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Weinberg's book on cosmology has explicit details on perturbation theory. In general, in cosmology, the way perturbed quantities are derived is simple but just computationally laborious. Assume that t …

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

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. …

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 …

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 …

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 …

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

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 …

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. …

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). …

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 …

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 …

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 …

$R_{\mu \nu}F^{\mu \nu}$ 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.

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 …