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The source of gravity in GR is not just mass, but the full energy-momentum tensor; this tensorial quantity is a measure of energy, momentum and stress, and applies to ALL forms of matter and all fields that are non-gravitational. Furthermore, there exists a quantity in differential geometry which is automatically conserved in a small neighbourhood on a ...

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In the solar system, there is only a weak gravitational field outside the sun. So for practical purpose, you can expand the metric to the first order in $\epsilon$(and I guess that why have this parameter in the definition), $$g_{\mu\nu} = e^{-\Phi }\eta_{\mu\nu} = e^{-\epsilon\phi} \eta_{\mu\nu} \approx -(1-\epsilon\phi) dt^2 + (1-\epsilon\phi ) (dx^2 ... 3 There is no mention of Euclidean space here, because the 2-spheres are not Euclidean. They're not going to be embedded in Euclidean space either. The Schwarzschild r coordinate is defined so as to make the areas of the family of 2-spheres 4\pi r^2. In other words, we're simply labeling the 2-spheres by the their areas. Euclidean space would ... 5 If you want, you can go and use the ansatz:$$ds^{2} = -A(r) dt^{2} + B(r) dr^{2} + 2C(r)\,dt\,dr + f(r)\left(d\theta^{2} + \sin^{2}\theta d\phi^{2}\right)$$Where the functions only depend on r due to the fact that t generates a symmetry of the spacetime -- you are assuming a static spacetime. Note, however, that you are free to arbitrarily rescale ... 0 General relativity is not a quantum theory of gravity, and as such, the application of Feynman diagrams is doubtful. One can try to quantize it, but this will lead to a non-renormalizable theory due to the dimension of the coupling constant. This is where other approaches of quantum gravity come in, e.g. string theory. You can, however, calculate ... 2 This isn't an answer, because I don't think your question has an answer, but it got too long to put in a comment. Anyhow, when physicists try to describe the universe we do it by constructing mathetical models. Then we use these models to calculate what will happen and do experiments to see if we we correct. If we got the correct answer it means our model ... 0 The galaxies and (and binded objects) are maintained by gravity. The rest is explained here, Whats left at the center of the Universe after Big bang? 0 Well, the singularity does not concern the differentiable structure: Even around the tip of a cone (including the tip) you can define a smooth differentiable structure (obviously this smooth structure cannot be induced by the natural one in R^3 when the cone is viewed as embedded in R^3). Here the singularity is metrical however! Consider a 2D smooth ... 2 Not only the position in the gravitational field is important, but also the velocity. Consider the Schwarzschild metric$$ \text{d}\tau^2 = \left(1 - \frac{2GM}{rc^2}\right)\text{d}t^2 - \frac{1}{c^2}\left(1 - \frac{2GM}{rc^2}\right)^{-1}\left(\text{d}x^2 + \text{d}y^2 +\text{d}z^2\right), $$where \text{d}\tau is the time measured by a moving clock at ... 1 The two don't work together, they are competing descriptions of gravity. From a QM perspective gravity is mediated by the graviton. Picture gravitons encountering photons, imparting the force of gravity thus changing the photons' paths. From a GR perspective mass warps spacetime, and any photons traveling through now have warped paths. So you could see ... 0 Clocks tick slower at lower altitudes. So 1. On the surface of the Earth will be the slowest. Now since the ISS has no way of knowing whether it is in orbit or in deep space, you might think that clock 2 and 3 should tick at the same rate. But instead clocks 2 and 3 will just feel like as if they were ticking at the same rate. Astronauts at 2 and 3 will not ... 1 There is a standard way to find out if the spacetime around you is curved. Surround yourself with a sphere of small test masses and wait and see what happens. If the sphere stays exactly the same shape you're in flat spacetime but if it changes shape or volume you're in a curved spacetime. In the case of the ISS the test masses nearer the Earth than you ... 