- #1

Alexroma

- 34

- 0

Imagine a spaceship coming from planet X to Earth with velocity

*V*. The time during which the spaceship can be observed from Earth equals

*t – t*, where

_{c}*t*is the travel time of the spaceship (the time it takes for the spaceship to come to Earth from planet X), and

*t*is the time it takes light to come to Earth from planet X; this expression of the observational time is explained by the fact that the spaceship is following behind its own light.

_{c}Alternatively, imagine a spaceship going from Earth to planet X with the same velocity. In this case, the time during which the spaceship can be observed from Earth equals

*t + t*, as it takes

_{c}*t*for the spaceship to go to planet X, plus it takes

*t*for the light to convey the image of its landing at that planet to observer on Earth.

_{c}So, we have two different observational times for incoming and receding objects travelling the same distance with the same velocity. In order to reconcile the difference between these two observational times, we determine the mean observational time (

*t*) as the geometric mean of them:

_{m}tm = √ (t – t

_{c})(t + t

_{c}) = √ t

^{2}– t

_{c}

^{2}

To find out the factor for time dilation, we divide the mean observational time by travel time:

t

_{m}/t = √ 1 – t

_{c}

^{2}/t

^{2}= √ 1 – V

^{2}/C

^{2}

Therefore, in result we have the same factor for the time dilation as the factor used in the special theory of relativity, i.e. the reciprocal of the Lorentz factor (√ 1 – V

^{2}/C

^{2}).