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I was just trying to calculate the time dilation for two sets of events using the Lorentz-Transformation. (I'm trying to understad the special theory of relativity at the moment - I mean what better could you do in this time of isolation?) But in one case I got the result $\Delta t'=\gamma\Delta t$ and in the other case the result $\Delta t'=\frac{\Delta t}{\gamma}$. And I don't know what's going on. Is this correct and I just need to understand how to interpret it or did I make a mistake?

So here are the two sets of events. Of course we have two observers $O$ and $O'$ witht heir frames of reference $S$ and $S'$ and relative to $O$ the observer $O'$ is moving along the x-axis with the velocity $v$.

For the first set of events $E_1$ and $E_2$ I assumed that for $O$ they happen on the same spot, i.e. in $S$ we have that $x_1=x_2$. I wanted to calculate $\Delta t'=t_2'-t_1'$. I just used the Lorentz-Transformation $t'=\gamma(t-\frac{v}{c^2}x)$ to substitute $t_2'$ and $t_1'$ and got the result $\Delta t'=\gamma\Delta t$. Straight forward enough I thought.

$\Delta t'=\gamma(t_2-\frac{v}{c^2}x_1)-\gamma(t_1-\frac{v}{c^2}x_1)$ (note that $x_1=x_2$)

$\Delta t'=\gamma(t_2-\frac{v}{c^2}x_1-t_1+\frac{v}{c^2}x_1)$

$\Delta t'=\gamma(t_2-t_1)$

$\Delta t'=\gamma\Delta t$

Next I took a different sets of events $E_3$ and $E_4$ for which I assumed that they both happened where the observer $O'$ was (at different times). That means for example that $x_3'=x_4'=0$ and more importantly that $x_3=vt_3$ and that $x_4=vt_4$. Again I substituted these fatcs and the Lorentz-transformation. But this time I got the seemingly conflicting result $\Delta t'=\frac{\Delta t}{\gamma}$.

$\Delta t'=\gamma(t_4-\frac{v}{c^2}vt_4)-\gamma(t_3-\frac{v}{c^2}vt_3)$

$\Delta t'=\gamma(t_4-\frac{v^2}{c^2}t_4-t_3+\frac{v^2}{c^2}t_3)$

$\Delta t'=\gamma(t_4(1-\frac{v^2}{c^2})-t_3(1-\frac{v^2}{c^2}))$

$\Delta t'=\frac{(1-\frac{v^2}{c^2})}{\sqrt{1-\frac{v^2}{c^2}}}(t_4-t_3)$

$\Delta t'=\frac{(1-\frac{v^2}{c^2})\sqrt{1-\frac{v^2}{c^2}}}{(1-\frac{v^2}{c^2})}(t_4-t_3)$

$\Delta t'=\sqrt{1-\frac{v^2}{c^2}}\Delta t$

$\Delta t'=\frac{\Delta t}{\gamma}$

Am I to believe then that $\Delta t'=\gamma\Delta t=\frac{\Delta t}{\gamma}$? I think not! But what is going on? I'd appreciate any help I can get.

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There is no contradiction, because $\Delta t$ and $\Delta t'$ mean different things in different situations! In the first case $\Delta t$ is the time elapsed in the frame where the events happen at the same place, while in the second case that would be $\Delta t'$. And the results are perfectly consistent: in both cases, the time measured by an observer that is moving with respect to the events is $\gamma$ times the time measured by an observer for whom the events happen at the same position.

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