1 This is my guess Think about curve space is like sound wave, When you quantize the wave to be point, like phonon of sound wave, you get graviton from gravity 3 At the level of understanding the data and observations we have up to now, General Relativity describes well what we perceive of the Cosmos and Quantum Field Theory what we observe in the microcosm of elementary particles and their interactions. The two have not been joined up to now, i.e. there is no accepted unified theory that joins smoothly these two ... 6 The curvature of the universe can be derived from the temperature fluctuations in the Cosmic Microwave Background. For a given amount of radiation, baryons, dark matter and dark energy in the universe, these temperature fluctuations can be calculated theoretically, and compared with observations, and so one searches for the values that yield the ... 1 Firstly, it should be made clear that being isotropic is a very special and rare property. (A spacetime can never be truly isotropic because no isometry can map spacelike vectors to timelike vectors, for example, so I'll talk about "space" being isotropic). There are very few spaces isotropic around every point, only very few spaces will even be isotropic ... 3 g^{\alpha\beta} is symmetric in \alpha and \beta, while R_{\alpha\beta\gamma\mu} is anti-symmetric in \alpha and \beta, so the contraction g^{\alpha\beta}R_{\alpha\beta\gamma\mu} is necessarily 0, and cannot be R_{\gamma\mu}. Moreover, it is not correct to say, that if the contraction of 2 tensors with another tensor (here the metric ... 1 the emitter and the receiver move with constant velocities relative to an inertial frame and v is the constant velocity of the receiver relative to the emitter and away from it. No, both the emitter and the receiver are accelerating, and the receiver has gained an extra velocity v between the time the photon was emitted and the time it was received. ... 1 The null curves have:$$ 0 = ds^2 = -dt^2 + e^t dx^2$$that gives you the null curves differential equation:$$ \frac{dx}{dt} = e^{-t/2} $$Solving it you get the equation of the null curves, and so the light cone. 0 I myself overlooked it too but wikipedia actually happens to have a great such list at https://en.wikipedia.org/wiki/Quantum_gravity#Points_of_tension There are other points of tension between quantum mechanics and general relativity. First, classical general relativity breaks down at singularities, and quantum mechanics becomes inconsistent ... 5 You're quite correct that the metaphor is misleading, and indeed you'll find professional relativists tend to be rather scornful of it. There are a number of problems with it, of which the problem you mention is just one. For example the diagram implies only space is bent, while the bending is of spacetime so time is bent as well. The diagram also implies ... 1 Partly this answer is just gathering together the comments above, though there are a couple of points that haven't been mentioned. Firstly, as mentioned in the comments electromagnetic waves do gravitate and the links in the comments cover this well. In the early universe (for the first 47,000 years after the Big Bang) EM radiation was the dominant ... 6 I believe you've already spotted the answer to your question with this sentence: And a black hole will shift the trajectory entirely. This is all dependent on the proximity to the source of gravity. You can "shift light" (bend its trajectory) as much as you want with as little mass as you want using a black hole. Just let the light get arbitrarily ... 3 In principle, yes, you can specify the stress tensor and solve the resulting equations, but in practice, this is hard to do because the field equations are non-linear PDEs...darn. The simplest possible example is the case in which the stress tensor vanishes; T_{\mu\nu} = 0 namely the vacuum equations. The field equations with vanishing cosmological ... 2 I really don’t like that diagram. No, REALLY. I think it conveys a bad intuition that may confuse you. I don’t like what it does with the connexion coefficients (Christoffel symbols). Here’s why. In a general manifold, tangent spaces of course are not comparable at different points. This is in contrast with the situation in Euclidean space, which can be ... 1 I'll try to explain parallel transport first: This image shows how a vector generally does not remain parallel to itself under parallel transport. The vector here is moved along the path A-N-B and is constrained to remain in the tangential plane to the spherical surface at all points of the path because the geometry does not know of the "third" dimension ... 