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Over the real numbers, any non-degenerate quadratic form is determined (up to a change of basis) by its signature, which consists entirely of $1$s and $-1$s.

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Another way of looking at user40330's answer is to think of the inverse metric as the map from the space of one-forms (or differentials, if you prefer) and mapping them to the space of vectors (or directional derivatives, if you prefer that language), and then thinking of the metric as the inverse of this map. Namely $$g^{-1}({ d}v)=g^{ab}v_{b} = v^{a} = ... 0 The tensor algebra is symmetric between one-forms and vectors. One could start with defining any of them first and then obtain the rest of the things. The inverse metric tensor is a linear map that takes two one forms on a manifold and maps into \mathbb{R}. g^{\mu \nu}: A_\mu,B_\nu \rightarrow \mathbb{R} It of course tranforms like a vector with ... 6 At the most basic level, you can just use the definition of the Christoffel symbols in terms of the metric: \Gamma^i_{jk} = g^{is} (\partial_j g_{sk} + \partial_k g_{sj} - \partial_s g_{jk}). Plugging this into the right-hand side of your expression will yield the left-hand side. However, one can obtain your expression directly from one of the ... 0 Providing a particular metric tensor, is it possible to know information about the stress-energy tensor? If it is your question then the answer is following: For a perfect fluid the stress-energy tensor is T^{ik}=(p+\rho c^2)u^iu^k-pg^{ik}. So you need to know the pressure p and the energy-density \rho of the fluid and the time-like velocities u^i ... 2 When we talk about the geometry of GR, it is understood that the manifold of spacetime is not a Riemannian one, but rather a Lorentzian manifold. This means that the metric is not positive definite. With this understanding, we call g(.,.):=\langle.,.\rangle an inner product as usual. This lack of positive definiteness has many consequences. It is the ... 2 Comments to the question (v2): On one hand, let there be given a configuration space (Q,g) endowed with a metric g. (As ACuriousMind points out in a comment, there is a 1-1 correspondence between a metric g and the kinetic term in a Lagrangian.) On the other hand, note that the canonical symplectic 2-form \omega on the cotangent bundle T^{\ast}Q ... 1 No, but you are most likely to get one from the kinetic term of the Lagrangian itself. In most cases one requires it to be a convex function in the \dot q variables. You then get a metric if such kinetic term is quadratic in \dot q (and of course sensible kinetic energy is positive-definite). The metric and symplectic structures on a manifold are ... 0 Mm.... At first I repeated your calculation and got the same answer as yours, then I checked the paper you gave and found it consist with (32 a) and it seems not a typo, so I read it from begining - oh brother it's not 2+1 gravity - -b it's 3+1 gravity and you should treat d\Omega^2 more carefully: r^2 d\Omega^2=r^2d\theta^2+r^2\sin^2\theta d\phi^2 so ... 0 You have the order wrong in which you "get" things. You get the energy-momentum tensor from your specific matter theory. You do not know what g_{ab} is. Then, given some general assumptions about your spacetime, you write down an ansatz for the metric. For Schwarzschild, we have time independence and \mathrm{SO}(3) isometry. Then you calculate the Ricci ... 3 I'll prove a formula that is probably easier to use for this. \begin{split} \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} g^{\mu\nu} \partial_\nu \phi \right) &= \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} \right) g^{\mu\nu} \partial_\nu \phi + \partial_\mu \left( g^{\mu\nu} \partial_\nu \phi \right) \\ &= ... 1 All timelike geodesics in Minkowski spacetime start at past timelike infinity and end at future timelike infinity. The worldlines of Rindler observers are not geodesics, whereas the worldlines of Minkowski observers are. Heuristically think of a flat Euclidean plane. There are plenty of inextendible curves that don't go to infinity, but all geodesics start ... 2 The kinetic term of the Lagrangian is proportional to$$g_{ij}v^iv^j$$where the vs are the generalised velocities. Writing them as the time derivative of the generalised coordinates, i.e. v^i\dot q^i, taking the square root, and multiplying by a small time lapse \epsilon you get$$\sqrt{g_{ij}\dot q^i\dot q^j}\epsilon,$$which is a first order ... 1 One thing you can use the curvature tensor for is to detect singularities in the spacetime. For the Schwarzschild solution, the simpler curvature scalars formed from the Ricci tensor, R and R_{ab} R^{ab} vanish everywhere due to the fact that this is a vacuum solution to the Einstein equations. But since you have the full Riemann tensor at your ... 0 If you are working with the Newman Penrose null tetrad, i.e. with respect to which the metric looks like g=\left(\begin{array}{cccc}0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & 1 & 0\end{array}\right) then the standard notation is \lbrace l,n,m,\overline{m}\rbrace, where l,n distinguish the ... 3 The issue here is that the Schwarzschild coordinates are divided into two disconnected patches by the coordinate singularity at r=2m. There is no physical connection and the two pieces can be viewed as separate solutions. The Eddington-Finkelstein coordinates take the outer solution and extend it beyond the horizon, but inside the horizon it is different ... 2 Relativists tend to use the proper time, d\tau, and the proper distance, ds, interchangably. If you're working with proper time you'd expect the equation for it to look like:$$ d\tau^2 = dt^2 + \text{other terms} $$while if you're working with proper distance you expect:$$ ds^2 = dx^2 + dy^2 + dz^2 + \text{other terms}  The sign problem comes ...

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Could you provide a simple reason for these two conventions? The reason behind the (-,+,+,+) convention (the "mostly plus metric") is that a positive length in 3 dimensional space (e.g., the distance from my head to my toes) should still be a positive length in 4 dimensional space-time. Why should the distance from my head to my toes all of a sudden ...

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related: http://math.stackexchange.com/q/160882/224026 see also: http://en.wikipedia.org/wiki/Metric_tensor: " From the coordinate-independent point of view, a metric tensor is defined to be a nondegenerate symmetric bilinear form on each tangent space that varies smoothly from point to point. "

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1- A degenerate matrix is a matrix whose rank is smaller thank its dimension. 2- A singular (non-invertible) matrix is one who has a vanishing determinant. Equivalence of the two : A matrix whose rank is smaller than it's dimension when diagonalized will have at least one zero eigenvalue, and consequently a vanishing determinant.

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If $\text{det } g = 0$, then $\text{ker } g \neq \{\vec{0}\}$, ie there is some vector $X \in \text{ker } g$, such that $g(X,\ast)$ gives zero 1-form, so $g(X,Y)=0$ for any $Y$.

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