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Let’s assume I measure time t in my rest frame - thus in a moving frame I have t’ corrected according to the Lorentz factor.

I have a moving primed frame such that t’= 0.5t for example.

I measure 1 second in my rest frame, which is observed as 0.5 second in my primed frame.

If I now consider my primed frame to be the rest frame, and my previous rest frame to now be my “primed” frame, and measure 0.5 seconds in my now rest frame, I get 0.25 seconds in my now moving frame. How does that work out? Shouldn’t I get my whole 1 second back?

It is safe to assume I could consider any possible frame of reference, so what detail is it that I’m missing?

Edit*: I have 2 pairs of frames:

Rest1: Let time be t Motion1: moving observer relative to Rest1 such that t’=.5t

Rest2: chosen such that Rest2 is at same speed as Motion1. Let time be T Motion2: chosen such that Motion2 is at same speed as Rest1. T’ = 0.5T

If t = 1, then t’=0.5 Since Rest2 and Motion 1 are at the same speed, then T = t’ = 0.5 Therefore T’ = 0.5*0.5 = 0.25

Motion2 and Rest1 are at the same speed, thus T’ = t, 1=0.25!?

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  • $\begingroup$ Joeseph123, while it's clear what "my rest frame" is, it isn't at all clear what "my primed frame" means or what "my now moving frame" is. Have you taken any time to look at the numerous questions (and their frequently asked duplicates) here regarding symmetrical time dilation for observers in relative uniform motion? $\endgroup$ – Alfred Centauri Sep 22 '19 at 0:14
  • $\begingroup$ primed as in the frame with the apostrophe’ like t’ or t prime in calculus. $\endgroup$ – Joeseph123 Sep 22 '19 at 0:18
  • $\begingroup$ All the questions I’ve seen about time dilation are not my question...I’m asking about how my understanding of Lorentz transformations is incorrect, when I haven’t violated any principles (as far as I understand.) This is Not the twin paradox either. $\endgroup$ – Joeseph123 Sep 22 '19 at 0:21
  • $\begingroup$ Alfred is implying that you can't be at rest in the unprimed & primed frames at the same time. Your thought experiment should have 2 different observers, one per frame. Otherwise, it gets confusing to think about, and to talk about. $\endgroup$ – PM 2Ring Sep 22 '19 at 0:35
  • $\begingroup$ Perhaps this analogy will help. Say you & I are facing each, 2m apart. We then both step 1m backwards, so we're now twice as far apart. So you now appear half as big to me, but I also appear half as big to you, not twice as big. $\endgroup$ – PM 2Ring Sep 22 '19 at 0:42
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I measure 1 second in my rest frame, which is observed as 0.5 second in my primed frame.

(1) this doesn't make much sense to me, and (2) it seems wrong in any sense I can think of.

Disregarding the unusual phrase "my primed frame", from any inertial reference frame moving relative to you, your 'wristwatch' runs slow. That is, any inertial observer will observe your 1 second elapsed time on your wristwatch to have taken longer than 1 second according to their system of synchronized clocks at rest.

Your question as written has so many problems that I believe I will regret posting this 'answer' (which is too long to be a comment).

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