Will O' Clocks Show Different Times?

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When observing a moving frame O', an observer in frame O will see that clocks at different positions in O' generally show different times due to the effects of relativity. Clocks situated on a plane perpendicular to the direction of motion will appear synchronized to the observer in O, while those aligned with the motion will not. The concept of simultaneity is relative, meaning observers in O and O' will disagree on the synchronization of clocks. The finite speed of light also introduces time delays in observations, complicating the interpretation of what is "seen" versus what is "observed." Ultimately, the discussion emphasizes the importance of understanding the relativistic effects on time measurement between different frames.
  • #31
Suppose O' is a frame that observes an infinite grid with all the clocks synchronized to O' itself.

Now, O is a frame (me) who has a relative speed with respect to O' (thus all the other clocks of the grid as well)

How will my observation of the times be affected by my motion?
Specifically, suppose the origin (both space and time) was the event when O' and O touched each other

Assuming O is going towards the x-axis of the grid, what will the observed time of a general grid point be according to O?
 
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  • #32
aberrated said:
Suppose O' is a frame that observes an infinite grid with all the clocks synchronized to O' itself.

Now, O is a frame (me) who has a relative speed with respect to O' (thus all the other clocks of the grid as well)

Further let the x-axes of the two grids be parallel, likewise for the y- and z-axes, and let the relative speed v of O' with respect to O be in the +x direction.

How will my observation of the times be affected by my motion?
Specifically, suppose the origin (both space and time) was the event when O' and O touched each other

I assume you mean that the origins (x=y=z=0 points) of the two grids coincide at t=t'=0.

Then two clocks whose x'-coordinates differ by \Delta x' in O' (and which are synchronized in O'), will be observed in O to be out of synchronization by an amount

\Delta t' = \frac{v \Delta x'}{c^2}

with the clock that is in the "forward" position (with respect to the motion of the clocks in O), having a reading that is behind the other one.

So if the clocks are 1 light-second apart in O', and are moving at 0.5c (0.5 light-seconds per second) in O, then in O they will be observed to have readings that are 0.5 second apart, and decreasing as you proceed in the +x direction. All O' clocks with the same x' but different y' and z' will have the same reading in O.
 

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