3 It seems that OP is pondering the following. What happens in a field theory [in OP's case: GR] if spacetime M has a non-empty boundary \partial M\neq \emptyset, and we don't impose pertinent (e.g. Dirichlet) boundary conditions (BC) on the fields \phi^{\alpha}(x) [in OP's case: the metric tensor g_{\mu\nu}(x)]? I) Firstly, it should stressed ... 2 For reference, Weinberg p. 378: A metric space is said to be homogeneous if there exist infinitesimal isometries (13.1.3) that carry any given point X into any other point in its immediate neighborhood. Equation (13.1.3) defines an infinitesimal transformation and (13.1.5) concludes the Killing equation \xi_{\sigma;\rho} + \xi_{\rho;\sigma} = 0 ... 3 This is a more complicated question that you probably realise. This first point to make is that the speed of light is always locally c, that is, if you measure the speed of light at your location you will always get the result c. The problem comes when you measure the speed of light at some location distant from you. To measure the speed of light ... 0 The following articles may be helpful. This is actually an active scientific topic. http://www.symmetrymagazine.org/breaking/2009/02/19/most-extreme-gamma-ray-blast-also-probes-quantum-gravity http://www.sciencemag.org/content/323/5922/1688.abstract http://www.sciencemag.org/content/early/2013/11/20/science.1242353.abstract 0 There is no experimental evidence on whether light travels slower in a gravity field. Some quantum gravity theories require light to be slower in an intensive gravitational field while others not so. So, it is to be determined by experiments or astronomical observations. Light travels in glass as fast as in vacuum. Because microscopically, glass is nothing ... 0 Why do you say that the box doesn't shrink in a uniform gravitational field? The photons outside the box also experience the field! Thus, the photons "falling" on top of the box are slightly blue-shifted while the photons hitting it from below are slightly red-shifted. And the box squeezes in the same amount as its rocket analogue. 1 This result follows from i) Uniformization theorem and ii) Gauss-Bonnet theorem in 2d. According to the statement of uniformization theorem from this wiki page : every connected Riemann surface X admits a unique complete 2-dimensional real Riemann metric with constant curvature −1, 0 or 1 inducing the same conformal structure On the other hand, ... 1 Why not use explicit construction for such a surface? From The Manifold Atlas: Any hyperbolic metric on a closed, orientable surface S_g of genus g\ge 2 is obtained by the following construction: choose a geodesic 4g-gon in the hyperbolic plane {\Bbb H}^2 with area 4(g-1)\pi. (This implies that the sum of interior angles is 2\pi.) Then ... 2 Of course, the metric \eta_{\mu\nu} is not a unique solution for Einstein vacuum equations compatible with your given initial data. And yes, we can interpret the alternatives as arising from coordinate functions. Let us take the simplest of such function: redefine time by introducing new 'time' variable \tau through a relation t=f(\tau) (spacial ... 5 Spin 2 just means that the gravitational field is given by a metric field and general covariance, which is the nonlinear expression of a massless spin 2 representation of the Poincare group. The latter appears when linearizing around the Minkowski metric and dropping all interactions. See the classical paper by S. Weinberg, Phys.Rev. 138 (1965), B988-B1002 ... 5 Once inside the Schwarzschild horizon, it is immaterial to observers outside whether what went in was "matter" (fermionic stuff with a rest mass) or "pure energy" (bosonic stuff): it all adds to the black hole's mass by the amount \sqrt{m_0+(p^2/c^2)} where m_0 is the rest mass and p its momentum (at infinity). After this, all the body's energy stays ... 1 I am assuming that matter is considered energy when it is broken down to its simple building blocks. If that's your criteria for being "considered energy", then at least classically the singularity of a black hole of any mass whatsoever will do this, simply because the gravitational tidal forces diverge to infinity near those types of (curvature) ... 