# Does a pendulum violate general relativity?

General relativity asserts that gravity slows down clocks. The greater the gravitational field, the slower the clock. But taking into account the formula of the pendulum, i.e. $$T=2π\times \sqrt{L/g}$$ we see that if gravity gets stronger the pendulum (which is the simplest clock) gets faster. We consider the formula in its domain with g≠0.

• Gravitational time dilation results from accelerating reference frame near a massive object. The period of oscillation of the pendulum decreases with more gravity, but that does not mean the pendulums clock will run differently with respect to some observer in such frame. See: gravitational time dilation on wikipedia Commented Jan 20, 2017 at 19:39
• @mfc: A pendulum clock doesn't define the passage of time... it measures the passage of time. A pendulum clock is designed to measure time semi-accurately when placed in a certain gravitational field. A pendulum clock will stop when in free fall, but that doesn't mean that time stops. Commented Jan 20, 2017 at 20:32
• @James: yes, the pendulum measures the passage of time. We know it works as a clock. If it is not so accurate it doesn't matter. If we cut power also your electronic clock will stop and this doesn't mean time stops.
– mfc
Commented Jan 20, 2017 at 20:37
• A pendulum doesn't actually measure time. It measures the number of cycles it has completed, which we then calculate time from. If you assume $g = 9.81 m/s^2$ and shorthand the equation to a simple x-number-of-cycles-per-minute form for convenience, the shorthand equation no longer holds when $g$ increases. Commented Jan 20, 2017 at 20:56
• I think your assertion that "we can't treat a pendulum differently of another kind of clock" is wrong. I can make a clock that tells time by counting the number of oxygen molecules that have bounced off it. Maybe it works perfectly well at sea level, but then slows down when I take it up a mountain or put it out in the cold. That doesn't mean time slowed down. Similarly, you chose to measure time in a bad way (with a pendulum) instead of in a good way (with cesium atoms), so you can't expect much. Commented Jan 20, 2017 at 22:25

This question is sort of interesting. There are two competing phenomena. One is the increased frequency of a pendulum with gravity and the other is the time dilation of such a pendulum as seen by a distant observer.

Let us look at the surface gravity. The Schwarzschild metric has the Killing vector $\xi_t~=~\sqrt{1~-~2m/r}$, $m~=~GM/c^2$. The gravity on a stationary surface held at a fixed radius is given by $$g^2~=~(\nabla_r\xi_t)(\nabla^r\xi^t)~=~\frac{1}{4}\frac{1}{(1~-~2m/r)^2}\frac{4m^2}{r^4}$$ or $$g~=~\frac{1}{1~-~2m/r}\frac{m}{r^2}~=~\frac{m}{r^2~-~2mr}.$$ What this says is that for $r$ large this is $g~=~GM/r^2$, which is the Newtonian result. For $r~\rightarrow~2m$, up to the event horizon of a black hole, the surface gravity $g~\rightarrow~\infty$.

Using this in a standard formula for a pendulum of length $\ell$ it indicates that the periodicity is $$T~=~2\pi\sqrt{\frac{\ell(r^2~-~2mr)}{m}}.$$ The periodicity $T~\rightarrow~0$ as $r~\rightarrow~2m$. This means the frequency diverges. So a pendulum held fixed on a constant surface just above the event horizon has a huge frequency.

Now let us ponder what a distant observer witnesses. The time dilation of the periodicity will be simply $$T'~=~\frac{1}{1~-~2m/r}T~=~2\pi\sqrt{\frac{\ell r^3}{m(r~-~2m)}}T.$$ The time dilation then means the distant observer will witness the pendulum with a huge periodicity very close to the event horizon. So the frequency $\nu~\rightarrow~0$. This means the pendulum will appear frozen when held fixed near the event horizon. Again, this holds of course for the pendulum on some fixed radius above the horizon and not for a freely falling one.

• I don't understand what you want to demonstrate. We know the parallelism with the Schwarzschild solution. Also: general relativity doesn't need event horizons to work and we don't know yet what black holes actually are. ANY clock's period diverges in such situation. So, let's consider the pendulum formula within its domain (g≠0) and without a diverging gravity. A local observer, if the planet increased its density, would measure a faster period (violation!) and a distant observer would measure a period difference inversely proportional to the difference (if there's) in gravity (violation!)
– mfc
Commented Jan 20, 2017 at 21:51
• I meant (not characters left) the distant observer measures a difference in the period according to the difference in gravity of his reference system with respect to the other one. If ∆g is negative he should measure a slower clock. But for the pendulum this doesn't occur (violation), since the greater is g, the faster is the pendulum. No theory is perfect (e.g. GR doesn't fit in QFT). Gravity changes the weight of a clock's components and "depending on how this fact acts on the specific mechanics of the clock-system", its period can get faster or slower as the pendulum proves.
– mfc
Commented Jan 20, 2017 at 22:10
• This is a rather canonical general relativity problem. The Killing vector field for the metric predicts the periodicity of oscillation will shorted due to increased gravity. However, a distant observer witnesses this slowed by time dilation. Commented Jan 20, 2017 at 22:42
• we have special relativity where the discourse on two different reference systems comes into play if they move the one with respect to the other. This is not the case. Also, the distance between two reference systems is not important (there may be a time delay for signals from a system to the other, but time delay is not time dilation). So, we come to "general" relativity. Where the distance still doesn't matter. Gravity matter. You have to consider the ∆g. And the problem with the pendulum is still there. If not maybe you want to be clearer with equations thanks.
– mfc
Commented Jan 20, 2017 at 22:52
• Also consider the Pound-Rebka experiment. A lower gravity accelerates clocks and a redshift is measured. This is also general relativity, i.e. the same issue (gravitational time dilation). Now, imagine to measure the frequency of the emitted light with a pendulum-based mechanism. You would measure a blueshift (violation of GR). What about this?
– mfc
Commented Jan 20, 2017 at 23:00

A rod is a pendulum. Is a rod a clock? No, I don't think so.

A pendulum clock is a clock. Do I need to justify that claim?

From the above we conclude that a pendulum is not a pendulum clock, and a pendulum clock is not a pendulum.

Now the question arises what is a pendulum clock?

Well, I would say that its a rod and a gravitating mass. And some support structure that keeps the rod and the mass separated.

This pendulum clock is such that it gets seriously messed up by nearby extra masses, it speeds up a lot or slows down a lot.

But if extra masses are placed symmetrically around the pendulum clock, then the clock slows down slightly as as predicted by general relativity. General relativity says that masses have that effect on clocks. That by the way is more correct statement than the "gravity has effect on clocks".

I believe you could see it as such, that the pendulum is capable of measuring 'time' in the configuration space, but is not a valid timekeeper in the spacetime manifold, which has 'time' as the fourth dimension in its tensors. These two 'times' are not congruent and contradicting, as you rightfully identified. This is exactly the problem between quantum mechanics (configuration space/ state space to be more precise) and the spacetime manifold. They haven't solved this, as this gets near the problem of quantum gravity.