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I first read about it on A Brief History of Time(Stephen Hawking). In 1962, a relativity experiment was executed: identical (classical) watches put on a water tower, one is on very high, other one is at the bottom. And of course, after a certain time, the one which is closer to ground was slower than the other.

What I want to learn is what happens if I add this:

sea tower experiment

and re-do the experiment? The stick bounds the wheels of the identical watches externally. It has no mass, it is non-flexible, it has no friction, strong enough etc. So, think it as an ideal stick for phsic exam question.

What happens then? If without stick, lower clock is 30 minutes late than higher one, with stick, will it be 15 mins further from reference clocks? Or something entirely different will happen? Thanks.

P.S: All the mechanism that clocks use are fully Analog, not stepped!

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The upper clock tries to force the lower clock to speed up, by exerting a force on the shaft, the force is produced by a motor.

The lower clock tries to force the upper clock to slow down, by exerting a force on the shaft, the force is produced by the mechanism that limits the clocks rate in normal use.

Which clock wins? The lower clock wins, except if the upper clock manages to break the lower clock's clock mechanism.

If the questioner does not like this kind of trivial answer, then the questioner should get rid of the clock mechanism of the lower clock.

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  • $\begingroup$ Why lower clock wins? Can you explain it in detail? ***Think all the hardwares ideal, no friction no breaking nothing unwanted happens. $\endgroup$ – Alper91 Dec 12 '15 at 9:29
  • $\begingroup$ There's an escape wheel and a pallet that keep the ticking rate constant regardless of how tight the mainspring is wound. Extra force does not cause extra speed of clock hands. $\endgroup$ – stuffu Dec 12 '15 at 9:43
  • $\begingroup$ In that case, I have to change my original statement to this: All the mechanism that clocks use are fully Analog, not stepped! $\endgroup$ – Alper91 Dec 12 '15 at 9:45
  • $\begingroup$ That's no clear enough. Do we perhaps just have two identical motors at both ends of a vertical shaft? $\endgroup$ – stuffu Dec 12 '15 at 9:51
  • $\begingroup$ I don't see the point, why would be a difference between mechanical and electrical force? After all, they both provide nothing but kinetic energy to the clocks. Ok, let go on your way, we have two identical motors at both ends of a vertical shaft. Then why? Why lower clock wins? And by 'win', how it will win? The time on both clocks will be same as other identical reference ground clocks? Or both will have a time slightly over by other reference ground clock? $\endgroup$ – Alper91 Dec 12 '15 at 10:01
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Let's say two identical steam engines are connected to a vertical shaft. Both engines are equipped with a very sensitive centrifugal regulator. What happens?

The lower engine turns itself off, as the rotation speed is too large, according to the regulator.

Let's say two identical electric motors are connected to a vertical shaft. Both motors are equipped with a very careful worker who adjusts the rotation speed according to his atomic clock. What happens?

The lower worker turns the lower motor off, as the rotation speed is too large, according to the his atomic clock.

(Is there a brake the worker is supposed to apply? How effective is the brake?)

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When two clocks wrestle, the stronger one wins.

As the upper clock is stronger it wins.

That's the answer, when we allow some sloppiness about the meaning of "clock".

The clocks were identical, and an upper thing always has an advantage in wrestling match, as proximity of large mass causes weakness.

"Proximity of large mass causes weakness" can be trivially deduced using the equivalence principle: A strong rope is lowered down from the rear of an accelerating rocket, the rope must break at a horizon, which horizon for the rocket staff may be equally well either a Rindler-horizon or an event horizon of a very large black hole.

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