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Warp drives are not allowed by the basic laws of physics, in particular the theory of relativity prohibits any superluminal motion or superluminal propagation of usable information. So whatever "exotic matter" or other wordings are proposed to justify the superluminal warp drives is banned, too. The typical "exotic matter" needed for warp drives would need ...


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Mathematically, that's due to superposition. Both masses produce some gravitational field, which add together to give the "net field". (The same goes for electromagnetism, where one may add electric/magnetic field strengths,electric potentials etcetera for every point in space.) Ever so slightly changing the field strength at the well. As gravity gets weaker ...


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Parity is a symmetry that we impose to the Dirac and Maxwell Lagrangian because we know experimentally that electromagnetism conserves parity. Demanding parity invariance will then tell you how Dirac spinors and vectors transform under such a symmetry. The correct transformation properties are: $\bullet$ Spinors: ...


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In other words, in order to connect an abstract mathematical description of motion, which in some sense can be transformed in any way, to physical observations, doesn't one have to anchor the mathematical reference somehow to physically meaningful concepts like unaccelerated frames? Take something simpler that General Relativity. Consider ...


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The positive energy theorem talks about the lower bound on the total energy/mass, like the ADM mass. To be able to define such a concept of the total energy/mass in general relativity, one needs some asymptotic region respecting a time-translational symmetry. That's the region where the gravitational potential (something like the deviation of $g_{00}$ from ...


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I wonder if your question is a form of the hole argument. It's true that you can't always find the geodesics (i.e., the unaccelerated worldlines) in certain coordinates given only the Riemann curvature in those coordinates, but that's just a gauge redundancy, not a physical ambiguity. There are up-to-diffeomorphism uniqueness results for the GR initial-value ...


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This is too long for a comment so I'll post it as an answer, even though this question is years old. If Alcubierre warp bubbles are physically possible, which is exceedingly unlikely, and if the equivalence principle is correct, you could definitely escape from a black hole in one, because there's nothing locally special about the event horizon. In a large ...


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The statement that space is Euclidean is a broad statement, not meant to hold near very massive bodies or in arbitary volumes. One sense in which it can be meant is to hold "on average" for the whole spacetime - the universe - as such. The best candidate for the overall metric of spacetime is the FRLW metric, which is the exact solution for a universe that ...


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It takes time for light to bend, i.e. the bending of light is a consequence of the curvature of geodesics in four dimensional space-time. On the other hand, a three dimensional snapshot of space appears to have geodesics that are straight lines (some specifics shown below). This is exactly why we thought space was Euclidean for so long and why all ...


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First off, the question focuses a lot on the issue of how to tell whether a particular set of coordinates is inertial. This is the wrong thing to focus on, and it wouldn't even normally occur to a relativist to ask whether a certain set of coordinates was inertial. Coordinate systems are global things, whereas an inertial frame of reference only exists ...


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We can't say much about closed timelike curves with any certainty; they are an artefact of the existence of solutions to the general relativity equations which allow them. It is possible (and quite a few physicists believe this) that a theory of quantum gravity may preclude CTC from occurring, or that CTC may occur but the information might be censored by ...


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If somehow, it is only possible to create one alcubierre drive, and it can never turn around, then you won't get closed timelike curves. Otherwise, any construction is going to have them. The reason is that you can "zoom out" far enough that the distortions to spacetime caused by the drive are no longer present, and the person flying the drive then just ...


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$$\nabla_{\mu}\left[U,V\right]_{\nu} = \nabla_{\mu}\left(g_{\nu\lambda}\left[U,V\right]^{\lambda}\right)$$ You have two terms inside the parentheses, and you have to apply the derivative to both of them. Myself, I'd just remember that: $$\left[U,V\right]^{a} = U^{b}\nabla_{b}V^{a} - V^{b}\nabla_{b}U^{a}$$, so, we have: $$\begin{align} ...


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Notice that $(\nabla_\alpha A^\mu) (\nabla_\beta B_\mu) = (\nabla_\alpha A_\mu) (\nabla_\beta B^\mu) $ and hence (looking only at the first row), the first term reads $$-(\nabla_\mu U^\tau)(\nabla_\nu V_\tau) = -(\nabla_\nu V_\tau)(\nabla_\mu U^\tau)= -(\nabla_\nu V^\tau)(\nabla_\mu U_\tau),$$ which is the symmetric of the forth and thus cancels with it. ...


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The structure of general relativity does not allow the theory itself to be classified in terms of global, discrete symmetries such as time-reversal. The Einstein field equations don't refer to a time coordinate; they're expressed tensorially, which means that they are completely independent of what coordinates you choose. Since there is no guarantee that you ...


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Astrophysically no, because the star has to go through supernova to form a blackhole. Some of the mass will be pushed away by the explosion, so the black hole will have a smaller mass than the star before supernova. If we ignore this and assume that mass stays the same then the object will feel the same gravity if it is at a safe distance, for the following ...



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