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 ("absolute") 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 relativistically more massive on the forward swing?)

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

Please note, gravity should not play much of a role here. I could ask the same question with an oscillating tuning fork to avoid most gravitational complications.

  • $\begingroup$ "pendulum" and "relativistic speeds" don't work together. At best you have a carousel. Unless your pendulum is hovering over a black hole? $\endgroup$ Commented Dec 19, 2016 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) $\endgroup$ Commented Dec 19, 2016 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? $\endgroup$ Commented Dec 19, 2016 at 22:16
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    $\begingroup$ @JanDvorak Perhaps. I used it as an example. You can pick any massive object to suit the purpose :) $\endgroup$ Commented Dec 19, 2016 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
    Commented Dec 19, 2016 at 22:23

2 Answers 2


Yes, although the observer travelling with the pendulum measures the same speed forwards and back, however in the inertial observer's frame, the speed of the pendulum relative to the moving frame varies between the forward swing and the back swing. You can verify this using the velocity addition formula used to calculate relative velocities (shown below using natural units, that is all velocities are expressed as a fraction of 1, where 1 is the speed of light).

$$u={v+u' \over 1+vu'}$$

where $v$ is the velocity of the system containing the pendulum, $u'$ is the velocity of the pendulum as measured in that system, and $u$ is the absolute velocity of the pendulum as measured in the inertial observers frame. The speed of the pendulum relative to the pivot as measured by the inertial observer is $|u - v|$ and you can calculate that forward swing $+u'$ yields a slower relative speed than $-u'$ on the back-swing.

Since the pendulum is swinging faster on the back-swing, the duration of its swing will be shorter and that will compensate for the fact that the clock is ticking faster (due to less time-dilation) and the number of ticks will add up exactly to the number of ticks on the forward-swing, which has a longer duration but more time dilation.

We can express this relationship algebraically, noting that the duration of a half-period in the inertial frame is $\Delta t = {L \over |u - v|}$ where $L$ is the distance traveled in a half-swing; and the formula for time elapsed in the clock's frame (i.e. the number of ticks) is $\Delta t' = \Delta t \sqrt {1 - u^2}$. Manipulating and simplifying the formula for the case where $u$ is positive (the forward swing) and negative (the back swing) will demonstrate that $\Delta t$ is the same for both even though they travel at different speeds.

(I have not yet explored the underlying mechanism for why the speed varies between the forward and back swing).


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? $\endgroup$ Commented Dec 22, 2016 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) $\endgroup$ Commented Dec 22, 2016 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
    Commented Dec 23, 2016 at 1:41
  • $\begingroup$ surely both observers agree how many ticks elapse in a half-period, if we want to keep reality consistent. But the problem (from my naive perspective) is that the inertial observer (who's not with the pendulum pivot) should expect more ticks on the back swing than on the forward swing due to reduced time dilation. The only way I can resolve this is by positing that the pendulum actually swings 'shorter' on the way back, compensating for the missing time dilation. But what causes this? Is there some other adjustment I'm missing? $\endgroup$ Commented Oct 20, 2019 at 9:16

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