4 If you have two points p,q spacelike separated in a spacetime M there is not anything like the shortest spacelike curve joining them! Any spacelike curve joining them can be continuously deformed closer and closer to a lightlike curve joining the same points. So the inf of the set of the lengths of spacelike curves joining the points is always zero and ... 1 Apart from the fact that the concept of relativistic mass is best avoided, as John Rennie mentioned, it is also a concept of special relativity: it can only be defined in an inertial frame (a Minkowski spacetime) where special relativity is valid. However, the expansion of space is a consequence of general relativity. There is no global inertial frame ... 3 You say: A distant quasar would be less massive in its frame of reference than our observations would suggest and this refers to the notorious expression for the relativistic mass:$$ m = \gamma m_0 $$The trouble is that relativistic mass is a troublesome concept that causes more problems than it solves. For example, the gravitational field of a ... 2 OK, let us start from your example. I think that it is too pathological to be considered as a safe starting point for this discussion, which is worth and interesting however. Nevertheless I would like to spend some words about this case since it permits to introduce some general issue useful in the second part of my answer. AdS_n is not globally hyperbolic. ... 2 I'd like to add a clarification to the other answers, some of which seem to imply that the precession of Mercury's orbital perehelion is owing to general relativistic frame dragging. In particular, the statement that the Sun drags the fabric of space time around with it could be, in my opinion, misleading because most of the precession is NOT owing to "frame ... -4 The solution of Einstein, contesting Newton´s laws, was challenged by several scientists including Dr. Thomas Van Flandern astronomer who worked at the U.S. Naval Observatory in Washington. According to them, Einstein would have gotten this information (43 "arc) and" adjusted "the arguments for the result of the equation, previously known, were achieved, ... 0 @AlfredCentauri The condition of locality conventionally only requires a locally constant gravitational field during the time the experiment runs. For a stationary rocket in the gravitational field of e.g. the earth, this condition is perfectly satisfied, without requiring the gravitational field to be uniform throughout space. Clearly, the stars are not ... 1 I think you should have a lowered index on the RHS:$$\begin{align} \frac{dh_{ab}}{dt} &= \frac{d}{dt}\left(h_{ma}h_{nb}h^{mn}\right)\\ &=\frac{dh_{ma}}{dt}\delta_{b}{}^{m} + \frac{dh_{nb}}{dt}\delta_{a}{}^{n} + h_{ma}h_{nb}\frac{dh^{mn}}{dt}\\ \frac{dh_{ab}}{dt}&= 2 \frac{dh_{ab}}{dt} + h_{ma}h_{nb}\frac{dh^{mn}}{dt}\\ \frac{dh_{ab}}{dt} &= ...

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When you write the Dirac equation in a curved spacetime, in the context of General Relativity (which allows curvature, but not torsion) , you have a spin connection : $$\nabla_\mu\psi=\left(\partial_{\mu}-\frac i4\omega_{\mu}^{IJ}\sigma_{IJ}\right)\psi$$ Now, the Einstein-Cartan theory is not General Relativity, because it allows curvature, but also ...

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In his original work, Fermi considered only vectors $f^{\mu}$ which are orthogonal to the curve $f^{\mu} v_{\mu} = 0$. His analysis is relevant to the spin or photon polarization vectors which are orthogonal to the four-velocity by definition. Walker generalized Fermi's work to vectors which are not necessarily orthogonal to the velocity. (Thus the ...

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You can derive the desired expression in the following way: \begin{align}\delta(R_{ab}R^{ab}) &=\delta R_{ab} R^{ab}+R_{ab}\delta R^{ab}\\ &=\delta R_{ab}R^{ab}+R_{ab}\delta R_{cd}g^{ca}g^{db}\\ &=\delta R_{ab}R^{ab}+R^{cd}\delta R_{cd}\\ &=2R^{ab}\delta R_{ab}\\ &=2R^{ab}\delta(R_{cadb}g^{cd})\\ ...

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However, I had difficulty understanding that answer and would like to understand how to do it this way. That is to say, I'd really like to know what property or identity that I'm missing before I can use use the Bianchi identities to show that it is manifestly zero. The other proof uses the first Bianchi identity. That's where the starting assumption ...

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