Suppose we have to observer $A$ and $B$ with their time given by $t_A$ and $t_B$.To synchronize their clocks observer $A$ send lights at $t_A$ towards $B$. $B$ receives the light at $t_B$ and send it back to $A$,and $A$ receives the light at $t_A'$.Finaly $A$ sends a message to $B$ with instructions that his clock should have $\frac{t_A+t_B}{2}$ in $t_B$.

Now suppose we have two events, $e_A$ in the worldline of $A$ and $e_B$ on the worldline of $B$. Suppose also that these two observers agree that these events are simultaneous, than they can synchronize their clocks by setting their clocks to zero for these events.

My question is, given another event $e_c$, and suppose that $e_c$ is simultaneous to $e_A$ from the point of view of $A$, are $e_A$ and $e_c$ simultaneous from the point of view of $B$?

  • $\begingroup$ With two observers only, remember that they will ways be on "the same plane" .This is basic geometry. $\endgroup$ – Sidarth May 29 at 5:19
  • $\begingroup$ @Sidarth Well, they aren't actually planes, they are spaces. But 4D diagrams aren't easy to draw or understand, so it's customary in Special Relativity diagrams to suppress 1 or 2 space dimensions, so we get lines or planes of simultaneity. Please see en.wikipedia.org/wiki/Relativity_of_simultaneity $\endgroup$ – PM 2Ring May 29 at 5:56
  • $\begingroup$ Are A & B at rest relative to each other? $\endgroup$ – PM 2Ring May 29 at 5:57
  • $\begingroup$ They are at rest in a uniform rotating frame $\endgroup$ – amilton moreira May 29 at 5:59
  • $\begingroup$ @amiltonmoreira If they are in a uniform rotating frame, then they are not inertial observers. So, you cannot blindly use special relativity $\endgroup$ – silverrahul May 29 at 6:21

If the two observers are moving relative to each other, then in general events that seem simultaneous to one will not seem simultaneous to the other. That is because their respective time axes will be tilted relative to each other, which means that surfaces of constant time in one frame will be out of alignment with surfaces of constant time in the other. It doesn't matter whether they synchronise the clocks they are holding (ie at the two positions in space where they are located at the time they perform the synchronisation)- everywhere else, time will be out of synch between the two frames.

  • $\begingroup$ I will edit my question $\endgroup$ – amilton moreira May 29 at 7:35

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