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Over the years I have seen this image which always confused me: Clock lattice diagram

(from Wikipedia Spacetime)

"In special relativity, the observer measures events against an infinite latticework of synchronized clocks." This sounds needlessly artificial and abstract.

Let me take a stab at what they are trying to say. Say the event is a star exploding which is 1 billion light years away. The light reaches my eyeballs 1 billion years after the event happened. Therefore, one of these synchronized clocks mis-times the event. Do these clocks account for the time taken for the light from events to reach our eyeballs?

In the exact same frame, I could have an observer 1 billion light years away right next to the star when it exploded registering a "more correct" time.

So the observers disagree on the timing of the events, even though they are in the same frame? Or does the far-away observer back-calculate the time of the event, knowing the distance to the star, and then they agree?

Now that I think about it, are all these clocks actually out of synch, but appear to be in synch by the time the light from all of them has reached the eyeballs of our observer?

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    $\begingroup$ The clocks are in synch and the observers are smart enough to correct for the travel time of light, so they all agree on when the explosion happened. $\endgroup$
    – WillO
    Aug 18 '21 at 4:47
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    $\begingroup$ They are in synch. $\endgroup$
    – WillO
    Aug 18 '21 at 4:50
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    $\begingroup$ I gave you the same answer three times. You keep insisting that it's wrong. If you're so sure of the answer, why did you ask the question? $\endgroup$
    – WillO
    Aug 18 '21 at 5:08
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    $\begingroup$ The fact that the clocks are in sync does not depend on you receiving photons from those clocks with your eyeballs. $\endgroup$ Aug 18 '21 at 5:20
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    $\begingroup$ The clocks are all in synch in the God-like perspective. So in the mortal perspective, if you look at a clock 10 light-seconds away, it's lagging by 10 seconds compared to the clock next to you. $\endgroup$
    – PM 2Ring
    Aug 18 '21 at 7:29
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All it means is that in your stationary frame you have a plane of simultaneity, so if it is 2:37pm where you are it is 2:37pm everywhere in your frame of reference.

For example, if it is 2:37pm where you are 'now' it is also 2:37pm 'now' on Jupiter, which is about 33 light minutes away. If a light from Jupiter arrives at you now at 2:37, then in your frame of reference it must have left Jupiter 33 minutes ago at 2:04pm.

So all it really means is that in your intertial rest frame it is the same time everywhere.

To take your exploding star example, if it is 2:37pm on August 18th 2021 when you see the explosion, then it is also 'now' the same time and date a billion light years away, and the star would have exploded a billion years earlier.

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  • $\begingroup$ Thank you. "If a light from Jupiter arrives at you now at 2:37, then in your frame of reference it must have left Jupiter 33 minutes ago at 2:04pm." And so one of the synchronized clocks located next to Jupiter to me on earth appears to read out 2:04pm, right? But if I were to be able to teleport instantaneously to that clock on Jupiter, it would read 2:37pm, correct? I understand now that the clocks in the Wikipedia diagram are not a display of what the observer sees. $\endgroup$
    – HelpMe
    Aug 18 '21 at 14:55
  • $\begingroup$ Indeed. If you were to look at the clock on Jupiter through a telescope, you would see it as reading 2:04 because the image you are seeing left the clock 33 minutes earlier. But the current time on the Jupiter clock would be 2:37, the same as your local clock. $\endgroup$ Aug 18 '21 at 15:08
  • $\begingroup$ Thanks, Marco!! $\endgroup$
    – HelpMe
    Aug 18 '21 at 15:09
  • $\begingroup$ My pleasure. Most of the paradoxes of SR and most of the questions about it on this site arise where the relativity of simultaneity hasn't been taken into account properly. The physics teaching community should really have a think about how to fix that. $\endgroup$ Aug 18 '21 at 15:24
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    $\begingroup$ @JEB I mean 'fix it' in the sense of placing more emphasis on the teaching of the relativity of simultaneity and less on time dilation. $\endgroup$ Aug 18 '21 at 19:29
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"In special relativity, the observer measures events against an infinite latticework of synchronized clocks." This sounds needlessly artificial and abstract.

It IS artificial and abstract, but it is far from needless. It serves a very good purpose to think of it along those lines.

