How to prove time dilatation from the Lorentz transform formula: $$ t' = \gamma\left(t-\frac{Ux}{c^2}\right) $$ (U: the velocity of the referential R' relative to R)
So far I've found this formula : $$ \Delta t' = \gamma\left(\Delta t-\frac{U\Delta x}{c^2}\right) $$
but I don't know how to handle the $ \Delta x $ from here.
I have seen in the literature that $ \Delta t' = \frac{\Delta t}{\gamma} $
but I clearly don't know how to infer this from the Lorentz Transform.
T.I.A.
The starting point is to be clear on what time dilat(at)ion means. Here it is...
The time interval between two events as found in an inertial frame of reference where the events occur in different places is greater than the time interval ($T$, say) in the inertial frame where they occur in the same place.
We assume that frames of reference are equipped with synchronised clocks everywhere, so that the time of an event can be registered at the place in the frame where it occurs.
With the Lorentz transform equations that you've quoted, it's easiest to use the undashed (unprimed) frame as the one in which the events occur in the same place and at time $T$ apart, and to make substitutions accordingly, remembering that $\Delta x$ is the spatial separation of the events.
The dilated value in the dashed frame of the time between the events follows almost immediately – remembering that 𝛾 > 1.
In the formula$$\Delta t'=\gamma\left(\Delta t-\frac{U\Delta x}{c^{2}}\right)$$ we suppose that two events occur in the same place, i.e.$\;\Delta x=0$
we find in the moving reference frame $\mathcal{R}'$ $$\Delta t'=\gamma\Delta t$$
The inverse Lorentz transformation gives $$\Delta t=\gamma\left(\Delta t'+\frac{U\Delta x'}{c^{2}}\right)$$ if in $\mathcal{R}'$ two events occur in the same place , i.e.$\;\Delta x'=0$ we have $$\Delta t=\gamma\Delta t'$$ or $$\Delta t'=\frac{\Delta t}{\gamma}$$
