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Let's recap: upper indices are vectors ($x^\mu$), the inner product on Minkowski space is given by $g_{\mu \nu}$ so "dual vectors" have lower indices $x_\nu = g_{\nu \rho} x^\rho.$ Then you see that a matrix (in the sense of linear map between vectors) has one upper and one lower index, because it maps a vector to another vector: $$x^\mu \mapsto ... 2 I) Pragmatically speaking, the most important property of \sqrt{-g} for model building purposes, is not per se the fact that \sqrt{-g}d^{4}x measures the volume element of a 4-dimensional Parallelepiped with infinitesimal edges dx^0, \ldots, dx^3. II) A more important property is that \sqrt{-g}d^{4}x transforms as a scalar (i.e. is invariant) under ... 4 A variation of a tensor is always a tensor and the formula for the value above doesn't show otherwise. What you probably find surprising is that \delta g_{\mu\nu} and \delta g^{\rho\sigma} are not related to each other by simply raising the indices \mu,\nu or lowering the indices \rho,\sigma. Indeed, they're not related in this way. In this case, ... 3 A perfect fluid is defined by the property that, in the local rest frame, it allows no energy fluxes and no anisotropic stresses. Thus, at a given space-time point, in the local rest frame [in which the components of the 4-velocity are u^{\alpha} = (1, 0, 0, 0)^{\mathsf{T}}], the energy momentum tensor components are T^{\alpha\beta} = \mathrm{diag}(e, p, ... 3 First off, please don't use units with c\ne 1 in GR. It makes everything horribly messy. What we normally think of as a ruler or clock measurement is represented in GR by an upper index quantity like \Delta x^\mu. Therefore in a Cartesian coordinate system in the fluid's rest frame, we are guaranteed that u^\mu=(1,0,0,0), not (-1,0,0,0). This is ... 1 As soon as you get something like \delta_{bd}, alarm bells should ring, as this is not a tensor. The inverse metric g^{ac} is defined by the identity$$ g^{ac}g_{cb} = \delta^a_b $$If you plug this into your expression (and use the fact that g is symmetric), you will obtain the correct equation. 0 It can be show easily by the next reasoning.$$ DA_{i} = g_{ik}DA^{k}, $$because DA_{i} is a vector (according to the definition of covariant derivative). On the other hand,$$ DA_{i} = D(g_{ik}A^{k}) = g_{ik}DA^{k} + A^{k}Dg_{ik}. $$So,$$ g_{ik}DA^{k} + A^{k}Dg_{ik} = g_{ik}DA^{k} \Rightarrow Dg_{ik} = 0.  So, it isn't a condition, it is a ...
The elements of this particular metric tensor do not depend on time. $dx^j$ and $dt$ are treated as constants when you take the time derivative. You should also have $\Gamma^{\lambda}_{00} = -\frac{1}{2} g^{\nu \lambda} \frac{\partial g_{00}}{\partial x^\nu}$.