How did we get the time dilation equation from this proper time equation?

Question

As you can see in the text book we get time dilation equation from equation number 1.9

I just can't see how he did it here:

For now, we define the proper time between events B and the origin to be the time ticked off by a clock which actually passes through both events. It is a directly measurable quantity, and it is closely related to the interval.

Let the clock be at rest in frame $\bar O$, so that the proper time $ \Delta \tau $ is the same as the coordinate time $\Delta \bar t $.

Then, since the clock is at rest in $\bar O $, we have:

$\Delta \bar x $ = $\Delta \bar y $ = $\Delta \bar z $ =0, (1.9) so:

$ \Delta S^2 $= $ -\Delta \bar t^2 $ =$ -\Delta \tau^2 $

The proper time is just the square root of the negative of the interval. By expressing the interval in terms of $O$ coordinates we get:

$\Delta \tau = {[(\Delta \tau^2)- (\Delta x^2) -(\Delta y^2)-(\Delta z^2)]}^{1/2}$

= $\Delta t {( 1- v^2)}^{1/2}$

This is the time dilation all over again.

Answer 1

Your thingy is traveling with velocity $v$. Which means that $$ \Delta \vec{r} = \vec{v} \Delta t,\quad \left| \Delta r \right| = v \Delta t $$ (the definition of velocity).

Square both sides: $$ \Delta r^2 = \Delta x^2 + \Delta y^2 + \Delta z^2 = v^2 \Delta t^2 $$

(the first equality is the 3d Pythagorean theorem). Substituting this into (1) will get you to (2).