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I'm reasonably familiar with special relativity and its effect such as time desynchronization, but I'm having trouble understanding how these effects come into play when we also consider the time for light to reach us. I'll describe my problem using this image:

desync

We will be observing 2 objects (the red vertical lines). There are also 2 observers: A, which is standing still and B which is moving towards the objects.

We should expect that B sees a time desync. If we look at the plane of simultaneity (black solid line) we see that the farther object is observed further in the future. However the actual photons reaching A and B (at the point they meet) are the same. How can B see 'further' in the future than A, but still receive the same photons?

To clarify I'm aware that B doesn't see the solid line, but I would expect B to see further into the future than A as special relativity predicts a time desync.

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    $\begingroup$ You never see into the future. What you see are photons arriving at your eye from objects they left in your past. $\endgroup$
    – Marco Ocram
    Commented Jan 15 at 8:01

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The red vertical lines represent $x=4$ and $x=3$ in A frame. The photon will reach them at times $t=-1$ and $t=0$ respectively. The corresponding values for B have different $x'$ and also different $t'$. The $t'$ will be more in the past, according to the simultaneity lines for B. Or, in other words, B will see A times as being 'in the future', in the meaning that they are bigger numerical values.

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  • $\begingroup$ If we draw another parallel line it will indeed reach A and B at different coordinate times, but it will still seem as if B sees no time desync right? My main problem is that A and B see the same photons even though B should see the farther red object further in the future than A would. If it's not clear what I mean I can add another picture. $\endgroup$
    – catmousedog
    Commented Jan 15 at 1:43
  • $\begingroup$ My previous answer was mentioning another spacial dimensions. I changed to focus on $x$ direction only, and the relations between its values for A and B, regarding space and time. $\endgroup$
    – Claudio Saspinski
    Commented Jan 15 at 3:33
  • $\begingroup$ So A and B will observe no time desync between the red objects, but it will find a time desync in the coordinates. So even though B will always receive the same photons as A (all light-like lines will see the farther red object as 1 second in the past) it experiences time desync in its coordinates? That would make the coordinates not seem very representative of what B is actually observing though right? $\endgroup$
    – catmousedog
    Commented Jan 15 at 11:53
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In your diagram the dashed line represents light arriving at A and B having earlier left the two objects. Your trouble in understanding might vanish if you consider, from the perspectives of A and B separately, what time it was when the light left each of the two objects. You will find that the times in B's frame differ from the times in A's, and that accounts for the lack of simultaneity.

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  • $\begingroup$ From this diagram it looks like both A and B observe the same simultaneity. They both receive the same photons. That is if there were two events on the red vertical lines at (ct, x)=(3,0) and (4,-1) then both A and B would receive these photons at the same time, even though I would expect B to receive them out of sync. $\endgroup$
    – catmousedog
    Commented Jan 15 at 11:43
  • $\begingroup$ You have not followed what I suggested, which was to consider the times, in each frame, when the light left each of the two objects. The events happen at 3,0 and 4,-1 only in the frame of A. They are not simultaneous in either A's frame or in B's. $\endgroup$
    – Marco Ocram
    Commented Jan 15 at 16:30
  • $\begingroup$ Say the objects on the red vertical lines only send a single photon each, for example at (3,0) and (4,-1) both objects explode. Both A and B receive these exact same photons without any delay in between them. So even though you could label them as being out of sync for B, B will still see them happen at the same time just like A? $\endgroup$
    – catmousedog
    Commented Jan 15 at 17:49
  • $\begingroup$ Yes, you are confusing two entirely separate issues. The two objects do not explode simultaneously in either frame. A and B see them simultaneously because A and B are at the same point in space when the photons arrive from the objects. $\endgroup$
    – Marco Ocram
    Commented Jan 15 at 20:25
  • $\begingroup$ I'm talking about what an observer actually sees. The plane of simultaneity predicts B sees a time desync, but if we use my diagram then B will always see the same desync as A. (They physically receive the same photons) If I draw two more photons coming from (4,0) and (3,1), then they will both still arrive without any additional desync at B. A and B receive these photons at different times because they are no longer at the same position. But they will still both see the objects explode at the "same" time, or rather with the same desync. $\endgroup$
    – catmousedog
    Commented Jan 16 at 1:08

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