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According to "Exact Solutions of the Einstein Field Equations", the most general cylindrically symmetric metric is $$ds^2 = e^{-2U} (\gamma_{MN} dx^M dx^N + W^2 d\phi^2) + e^{2U} (dz + A d\phi)^2$$ With Killing vectors $\eta = \partial_\phi$ and $\zeta = \partial_z$, and all functions independant of $z$ and $\phi$. The other two ...

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I think your last sentence shows you're on the right track. A one form is a linear functional that maps vectors to real numbers. You give it a vector as an input, and, as you say, it returns the rate of change in the implied direction. Let's say we have a path whose tangent at a point is defined by the vector $v^j\,\partial_j$ - the differential operator ...

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1. Void vs Vacuum The first thing that needs to be done is to distinguish between void and space (ie vacuum). Space is not nothing, because you can move things in it; think of it as the medium in which particles can move. For if space was exactly nothing; then where could you put a particle? There is no place you can put it. 2. Crumpling space The ...

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General relativity in low dimensions gets progressively simpler, making it difficult to really make interesting statements on it. Here is the situation : 2+1 dimensions : In 2+1 dimensions, the Riemann tensor depends only on the Ricci tensor. So if spacetime is Ricci flat at a point, it is totally flat. This means that when there is a 0 stress energy ...

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The Killing vector that generates dilatations is $$\xi^a = \left( x^\mu , z \right)$$ The norm of this is $$\| \xi \| = g_{ab} \xi^a \xi^b = g_{\mu\nu} \xi^\mu \xi^\nu + g_{zz} \xi^z \xi^z = \frac{L^2}{z^2} \left( \eta_{\mu\nu} x^\mu x^\nu + z^2 \right)$$

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Describing ´space´ as ´empty´ does very little when you are juxtaposing that with a field.. Spacetime is not ´nothing,´ (though its emptiness can be used to describe it as empty at a localized point given[quantum fluctuations notwithstanding]no actual matter is within whatever frame of reference you are using)it is a field like the Higgs field you ...

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Yes. You can use any coordinate system. And then the transformation between two coordinate systems can be rather complex. In general relativity in fact, there aren't global inertial frames, so you are forced to either use general coordinate systems or else to use frames locally and patch the results together. The former is often easiest after you've put in ...

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To answer the question in your title, resolution of vectors into unique components depends only (1) on the fact of the vector space over the underlying field; as long as one has a basis (any $N$ linearly independent vectors where $N$ is the space's dimension), unique resolution of a vector into components with respect to this basis is defined. However, ...

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Start by rewriting the scalar product as a covariant-contravariant contraction, like so: $${\bf u}\cdot{\bf v} = g_{ij}u^iv^j = (g_{ij}u^i)v^j = u_jv^j$$ Now transform the components with your $S$ and $T$ matrices,  u_jv^j = \left( S_j^a {\bar u}_a \right) \left( T^j_b {\bar v}^b \right) = (S_j^a T^j_b) {\bar u}_a {\bar v}^b = \delta^a_b {\bar u}_a {\bar ...

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