most misleading notion of GR. Only for your personal MISconceptions about GR, about geometry, and.... Tom Roberts tjroberts(a)lucent.com |