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One obvious subset of solutions is the Einstein manifold where the Ricci tensor is proportional to the metric tensor $R_{\mu \nu} = \frac{R}{4} g_{\mu \nu}$. If you then consider a diagonal metric wit …

Just substitute when in doubt: $$\nabla^a G_{ac} = \nabla^a\left(R_{ac} - \frac{1}{2}Rg_{ac}\right) =\nabla^a R_{ac} - \frac{1}{2}\nabla_c R $$ So $(3.2.30)$ is: $$2\nabla^a G_{ac}=0$$ Metric com …

In fact, it has a non-zero Weyl tensor as the only nonzero curvature (after Riemann tensor is decomposed into Weyl tensor, Ricci scalar and Ricci tensor). …

You can put $\xi=0$ and you'll get a minimally-coupled massive scalar in curved spacetime. Nothing prevents you from doing that. The reason you usually see the $\xi R \psi$ term in the equations of mo …

No, the dimensions do not curve. It's the metric that changes, and induces curves and bends in the spacetime geometry, and consequently the geodesics as well. The Einstein's equation is $G_{\mu \nu} …

You can't define a trace on $2$ similar indices (both up or both down). Evaluating a trace implies contraction of indices in a coordinate invariant way (just what you said). Remember to follow Einste …

$R_{\mu \nu}F^{\mu \nu}$ is simply zero. No computations are needed to see this. Just note that $R_{\mu \nu}$ is a symmetric tensor while $F^{\mu \nu}$ is antisymmetric.

fields) $T_{\mu \nu}$ is something that you have to put in by hand in Einstein's equations: $$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \frac{8\pi G}{c^4} T_{\mu \nu}$$ to see how it determines the curvature …

I think your question is good because it asks about the fundamental nature of gravity seen from both the conventionally geometric approach and QFT approach. First, the quoted text from Wikipedia is n …

For the same reason that given a function $f(x)$ which zero at some $x=x_0$, it does not imply that $f'(x_0)=0$.

TL;DR- General relativity (GR) is based upon general coordinate invariance; this means that physics is invariant under a general coordinate transformation (GCT). This invariance implies the contracted …

The first one is often used as a measure of curvature. Another interesting scalar that you can create from a combination of these is the Gauss-Bonnet term. …

The easiest way to go about this is to think of variation in terms of the background field method. Here, you write the metric as some background plus perturbation: \begin{equation} g_{\mu \nu} = \bar …

Otherwise, contracting indices between just curvature tensors leads to different objects. … This is because you can have many distinct possible contractions between curvature tensors. …

Also, the correct measure of curvature is not Ricci tensor but the Riemann tensor $R_{\mu \nu \rho \sigma}$. This is non-zero for Schwarzschild everywhere outside the point source/spherical object. … And lastly, it is entirely possible for spacetime to have curvature without any matter: freely propagating gravitational waves carry energy and momentum themselves, producing non-zero curvature. …