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You have a good intuition on the possible answer because it does involve oscillations, but you need to visualize the string differently. The origin of the conundrum is not quantum, but relativistic. Here's why: Consider an inertial frame O and set up a string spanning the entire $x$ axis, stretching from $x \rightarrow -\infty$ to $x \rightarrow \infty$. ...


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My reasoning being that due to Lorentz invariance, if two objects are located at time-like separated spacetime points, then it is possible to find a frame in which they are located at the same spatial point, in which case any direct interaction between the two objects would violate causality (one object would [instantaneously] affect the other despite the ...


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//Answering my own question with a modified version of my comment The key is that the devices are defined to only explode when hit simultaneously from the reference frame of the moving train. However, what is simultaneous inside the train is not necessarily simultaneous to an outside observer. Hence to an outside observer the device will appear to be ...


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Assume you have a set $F\subset \mathcal M$ called a future set which is the chronological future of some set $S\subset \mathcal M$, ie., $F=I^+[S].$ Where $\mathcal M$ is your spacetime. Similarly, assume you have a past set $P=I^-[S']$ for some $S'\subset\mathcal M$. So that we are clear, $F=I^+[S]$ means that $F$ is exactly those events $f$ where there ...


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Indeed, the QFT notion of locality is that observables at space-like separation commute, i.e. $$ (x - y)^2 < 0 \implies [\mathcal{O}_1(x),\mathcal{O}_2(y)] = 0 $$ for all local observables $\mathcal{O}_1,\mathcal{O}_2$, which are generically polynomials in the fields and their derivatives. This is our notion of locality because, classically, we know that ...


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It is not the definition of an event horizon, and in fact you can choose coordinates that are regular near the event horizon. A common reason for coordinates that are irregular at the horizon is if the coordinate is primarily made to represent time far away. In that case, a timelike curve has a negative interval in your convention, so you can have time ...



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