How does time dilation work for a mechanical clock?

Question

I understand that time dilation is when the gravitational pull will change the distance the speed of light is traveling. However, how would this work for a mechanical clock? Say the arms of the clock act as the speed of light and it sits on a strong gravitational pull. Since the arms travel in a circle it would be faster going from the 12 spot to the 6 (since it is moving toward the pull) but slower going from the 6 to the 12 (since it is moving away from pull)? Therefore, evening itself out and there would be no time change, right?

Answer 1

No, I don't think you quite understand how time dilation works.

The effect has absolutely nothing to do with the mechanism of the clock itself. In your case turning the clock by 90 degrees would eliminate the effect you are discussing.

It has everything to do with what happens to the clock ticks you send from the rest frame of the clock to the rest frame of the observer using light or anything else as a signal. How you make those ticks is irrelevant, as long as they are proportional to local time (which they would not be if one followed your mechanical example in detail).

Of course you will screw up any practical experiment by using a clock which explicitly depends on the local gravitational field like an hourglass filled with sand. But then, Einstein was not trying to explain how to screw up experiments but he did explain what the irreducible effects in perfect experiments would be. And that is just what we see in real world experiments done by people who know how to build really good clocks that are independent of pretty much anything.

Try reading Einstein's paper again. He made it very, very clear what the necessities are to connect time measurements in different rest frames. And why those necessities do not lead to a universal definition of time but only to a relative definition of time. The clocks are just utensils. To analyze their internals adds nothing to the physics.

:-)