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Mathematical discipline which studies some properties of smooth manifolds, which allow to generalize calculus to beyond $\mathbb{R}^n$. General relativity is written in this language.

2 votes
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Is the Christoffel connection in 2D a total derivative?

Simple counterexample. Consider the following metric in $(t,x)$ coordinates \begin{equation} ds^{2} = -a(x) dt^{2} + dx^{2}, \end{equation} then \begin{equation} \Gamma_{tx}^{t} = \frac{1}{2}g^{tt}\pa …
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4 votes
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Building a Lorentzian metric from an induced metric in Euclidean space

The Riemannian metric is not sufficient. You need an additional line field $v_{a}$, and if you want your Lorentzian metric to be non-degenerate, this line field should be everywhere non-zero. Choosing …
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4d Riemannian Manifold as Part of 4d Lorentzian Manifold

The 4-Riemannian manifold is a useful starting point, because a Riemannian metric always exists given that it only requires paracompactnes (contrary to the Lorentzian metric that has topological obstr …
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5 votes
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Why in physics we work with metric without defining topology/smooth structure first?

For most applications, you can forget about the set $M$ and the topology on $M$ and work only on one of the charts. In this perspective, the metric has a predominant role since the (pseudo)Riemannian …
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2 votes
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Derivation of the Schwarzschild metric: why are $g_{22}$ and $g_{33}$ the same as for flat s...

The Schwarzschild solution is a spherically symmetric solution produced by a central source. This means that at $t = \mathrm{const}$ the metric should be invariant under rotations. \begin{equation} d …
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2 votes
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A question about the topology of spacetime and the existence of CTCs

What do you mean by "how far this mismatch of topologies can go while avoiding CTCs"? You have already noticed that the two topologies coincide if and only if the spacetime is strongly causal. It is t …
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1 vote

Derivation of metric flatness locally

For the first question, because $K$ is a general matrix with $D^2$ elements but $g$ has only $\frac{1}{2}D(D+1)$ independent elements, in order to diagonalize $g$ you only need $\frac{1}{2}D(D+1)$ ele …
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