Question

Consider a mirror clock and a light pulse is released inside of the clock. The clock is then move by someone at speed v. Shows that time dilation, t’, in a moving frame as viewed by someone at rest as

t' = t/√(1-v^2/c^2)

Answer 1

If the clock doesn't move, the time required to pass the distance L is

$$t=2L/c. \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(1)$$

However, if they are moving with constant speed v, the time will increase because the distance will increase, instead of 2L the distance becomes

$$d=2\sqrt{L^2+(vt'/2)^2},$$

and for the light the time required to pass that distance increases to t', which we will find below:

$$t'=d/c=\frac{2}{c}\sqrt{L^2+(vt'/2)^2},\\ \space\\ t'^2=4L^2/c^2+v^2t'^2/c^2.$$

Make substitution using equation (1):

$$t'^2=t^2+v^2t'^2/c^2,\\ t'=\frac{t}{\sqrt{1-v^2/c^2}}.$$

That's it.