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Can anybody give me an explanation about non coordinate bases and its importance?
Let's take a very simple example. Suppose you have polar coordinates, r and [itex]\theta[/itex].matness said:Can anybody give me an explanation about non coordinate bases and its importance?
Weren't one-forms covariant tensors?Such an orthonormal basis can be defined in general by specifying an array of n one-forms (aka contravariant tensors) that map the vectors of the tangent space to n scalars.
Probably <checking>, yes, it appears I got that backwards :-(Ratzinger said:Weren't one-forms covariant tensors?
The theory itself allows the possibility for 4-manifolds other than R^{4}. Indeed, there are interesting solutions that don't have R^{4} as the base manifold. An interesting related question is "How can we determine the topology of spacetime?"bchui said:In General Relativity, the manifold [tex]M[/tex] is 4-dimensional, so why don;t we simply take [tex]M=\Re^4[/tex] and use much simplier symbols in the differential geometry?
note that manifold is only required to look locally like R^n, it may look globally different from R^nIn General Relativity, the manifold is 4-dimensional, so why don;t we simply take and use much simplier symbols in the differential geometry?
[itex] \left\{ r, \theta, \phi \right\}[/itex] is not a coordinate system for [itex]S^3[/itex]. [itex]S^3[/itex] is the 3-dimensional surface of a 4-dimensional ball; [itex]S^3[/itex] is not a 3-dimensional ball.bchui said:Hold on, should Schwarzschild be [tex]S^3\times\Re [/tex] instead, for [tex](r,\theta,\varphi)[/tex] is 3-dimensional spherical plus [tex]t\in\Re[/tex]?
One has to be a little careful here, but basically yes.bchui said:Schwarzschild is a metric on the manifold
[tex]M=S^2\times {\bf R}^+\times {\bf R}[/tex], for we have [tex]r>0[/tex] and [tex]t\in {\bf R}[/tex]
No, an n-dimensional manifold is that looks locally like [itex]\mathbb{R}^n[/itex], but not necessarily globally. The "actual space" for Schwarzschild is [itex]M[/itex], not [itex]\mathbb{R}^4[/itex].So, [tex]M[/tex] is actually the domain of the parameter values, not the "actual space" we want to describe? For, the "actual space" we want to describe is [tex]{\bf R}^4[/tex]?
The open Friedman-Robertson-Walker universes have underlying manifold [itex]\mathbb{R}^4[/itex], while the closed Friedman-Robertson-Walker universes have underlying manifold [itex]\mathbb{R} \times S^{3}[/itex].Does that applies to Robertson-Walker metric and many others?