- #1
Karl Coryat
- 104
- 3
[later edit: Sorry for the title -- I am glad to see this was interesting enough to spark discussion.]
Twin A takes off and leaves Twin B behind. Rather than switching on the reverse-thrust, slowing down, and beginning the journey back home (an acceleration that would be distinctly detectable onboard), Twin A has a close encounter with a star, which gravitationally slingshots the ship around the horn and back toward Twin B.
Of course, this acceleration breaks the symmetry of the situation for the twins. But, now Twin A's ship is undergoing a free-fall, weightless acceleration during the turn-around. As I understand it, if the ship were sealed shut, Twin A and his clocks and instruments could just as well think they were in an inertial frame for the entire trip -- only to find themselves back where they started, Twin B having aged more. How do the onboard clocks "know" that they are undergoing an acceleration or are in a gravitational well during the turn-around, so as to tick more slowly than if the star weren't there at all?
I'm sure this has a simple solution, but it isn't coming to me. Thanks for your help.
Twin A takes off and leaves Twin B behind. Rather than switching on the reverse-thrust, slowing down, and beginning the journey back home (an acceleration that would be distinctly detectable onboard), Twin A has a close encounter with a star, which gravitationally slingshots the ship around the horn and back toward Twin B.
Of course, this acceleration breaks the symmetry of the situation for the twins. But, now Twin A's ship is undergoing a free-fall, weightless acceleration during the turn-around. As I understand it, if the ship were sealed shut, Twin A and his clocks and instruments could just as well think they were in an inertial frame for the entire trip -- only to find themselves back where they started, Twin B having aged more. How do the onboard clocks "know" that they are undergoing an acceleration or are in a gravitational well during the turn-around, so as to tick more slowly than if the star weren't there at all?
I'm sure this has a simple solution, but it isn't coming to me. Thanks for your help.
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