0

I'll preface by saying I'm fairly new to physics - was an English major back in college :-)

I'm learning time dilation. As I understand it, if a rocket is travelling away and then to from Earth at velocity v, the amount of time experienced will be dilated by a factor $\gamma$, which is equal to $$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$

However, consider the following scenario. We have a really tall elevator going from Earth up into space. At time 0, we shoot the elevator upwards, at velocity $v$. At the same time, a car drops from the elevator at constant velocity $v$.

As I see it, the elevator car is stationary with respect to Earth. On the other hand, it seems like it experiences time dilation twice: first from the movement relative to the elevator, and then from the movement relative to the Earth. I understand that this is crazy, since it has to be stationary w.r.t the Earth, but it should have time dilation with respect to the flying elevator.

I'm sure the resolution to this lies in its net velocity w.r.t. the Earth being zero, but still don't understand how the time it experiences w.r.t the time measured on Earth doesn't have some $\gamma^2$ term.

Any help is appreciated, thanks!

Hank Lyensen
  • 1

1 Answers1

0

You are mixing 3 different reference frames here. Time dilation is how a specific observer sees the speed of the clock of another observer moving at v relative to him. If different observers look at a moving object A, but these observers are moving relative to each other, each one will see a different time dilation for the same object A, given by the equation you wrote (with a different speed for each observer). If the car is at rest relative to earth, it will experience no time dilation relative to earth, and will see that the rocket experiences the same time dilation than an observer at earth sees.

Pato Galmarini
  • 2,675