curvature of spacetime
in [Relativity]
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From: G=EMC^2 Glazier on 23 Mar 2006 18:31 My curvature of spacetime theory has space concave,and convex. It is that simple. It answers many hard questions,and that makes it good. It fits well with my equation for the universe. Einstien would love it.,for he gave me the concave spacetime,and that's the front half TreBert
From: GSS on 24 Mar 2006 12:37 JanPB wrote: > GSS wrote: > > > > From equations (5) we can compute the components of the Riemann tensor > > R_ijkl for the Euclidean 3-d space which turn out to be all zero. When > > we compute the components of the Riemann tensor R_ijkl for the > > Riemannian 3-d space from equations (5) and (6), just two components > > namely R_1212 and R_1313 turn out to be non-zero and are given below. > > > > R_1212 = (GM/c^2 r).(1/(1 - 2GM/c^2 r)) ....... (7) > > R_1313 = (GM.Sin^2 (q)/c^2 r).(1/(1 - 2GM/c^2 r)^2) .... (8) > > This doesn't look quite right. Denoting the coordinates (r,theta,phi) > by indices 1,2,3 and using G = c = 1 I get: > > R_1212 = -M/r . (1/(1 - 2M/r)) > R_1313 = -M sin^2(theta)/r . (1/(1 - 2M/r)) > R_2323 = 2Mr sin^2(theta) Yes you are right except that R_1313 should still be with + sign. > > Hence the intrinsic curvature of Riemannian 3-d space, associated with > > static spherically symmetric gravitational field, is given by just two > > three > > > components of the Riemann tensor given by equations (7) and (8) above. > > > > Now in order to visualize this intrinsic curvature of Riemannian 3-d > > space we must be able to visualize the existence of specific 2-d > > surfaces with non-zero Gaussian curvature which were plane surfaces in > > the initial Euclidean space. > > One can consider the three planes (r,theta), (r,phi), (theta,phi) and Here (r,theta) will represent a 'plane' only if we take phi=0. Otherwise for phi = phi1 etc it can represent an infinite number of planes passing through the theta =0 axis. However, (r,phi) and (theta,phi) do not represent plane surfaces. You may therefore reconsider your subsequent remarks. GSS ------------------------------------------------------------------- > in each case look at the 2D submanifold made of geodesics emanating in > directions contained in those planes. Gaussian curvatures of these 3 > surfaces are the three sectional curvatures which easily follow from > the R components above: > > S_12 = R_1212/g_rr . g_theta theta = -M/r^3 > S_13 = R_1313/g_rr . g_phi phi = -M/r^3 > S_23 = R_2323/g_theta theta . g_phi phi = 2M/r^3 > > The equality of S_12 and S_13 is to be expected by the rotational > symmetry of the situation. The two 2D surfaces obtained from geodesics > in the (r,theta)- and (r,phi)-direction look like planes in the > standard 3D drawing of the situation (since by symmetry geodesics > emanating in - say - the (r,theta)-plane must stay in that plane), > these "planes" are really strongly negatively curved (-M/r^3) and when > embedded in the Euclidean space they assume the familiar funnel shape. > > The third 2D surface, i.e. the one obtained from geodesics emanating in > the (theta,phi)-direction "looks" positively curved in the drawing > (since the geodesics immediately begin deviating towards the origin) > and is in fact truly positively curved (2M/r^3). > > > You are therefore requested to clarify whether, > > > > (a) All possible plane surfaces in the original Euclidean 3-d space > > (equations (1)) get curved with non-zero Gaussian curvature in the > > Riemannian 3-d space (equations (3)). > > So far we've seen that all planes through the origin become "funnels". > > > (b) Only a few specific identifiable plane surfaces in the original > > Euclidean 3-d space (equations (1)) get curved with non-zero Gaussian > > curvature in the Riemannian 3-d space (equations (3)). If so, then can > > we identify these plane surfaces in the Euclidean space and compute the > > Gaussian curvature of the corresponding curved surfaces in the > > Riemannian space? > > It doesn't look like any plane "stays" flat in this sense. > > -- > Jan Bielawski
