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So I look at the Lorentz transformation in the usual one dimensional case. Let's say the transformations are:

$$t' = \gamma(t-vx/c^2) \\ x'=\gamma(x-vt)$$

If I want to "track" the origin of the prime frame in terms of non-primed coordinates, then I need to ensure that $x=vt$ which gives me $t'=t/\gamma$. We can see that the primed clock is slowing down.

However, if I want to target some fixed non-zero location in the primed frame, say $x_0'$, then I need to have $x=vt+x_0'/\gamma$. That means the transformed time of the clock standing at the $x_0'$ position in primed frame has reading $$ t'=t/\gamma - \frac{v}{c^2\gamma^2}x_0'$$

There could be some further simplifications probably. But why is it that, depending on the fixed position in the primed frame, the time in the unprimed frame has a different offset?

I understand this has no affect on the time "ticking rate" of the primed frame, from the POV of the unprimed frame, because the ticking rate is not affected by the time offset.

In some sense, if you freeze time "t" in the unprimed frame, and "walk" around the universe, you would actually be moving through time in the primed frame!

I drew a x-t plot and understand that this non-uniform (in space) offsetting of $t'$ helps to keep the $x=ct$ line invariant in both frames. But is there a more intuitive physics explanation?

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  • $\begingroup$ don't you mean $x'_0$? $\endgroup$ – JEB Jun 5 at 4:20
  • $\begingroup$ Yes. Thanks, fixed. $\endgroup$ – Shuheng Zheng Jun 5 at 5:52
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"In some sense, if you freeze time "t" in the unprimed frame, and "walk" around the universe, you would actually be moving through time in the primed frame!"

You have already stated the intuitive reason: simultaneity is relative, so by shifting $x'_0$ (I refuse to call an $x'$-coordinate $x$-anything without a prime) at fixed $t'=0$, the time in the $x$-frame changes. Or, moving observers have different hyperplanes of simultaneity, which in layman's terms is: their definitions of "now" differ, and the size of disagreement depends on position.

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  • $\begingroup$ I always thought the relativity of simultaneity is due to the speed of light taking some time before information could reach the observer. $\endgroup$ – Shuheng Zheng Jun 5 at 6:11
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    $\begingroup$ No, light delay is an additional effect (on top of the Lorentz transform), and it is necessary to take that delay into account to predict what you actually see. $\endgroup$ – m4r35n357 Jun 5 at 9:29
  • $\begingroup$ @Shuheng Check out this gif anim from Wikipedia. $\endgroup$ – PM 2Ring Jun 5 at 12:12
  • $\begingroup$ Is there a more direct way to conclude, at least heuristically, why relativity of simultaneousness is a consequence of the postulates of special relativity? $\endgroup$ – Shuheng Zheng Jun 5 at 13:20
  • $\begingroup$ The lightning flash at 2 ends of the moving train. If they are simultaneously on the platform, the one at the front of the car is 1st on the train. Now we use light propagation to derive this, but you can do the experiment with synchronized observers spaced every foot along the track, and similarly inside the car. They then compare there notes after it's all done: the track people all conclude simultaneous flashes, and all train observers say the from strike was first, even though all didn't see it 1st. $\endgroup$ – JEB Jun 6 at 2:55
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Okay, I actually read your whole post and I understand your quandary. Even if you were to make static the universe,( even freeze time) you will move thru time as you browse around in that frozen universe. That one will take some more thought.

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