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I am taking a course in introductory general relativity and came up on this question, which a google search didn't answer. The rotation of the Earth can be measured using a Foucault pendulum. The moon also rotates to always keep the same side facing the Earth, and it seems therefore that we should be able to measure this rotation using a pendulum on the moon. However, I have just learnt that a particle falling freely in a gravitational field is really following a geodesic curve through space-time. This got me wondering whether the apparent rotation of the moon is an artifact of the curvature of space-time, or if it is a real physical effect. I think my question can be phrased as in the title - would a Foucault pendulum rotate on the moon?

Another way of phrasing the question which may be clearer: If a (non-rotating) satellite were held still above the surface of the Earth, and then given a sudden sideways kick, putting it into orbit: would its orientation remain fixed relative to the stars, or would it rotate to always face the same side towards the Earth?

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Simplifying assumptions: The Earth is a perfect gravitational point source, and the orbits are perfectly circular.

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    $\begingroup$ Moon: "hip hip hooray I got coriolis effect!" $\endgroup$
    – user6760
    Feb 20, 2021 at 10:21

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There is no significance in the fact that the same face of the Moon is always oriented towards the Earth. It simply means that the rotation of the Moon around its own axis has the same period as the orbit of the Moon around the Earth.

(This wasn't always so. It can be reconstructed that the Moon previously had a faster rotation rate than it has now. Celestial bodies have a gravitational interaction (involving tidal effect) that tends to slow rotation down. When the Moon rotation was slowed down to once per month the rotation of the Moon went into tidal lock.)

Any satellite orbiting any celestial body can have any rotation rate of its own.

Some of the Earth orbiting artificial satellites have a design where the satellite extends a pole after it has inserted into orbit. The purpose of the pole is to facilitate tidal lock of the satellite.

Uneven pressure from solar wind tends to slowly change the orientation of a satellite. The tidal effect from the pole extending from the satellite is very small, but the effect from the solar wind is very, very small, so very little means are sufficient to counteract it.

There is no connection between tidal lock and relativistic physics. Tidal lock is accounted for in terms of Newtonian dynamics. There is a theoretical relativistic correction, but the difference is so small that it's totally negligible.

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  • $\begingroup$ Thank you, this is a very clear answer. In terms of Newtonian mechanics, this makes a lot of sense to me. $\endgroup$
    – Jakob KS
    Feb 20, 2021 at 11:04
  • $\begingroup$ However, in terms of relativity, I find it surprising. If a satellite simply follows a straight line through space-time, how does it "know" to keep its orientation fixed relative to the stars? I would have thought that its orientation would somehow follow the curvature of space-time, and this would cause it to rotate. $\endgroup$
    – Jakob KS
    Feb 20, 2021 at 11:16
  • $\begingroup$ That's because in general relativity, a "straight line" is a geodesic which accounts for curvature due to gravitation. $\endgroup$ Feb 20, 2021 at 17:19
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    $\begingroup$ @JakobKS Objects follow straight paths through spacetime, not through space. How Gravity Makes Things Fall has a great visual demonstration of this fact that is better than the typical "heavy ball on a rubber sheet" metaphor. $\endgroup$
    – trent
    Feb 20, 2021 at 17:29
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    $\begingroup$ @JakobKS Imagine a game of curling on an ice surface that is not perfectly flat, but instead a very shallow dish. Whether on a flat ice rink or a curved one: a curling stone that is released without initial rotation will not subsequently rotate a bit. The slope will bend the trajectory, but no rotation will be imparted to the stone. Over the size of a curling stone gravity is uniform, so no torque on the stone arises. Because gravity is uniform we must think of gravity as acting upon the center of mass. $\endgroup$
    – Cleonis
    Feb 20, 2021 at 17:30
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Foucault's pendulum would, indeed, work on the Moon - though because the rotation is much slower (a period of about 2360 ks versus about 86 ks), the pendulum will also precess that much slower. (It will also swing with a longer period, due to the reduced gravity, but that's a different matter.)

Regarding your question about whether the rotation is "real" or not in the context of general relativity, yes it is real: Foucault's pendulum is, in fact, not a free-falling body in a gravitational field. As a pendulum, it is anchored at one end, and a string connects the pendulum bob to a hook. This string exerts a real force on that bob, and if it's moving under a force, it cannot be in free fall, by definition. As the Moon rotates, the anchored tip of the string is steadily dragged ever so slightly, which pulls the bob sideways which, in turn, causes the shift in its swinging motion.

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  • $\begingroup$ In my opinion, the apparent rotation of the plane of the swing of a pendulum has nothing to do with 'the bob being pulled sideways'. The pendulum swings in a plane that is fixed 'with respect to the fixed stars', or to the mean distribution of mass in the universe. A rotating wheel behaves the same way as the pendulum... $\endgroup$
    – user124096
    Feb 21, 2021 at 18:15
  • $\begingroup$ The plane should be fixed except for external forces. The tidal acceleration at the Moon due to the Earth is 81 times of what it is here due to the Moon (mass ratio). And it has 27 times the duration to have an effect (sideral period of rotation ratio). And to keep swinging for many days, the pendulum has to be huge -- don't know whether this matters. $\endgroup$
    – Rainald62
    Feb 22, 2021 at 0:40
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I think I understand your question. It's not so much about Foucault's Pendulum as it is about frames of reference. I believe this is what you are thinking: If you ignore the rotation of the moon and just think of it as an orbiting point, then an obvious way to use it to generate a coordinate system that follows it is to have one axis tangent to the orbit, and the other two always perpendicular to that one. It is natural to think that, if the moon is just following a geodesic, then that coordinate system that has one axis tangent to the orbit is the natural coordinate system associated to that geodesic, and in that coordinate system the moon isn't rotating.

But it doesn't work that way. I think a roughly correct way to get the coordinate system that follows the moon is to imagine replacing the moon with a family of small particles. The ones nearer the earth orbit faster, and the ones farther away orbit slower. But the closer they get to the particle at the center of where the moon was, the closer they get to moving at the same speed. The coordinate system comes from taking the limit, and it will not have an axis tangent to the orbit, nor will it keep an axis pointing towards the earth. It will in fact maintain the same orientation as it orbits.

Really, the best way to think of the geodesic path is to drop one of the space dimensions and model the earth as a disk for each slice of spacetime, combining them to make a cylinder running from the past to the future. Then the geodesic path of the moon is a spiral around that cylinder, and the full coordinate system has a $t$ axis, an $x$ axis, and a $y$ axis, plus a $z$ axis that we are ignoring in this mental model with 2D of space and 1D of time. In general, when you project from spacetime down to space, the axes of the coordinate system won't turn to follow the direction of the track. They'll keep the same orientation like one of those gyroscope balls some people used to have in their cars.

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  • $\begingroup$ Thank you! This was precisely the question I was really trying to ask - I apologize for the lack of clarity in my original post. GR is really super interesting, but also so hard to wrap your head around. I think you hit the nail on the head in that I don't understand how to project from spacetime down to space, so I'll try to look up some resources on that. $\endgroup$
    – Jakob KS
    Feb 21, 2021 at 8:20
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    $\begingroup$ More generally, GR doesn't invalidate Newtonian mechanics, it explains it. There are some differences in extreme situations (e.g. Einstein explained the anomoly in Mercury's orbital period), but most ordinary phenomena can be calculated purely using Newton's laws (rocket scientists generally just use Newton). $\endgroup$
    – Barmar
    Feb 21, 2021 at 14:39
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Yes, it would work. Of course, you should take into account that gravity at the Moon's surface is only 1/6 of the value on Earth and that a Moon (sidereal) day is 27 Earth days.

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