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How would a “loop-hole free” Bell test appear to an observer moving at relativistic speeds compared to Alice and Bob?

In most such tests, Alice and Bob are 1km+ away from each other to prevent information (e.g. info about their polarizer settings) from being able to influence the result. Also, Alice and Bob make their observations at slightly different times due to one of them being slightly farther away from the particle source. (Hensen, et al., 2015; see summary with diagrams here).

However, it occurred to me that due to the short separation of mere microseconds between the measurements conducted by Alice and Bob, could an observer travelling at relativistic speeds, receiving radio transmissions of the detection results in real-time, perceive the sequence of measurements to have occurred in a different order than the order in which they would appear to a stationary observer to have occurred?

This question is based on Einstein’s paper on special relativity, particularly where he writes:

(...) two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system. (A. Einstein, “On The Electrodynamics of Moving Bodies”, 1905)

If I understand it correctly, this also means that two non-simultaneous events A and B, which appear to a stationary observer to occur sequentially with A happening an instant before B, can meanwhile appear to a moving observer to occur exactly simultaneously, or even to appear to happen in the opposite sequence, with B happening before A.

If indeed the sequence in which the measurements by Alice and Bob take place is merely an artifact of these two observers being stationary relative to one another, and if indeed another observer might see the measurements take place in reverse order, then what does this mean for any non-local effects we assume that Alice’s measurements have upon Bob’s measurements due to the effects of quantum entanglement?

If I am not ubderstanding it correctly then what am I getting wrong?

(Thank you, I am just an amateur physics buff trying to better understand these ideas.)

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    $\begingroup$ The time-, null-, or space-like nature of separations is invariant. If the test is space-like in the lab frame it is space-like to all observers. $\endgroup$ – dmckee Sep 12 '18 at 19:53
  • $\begingroup$ @dmckee your statement is correct, but I don't see how it addresses this question. The separation of A and B is space-like to all observers. This space-like separation is in fact the very property which allows the ordering of events A and B to be reversed depending on the frame of reference. If A and B were time-like separated then their ordering can not be reversed no matter the Lorentz transformation performed. $\endgroup$ – enumaris Sep 12 '18 at 20:25
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    $\begingroup$ I'm not an expert, but don't "loophole free" tests work with space-like separated end stations? If so the question of temporal order is meaningless from the get-go, because it is an observer artifact that can have no effect on the physics. Perhaps I should have made that an answer, because your question seems to be "what effect does changing the observed order of space-like separated effect have, and the answer is "none". The events can not classically affect one another. $\endgroup$ – dmckee Sep 13 '18 at 14:51

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