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## about time measurement vs measurement of movement

 Quote by DaleSpam I'm sorry, but I don't understand how the concept you describe here differs from Lorentz's aether theory. Can you explain a bit (preferably avoiding gravity if possible so that we can stick to SR)?
The gravity example is the easiest because it is true (which helps to confuse crackpots), whereas the argument's I've heard from crackpots about SR are false. I've heard crackpots argue that there is something about motion that makes certain types of clocks tick slower than others. Ie, if you sent a different type of clock up in a GPS satellite, it might act differently. This differs from Lorentz aether theory (or relativity) because in reality all clocks tick at the same rate (in their own frame) regardless of their speed relative to another object.

I say this only because the OP's line of questioning seemed to be going in that direction - implying to me that if speed affects clocks, that it might affect different clocks differently.

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 Quote by russ_watters The gravity example is the easiest because it is true (which helps to confuse crackpots), whereas the argument's I've heard from crackpots about SR are false.
A pendulum clock on the earth does tick at a different rate than a pendulum clock on the moon, but a pendulum clock at rest on the surface of the earth ticks at the same rate as a pendulum clock moving at a uniform (non-relativistic) velocity wrt the earth.

 Quote by DaleSpam A pendulum clock on the earth does tick at a different rate than a pendulum clock on the moon, but a pendulum clock at rest on the surface of the earth ticks at the same rate as a pendulum clock moving at a uniform (non-relativistic) velocity wrt the earth.
I think I see a problem with this, Dale. The rate of clock on the Earth would be the same as the rate of a clock at assymtotic infinity. The rate of the clock at assymtotic infinity would be the same as the rate of a clock on the moon...

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 Quote by Phrak The rate of clock on the Earth would be the same as the rate of a clock at assymtotic infinity. The rate of the clock at assymtotic infinity would be the same as the rate of a clock on the moon...
No, an inertially moving or resting pendulum clock at asymptotic infinity would have a rate of 0:
$$\omega =\sqrt{\frac{g}{L}} = \sqrt{\frac{0}{L}} = 0$$

 Quote by DaleSpam No, an inertially moving or resting pendulum clock at asymptotic infinity would have a rate of 0: $$\omega =\sqrt{\frac{g}{L}} = \sqrt{\frac{0}{L}} = 0$$
Oh, a pendulum clock! Good grief, how did I miss that.

 Tags special relativity., time measurement