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

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.

• Hi, I changed your image to text and used mathjax to format it. Please feel free to edit any mistakes I might have made. If you are happy with the answer you got, as well as accepting it, you can also upvote it. Regards – user154420 Jun 29 '17 at 8:59
• Is there a typo in the second last equation? Also, have you considered factoring out a $\delta t ^2$ to get the desired result? – Rumplestillskin Jun 30 '17 at 0:55

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