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It is an extremely complicated problem. Imposing restrictions on the stress energy tensor will give you different possible spacetimes (which spacetimes is also quite a complex problem), but if those restriction apply in all cases is quite hard to prove. The stress energy tensor still has to obey the field equations of the various matter fields that exist, ...


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Usually, the variation of a filed $\delta\phi$ is defined to be \begin{equation} \delta\phi\left(x\right)=\phi^{'}\left(x\right)-\phi\left(x\right) \end{equation} where the new field $\phi^{'}$ and old field $\phi$ are evaluated at the same point, if we take a active transformation point of view. So I think $\delta g_{\mu\nu}$ should be ...


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Classical Lagrangian field theory deals with fields $\phi: M \to N$, where $M$ is spacetime and $N$ is the target-space of the fields. We shall for convenience call $M$ and $N$ the horizontal and the vertical space, respectively. The metric $g$ can be viewed as a classical field of this kind. OP is asking about finding the Euler-Lagrange equations. In that ...


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One of the basic principles in relativity is that spacetime always looks locally flat. By this I mean that if you restrict your observations to a small region surrounding you the curvature becomes negligable. You can make the effects of curvature arbitrarily small by making the region you consider arbitrarily small. The point of this is that for your clock ...


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Actually the clocks tick at the same rate. You probably forgot that distances do not stay the same when you rotate the clock. (In non-Euclidean space turning a rod changes its length)


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No, not in general. Things are only negligible for certain purposes. For example, a field that obeys the Klein-Gordon equation has the stress-energy tensor:$$T^{\mu\nu} = \frac{\hbar^2}{m} \left (g^{\mu \alpha} g^{\nu \beta} + g^{\mu \beta} g^{\nu \alpha} - g^{\mu\nu} g^{\alpha \beta} \right ) \partial_{\alpha}\bar\psi \partial_{\beta}\psi - g^{\mu\nu} m c^2 ...


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The metric tensor is unrelated to the topology of spacetime, as it does not actually qualify as a metric in the topological sense : it is not positive definite, nor does it apply to spacetime points (at least not directly : you can always find the length between two spacetime points by integrating the tangent vector of the path between them). The topology of ...


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There is no general solution for the two body problem in general relativity. But! There are a few solutions for specific two body problems. These include the Curzon-Chazy metric (Two particles on a cylindrically symmetric axis) $ds^2 = e^{-2\psi} dt^2 - e^{2(\psi - \gamma)} (d\rho^2 + dz^2) - e^{2\psi} \rho^2 d\phi^2$ and the Israel-Khan metric ("two ...


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Under the GR framework there is no known solution. Because the stars and planets are evolving we can say for sure that there is a well-defined 2-body problem and solution. GR was invented to describe gravity at large ant the problem starts with the temptation to use GR in local fields. The contradiction between 'space expands' and 'orbits do not expand' ...


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There is simply no closed-form 2-body problem in GR. The reason is as follows: The governing equations of GR are the Einstein field equations. One obtains the metric tensor as a solution to the field equations which describes local geometry of spacetime, which itself is determined by the local energy-momentum tensor which induces spacetime curvature, etc. ...


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I do not agree with the answer given by @ACuriousMind. @Scardenalli has asked for a compact Ricci-flat Riemannian manifold $M$ having as isometry group $U(1)\times SU(2)\times SU(3)$. This does not imply that $M$ must be a symmetric Riemannian manifold. However, the answer to @Scardenalli's question is still no, and it follows from a classical result in ...


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The Schwarzschild radius $r$ is not simply the distance to the centre of the black hole. If you measured that distance by letting down a tape measure you'd find the distance was substantially greater than $r$. See for example my answer to How much extra distance to an event horizon?, where I do this calculation. We actually define $r$ to be the ...


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In ordinary vector spaces, the dot product $\cdot$ is a binary operator which takes a pair of vectors $(A,B)$ in the space to the field over which the space is defined. Formally, for a vector space $V$ over a field $K$, the dot product $(\ \ , \ )$ is a bilinear map $$(\ \ , \ ): V \times V \to K.$$ The inner product only has assumes the standard meaning ...



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