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I think you didn't understand perturbation theory in the context of GR. You split the metric tensor into an exact solution ($\eta$) of Einstein's equations, and a perturbation ($h$) as: $$g_{\mu \nu} …

You can have some $g_{00}(t)$, but you can always redefine your time coordinate so that you recover the (cosmic time) FLRW form of metric (where $g_{00}= $ constant). Say you have $g_{00} = f(t)$ and …

1) You can choose whichever coordinates you like in GR, in order to aid physical intuition and/or to ease calculations, etc. In this case, the manifold is sliced into 2-spheres which is done by choosi …

Because the definition of box operator is $g^{\mu \nu} \nabla_\mu \nabla_\nu$ and not $g^{\mu \nu} \partial_\mu \partial_\nu$. You need to use covariant derivatives instead of partial derivatives. You …

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 …

Expansions around an arbitrary background. You can later put background curvatures to zero to get the expansion around flat space. Ricci scalar up to $\mathcal{O}(h^3)$: $$R + \epsilon (- h^{\alpha …

You should have the same kind of tensor on both sides of the equation. By 'kind', I mean the rank of a tensor, which is specified by two numbers $(m,n)$ where $m=$ number of (free, not dummy) contrava …

$\delta g^{\mu \nu}$ is definitely a tensor, but you have to be careful when you have variations of a tensor. To be clear, it is $\delta (g^{\mu \nu})$, not $(\delta g)^{\mu \nu}$. Then you must expre …

The proper time is the time experienced by the particle itself. When you thus choose your reference frame as that of the particle itself, you fix your spatial coordinates at the particle's location. S …

I am late, but I'll answer anyway. In linearized gravity, $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$. We wish to find the expression for $g^{\mu \nu}$ up to linear order, which in general, must be s …

First of all, your counting of unknowns and equations is incorrect. There are $10$ unknowns (inverse metric components) and $10$ equations. The second equation in your question can be split into $6$ i …

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} …

The standard (QFT) way of quantizing gravity is by applying what is known as the background field method. Here, you write the metric tensor $g_{\mu \nu}$ as a sum of some classical background spacetim …

The spatial part of the metric is important when test particles have relativistic velocities, like photons. So you need this form of the metric when studying deflection of light by the sun, which was …

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 …