What's more general than Riemann space?
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From: Daryl McCullough on 21 Jun 2010 13:18 Edward Green says... >D.F. Lawden, "Introduction to Tensor Calculus, Relativity and >Cosmology", p.89 > >"It will be proved in Chapter 6 that, in the presence of a >gravitational field, space-time ceases to be Euclidean in Minkowski"s >sense and becomes an R_4. This is our chief reason for considering >such spaces. However, we can generalize the concept of the space in >which our tensors are to be defined yet further. Until section 37 is >reached, we shall make no further reference to the metric. This >implies that the theory of tensors, as developed thus far, is >applicable in a very general N-dimensional space in which it is >assumed it is possible to set up a coordinate frame but which is not >assumed to possess a metric. In such a hypothetical space, the >distance between two points is not even defined. It will be referred >to as #_N. R_N is a particular #_N for which a metric is specified." > ># was since identified as a script S. It's clear that he's talking about generalizing Riemannian geometry by dropping the requirement that there is a metric tensor. Galilean spacetime is already an example of such a more general geometry. There is no metric that allows you to compute the distance between two events that take place at different times. -- Daryl McCullough Ithaca, NY
From: Rock Brentwood on 21 Jun 2010 18:35 On Jun 12, 6:04 pm, Edward Green <spamspamsp...(a)netzero.com> wrote: > What's more general than Riemann space? A Riemann-Cartan space is more general. In it, the connection and metric are independent structures, while in a Riemann space one uses the Levi-Civita connection associated with the metric -- which requires the metric to be non-degenerate. A generalization of Riemann and Riemann-Cartan spaces is one which admits degenerate metrics. Then, the appropriate way to go (i.e. a way that encapsulates the Newton-Cartan spaces of non-relativistic space- time geometry) is to take BOTH the covariant and contravariant metrics as independent. Subject them to the condition that for an n- dimensional manifold the symmetry group that preserves both metrics be n(n-1)/2 dimensional. An additional condition is required (which I'll explain below) to get sensible results. To make this precise, define the following: V = span (e_i: i = 0, ..., n-1) --- the tangent vectors V' = span (E^i: i = 0, ..., n-1) --- the cotangent vectors with (e_i) and (E^i) being dual. Then the metric has the form g = sum g_{ij} E^i * E^j the dual metric the form g' = sum g^{ij} e_i * e_j and the "contact form" is I = sum e_i * E^i, where I'm using ()*() to denote tensor products. The requirements are then that the most general symmetry that preserves g, g' and I should have n(n-1)/2 independent degrees of freedom. The additional condition is that there the symmetry group should not leave any of the dimensions fixed. More precisely, that there be no decomposition V = V1 + V2, V' = V1' + V2' such that V1 is orthogonal to V2', V1' is orthogonal to V2, V1 and V1' are both invariant. This geometry does not have a name. The result of these conditions are that the most general form the two metrics can have are either: (a) both g and g' are non-degenerate and proportional to one anothers' inverses (b) one of g or g' is non-degenerate and the other is 0 (c) both metrics are degenerate and act on orthogonal spaces, i.e., g = sum_{i,j<A} g_{ij} E^i * E^j g' = sum_{i,j >= A} g^{ij} e_i * e_j for some A between 1 and n-2. Cases (a) and (b) can be combined into one, as follows: g = k g_0, g' = K g_0^{-1} for some non-degenerate metric g_0, where k and K are scaling coefficients, not both equal to 0. For n = 4, this classification leads to the following. Let (w,x,y,z) denote a tangent vector basis, (W,X,Y,Z) its dual cotangent vector basis, with the two metrics diagonal with respect to both. Then the cases are: Non-degenerate metrics: (1) g_0 = W*W + X*X + Y*Y + Z*Z -- 4-D