Time dilation formula and example What is the time dilation formula by constant velocity?
Is the right formula 
$$t'=t\times \sqrt{1-v^2/c^2}$$ or is it 
$$t'=\frac{t}{\sqrt{1-v^2/c^2}}$$ ?
And can someone show me a computational example regarding the time dilation formula?
Thank you in advance.
 A: In words, remember this:  the relatively moving clock 'ticks' slower.
So, if $t$ denotes the elapsed time according to your clock (which is at rest with respect to you), and $t'$ denotes the elapsed time according to a relatively moving clock, $t > t'$; your clock shows a greater elapsed time that the relatively moving clock.
Since $\sqrt{1 - \frac{v^2}{c^2}}$ is less than one for $v > 0$ (with $v$ the relative speed), the time dilation formula must be (for the chosen notation) 
$$t' = t\sqrt{1 - \frac{v^2}{c^2}} $$
So, for example, if your clock shows an elapsed time of $1s$, a clock relatively moving with a speed $0.5c$ will show an elapsed time of
$$\sqrt{1 - \frac{1}{4}}  = 0.866s$$
Now, you are free to choose the notation by which you keep track of the elapsed times according to two relatively moving clocks.  So, to keep things straight, remember the relatively moving clock 'ticks' slower.
This is, by the way, perfectly symmetrical.  To an observer at rest with the primed clock, it is your clock that 'ticks' slower.
A: Suppose you see a moving spaceship. For clarity we use the notation:


*

*$t'$ is the time in the moving frame (in the spaceship).

*$t $ is the time in the stationary frame (on Earth).


When the two clocks are compared one will find that the Earth has aged more than the spaceship. Thus we always have, $t>t'$. Since $1/\sqrt{1-v^2/c^2} >1$ for any nonzero $v$, we must have,
\begin{equation}
 t = \frac{t'}{ \sqrt{1-v^2/c^2}}
\end{equation}
A computational example is very straightfoward. Suppose a ship is moving at $v=4c/5$ then 
\begin{equation}
 t = \frac{t'}{ \sqrt{9/25}} = \frac{5}{3}t'
\end{equation}
So one second in the spaceship will correspond to $1.67s$ on Earth.
A: It depends on which inertial frame you're referring to - there isn't one stationary frame and one moving frame, each inertial frame is moving relative to the other (I think the notation of $t$ and $t'$ is often confused for this reason and will use something a bit more verbose).
Specifically, if you are stationary in the first frame, $S (x, t)$, then the clock of the other inertial frame $S' (x', t')$ will be moving away from you and ticking slowly, so that for every one tick of your own clock in $S$, there is less than one tick of the moving clock from $S'$:
$1\ tick\ of\ stationary\ clock\ in\ S = \sqrt{1-v^2/c^2} \times 1\ tick\ of\ moving\ clock\ from\ S'$
On the other hand, if you are stationary in the other inertial frame, $S'$, then the opposite will be true - the clock of the other inertial frame $S$ will be moving away from you and ticking slowly, so that for every one tick of your own clock in $S'$, there is less than one tick of the moving clock from $S$:
$1\ tick\ of\ stationary\ clock\ in\ S' = \sqrt{1-v^2/c^2} \times 1\ tick\ of\ moving\ clock\ from\ S$
As an example of using the equation, if $v = \frac{\sqrt{3}}{2}c$, then $\sqrt{1-v^2/c^2} = \sqrt{1-3/4} = \frac{1}{2}$, so that you have two ticks of the stationary clock for every one tick of the moving clock.
Regarding understanding the equation, I think that's best done with interactive visuals, which I can't include here (SVG's with animations), but have tried to do so with the geometry of time-dilation, hope it's helpful.
A: Which is which is always confusing.  I like the t>t' basis to keep things straight where t' is the elapsed time in the moving frame and t the elapsed time in the "stationary" frame.
I cannot do the math, but it has been shown that if a rocket ship blasts to a point X in the universe and returns to Earth at near the speed of light,it is always the elapsed time on Earth which is greater, even if you look at it from the point of view of the rocket ship.  That's because the stationary frame of reference is the initial frame of the rocket ship going out.  When the rocket ship "turns around" and goes back to Earth, the relativistic speeds of the still receding Earth and the double speed of the rocket ship catching up to Earth are so great that even though initially the time elapses more quickly on the rocket ship relative to Earth, the time on the rocket ship when it turns around slows down so much that the elapsed time on Earth catches up and surpasses the elapsed time on the rocket ship as if the original frame of reference was that of Earth.