From: JanPB on 24 Mar 2006 17:51 GSS wrote: > JanPB wrote: > > GSS wrote: > > > > > > From equations (5) we can compute the components of the Riemann tensor > > > R_ijkl for the Euclidean 3-d space which turn out to be all zero. When > > > we compute the components of the Riemann tensor R_ijkl for the > > > Riemannian 3-d space from equations (5) and (6), just two components > > > namely R_1212 and R_1313 turn out to be non-zero and are given below. > > > > > > R_1212 = (GM/c^2 r).(1/(1 - 2GM/c^2 r)) ....... (7) > > > R_1313 = (GM.Sin^2 (q)/c^2 r).(1/(1 - 2GM/c^2 r)^2) .... (8) > > > > This doesn't look quite right. Denoting the coordinates (r,theta,phi) > > by indices 1,2,3 and using G = c = 1 I get: > > > > R_1212 = -M/r . (1/(1 - 2M/r)) > > R_1313 = -M sin^2(theta)/r . (1/(1 - 2M/r)) > > R_2323 = 2Mr sin^2(theta) > > Yes you are right except that R_1313 should still be with + sign. It must be negative because the sectional curvature S_13 must be equal to S_12 (which is negative) by the spherical symmetry, and the denominator of S_13 is positive (since both g_rr and g_phi phi are). > > > Hence the intrinsic curvature of Riemannian 3-d space, associated with > > > static spherically symmetric gravitational field, is given by just two > > > > three > > > > > components of the Riemann tensor given by equations (7) and (8) above. > > > > > > Now in order to visualize this intrinsic curvature of Riemannian 3-d > > > space we must be able to visualize the existence of specific 2-d > > > surfaces with non-zero Gaussian curvature which were plane surfaces in > > > the initial Euclidean space. > > > > One can consider the three planes (r,theta), (r,phi), (theta,phi) and > > Here (r,theta) will represent a 'plane' only if we take phi=0. > Otherwise for phi = phi1 etc it can represent an infinite number of > planes passing through the theta =0 axis. > > However, (r,phi) and (theta,phi) do not represent plane surfaces. > > You may therefore reconsider your subsequent remarks. Sorry, my sloppiness. I meant planes spanned by (d/dr, d/d theta), (d/dr, d/d phi), and (d/d theta, d/d phi) *at* some point (r0, theta0, phi0). -- Jan Bielawski
From: Hexenmeister on 25 Mar 2006 07:03 "JanPB" <filmart(a)gmail.com> wrote in message news:1143240686.116255.303260(a)t31g2000cwb.googlegroups.com... | GSS wrote: | > JanPB wrote: | > > GSS wrote: | > > > | > > > From equations (5) we can compute the components of the Riemann tensor | > > > R_ijkl for the Euclidean 3-d space which turn out to be all zero. When | > > > we compute the components of the Riemann tensor R_ijkl for the | > > > Riemannian 3-d space from equations (5) and (6), just two components | > > > namely R_1212 and R_1313 turn out to be non-zero and are given below. | > > > | > > > R_1212 = (GM/c^2 r).(1/(1 - 2GM/c^2 r)) ....... (7) | > > > R_1313 = (GM.Sin^2 (q)/c^2 r).(1/(1 - 2GM/c^2 r)^2) .... (8) | > > | > > This doesn't look quite right. Denoting the coordinates (r,theta,phi) | > > by indices 1,2,3 and using G = c = 1 I get: | > > | > > R_1212 = -M/r . (1/(1 - 2M/r)) | > > R_1313 = -M sin^2(theta)/r . (1/(1 - 2M/r)) | > > R_2323 = 2Mr sin^2(theta) | > | > Yes you are right except that R_1313 should still be with + sign. | | It must be negative because the sectional curvature S_13 must be equal | to S_12 (which is negative) by the spherical symmetry, and the | denominator of S_13 is positive (since both g_rr and g_phi phi are). | | > > > Hence the intrinsic curvature of Riemannian 3-d space, associated with | > > > static spherically symmetric gravitational field, is given by just two | > > | > > three | > > | > > > components of the Riemann tensor given by equations (7) and (8) above. | > > > | > > > Now in order to visualize this intrinsic curvature of Riemannian 3-d | > > > space we must be able to visualize the existence of specific 2-d | > > > surfaces with non-zero Gaussian curvature which were plane surfaces in | > > > the initial Euclidean space. | > > | > > One can consider the three planes (r,theta), (r,phi), (theta,phi) and | > | > Here (r,theta) will represent a 'plane' only if we take phi=0. | > Otherwise for phi = phi1 etc it can represent an infinite number of | > planes passing through the theta =0 axis. | > | > However, (r,phi) and (theta,phi) do not represent plane surfaces. | > | > You may therefore reconsider your subsequent remarks. | | Sorry, my sloppiness. Yep. I'm not interested in discussing these issues with you anymore because the only thing you can discuss is your self-aggrandisation dreamland. Androcles.
From: Nick on 25 Mar 2006 21:55
Electrons and protons are electric and magnetic. How does the electromagnetic force hold electrons to the surface of an atom in precise shells? How does the electric field contribute? How does the magnetic field contribute? |