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.

  • $\begingroup$ 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 $\endgroup$
    – user154420
    Jun 29 '17 at 8:59
  • $\begingroup$ Is there a typo in the second last equation? Also, have you considered factoring out a $\delta t ^2$ to get the desired result? $\endgroup$ 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: $$ \Delta r^2 = \Delta x^2 + \Delta y^2 + \Delta z^2 = v^2 \Delta t^2 $$

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

  • $\begingroup$ what is delta r @SolenodonParadoxus $\endgroup$ Jun 29 '17 at 6:14
  • 1
    $\begingroup$ Distance travelled is equal to velocity by time. $\endgroup$
    – user154420
    Jun 29 '17 at 6:24

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