" Therefore, one of these synchronized clocks mis-times the event. Do these clocks account for the time taken for the light from events to reach our eyeballs? "

No. The clocks do not account for time taken for light to reach eyeballs. Because, the clocks are not concerned with when the light from the event reached your eyeball. The clocks are concerned with when the event OCCURED , IN YOUR FRAME.

" In the exact same frame, I could have an observer 1 billion light years away right next to the star when it exploded registering a "more correct" time. "

In the same frame, all the clocks in the lattice would display the exact same time, when that event occurs. For an observer , 1 billion light years away, he would have his OWN lattice of clocks, and that lattice would also display some time on those clocks when that event happens, IN HIS FRAME. The time displayed on his clocks when the event happens in his frame, would NOT necessarily be the same as the time displayed on your lattice of clocks when the event happens in your frame.

" So the observers disagree on the timing of the events, even though they are in the same frame? Or does the far-away observer back-calculate the time of the event, knowing the distance to the star, and then they agree?"

Whether the observers clocks show the same time or not, depend on the specifics of the example. Specifically, whether their clocks were synced with each other some time in the past, and what were the sequence of " their motions " from the event of that syncing of clocks till now.

" are all these clocks actually out of synch, but appear to be in synch by the time the light from all of them has reached the eyeballs of our observer? "

No, in your frame, your lattice of clocks are always in sync. That is the premise of this system of infinite lattice of clocks. They are all in sync in your frame.

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  • $\begingroup$ Awesome, so you agree that the diagram is misleading in that it does not display what an observer would see when he looks out at his infinite lattice of synchronized clocks, looking at some clocks far away. Rather, it is the time you would read if you could teleport instantaneously to an arbitrary clock in his frame. $\endgroup$
    – HelpMe
    Aug 18 '21 at 14:49
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    $\begingroup$ @HelpMe I would not say that the diagram is misleading. Rather , i would say that you misinterpreted the diagram. The diagram never claims to show what an observer would "see" as the reading on the faraway clock. It always means what the observer would read if he could teleport instantaneously to the faraway clock. Since, teleportation is not real, the more accurate way to put it would be, the diagram shows what is the time at the faraway clock in the frame of the observer. $\endgroup$ Aug 18 '21 at 15:21
  • $\begingroup$ Thank you for clarifying that for me. $\endgroup$
    – HelpMe
    Aug 18 '21 at 16:47
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Now that I think about it, are all these clocks actually out of synch, but appear to be in synch by the time the light from all of them has reached the eyeballs of our observer?

The opposite is true. Here's how you synchronize two clocks at positions $\mathbf x_A$ and $\mathbf x_B$.

  1. At time $t_1$ (as measured on clock $A$), a light pulse is emitted from the point $\mathbf x_A$ toward the point $\mathbf x_B$.
  2. At time $t_2$ (as measured on clock $B$), the pulse is received at $\mathbf x_B$ and sent back toward $\mathbf x_A$.
  3. At time $t_3$ (as measured on clock $A$), the reflected pulse arrives back at $\mathbf x_A$.

If $t_2 = (t_1+t_3)/2$, then we say the clocks at $A$ and $B$ are synchronized. Keeping this procedure in mind, it should be clear that an observer looking out at the lattice of clocks would see (with her eyes) the more distant clocks to be lagging because of the finite speed of light delay; however, she could compensate for this by mentally adding $d/c$ to each clock time (where $d$ is the distance to the clock) and if she did so, all of these naive time-of-flight discrepancies would disappear.


"In special relativity, the observer measures events against an infinite latticework of synchronized clocks." This sounds needlessly artificial and abstract.

The crucial point is that when we specify the time at which some event occurs, then that time should be understood as being measured on a clock at the location of the event. For example, when we say "a firecracker exploded at location $\mathbf x$ at 2:30 PM," what we mean is that a clock sitting at $\mathbf x$ read 2:30 PM when the firecracker exploded.

In Newtonian physics, we implicitly label events with a location and time, but that time is not localized to any particular position - we simply say that it is the time. As a result, we mathematically equate an observer's reference frame with a lattice of measuring sticks (against which one measures position) and a single clock which we could imagine sitting on the observer's wrist. In contrast, the perspective presented in special relativity is that we should make the idea of time a local one, by placing a clock at each lattice point.

