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When MTW say the universe is isotropic, they mean it is isotropic everywhere i.e. at all points in the universe. It's easy to construct universes that are isotropic at a single point and not homogeneous, for example CuriousOne's suggestion of a ball with density that is a function of distance from the centre. However this ball is only isotropic if you are ...


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To know what a closed timelike curve looks like, you just do like every spacetime metric. You compute geodesics and field equations and all of that. Unfortunately, things start getting complicated. Closed timelike curves have a lot of weird behaviours, especially when it comes to matter fields upon them. They may not have a properly defined Cauchy problem, ...


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The idea behind that quote is that you can't really separate space and time in General Relativity, which is the most complete scientific theory concerning the geometry of space and time. Instead, it works best to consider them as one integrated thing, called spacetime. To go into a little more detail, first consider galilean spacetime. Here you can think ...


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As far as I understood from my so far cursory look into a living review article by Poisson, Pound and Vega on The Motion of Point Particles in Curved Spacetime, it's a bit messy. But I think if you manage to go through GR, this should be manageable, as well. It will probably help if you've dealt with Green's functions before and even better if you've seen ...


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You never have to make energy come from nowhere, the fact that the Einstein tensor has zero divergence means thavyiu can write any spacetime and the corresponding stress-energy tensor will have zero divergence. And zero divergence means that changes in energy (or momentum) are effected only by energy (or momentum respectively) net flowing in or out of a ...


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Really, to answer this carefully, we have to really think through what a horizon is. And for a general spacetime, there are several different notions of horizon, and "event horizon" is probably the most difficult of them to work with. The formal definition of "event horizon" says "Let's go to the distant future, take every freely-falling path that ...


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You ask: If the clock is running slowly compared to a distant clock is this equivalent to the clock having a lower energy compared to a distant clock? but you have to very careful what you mean by energy in general relativity. As it stands your question too vague to be usefully answered. However in the weak field limit there is a sense in which time ...


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I agree with everything John Rennie said but let me just take a slightly different direction. Note that Schwarzschild space-time has a time-like Killing field $\xi^{\mu}$. In a stationary space-time such as this, one can define the Newtonian analog of gravitational potential by $\phi \equiv \frac{1}{2}\log(-\xi_{\mu}\xi^{\mu})$. One can then easily show ...


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There are a few misconceptions in your scenario that cause the misunderstanding. First of all, by definition, causality means that if the time interval between two events is positive in one reference frame, then it is positive in any other reference frame of your choice and viceversa, provided the velocity the events propagate to be smaller than $c$. If, on ...


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$\require{cancel}I) $OP's is considering Dirac fermions in a curved spacetime. OP's action has various shortcomings. The correct action reads$^1$ $$ S~=~\int\!d^nx~ {\cal L}, \qquad {\cal L} ~=~e L, \qquad L~=~T-V,\qquad e~:=~\det(e^a{}_{\mu})~=~\sqrt{|g|}, $$ $$ T~=~\frac{i}{2} \bar{\psi} \stackrel{\leftrightarrow}{\cancel{\nabla}} \psi, \qquad V~=~ ...


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I got the answer reading a book from E. Poisson, what I was doing was indeed wrong, you have to start with the induced metric given by $$ h_{ab}= g_{\mu\nu}e^{\mu}_a e^{\nu}_b $$ where $$e^{\mu}_a=\frac{\partial x^{\mu}}{\partial y^a}$$ are the tangent vectors to curves of the hypersurface. Then, you just replace $g$ by $h$ in the usual relation ...


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It appears that all you have to do is integrate (6) to obtain: $$\alpha = \int dx \frac{2GM}{c^2}\frac{y}{(x^2+y^2)^\frac{3}{2}} = \frac{2GM}{c^2}\frac{xy}{y^2\sqrt{x^2+y^2}}=\frac{2GM}{c^2}\frac{x}{y\sqrt{x^2+y^2}} \equiv \frac{4GM}{c^2R}$$ So we obtain: $$R=\frac{2y\sqrt{x^2+y^2}}{x}$$


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Due to relativity, the clocks on the GPS satellites move fast by about 38 ┬Ás per day. Which would be a problem, but not that big a problem because they all move fast by the same amount. Still you'd need to synchronize the clocks from time to time, because the satellite's position in space also depends on the clock. HOWEVER, if you do that once a week, ...



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