If a clock is swinging on a pendulum (assume it's on some massive object with gravity) what happens when that system is moving at relativistic speeds? To an inertial observer, on the back-swing (i.e. opposite the direction of movement of the system) the pendulum has a lower velocity than on the forward swing. The clock should experience less time dilation on the back-swing, and that should register as a different number of 'ticks' on the clock, but that can't be because then the 'moving' observer would notice that too! Is the length (duration) of each swing different to cancel out that effect, and what causes the change in duration (is it because the weight/bob is realistically more massive on the forward swing?)

I don't know too much, so please keep the explanation simple!

  • $\begingroup$ "pendulum" and "relativistic speeds" don't work together. At best you have a carousel. Unless your pendulum is hovering over a black hole? $\endgroup$ – John Dvorak Dec 19 '16 at 22:08
  • $\begingroup$ @JanDvorak what do you mean? if a asteroid is moving past at high speed, a pendulum on it won't swing in a regular fashion? (I'm talking about a pendulum swinging parallel to the direction of motion) $\endgroup$ – afuna Dec 19 '16 at 22:14
  • $\begingroup$ Sorry. I misread, I thought the mass of the pendulum was moving at relativistic speeds relative to its hinge. But, doesn't an asteroid have a bit too low a mass for a pendulum? $\endgroup$ – John Dvorak Dec 19 '16 at 22:16
  • $\begingroup$ @JanDvorak Perhaps. I used it as an example. You can pick any massive object to suit the purpose :) $\endgroup$ – afuna Dec 19 '16 at 22:16
  • $\begingroup$ It's kinda the same idea, I think.....have you studied the revolving wheel and spoke example?spacetimetravel.org/tompkins/node7.html or the Ehrenfest Paradox. It's related but maybe not close enough, sorry $\endgroup$ – user139561 Dec 19 '16 at 22:23

I see two main issues with your scenario:

1) A pendulum is not an inertial reference frame, so special relativity does not really work here.

2) If you took the problem to be a bunch of instantaneous inertial reference frames, as one does in the classic relativistic rocket problem, you will find that your assumption of how time is measured is incorrect: the two observers will not observe a different number of ticks on the front and back swings, because the number of ticks is dilated from the perspective of a stationary observer as well. So it might take six ticks for the pendulum to go forward, and six ticks to go back, but to the stationary observer, the six forward-moving ticks will happen more slowly than the ones on the backswing. An observer sitting on the pendulum would see them all happen at the same rate, and an observer sitting on the (constant-velocity) pivot point of the pendulum would see symmetric time dilation for the front and back swings.

A pendulum basically is a clock, so if we ignore the problem of inertial reference frames, any relativistic effects that affect the pendulum will similarly affect the clock.

  • $\begingroup$ 'six ticks' more slowly means a longer time as measured on the inertial clock. i.e. the pendulum swings 'slower' on the way back - that is what I suggested at the end of my question. But what is the interpretation of that? What translates the potential at the beginning of the forward swing into less velocity: because it is relativistically more massive? $\endgroup$ – afuna Dec 22 '16 at 21:33
  • $\begingroup$ (n.b. I hadn't meant to analyze an observer on the pendulum, rather to compare with an observer moving alongside the pendulum's pivot) $\endgroup$ – afuna Dec 22 '16 at 21:34
  • $\begingroup$ I'm not quite sure what you're getting at. An observer on the pendulum's pivot will see things exactly as if he/she and the pendulum were both stationary. An observer on the "ground" will see time dilation, varying with the relative velocity between the pendulum head and the ground (assuming the clock is on the moving end of the pendulum). $\endgroup$ – tdinap Dec 23 '16 at 1:41

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