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4

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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A closed timelike curve wouldn't actually "look" like anything because it's an abstract thing. You can't actually see any lightcones or worldlines. A metric is an abstract thing too, to do with your measurements of distance and time, typically made using the motion of light. And the crucial point is this: you don't travel along your worldline. You move ...


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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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Light will not travel on a geodesic in the background spacetime, as light is an electromagnetic field and has stress-energy and so itself warps the spacetime around it. However if the light induced curvature is small compared to the curvature already there, then it won't change things much, so will go on an approximately geodesic route. Same for you ...


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I"m pretty sure that this discussion does appear in Hawking and Ellis, though I admit that it's been a while since I looked. It's not done through a Penrose diagram, though. The argument really comes down to the fact that for sufficiently small $r$, $d\phi$ is timelike. But, by construction, the orbits of $\phi$ are closed curves. When $d\phi$ is ...


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This is really one google search away, see e.g. page 26 (marked 64) here. As already noted by John Rennie, Penrose diagrams are not suited for the analysis of Kerr CTCs because they show a $\phi = const., \theta=\pi/2$ slice of the global structure. The $r<0$ region is however accessible only through $\theta \neq \pi/2$. The Boyer-Lindquist coordinates ...


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Timaeus : I can't answer your question, but would like to comment. However what I want to say is to big for a comment, so I'm using the answer facility. Apologies in advance, feel free to downvote. The region beyond the ring singularity in the maximal Kerr spacetime is described as having closed timelike curves. Let's imagine we're out in space, at a safe ...


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You may be inquiring about a mechanism to separate sound vibrations which originated yesterday from sound vibrations which originated just an instant ago. If you stood inside the Capitol dome with sensitive enough apparatus, could you hear the residual vibrations of voices which spoke yesterday or a hundred years ago? The question is (1) whether vibrations ...


3

Time is a continous flux that goes from the past to the future; you can't stop it, you can only accelerate or decellerate it. From relative theories and from quantum mechanics we know that, in the most of case, the max speed of informations is the speed of light. So we can see events from the past: we can observe the light of a star that, now, is died. Only ...


2

Here is one solution to Einstein's field equations: $$ds^2=dt^2-dx^2-dy^2-dz^2 \text{ on } \{(x,y,z,t):x,y,z,t\in\mathbb R\}.$$ It has a global coordinate system, a global reference frame, no closed time like curves, and a famous name, Minkowski space. Our second solution is a different manifold $\mathbb R^3\times \mathbb S$, which can (but doesn't have ...


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Calculating the power emitted as gravitational waves is relatively straightforward, and you'll find it described in any advanced work on GR. I found a nice description in Gravitational Waves: Sources, Detectors and Searches. To summarise an awful lot of algebra, the power emitted as gravitational waves by a rotating object is approximately: $$ P = ...


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Gravitational waves are weak. The reason that we try to find them coming from huge astronomy bodies is because we need a lot of mass to produce a measurable gravitational wave. As you can see, gravity is very weak compared to the other fundamental forces. This is why we need need huge masses to measure them. You could not make a time machine with a ...



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