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I want to show that ,for me at rest on earth, one clock in movement with speed V ticks slower. The proper time is the time of the clock at movement, so i do: $\Delta t=\gamma\Delta t'+\frac {V}{c^2}\gamma \Delta x' $ but $\Delta x'=0$ so its done... My question is: Why cant i convert from $\Delta t'=\gamma \Delta t -\frac{V}{c^2}\gamma \Delta x$ like this. In my conception the transformations work both ways.

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  • $\begingroup$ I'm not sure I understand your question. Are $\Delta t$ and $\Delta x$ supposed to be in your frame, while $\Delta t'$ and $\Delta x'$ are in the rest frame of the clock? $\endgroup$
    – J. Murray
    Commented Mar 4, 2020 at 3:03

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I cannot comment because of my low reputation hence I am writing this as an answer
I must say that in the latter transformation $$\Delta t'=\gamma \Delta t -\frac{V}{c^2}\gamma \Delta x$$ where $\Delta x=V \Delta t$ (The event describing the presence of observer in motion at time = $\Delta t$) this implies
$$\Delta t'=\gamma \Delta t -\frac{V^2}{c^2}\gamma \Delta t \implies \Delta t'=\gamma(1-\frac{V^2}{c^2}) \Delta t \implies \Delta t'=\frac{\Delta t}{\gamma} $$ which yields the same result

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  • $\begingroup$ I assume these were the notations you meant to use $\endgroup$
    – GKI
    Commented Mar 4, 2020 at 13:57

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