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neutrino
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This is something that arose out of a section in Richard Mould's Basic Relativity.
He begins SR with his so-called "Physical Threorms (PT)," which are gedanken experiments used to show the effects time-dilation, length-contraction, and the concept of simultaneity in relativity, in that order.
I'll give a gist of his second PT: An observer, in whose frame a rod of length L is at rest, obserevs a clock that reads 5:00 as it passes the left end of the rod (event A) with velocity v to the right. By the time (say, an hour) it reaches the other end, the moving clock reads 5:\frac{L}{\gamma v} due to time dilation(event B). Both observers must agree on the "facts" (events A and B).
Now, the second observer agrees that his clock does read 5:\frac{L}{\gamma v} when the moving rod's end B passes him. But according to him, it is because a rod of length \frac{L}{\gamma } is moving with a velocity v to his right.
Mould does not talk about how the first observer's clock appears to the second.
Now here's my question: If I were to replace the rod with two asteroids (assumed to be moving at the same uniform velocity) separated by a distance L, with event A and B corresponding to passing asteroid 1 and 2, respectively. How would I explain the fact that both observers agree that the second observer's clock reads 5:\frac{L}{\gamma v} when passing asteroid two, since there is no contraction of length involved.
I hope my question is clear.
He begins SR with his so-called "Physical Threorms (PT)," which are gedanken experiments used to show the effects time-dilation, length-contraction, and the concept of simultaneity in relativity, in that order.
I'll give a gist of his second PT: An observer, in whose frame a rod of length L is at rest, obserevs a clock that reads 5:00 as it passes the left end of the rod (event A) with velocity v to the right. By the time (say, an hour) it reaches the other end, the moving clock reads 5:\frac{L}{\gamma v} due to time dilation(event B). Both observers must agree on the "facts" (events A and B).
Now, the second observer agrees that his clock does read 5:\frac{L}{\gamma v} when the moving rod's end B passes him. But according to him, it is because a rod of length \frac{L}{\gamma } is moving with a velocity v to his right.
Mould does not talk about how the first observer's clock appears to the second.
Now here's my question: If I were to replace the rod with two asteroids (assumed to be moving at the same uniform velocity) separated by a distance L, with event A and B corresponding to passing asteroid 1 and 2, respectively. How would I explain the fact that both observers agree that the second observer's clock reads 5:\frac{L}{\gamma v} when passing asteroid two, since there is no contraction of length involved.
I hope my question is clear.
