# Confusion with the Lorentz transformation

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.

• 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? Commented Mar 4, 2020 at 3:03

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