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The singularity comes from the scale factor $a(t)$: $$ds^2 = -dt^2 + [a(t)]^2 ( dr^2 + r^2 d \Omega^2)$$ By solving the Friedmann equations for the scale factor we know that: $$a(t) = a_0 t^{\lambda}$$ where $\lambda$ is some positive number that depends on the matter-radiation ratio of the universe. At $t=0$ the scale factor becomes $a(0)=0$. So at ...


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Wald is a first rate relativist, and as such he is phrasing the concept of general covariance in terms of purely geometrical quantities, rather than resorting to the somewhat imprecise notion of coordinate transformations. In the discussion on pg. 57, he goes on to give an example of what it means to violate the principle of general covariance. In his ...


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All curves can be parameterised by an affine parameter (commonly written as $\lambda$ in the GR books I have). However only for timelike curves does the parameter $\lambda$ have a physical meaning i.e. it's the elapsed time $\tau$ shown on a clock by the observer following the curve. So there's no mathematical difference between parameterising a timelike ...


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Yes, the proper time along a timelike curve in relativity is very much analogous to the length of a curve. Just as the length of a curve is invariant under rotations, the proper time along a curve is invariant under Lorentz transformations. One difference with conventional length is that although a straight line is the shortest length between two points, a ...


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General covariance basically means you can change your coordinate system arbitrarily and express the laws of physics in the new coordinates. Because of this freedom, the relationship between coordinate distances, angles, etc. and physical distances, angles, etc. is variable and is expressed by the metric. So the quoted statement is basically saying that ...


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I will do this explicitly. In $AdS_3$, you can see we must have $z=0$ ($x,y,z$ are symmetric so just choose $z$). For general cases, we must have $u,v,x,y \neq 0$, so we must have: $\sin t \neq 0, \cos t \neq 0, \cosh \rho \neq 0, \sinh \rho \neq0, \sin \theta \neq 0, \cos \theta \neq 0, \cos \phi \neq 0$. Thus we have $\sin \phi =0$, i.e, $\phi =0$ ($n ...


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I lost one term, which is the one containing $\nabla^t(\partial_t)^r=g^{tt}\Gamma^r_{tt}=-M/r^2$, so this term is $$ -\frac{1}{8\pi}\int r^2\sin\theta \nabla^t(\partial_t)^r d\theta d\phi=\frac{M}{8\pi}\int \sin\theta d\theta d\phi=M/2 $$ Add this term to previous one and correct!


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Consider a contraction between $R_{\mu\nu\rho\sigma}$ and any number of $g$:s, such that the resulting tensor has two indices. Without loss of generality we can assume that each $g$ is contracted with $\text{Riemann}$ on both indices or not at all, and that no $g$ is contracted with another $g$. The first is because contracting once just raises an index, the ...


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I could give an example of what people mean when they "say": ... metric tensor depend on the local coordinate system and therefore are not intrinsic to the surface Take for example the Schwarzschild metric. We have $$ds^2 = -\left(1-\frac{2m}{r}\right)dt^2 + \left(1-\frac{2m}{r}\right)^{-1}dr^2 +r^2(d\theta^2 +\sin^2\theta d\phi^2) $$ If you read this ...


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The identity in question is given as, $$(\nabla_\beta u^\alpha) u_\alpha=(\nabla_\beta u_\alpha)u^\alpha$$ Expanding the left-hand side, we find, $$(\nabla_{\beta}u^\alpha)u_\alpha = (\nabla_\beta g^{\alpha \delta}u_\delta)u_\alpha = (u_\delta \nabla_\beta g^{\alpha\delta} + g^{\alpha\delta}\nabla_\beta u_\delta)u_\alpha$$ The Levi-Civita connection is ...


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Eddington–Finkelstein coordinates use the same position coordinates as Schwarzschild coordinates, only the time coordinate is transformed, so first consider how to define Schwarzschild coordinates in a physical way. This pdf explains a way of defining the position coordinates in section 9.1.1: • We may assign a practical definition to the radial ...


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Coordinates are not physical. They are entirely up to you. In the Eddington Finkelstein coordinates, the lines of constant u theta and phi are null radial lines going out to infinity while r is the coordinate such that the surfaces swept out by the spherical symmetry have an area of 4 pi r^2. (Ie, take a point of the spacetime. Apply a spherical symmetery ...


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In Riemann Normal Coordinates; you can take advantage of the special symmetry \begin{equation} g_{ab, cd} = g_{cd, ab} \end{equation} which allows you to express the curvature (very simply) as \begin{equation} R_{\mu \nu \alpha \beta} = g_{\mu \beta, \alpha \nu} - g_{\mu \alpha, \nu \beta} \end{equation} If you transpose the first two indices of $R_{mnab}$ ...


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What are their purpose? The "purposes" of Hyperbolic geometries are many and varied in mathematics, but one stands out far beyond all others, at least historically as the purpose. Hyperbolic geometries were constructed to prove that the Euclid parallel postulate (see "Parallel Postulate" Wiki page) was logically independent of Euclid's other axioms of ...


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The purpose of any model of the hyperbolic plane is that some aspect in it will be easy to work with computationally or intuitively, e.g. writing out certain isometries, identifying geodesics, computing volumes, etc. Hyperbolic space is unbounded, a hyperbolic manifold can be bounded. I don't know Yes If you are not using the terms bounded and infinite to ...


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If you solve the geodesic equation in Schwarzschild space-time then you will obtain, for the freely falling particle, the coordinates of its worldline $x^{\mu}(\lambda)$ in say the (global) Schwarzschild coordinate system; here $\lambda$ is an affine parameter such as proper time $\tau$ for a massive particle. The $x^i(\lambda)$ will be the spatial ...


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The line element in terms of the metric $g_{\mu\nu}$ is given by, $$ds^2=g_{\mu\nu}dx^\mu dx^\nu$$ As you haven't provided the entire line element in the post, I can only say $g_{tt} =a$ and $g_{rr}=b$. If it is the case that $a,b$ are constants, as well as any other components of the metric, then trivially $R_{\mu\nu}=R=0$. Otherwise, you pretty much ...


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The metric of the Schwarzschild black hole is $$ ds^2=\left(1-\frac{2M}{r}\right)dt^2-\left(1-\frac{2M}{r}\right)^{-1}dr^2-r^2(d\theta^2+\sin^2\theta d\phi^2). $$ Therefore, the Lagrangian of a particle in this background is (with proper time $\tau$) $$ ...


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I) First recall the fact that $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ is (the double cover of) the identity component $SO^{+}(2,2;\mathbb{R})$ of the split orthogonal group $O(2,2;\mathbb{R})$. This follows partly because: There is a bijective isometry from the split real space $(\mathbb{R}^{2,2},||\cdot||^2)$ to the space of $2\times2 $ real ...



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