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Alice and Bob's spaceships are converging inertially at 0.99c. Alice knows enough about Bob's clock that she can calculate for any time of her clock what his clock would show "at that time" in her frame. Bob is similarly knowledgeable about Alice's clock. When Alice's clock shows Ta, Alice calculates Tb = what Bob's clock shows at that time in Alice's frame. When Bob's clock shows Tb, Bob calculates Tc = what Alice's clock shows at that time in Bob's frame. Question: does Ta = Tc ?

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closed as off-topic by AccidentalFourierTransform, DilithiumMatrix, Bill N, Jon Custer, Rory Alsop Jan 15 '17 at 19:18

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    $\begingroup$ Keep in mind that you can jigger the origins to make this true of any one time without regard to the general answer. In most examples that one time is taken to be $t=0$ and when they pass one another. $\endgroup$ – dmckee Jan 14 '17 at 20:28
  • $\begingroup$ To those who voted this as off-topic: this is no homework question, though I am not competent to answer it with assurance. I constructed the case to ask whether "now" as a concept is inconsistent. If Alice is thinking "I wonder if Bob is thinking of me now?" and exactly at that moment (according to Alice's calculations) Bob is thinking "I wonder if Alice is thinking of me now?", and exactly at that moment (according to Bob's calculations) the answer (as given by udrv) is "no", then "now" is meaningless. This is beyond the relativity of simultaneity—it's not even symmetrical! $\endgroup$ – John Fairfield Jan 19 '17 at 23:14
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Short answer: No. If they did, there would be no relativity of simultaneity.

The simplest way to track the conflict is to add a buoy, say in Alice's frame, at some distance from Alice's own location. Let the buoy carry its own clock, synchronized to that of Alice, such that when Bob passes the buoy location, the buoy clock shows time $T_A$, just like Alice's clock. This means Alice observes 2 simultaneous events at non-coincident locations: her clock beating time $T_A$ at her location, and the buoy clock beating time $T_A$ at its own location.

Now look at Bob as he passes the buoy at his time $T_B$ and buoy time (Alice's time) $T_A$. Relativity of simultaneity means that Bob cannot see both Alice and the buoy marking time $T_A$ at the same moment of his time: spatially separated events that look simultaneous to Alice never appear simultaneous to Bob. So although he does see the buoy clock showing $T_A$ at his time $T_B$, he finds that Alice's clock shows a different time ${\tilde T}_A$.

Fun exercise: Apply the same reasoning to a symmetrical setup using buoys in both Alice's and Bob's frame.

Note following comments: As @dmckee observed, the phenomenon is independent of a particular choice of time origin, in either frame. Alice and Bob may choose their individual $t=0$ marks arbitrarily, yet they always arrive at the same conclusion regarding the specific time intervals. In more formal if obvious terms, time intervals are invariant under time translations in any frame.

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  • $\begingroup$ Are you a teacher? That's a really grand exercise to demonstrate the ideas. I think you should probably include Dmckee's comment that one can make the assertion true of any one particular time through choice of origin. $\endgroup$ – WetSavannaAnimal Jan 15 '17 at 0:32
  • $\begingroup$ Are you a teacher? That's a really grand exercise to demonstrate the ideas. I think you should probably include Dmckee's comment that one can make the assertion true of any one particular time through choice of origin. $\endgroup$ – WetSavannaAnimal Jan 15 '17 at 0:32
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    $\begingroup$ Thanks, and no, unfortunately, in any case not at the moment. I added a few words as you suggested, good point. $\endgroup$ – udrv Jan 15 '17 at 2:32

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