Re: curvature of spacetimew


"JanPB" <filmart@xxxxxxxxx> wrote in message
news:1143202470.795415.4680@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| 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)
|
| > 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
| 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.

It doesn't look like any plane "stays" flat in this nonsense either.

Androcles.


| --
| Jan Bielawski
|


.

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