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Certain condensed matter systems show emergent behavior that is similar to general relativity: see this for example. Also, in fluid mechanics, sound waves can become trapped behind an "event horizon" called an acoustic black hole. Finally, the Einstein field equations are essentially the only possible classical equations of motion for a massless spin-two ...

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General relativity is a theory of gravity; as such, it makes predictions about gravity. However, general relativity does make predictions about time and physical entities such as black holes. Some of the predictions general relativity did make: Gravitational waves exist (proved by LIGO last year) Black holes exist Light bends (proved in 1919 by an ...

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Comments to the post (v2): Ref. 1 is considering the $d$-dimensional real Euclidean space $(\mathbb{R}^d,|\cdot|^2)$ with the standard norm $$|x|^2~:=~\sum_{\mu=1}^d (x^{\mu})^2~=~\sum_{\mu,\nu=1}^d x^{\mu}\eta_{\mu\nu}x^{\nu}, \qquad \eta_{\mu\nu} ~=~{\rm diag}(1,\ldots, 1),\tag{A}$$ and inner product $$\langle x ,y\rangle~:=~\sum_{\mu,\nu=1}^d x^{\mu}\... 2 I can answer to the first part of your question. A metric with harmonic coefficients is for example the FLRW metric for an universe with positive curvature. In this case the metric takes the form (c=1):$$ds^2 = dt^2 - a^2(t) \left(dr^2 + \frac{1}{\sqrt{k}} \sin(r\sqrt{k}) d\Omega^2 \right)$$0 for getting Christofel symbols we should notice that if two vectors are parallel transported along any curve then the inner product between them remains constant under parallel transport. so you should write your sentence for 3 parameter a, lambda and nu cyclical,and then you obtain the correct Christofel. 4 I wonder of you are overthinking this. Wald says: If the universe had always expanded at its present rate that is, \dot{a} is a constant and independent of time. In that case the value of a at time t after the Big Bang is simply:$$ a = \dot{a} t $$So if you define T by T = a/\dot{a} then T is necessarily the age of the universe. 2 A spin-2 field in 4D is not five-dimensional. The standard spin-2 object transforms in the (1,1)-representations of the Lorentz group where the (m,n)-labels are half-integers labelling an equivalent representation of \mathfrak{su}(2)\oplus\mathfrak{su}(2), see the Wikipedia article on representations of the Lorentz group for more information. The spin ... 0 Lower index is a tool to map upper index to a real number (W dot V for example). So to define a lower index, you need g(v,w)v (with one slot waiting for w to fill in order to spit out the dot product. So you can think g(v,w) as a tool to make two vectors dot with each other. Now you have v available, if you encounter a w later g(v,w)v is a tool to map w to ... 0 The energy h\nu of a photon is simply the contraction$$ h\nu = g_{\alpha\beta} k^\alpha u^\beta $$of its momentum k^\mu with the frame's velocity u^\mu. The frequency shift is then given by$$ 1 + z = \frac{\nu_S}{\nu_O} = \frac{(g_{\alpha\beta} k^\alpha u^\beta)_S}{(g_{\sigma\rho} k^\sigma u^\rho)_O} $$where we denote the velocities of both source ... 3 Given a metric$$ \mathrm{d}s^2 = g_{MN}\mathrm{d}x^M\mathrm{d}x^N$$in n dimensions you find the decomposition down to d dimensions by rewriting the n-dimensional metric in terms of objects with no, one, and two d-dimensional indices (in the following indicated by Greek letters):$$ \mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu + 2 A_{\...

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A metric defines a flat space within an open neighborhood $U$ if and only if the Riemann tensor $R$ vanishes over that neighborhood. So you simply have to calculate $R$ and check that it vanishes in the neighborhood. The only if part of the assertion is clear. The if part is probably more interesting to you and indeed a simple proof of the if shows you (in ...

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My understanding (which is based somewhat on Jackson's chapter on SR in Classical Elecrodynamics) is that the invariance of the interval is not enough to derive the Lorentz transformations - you also need the second postulate (that the speed of light is constant in all frames). The invariance of the interval follows from the fact that spherical light waves ...

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To some extent Your answered the question already! Look at the basic postulates of cosmology, these are homogeneity and isotropy of space-time. Isotropy implies three Killing vectors (SO(3)) and homogeneity gives another three killing vectors (for translation in three spatial direction). Therefore altogether six Killing vectors. Remember we not considering ...

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Perhaps this is an easy way to see it. To first order in $h$, any tensor $T^{\mu \nu}$ can be written $T^{\mu \nu} = t^{\mu \nu} + \lambda h^{\mu \nu}$ for some $\lambda$ and tensor $\boldsymbol{t}$. Now, for the metric tensor, we must have that $g^{\mu \nu} = \eta^{\mu \nu} + \lambda h^{\mu \nu}$ by matching with the $\mathcal{O}(0)$ term, i.e. $g^{\mu \nu}... 1 Since no answers have been forthcoming I will summarise what has been discussed in the comments. The usual analysis of geodesic motion around a spherical mass assumes that the spacetime geometry is described by the Schwarzschild metric, and that the test mass is too small to perturb this metric to any significant extent. In that case the motion can be ... 1 On the contrary, it is even more important to be careful "how we define and measure time" in general relativity than it is in special relativity. In special relativity, the speed of the clocks affects the rate – as seen from another coordinate system (it's the time dilation). In general relativity, the gravitational potential influences the rate (... 0 If one allows an arbitrary metric$g_{\rho\sigma}$and just uses the symbol$\tau$for one of the coordinates, i.e.$\sigma^0$, then$g_{\tau\tau}=0$doesn't imply anything special about the point. It just says that the vectors purely in the$\tau$direction are null. But in the Minkowski signature, there are many null vectors at each point, anyway. So the ... 3 The Poincaré group is the semi-direct product of the six-dimensional Lorentz group and the four-dimensional translations and hence ten-dimensional (or "has ten parameters" is less precise diction). Since in a global inertial coordinate system you have to have the Minkowski metric by definition, only those transformations (diffeomorphisms) which preserve the ... 1 You are reading the proper time (which is, of course, the value of one of many possible coordinates). 0 It just means that the energy (e.g. in the GR language, the ADM energy) is minimized among all configurations with the same boundary conditions. It means that there are no gravitational or electromagnetic or other waves inside the space. In practice, it just means that the geometry is a Cartesian product$M^4\times Y$where$Y$is the manifold of compact ... 1 Time stops on the event horizon of a black hole according to a distant observer. One might consider Einstein'r original insight into relativity, where he realized on taking a street car that if the car were moving the speed of light he would never see a clock he was moving away from tick off its next increment of time. Something similar happens as one ... 0 Given mild differentiability conditions on the metric (I thought this might be$C^\infty$which is not very mild, but see comments to this answer by 0celo7 below) then, for any point$p$, you can always pick a coordinate system$\left\{x^i\right\}$which is locally flat, which means that$g_{ij}(p) = \pm\delta_{ij}\$ -- tangent vectors along coordinate ...

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The spacetime geometry may be indirectly read from the differential operators in the equations controlling other objects – particles and (matter) fields. Schrödinger's equation contains the differential operator $$\Delta = \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$$ which is the operator ...

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They are not always "square." Orthogonal bases like you describe are convenient for many reasons, such as the fact that a "length" can be described in easy terms, and that there is only one way to notate any given point. There are others, such as the polar coordinate system which are different. The polar system describes 2 dimensions, one linear and one ...

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