Euclidean signature (2) g_0 = W*W + X*X + Y*Y - Z*Z -- 3+1 Lorentzian signature (3) g_0 = W*W + X*X - Y*Y - Z*Z -- 2+2 "Penrosian" signature Degenerate metrics: (4) g = W*W, g' = x*x + y*y + z*z -- Galilean (5) g = W*W, g' = x*x + y*y - z*z -- Quasi-Galilean (6) g = W*W + X*X, g' = y*y + z*z (7) g = W*W + X*X, g' = y*y - z*z (8) g = W*W - X*X, g' = y*y + z*z (9) g = W*W - X*X, g' = y*y - z*z (10) g = W*W + X*X + Y*Y, g' = z*z -- "Archimedean" (11) g = W*W + X*X - Y*Y, g' = z*z -- Quasi-Archimedean Case (1) has only 1 type of dimension. Cases (2), (3), (4), (6) and (10) each have two types of dimensions. One can be called "spacelike", the other "timelike". The novelty of the generalization is that cases (5), (7), (8) and (11) each have THREE types of dimensions, while case (9) has FOUR. That is: space-like, time-like, (something else)-like and (something else yet)- like. In all, 4-dimensions has 11 types of signatures, with 3 of them being non-degenerate. The other 8 fall into the generalized geometry that goes beyond Riemann or Riemann-Cartan. Note, by the way, that in none of the degenerate cases can the analogue of a Levi-Civita connection be uniquely defined. This is actually one of the main distinguishing features of non-relativistic space-time. Instead, the closest you get to an analogue of "Levi- Civita" is to require that the covariant derivatives g, g' and I all be 0. Another feature that's novel has to do with the natural of conformal rescaling. Whereas in a Riemann or Riemann-Cartan space, a conformal rescaling has the form g -> k g, g' -> (1/k) g', I -> I by virtue of the fact that g' = g^{-1} in these geometries; in the more general geometry, there is a SECOND degree of conformal rescaling that is invisible to Riemann(-Cartan) eyes: g -> k g, g' -> k g', I -> I.
From: Rock Brentwood on 21 Jun 2010 18:38 On Jun 12, 7:04 pm, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote: > news:4c04f840-cc88-437e-b4e0-ffe904ca73fc(a)35g2000vbj.googlegroups.com... > | What's more general than Riemann space? > Euclidean space. No. Euclidean spaces are Riemannian geometries. Projective spaces are also Riemannian geometries. The example of interest to you of what is NOT a Riemannian (or Riemann- Cartan) geometry are Newton-Cartan spaces; in particular, the flat Newton-Cartan spacetime of classical non-relativistic physics.
From: Rock Brentwood on 21 Jun 2010 18:43 On Jun 13, 9:49 am, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote: > BUZZ!!! Euclidean space is a subset of Riemann space. No cookie for > you. > ================================================= > BUZZ!!! > Riemann space has a different postulate [structure] to Euclidean space. > http://en.wikipedia.org/wiki/Parallel_postulate (correction inserted) It has a *more general* structure than Euclidean space; but Euclidean spaces are special cases of Riemannian geometries; namely, those Riemannian geometries which possess constant positive definite (or negative definite) metrics. > I would claim that Euclidean space was a subset of > Lobachevsky space if I was [were] as ignorant as you. (Correction inserted) Euclidean spaces, however, are subsets Projective spaces. > You couldn't even buy a clue, let alone a cookie. Senile old men (never mind, former engineers) who spend all their hours on the USENET (with profiles listing as many as 1000 articles in some months) are in no position to be arguing anything with authority.
From: Androcles on 21 Jun 2010 18:54
"Rock Brentwood" <federation2005(a)netzero.com> wrote in message news:32efe3c7-96c1-4b8b-9915-edd08b26ba01(a)z8g2000yqz.googlegroups.com... On Jun 13, 9:49 am, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote: > BUZZ!!! Euclidean space is a subset of Riemann space. No cookie for > you. > ================================================= > BUZZ!!! > Riemann space has a different postulate to Euclidean space. > http://en.wikipedia.org/wiki/Parallel_postulate (correction inserted) ========================================= (incorrection deleted) |