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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
Accepted
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 …
4
votes
Accepted
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 …
0
votes
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 …
5
votes
Accepted
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 …
2
votes
Accepted
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 …
2
votes
Accepted
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 …
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 …