This is something that arose out of a section in Richard Mould's(adsbygoogle = window.adsbygoogle || []).push({}); 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 [itex]5:\frac{L}{\gamma v}[/itex] 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 [itex]5:\frac{L}{\gamma v}[/itex] when the moving rod's end B passes him. But according to him, it is because a rod of length [itex]\frac{L}{\gamma }[/itex] 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) seperated 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 [itex]5:\frac{L}{\gamma v}[/itex] when passing asteroid two, since there is no contraction of length involved.

I hope my question is clear.

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# Yet another question that is supposed to turn SR on its head

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