# Tag Info

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This isn't an answer, because I don't think your question has an answer, but it got too long to put in a comment. Anyhow, when physicists try to describe the universe we do it by constructing mathetical models. Then we use these models to calculate what will happen and do experiments to see if we we correct. If we got the correct answer it means our model ...

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The galaxies and (and binded objects) are maintained by gravity. The rest is explained here, Whats left at the center of the Universe after Big bang?

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Firstly, it should be made clear that being isotropic is a very special and rare property. (A spacetime can never be truly isotropic because no isometry can map spacelike vectors to timelike vectors, for example, so I'll talk about "space" being isotropic). There are very few spaces isotropic around every point, only very few spaces will even be isotropic ...

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For reference, Weinberg p. 378: A metric space is said to be homogeneous if there exist infinitesimal isometries (13.1.3) that carry any given point $X$ into any other point in its immediate neighborhood. Equation (13.1.3) defines an infinitesimal transformation and (13.1.5) concludes the Killing equation $\xi_{\sigma;\rho} + \xi_{\rho;\sigma} = 0$ ...

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It means that the metric and its inverse are well-defined everywhere in the space-time.

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Is teleparallelism an alternative to the introduction of a metric? Teleparallel gravity still comes with a metric - just take the tetrad field as orthonormal basis and there it is. The main difference between GR and teleparallelism is that the former uses curvature, the latter torsion to model gravity. According to Kleinert, there's actually a type of ...

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There is a sense in which metric theories of spacetime are "general". I simplify to four dimensions, but the argument generalizes to higher dimensions. Consider a particle whose path is parameterized by four coordinates $x^{a} = (t(s),x(s),y(s),z(s))$. We wish to describe the motion of the particle, given that at s=0, each of these functions has a known ...

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Conformal transformations are used in order to analyze the structure of a given spacetime. One can map a formally infinite spacetime to a compact interval, and study its properties there. This process is referred to as "conformal compactification", and enables one to draw Penrose diagrams. They serve to identify and classify horizons, infinities and ...

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I would like to add that the geometric picture and relationship between Nambu-Goto and Polyakov actions are only hints and heuristics. Specifically, string scattering amplitudes are computed in a Lorentzian space, but the worldsheets are Euclidean. One way to see it is that topology changes don't respect causality, so branching worldsheets are problematic ...

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As suggested in the comments, "lowering an index" is just coordinate notation for the isomorphism $\flat:TM\to T^*M$ between defined by $$X^\flat (Y) = \langle X,Y\rangle\text{,}$$ where $X$ and $Y$ are arbitrary vectors. I've tried using the definition of the metric ${g_{\alpha\beta}=\hat{\mathbf{e}}_\alpha\cdot\hat{\mathbf{e}}_\beta}$ where ...

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Also I've searched for it in books like Carroll's or Lawden's, but it's given pretty much as if it would be a definition. Because it is. No need for differential geometry, linear algebra is sufficient here: At a given point of space-time, the tangent space is just a vector space, the cotangent space its dual (ie the space of real-valued linear ...

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The Weyl tensor contains the information necessary to describe solutions of the Einstein equations in vacuum, given by $$R_{\mu\nu}=0.$$ From this we can deduce that the trace part of the Riemann tensor vanishes, but not its traceless part, which is given by the Weyl tensor. The latter therefore describes curvature phenomena in the absence of matter, like ...

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The Riemann tensor encapsulates all information about the 4-dimensional space-time. This information can generally divided into two sectors: Information about the curvature of space-time due to the existence of matter. This is given by the Ricci tensor according to the Einstein equation $$R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = 8 \pi G T_{\mu\nu}$$ ...

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Since the Lorentz transformation is valid for any $x\in M_{4}$, it can be rewritten as $\Lambda_{\rho}^{\mu}\eta_{\mu\nu}\Lambda_{\sigma}^{\nu}=\eta_{\rho\sigma}$. Substituting the infinitesimal form of the Lorentz transformation into the previous formula we get ...

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Note that if you lower an index of the Kronecker delta, it becomes the metric: $\eta_{\mu\nu}\delta^{\mu}_{\rho}=\delta_{\nu\rho}=\eta_{\nu\rho}$ And in your last step you got a wrong index. It should be $\omega_{\rho\sigma}$, not $\omega^{\rho}_{\sigma}$. Then, the metric terms cancel and you neglect cuadratic terms. That should be enough to solve it.

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Yes, you can regard the expression $$-\frac12m^2\phi^2-\frac16R\phi^2$$ as potential energy. Compare it to the harmonic oscillator: its potential energy is given by a term quadratic in position. The Ricci scalar can then be interpreted as simply contributing to the square of the mass of the scalar. To answer your question whether this is possible or not: ...

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