0
$\begingroup$

I'm trying to get my head around relativity and time dilation. What I can't figure out is how time can dilate for fast-moving observers if all velocity is relative. Here's a scenario:

Alice and Bob are both standing on the surface of a planet. Bob knows the planet is orbiting its star at 107,000 kph, so he gets into a rocket and accelerates to 107,000 kph in the opposite direction. He reasons that since he's actually decelerating to 0, time will start moving faster for him. But the star is moving through the Milky Way at 828,000 kph in the opposite direction to the planet's orbit, so he's actually speeding up. But the Milky Way is flying towards the Great Attractor at 2 million kph in the other direction, so he's actually slowing down.

Alice and Bob send each other light pulses to see whose clock is moving faster. Surely either Alice or Bob will experience time dilation relative to the other; what I can't figure out is which one. Alice thinks Bob accelerated to 107,000 kph and flew away, while Bob insists he slowed to a stop. The Milky Way thinks they're both wrong. Whose clock moves slower, and why?

Improve this question
$\endgroup$
2
  • $\begingroup$ You need to precisely define in terms of physics what it means to "know" and "decide". Until then it is meaningless. Reason with actual established physics principles and not colloquial language. If you start from false premises you will very well derive contradictions from them. $\endgroup$
    – Vincent Thacker
    Commented May 27 at 20:30
  • $\begingroup$ @VincentThacker edited. $\endgroup$
    – Justin Morgan
    Commented May 28 at 2:25

1 Answer 1

Reset to default
-1
$\begingroup$

If kinetic energy has been stored in Bob, then Bob's aging has decreased.

How to decide if kinetic energy has been stored in Bob? Imagine a flywheel on which Bob is attached. If Bob's motion was such that the wheel was spinning, then kinetic energy was stored in the wheel, of which Bob was a part.

The above works well, if Bob makes one or more rounds, otherwise not so well.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.