Though it may seem artificial and unnecessary, the distinction is shown to be crucial by a straightforward consequence of the postulates of special relativity (specifically the constancy of the speed of light). Imagine two observers - each with their own synchronized lattice of clocks - at the same point, but with one moving relative to the other. The clocks at their respective coordinate origins will agree, but the more distant clocks will not.

More concretely, consider an observer who measures a firecracker exploding at position $\mathbf x = (10\ \mathrm m, 0, 0)$ at time $t_\mathbf x = 0$. For clarity, I mean that the light from the firecracker arrived at the origin $\mathcal O$ at time $t_\mathcal O = (10\ \mathrm m) / c$, and the appropriate time-of-flight was compensated for. Then - by straightforward consequence of the postulates of relativity - a second observer at the position as the first, and whose clock (at the origin) agrees with that of the first, but who is moving with velocity $\mathbf v = v\hat x$ relative to the first would measures the time at which the firecracker explodes to be $t'_\mathbf x=-\gamma vx/c^2$.

As a result, it is strictly necessary to understand time as a local phenomenon as a consequence of the postulates of special relativity. The time at which a distant event occurs is entirely dependent on the reference frame of the observer and as such is not a meaningful notion in the absence of further context; that is, the notion of a single, global time is simply incompatible with the constancy of the speed of light.

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  • $\begingroup$ Thank you, that was very helpful. $\endgroup$
    – HelpMe
    Aug 18 '21 at 16:51
  • $\begingroup$ @HelpMe BTW, the procedure mentioned at the start of this answer is known as Einstein synchronisation. $\endgroup$
    – PM 2Ring
    Aug 19 '21 at 0:03
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"In special relativity, the observer measures events against an infinite latticework of synchronized clocks."

A more precise statement is that in special relativity, spacetime is modeled as if there were an infinite latticework of clocks. The clocks don't actually exist, and also the term "synchronized" is problematic. Synchronicity is defined in terms of these imaginary clocks, so it's a bit circular to assert that they are synchronous.

Say the event is a star exploding which is 1 billion light years away. The light reaches my eyeballs 1 billion years after the event happened. Therefore, one of these synchronized clocks mis-times the event.

In relativity, "event" refers to a point in four-dimensional spacetime. The idea with the synchronous clocks is that the time of an event is defined by the time shown by the clock at that event. The star exploding, and the light from the star's explosion reaching your eyeballs, are two different events. The clock next to the star correctly times the explosion, and the clock next to your eyeballs correctly times the light reaching you.

Compare it a video that has an on-screen display of the time. This is a record of the time when that moment of the video was recorded. When you actually watch it, the time will be different from the current time, but that doesn't mean the clock on the video mistimed the events. It correctly recorded the time at the time of the events. Similarly, if there were a lattice of clocks, then when you see the star exploding, you would see a clock next to it, and that clock would correctly record the time of the explosion.

Or does the far-away observer back-calculate the time of the event, knowing the distance to the star, and then they agree?

If the clocks were actually there, we would not have to back-calculate the time. We could just look at the clock and see what time the event was. But since we don't have the clocks, we calculate what the clock would say if it were there.

Now that I think about it, are all these clocks actually out of synch, but appear to be in synch by the time the light from all of them has reached the eyeballs of our observer?

There is no objective basis for whether the clocks are in synch or not. The clocks are simply a mental model for the time coordinate. We assign each event a time coordinate, and then we imagine a clock there showing that time coordinate. Someone in a different reference frame would have different times on their imaginary clocks. We could have a coordinate system such that by time the light reaches us, they are showing the same time as a clock next to us. That is an anisotropic coordinate system, and it violates the Copernican principle, but it is a valid coordinate system.

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The simple answer is this:

  1. If the observers are in the SAME frame of reference, then they will see the event at the exact SAME time (even when they are a billion light years apart).
  2. If the observers are in a DIFFERENT frame of reference, then they will see the event at DIFFERENT times.

Notes:

  • This is a thought experiment. In reality it is nearly impossible to be a billion light years apart and still be in the same frame of reference because galaxies are constantly moving.
  • The universe is expanding, which means that the same frame of reference at these large scales are not possible.
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