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We have $$ \Gamma^c_{ab} = \frac{1}{2} g^{cd} \left[ \partial_a g_{bd} + \partial_b g_{ad} -\partial_d g_{ab} \right] . $$ Varying this, we find \begin{align} \delta \Gamma^c_{ab} = \frac{1}{2} g^{cd} \left[ \partial_a h_{bd} + \partial_b h_{ad} -\partial_d h_{ab} \right] - \frac{1}{2} h^{cd} \left[ \partial_a g_{bd} + \partial_b g_{ad} -\partial_d ...


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The connection and the metric of a manifold are technically two independent quantities. If we allow both of those to be independent, then making a flat manifold with torsion is fairly trivial, for instance \begin{eqnarray} ds^2 &=& -dt^2 + dx^idx_i\\ {\Gamma^\alpha}_{\mu\nu} &=& g^{\alpha\beta} \varepsilon_{\beta\mu\nu} \end{eqnarray} This ...


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I see, though, that this breaks down at the pole θ=0, because all points with θ=0 are identical. But is this a problem? Yes, it is a problem, because if you omit the poles from a sphere, then it has the topology of a cylinder, not a sphere. There is a theorem in differential geometry called the Myers theorem. This is a theorem in Riemannian geometry (not ...


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In $n$ dimensions, the Ricci scalar curvature $R$ at a point measures how much the volume $V$ of a small $n$-dimensional ball of radius $\epsilon$ around that point differs from the Euclidean value $V_E$: $$\frac{V}{V_E}=1-\frac{R}{6(n+2)}\epsilon^2+O(\epsilon^4).$$ I am not aware of similar formulas for higher-order invariants, but that doesn’t mean they ...


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We observe phenomena consistent with mass distorting spacetime, and therefore we assume it does. Empirical observation is the ultimate master of all physics. Some of the evidence: https://en.wikipedia.org/wiki/Tests_of_general_relativity The "How does the mass of an object affect space and time", part of your question could mean one of two things. I will ...


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The $n^2(n^2-1)/12$ comes from the symmetries of the Riemann tensor and the algebraic Bianchi identity. $R_{abcd}$ is antisymmetric in $ab$ and in $cd$. This means that these pairs of indices can take $$m=\binom{n}{2}=\frac{n(n-1)}{2}$$ different values. For example, for $n=4$, they take the values $01,02,03,12,13,23$; for other values like $11$, the ...


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That depends on the rigidity of the stick, but since all the tidal forces are finite in the case of a nonrotating black hole in theory it should be possible to have a straight stick around it (in principle just like around the earth, where light is also slightly curved, but it is still possible to compensate for that on a stick). One thing you'll have to ...


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