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I'm taking a Special Relativity course and lately I've been wondering about this thing for some time. Imagine I'm at the origin of my spacetime lattice (hope I can say this), and I've synchronized all my clocks so that when light passes they start running synchronized.enter image description here

At a time T2, light has reached some clocks and they start ticking.
Let's suppose that at this time there's an event far away where the light has not reached the clocks in space. This the same situation of an event occurring at time T2 outside your light cone:enter image description here

  1. Can I make any assumption about this event at time T2?
  2. How much do I have to wait to see it?
  3. Do we say this event is too far in space, time or spacetime?

Please address any misconception (if there is) in comment, I will be glad to know them and edit my question.

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closed as unclear what you're asking by WillO, ZeroTheHero, sammy gerbil, Jon Custer, Michael Seifert Jun 9 '17 at 14:04

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ (2): Are you asking, when you see the clocks starting to tick? In $2*T_2-T_1$. You can see them. But you will see them only after the light from the clock arrived to you. If you ask, when you will see A to happen, it is independent from your other clocks. It will happen in $T_1 + \frac{dist(A)}{c}$. $\endgroup$ – peterh Jun 8 '17 at 18:59
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    $\begingroup$ Well you could make any assumption about any event outside your light cone, but it doesn't mean it's going to happen or is correct ;) $\endgroup$ – Kyle Kanos Jun 9 '17 at 2:21
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I do not understand why you relate clock synchronization with seeing event A. You will see event A when the light from it reaches you. You can draw the light leaving from A as two 45 deg lines. You will see A when the left line intersects the time axis. See light cone

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  • $\begingroup$ First of All, I wanna thank you both Wilyy Billy Williams and @peterh, thanks for the effort you make in trying to understand my strange questions and my poor language! You really are amazing! A little foreword: Since I've started studying special relativity, every time I get into a new problem, I've get used to picture myself putting each clock in space very slowly so its synchronized with the others and shooting a light sphere from the origin to all directions as soon as I start the experiment/observation and time starts to flow. This helps me more when there are multiple observer/frames. $\endgroup$ – Gabriele Scarlatti Jun 8 '17 at 20:21
  • $\begingroup$ I'm gonna read the light cone and then try to finish my explanation, and eventually edit the question to make it celarer! I need a little time! $\endgroup$ – Gabriele Scarlatti Jun 8 '17 at 20:29
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    $\begingroup$ Gabrielle Scarlatti, nothing wrong with your circles. You're doing it all in one reference frame, at two times. No problem. The clocks are in the same reference frame, they'll tick at the same rate. And you can use them to 'stay synchronized' to your light source since you can program in their distance from you. When they see A of course they'll think it's your light source, they'll correct for their distance to you, and say transmit the fact that they saw that light to you. You just have to do the different adjustments. Since you don't know where A is. ....... Continued $\endgroup$ – Bob Bee Jun 9 '17 at 5:04
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    $\begingroup$ ... Continued... but you know where all the clocks are, you just subtract the distance from you in light time (or twice the distance to account for thesis message travel time to you, and you get the time when they actually saw A. Then you just do geometrical triangulation. And you've located A. Just a little geometry and calculation, no need for abstract light cones. Do it simply with 2 or 3 clocks and it's easy. No need to learn about light cones. You can do light cones starting from simpler examples. $\endgroup$ – Bob Bee Jun 9 '17 at 5:08
  • $\begingroup$ Thanks @Bob Beee and all the others, reading your comments clarified a lot to me, also about things I did not question myself. I tried to edit the quesion to make it more appealing and clearer, hoping someone will find the same amount of help I've in reading your comments. Most of my question have been adressed in the comments, if someone think that now question is clearer and want to give a clean answer I will be happy to give Best Anser to someone! $\endgroup$ – Gabriele Scarlatti Jun 9 '17 at 8:28

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