- #1
Nantes
- 54
- 5
Hi guys! I'm a pharmacist who has been trying to understand how time dilation and Lorentz contraction and etc. work, out of pure curiosity. I have been reading a course by Michael Fowler, which I link the 4th section of: http://galileoandeinstein.physics.virginia.edu/lectures/sreltwins.html
In this section, he presents the problem of twins, one of which travels to Alpha Centauri (which is 4 light-years away) and back at relativistic speeds, and comes back younger than the other twin. It starts from the "How to give twins very different birthdays" part, in case you guys want to read the whole problem.
I'm puzzled by this declaration in bold in "What does he see?":
I can understand that, because at the moment the female twin in the spaceship turns around, the visual information of her turning around will take 4 years to reach Earth, during which time she will have traveled 60% of the distance this light traveled (since her speed is 60% that of light) and thus be 60% of the way back home already at the moment her earthbound twin sees her turning around. But then, suppose that, from the moment he sees her turning around, he doesn't do anything else but watch her coming back. At some point, when he's still seeing her part way towards Earth, her real self will reach the planet (since she's ahead of the light by a good margin). So he'd be able to meet with his sister face to face, but if he looked through his telescope, he would still be seeing her as being inside the spaceship on her way back... but that sounds absurd! Where is the flaw in my reasoning?
Thanks!
In this section, he presents the problem of twins, one of which travels to Alpha Centauri (which is 4 light-years away) and back at relativistic speeds, and comes back younger than the other twin. It starts from the "How to give twins very different birthdays" part, in case you guys want to read the whole problem.
I'm puzzled by this declaration in bold in "What does he see?":
"To him, her outward journey of 4 light years’ distance at a speed of 0.6c takes her 4/0.6 years, or 80 months. BUT he doesn’t see her turn around until 4 years later, because of the time light takes to get back to Earth from alpha-centauri! In other words, he will actually see her aging at half his rate for 80 + 48 = 128 months, during which time he will see 64 flashes.
When he sees his sister turn around, she is already more than half way back!"
I can understand that, because at the moment the female twin in the spaceship turns around, the visual information of her turning around will take 4 years to reach Earth, during which time she will have traveled 60% of the distance this light traveled (since her speed is 60% that of light) and thus be 60% of the way back home already at the moment her earthbound twin sees her turning around. But then, suppose that, from the moment he sees her turning around, he doesn't do anything else but watch her coming back. At some point, when he's still seeing her part way towards Earth, her real self will reach the planet (since she's ahead of the light by a good margin). So he'd be able to meet with his sister face to face, but if he looked through his telescope, he would still be seeing her as being inside the spaceship on her way back... but that sounds absurd! Where is the flaw in my reasoning?
Thanks!