# How does a Foucault pendulum work?

How exactly does a Foucault pendulum work? The usual explanation says that the plane of the oscillation of the pendulum is fixed while the earth rotates underneath. On Wikipedia, there is a demonstration of this effect, showing what it's like on the north pole. But surely that can't be right, for this gets at the heart of the hovering helicopter argument, which says that a helicopter hovering above the surface could wait for its destination to arrive (because the earth rotates underneath). But as we know, this doesn't work because of conservation of momentum. So surely the same must apply to the Foucault pendulum?

Now, I suspect the phenomenon has something to do with the Coriolis effect, but I can't really understand how. Wouldn't the Coriolis force just be too miniscule? Also, what if we set the pendulum oscillating in the east-west direction (at some point on the northern hemisphere)? Then surely there will be no Coriolis force?

• Aug 29, 2020 at 14:46
• On the observability of the Coriolis effect in meter-scale systems, see these cute videos.
– rob
Aug 29, 2020 at 15:04

## 2 Answers

The statement "the plane of the oscillation of the pendulum is fixed while the earth rotates underneath." is applicable only for a polar pendulum. On every other latitude there is more going on.

To drive that point home let me go over the following case: at the latitude of Paris it takes 32 hours for the plane of swing to go through a complete rotation. Of course that means that in 24 hours the plane of swing goes through 3/4 of a complete rotation. Hence if at t=0 the plane of swing is north-south then 24 hours later, when the Earth is back in the same orientation as 24 hours before, the plane of zwing is east-west.

The thing that has happened in those 24 hours is that the swinging pendulum bob has been exchanging momentum with the Earth. Of course, the change of momentum of the Earth is negligable, but that does not change the fact that exchange of momentum has taken place.

To make that exchange of momentum vivid I give the following thought demonstration. Imagine a small celestial body, say some asteroid, roughly spherical. Let that sphere be rotating. Build a structure such that a pendulum bob can swing, make that structure strong enough that the mass of the pendulum bob can be significant percentage of the mass of the sphere. Let's say the mass of the pendulum bob is 5% of the mass of the sphere.

With such a setup the swing of the pendulum bob will have a measurable effect on the orientation of the sphere. The sphere and the bob are exchanging momentum. That is the purpose of this thought demonstration: the same reasoning extends to a large celestial body: as a pendulum is swinging it is exchanging momentum with the celestial body.

For a pendulum not located on either of the poles that is why the plane of swing is rotating at a smaller angular velocity than the angular velocity of the celestial body itself.

Indeed it appears as if this dynamic effect cannot be strong enough. Then again, it is cumulative, that makes all the difference

Now, most derivations that you will encounter do not describe the dynamics in these terms. Yet they arrive at the actually observed rate at which the plane of swing is precessing. How does that happen?

Well, generally derivations are constructed in reverse. The construction of the derivation works backwards from the known observation.
Then in the final write-up the derivation is presented in forward steps.
The derivation then contains assumptions that produce the desired result, and these assumptions look plausible, but they are not necessarily physically justified assumptions.

Next, let me take the case of the actual Foucault pendulum in the Pantheon. Foucault describes that on rare occasions there was time for long uninterrupted runs. Over the course of such a long run the amplitude of the swing would decay to a mere 10 centimeters, but Foucault reports that the plane of swing was still rotating at the same rate.

The Foucault pendulum in the Pantheon has a 67 meter cable. At the latitude of Paris the distance to the Earth's axis is such that the required centripetal force to sustain circumnavigation of the Earth's axis corresponds to a weight on a 67 meter cable to be displaced about 10 centimeter.

Two jobs
So: the suspending cable has two jobs to do: it must provide the required centripetal force to sustain circling the Earth's axis, and it provides the restoring force for the swinging motion.

Let me now discuss 4 cases:

• the pendulum bob swings south to north:
The centripetal force is doing work, increasing the bob's angular velocity

• the pendulum bob swings north to south:
The centripetal force is doing negative work, decreasing the bob's angular velocity

• the pendulum bob swings west to east:
Now the pendulum bob is circumnavigating the Earth's axis faster than the Earth itself. Because of that velocity surplus the pendulum bob will swing wide.

• the pendulum bob swings east to west:
During that part of the swing the pendulum bob is circumnavigating the Earth's axis slower than the Earth itself. Because of that velocity deficit the pendulum bob is pulled closer to the Earth's axis

Interestingly, to a very acceptable approximation the rotation-of-Earth-effect is equally strong in every direction of the plane of swing.

Importantly, if there would not be a rotation-of-Earth-effect for east-west swing then the rotation of the plane of swing would stall there. We know it doesn't stall, hence any explanation that fails to predict/explain the effect for east-west swing is definitely wrong.

The image below gives a schematic presentation of the cumulative effect.

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Yes, the point is Coriolis force. If you want to study the motion of the Foucault's pendulum you have to consider the fact that it oscilates in a non inertial frame, which is Earth's surface, so apparent forces have to be taken into account.

Wouldn't the Coriolis force just be too miniscule?

If you want to restrict your study to few oscilations, then it's definitely minuscle, and you can set it to zero, but you want to go beyond this approximation and explain why the oscilation plane rotates, then the Coriolis force is the first order perturbation you have to account for.

what if we set the pendulum oscillating in the east-west direction (at some point on the northern hemisphere)? Then surely there will be no Coriolis force?

In this case still you have a non vanishing Coriolis force because it vanishes when the pendulum velocity vector $$\vec{v}$$ and the Earth's angular velocity $$\vec{\omega}$$ are parallel. In the case of East-West motion, $$\vec{v}$$ and $$\vec{\omega}$$ are perpendicular. They are parallel in a North-South oscillation at the equator.

• Yeah, I found a really good pdf article on the Foucault pendulum which went through the math, and then I saw how the Coriolis force was just given by twice the cross product between omega and the velocity of an object. And as you say, this means there is a Coriolis force when something moves east-west. It's just that I thought that there would be no force in this case because all I really knew of Coriolis was the demonstration with a canon ball getting shot due north from the equator, and how that resulted in a deflection. I imagined this wouldn't happen if something got shot east-west. Aug 29, 2020 at 22:40
• @FelisSuper About the cannon ball idea. Let's make that cannon ever larger. Set up a supercannon on, say, 45 degrees latitude. Fire to the west, at such velocity that the cannonball stays in flight for hours. During the flight the cannonball is effectively in a low orbit. Orbital motion is along a great circle. This shows that a westward travelling cannonball will proceed towards the Equator. That is: the cannonball idea predicts that the plane of swing turning will be different when swinging east-west. Observation: plane of swing turns uniformly. Aug 30, 2020 at 5:18
• @Cleonis Right, I did know that, but I thought that the amount of deflection would be too small as the bob on the pendulum moved just a tiny amount east-west. But as Matteo points out, the Coriolis force is indeed small (that's why we usually don't have to think about it), but during a long time span, that does lead to some deflection. Aug 30, 2020 at 8:50
• @Cleonis Btw, thank you for your answer to an earlier question of mine related to the anti-gravity wheel by Veritasium, it was indeed really helpful. Aug 30, 2020 at 8:55
• Just to note for those reading Cleonis' comment... his visual of a giant supercannon doesn't speak to Coriolis (indeed a westward shot in the NH would mean a deflection away from the equator (towards the north)) Aug 24, 2021 at 9:44