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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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Recall that the (global) conformal group is given by $${\rm Conf}(p,q)~\cong~O(p\!+\!1,q\!+\!1)/\{\pm {\bf 1} \},\tag{1}$$ cf. e.g. this Phys.SE post. Using the embedding $\imath: \mathbb{R}^{p,q}\hookrightarrow \overline{\mathbb{R}^{p,q}}$ into the conformal compactifification $\overline{\mathbb{R}^{p,q}}$, one may show after a short calculation that the ...

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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 Since you mention the following in one of your comments I'm less interested in Einsteins historical struggles and would love a more modern perspective on how to get to this insight. I hereby unashamedly ignore history, and offer instead a quick plausibility argument. Let's start with the equivalence principle which, loosely speaking, says that a (... 0 I have communicated with both of these fellows. The mathematics is based on Donaldson's theorem that in four dimensions there exists an infinite number of atlases of charts on a manifold that are homeomorphic but not diffeomorphic. I am not able to go into the mathematics, for it is pretty deep. It centers around the moduli space of self-dual connections SD/... 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. 0 Conceptually, the idea is that planes and spheres are equivalent from the point of conformal geometry. Conformal transformations map {planes, spheres} to {planes, spheres}, and in fact do this transitively -- any object in the set {planes, spheres} can be obtained from any other by a conformal transformation. Inversion is a reflection against a sphere, and ... 0 The Chern number you mention is the thing you get when you integrate a particular two-form over a surface. It turns out that this two form represents the first Chern class of the system (the system, in this case, consists of the parameter space and a line bundle describing the relative Berry phase along paths in the parameter space). The most important ... 0 Thinking of the sphere S^n as the one-point compactification of \mathbb{R}^n, we can consider the stereographic projection from the plane defined by x^0 = 0 to the unit sphere \{x\in \mathbb{R}^n\,:\,|x| = 1\}. This map is actually defined on \mathbb{R}^n\cup\{\infty\}, it takes the point \infty to the north pole of the unit sphere. Moreover, ... 2 I think you are supposed to simply argue that since the metric is "constant" in the sense of Eq. (C.2.3) on \Sigma_t, the curvature should be a constant as well. However, here is a more sophisticated point of view. Our definition of homogeneity is that the isometry group \mathrm{Iso}(\Sigma_t,h) is transitive, i.e. given p,q\in\Sigma_t there exists \... 1 The curvature in general relativity is determined completely by the metric (since the metric determines the Levi-Civita connection). Since the metric is constant on the homogeneous space so too must the curvature be. 1 Consider some path \gamma^{\mu}(\tau), and some vector x^{\mu}. Parallel transport is the condition when \gamma^{a}\nabla_{a}x^{b} = 0 Torsion is present if, for two paths \gamma^{a} and \delta^{a} that satisfy \partial_{a}\gamma^{b} = \partial_{a}\delta^{b} = 0, it is the case that parallel transport of \gamma along \delta produces a ... 1 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 ... 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 ... 0 Perhaps the most handy way is to proceed as follows: up to \tilde\Gamma^\Omega_{ab}=0 everything is fine. Now, write down this identity at \mathscr I^+ to get, since \tilde g^{\Omega a}=\delta^c_u and \partial_u\tilde g_{ab}=0,$$ \frac{1}{2}\left(\partial_a\tilde g_{ub}+\partial_b\tilde g_{ua}\right)=0. $$Now, fix a=\Omega and pick b=u,\theta,\... 0 I'll give you a few hints for you to deduce the answer. First, the \epsilon symbol is defined so that$$ \epsilon_{tr\theta\phi} = 1 $$Thus, Padmanabhan's equation is$$ d\Sigma_{tr} = r^2 \sin\theta d\theta d\phi $$This is what I wish to verify. Now,$$ d\Sigma_{tr} = \frac{\partial x^a}{\partial r} d\Sigma_{ta} = \frac{ x^a}{r} d\Sigma_{ta} = \frac{x^1}...

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They do decompose as symmetrized tensor products of Killing vectors. So nothing new here. I don't really remember the details but there are inequalities which tell how many "Killing objects" can a spacetime have, and this number is saturated for Minkowski and other maximally symmetric spaces. Class of spacetimes where you do expect to see nontrivial Killing ...

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You are presumably thinking of the FLRW metric for a universe with greater than critical density i.e. a closed universe. We normally use comoving coordinates to describe this, in which case the time coordinate is not curved and at every point along this time coordinate the three spatial coordinates have the topology of a 3-sphere. That is, if we draw a ...

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There are different descriptions of Spacetime according to General Relativity. Look at the De-Sitter-Space. It is a mathematical concept of Spacetime with a positive curvature. It is a submanifold of Minkowski-Space. there is also an Anti-De-Sitter-Space, which has a negative curvature. It plays a role in some cosmological theories (like Inflation). ...

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The usage of tensors in Physics, came about as a fortuitous "adaptation" to deal with the increased complexity of the Physics problems being investigated. The need to express more and more complex equations as succinctly as possible required a "shorthand" - and tensors came to the rescue.

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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 ...

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To add to the many good points already made, I think I agree with #Steve that what you are concerned about is the conundrum that an empty space seems to reflect the presence of matter. The key point is that we are dealing with a field theory, and the solution only describes the region of space outside of the mass, which is modelled as a single point as which ...

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I misunderstood what Menzel had intended by "covariant nature of the differential operator". He did not mean that the differential operation is synonymous with covariant differentiation. As Menzel is wont to do, he proceeded to expose a series of non-trivial equivalences, and then tacked on a final equivalence which does not obviously follow from the ...

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I think your mistake is in assuming that there are other terms. The way I see it, your only vector quantity is $w$, $u_{c} \nabla^{c}$ for example is a sum/scalar and doesn't have any vector character or components, thus does not need to be expanded into base vectors.

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Note that $\delta^i{}_j=\delta^i{}_k\delta^k{}_j$. Then $$\nabla_l\delta^i{}_j=\nabla_l(\delta^i{}_k\delta^k{}_j)=C(k,m)\nabla_l(\delta^i{}_k\delta^m{}_j)=C(k,m)[\delta^i{}_k\nabla_l\delta^m{}_j+\delta^m{}_j\nabla_l\delta^i{}_k]=2\nabla_l\delta^i{}_j$$ where $C(k,m)$ is the contraction of the $k$ and $m$ indices. Thus $\nabla_l\delta^i{}_j=0$.

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The partial derivatives $\partial_\lambda \delta^\mu_\nu$ are clearly zero because the components of the Kronecker delta are constant functions of spacetime coordinates (one or zero). One may always go to a locally Minkowski frame where the Christoffel symbols vanish and there, the covariant derivative is equal to the partial one and vanishes, too. Because ...

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You can write the Kronecker delta tensor as a product of the metric tensor $$\nabla_a(\delta^a_b) = \nabla_a (g_{bc} g^{ac}) = \nabla_a (g_{bc} g^{ac}) = g_{bc} \nabla_a g^{ac} + g^{ac}\nabla_a g_{bc}$$ As you may recall, the covariant derivative of the metric tensor is $0$ in general relativity. Version without using the metricity of the connection : ...

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A manifold $M$ is geodesically complete (i.e. has the property of "geodesic completeness") if for every point $P\in M$ and every direction $dx^\mu$ from that point, the infinitesimal line interval $dx^\mu$ away from $P$ may be transported by parallel transport by any amount $K\in(-\infty,+\infty)$ in both directions, i.e. if the geodesics (maximally straight ...

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