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The most compelling evidence of GR in presence of matter is, in my opinion, in neutron stars. These objects have a surface gravity $SG$ that is (geometric units): $SG_{NS}=GM/c^2 R \simeq 0.1$ This value is telling us that we can't use Newtonian gravity because we are in the strong field limit. For comparison, the sun has $SG_{SUN}=GM/c^2 R \simeq 10^{-5} ... 8 "According to Newton's law the negative mass should be repelled" -- Nope, in both Newtonian physics and in general relativity, negative mass would be attracted gravitationally to positive mass, although negative mass would exert a repulsive gravitational effect on positive mass (but if the negative mass is small compared to the mass of the black hole this ... 6 Frame-dragging effects are dependent on the spin of the central object, have been measured by experiments such as Gravity Probe B, and are definitely not dependent on the central metric. Also, any effects on a galactic scale are best quantified in terms of a continuous matter distribution, since the central black hole is a small fraction of the galaxy's ... 4 First of all, there is a quantum field theory on a curved background, even though it is not perfect. There are problems with global definitions of spinors, vacua, particle numbers etc. and this all seems to be a consequence of the core properties of GR such as no privileged definition of time or "global God observer". But the main issue of the theory is ... 4 In this link, the contradiction is "explained": The tremendous expansion greatly dilutes any initial curvature. Think, for example, of standing on a basketball. It would be obvious that you are standing on a (2-dimensional) curved surface. Now imagine expanding the basketball to the size of the Earth. As you stand on it now, it will appear to be flat ... 4 Energy conservation stems from Noether's theorem applied to time (i.e., time-invariance leads to energy conservation, similarly to how spatial-invariance leads to momentum conservation). Since the universe is expanding (and accelerating at that), the state of the universe today is different than it was yesterday and will be tomorrow, hence energy ... 3 Classically they are clearly topological. The metric does not appear, and you don't need a metric for integration on manifolds to make sense. Now in dimension 3 you can cast the Einstein-Hilbert action into a Chern-Simons theory as you say. The connection takes it values in the Lie algebra of the Poincare group. In higher dimensions you need to use higher ... 3 It doesn't. The covariant derivative is a map from$(k,l)$tensors to$(k,l+1)$tensors that satisfies certain basic properties. As such it cannot act on anything except tensors. The collection of components$\Gamma^a_{bc}$does not constitute a tensor. If you got to this expression via something like $$\nabla_d(\nabla_b A^a) = \nabla_d(\partial_b A^a) + ... 3 Your starting point is incorrect. You say: The point is, Rindler's observer shows us that the "action" of an accelerated observer on space-time is non trivial (there exists a black hole behind a uniformly accelerated observer). You're correct that there is a singularity, but it is only a coordinate singularity. The Riemann tensor is everywhere zero in ... 2 In relativity there's no objective frame-independent way to compare the rate two clocks at different locations are ticking--different coordinate systems can give different answers (ultimately this is due to the relativity of simultaneity). There is also no frame-independent notion of speed, so you can't say in any objective sense that clocks moving at high ... 2 Whilst the question is not a resource request, I would recommend Edward Witten's paper on the topic published in 1988, titled, 2+1 Dimensional Gravity as a Soluble System. In the paper, Witten shows: 2+1 dimensional gravity with or without \Lambda is soluble classically and at the quantum level 2+1 dimensional gravity is related to a Yang-Mills ... 2 The Lorentz group O(3,1) has spinor representations (actually SL(2,\mathbb C), that is the universal cover of O(3,1)), as well known. The problem is that now, in general relativity, we want to deal with generic transformations. So we are working with GL(4). Roughly speaking, the associated Lie Algebra \mathfrak{gl}(4) doesn't admit spinor ... 2 The gravitational Chern-Simons action is topological, yes. The gauge connection encodes the field of gravity and since it is being integrated over, the result does not depend on a metric. (In the expressions you write maybe the vielbein contribution is missing? Or maybe you mean to have absorbed it in the notation.) Notice that it's just the usual ... 2 I think it may have been Witten (perhaps somebody can correct me) who suggested that a TEO might not actually exist. Just as we might never, even in principle, posses the mathematics that can interpolate between the strongly coupled and free limits of QCD, it might not be possible to write down one set of equations that describes one corner of a TEO, say, ... 1 Relativistic time dilation and gravitational time dilation are two different coexisting phenomena. At speeds much less than c, the gravitational time dilation effect is much more pronounced, and is the reason that clocks on satellites must run slightly slower than earth time. At speeds approaching c, the relativistic time dilation effect is more ... 1 Although I think you may get the solution already, let me write answer as I thought. First, as you wrote the energy E can be written as $$E=m(-\xi_\mu\xi^\mu)^{1/2}=mV.$$ Because E is conserved, V is also conserved. Second, we are now in asymptotically flat spacetime. This means that at infinity you get V=1. Therefore ... 1 Yes, the observers on Earth will measure the astronaut's clocks to be be running slow while the astronauts will measure the clocks on Earth to be running fast. So the situation is asymmetric. The situation is asymmetric because the two sets of clocks are in different environments. Specifically there is a (gravitational) potential energy difference between ... 1 For an intuitive understanding of this, let me remind you that we know from special relativity that the proper time of an uniformly accelerated observer is "absolutely" slower than the proper time of an inertial guy (supposing clock hypothesis). The more the (proper) acceleration, the slower proper time of an accelerated observer is. Now, thanks to strong ... 1 Your claim that nothing will happen because they initially have no velocity is where things start to go wrong. In fact everything in relativity has the same speed (the speed of light). A particle that looks to be "at rest" in some reference frame simply has all of its velocity pointing in the "time direction". This is an intuitive reason why you would see ... 1 Yes, it will expand spacetime, thus time is also included 1 It is not true that these theories cannot coexist. To put things in context: Ever since Newton's time we have been thinking of things taking place in locations in space and time. Special relativity (SR) showed us that these are connected, and we really should be thinking of spacetime as the theater in which we live. General relativity (GR) simply gave some ... 1 There is no problem in treat Cristoffel symbols as tensors, because in some definitions they actually are tensors. If one defines abstractly a covariant derivative as an operator over tensors with the following properties: Linearity:$$ \nabla_c \left( \alpha A^{a_1,\dots,a_k}_{b_1,\dots,b_l} + \beta B^{a_1,\dots,a_k}_{b_1,\dots,b_l} \right)= \alpha ... 1 The Lagrangian for GR is $$L \propto \int R \sqrt{-g} \, d^4 x$$ where$R$is the Ricci scalar $$R = R^\mu_\mu = R^{\mu \nu}_{\mu \nu}$$ So, this is a scalar which is related linearly to all the components of the Riemann tensor, and is a second-order differential of the metric$g$of the form $$R \sim g \partial^2 g + (\partial g)^2$$ This is ... 1 What we like to call the energy, i.e., the total matter/energy content of space-time, might not be conserved. However, there is a lot of reason to suspect that fundamentally the universe is some big quantum system, and that space-time and particles and fields are emergent from this underlying idea. In that case, we expect there to be a Hamiltonian$H$and ... 1 In the Schwarzschild geometry, the Schwarzschild radius breaks naive dilation symmetry. In the simple case of a radial dilation$r \to \lambda r$, the geometry is only preserved by$R_S \to \lambda R_S\$. So, it naively seems like it would be difficult to find a working dilation, even just a radial dilation. I went to some effort (as an exercise for myself) ...