Why t=1/(1-v^2/c^2)^(-0.5)?

Question

V: velocity of the object
t :time which is measured by a steady observer
c:velocity of light
^:sign used for showing power

Answer 1

It is a consequence of the statement that the speed of light must be constant for all observers. Imagine this "light-clock": a meter stick with mirrors on either end, and a pulse of light passing back and forth between them. A round trip between the mirrors is one tick of this clock.

Now imagine one of these clocks in motion (in the direction perpendicular to the length of the meter stick). According to someone "at rest," the path of the light of the moving stick over one tick of the moving stick is two sides of an isosceles triangle. These are necessarily longer than one meter apiece, and since light travels at a constant speed, the tick of the moving clock appears to last longer than the tick of a stationary clock. This is the apparent "time dilation". The moving clock appears to run more slowly, even though the clocks are identically constructed.

If you work out the time-of-flight, the as compared to the time-of-flight for a stationary tick, you get this formula.

Answer 2

Because the term √(1-v²/c²) is the measure of the time dilation at relativistic velocities. The same effect happens also to mass, and length is effected by the 'Lorentz-Fitzgerald' contraction in exactly the same way.


Doug

Answer 3

this formula needs correction:

delta t' =1/(1-v^2/c^2)^0.5 * delta t

where delta t and delta t' denote time intervals in the 2 coordinate systems

notice also the + sign of 0.5