Why the Moon always shows the same face

[snorkack]

No asymmetry in the Moon itself is necessary for the Moon to librate. The effects of the asymmetries due to the eccentricity and inclination of the Moon's orbit, plus the asymmetry between the Moon's period of rotation and the Earth's, are sufficient.

That´´ s forced libration, not free. And has nothing to do with Moon showing the same face to Earth.
Suppose that Moon were a perfectly elastic perfect sphere of cheese, with craters and seas painted on with weightless paint, exactly equal density black and white, for markers.
Then Moon would have forced libration due to eccentricity - the orbital movement would be nonuniform due to eccentricity, but rotation uniform. Craters and seas would rock back and forth over the edge of Moon... but there would be no free libration. Because no restoring force. Nothing would stop Moon´´ s rotation period from being just slightly different from orbital period, causing the craters and seas to slowly drift from near side to edge, vanish to far side after several times of rocking back and forth and eventually cross far side and return from the other edge.

Some asymmetry of Moon must allow free libration, and therefore near side staying near side.

 

[PeterDonis]

That´´ s forced libration, not free.

Forced or free, the causes I gave are the ones that account for the observed libration of the Moon. So if there is any other kind, its magnitude must be comparatively very small.

And has nothing to do with Moon showing the same face to Earth.

The causes I gave certainly do. See below.

Because no restoring force.

Yes, there is: tidal gravity. See my explanation of tidal locking in post #30. That explanation applies perfectly well to a Moon that would be spherically symmetric in the absence of tidal gravity, because in the presence of tidal gravity, it is not spherically symmetric: there are tidal bulges on it, and if the Moon's period of rotation is different from its period of revolution around the Earth, those tidal bulges will move, which will create a restoring force tending to make the two periods the same.

It is true that, with a Moon that would be slightly asymmetric in the absence of tidal gravity, there will be an additional force tending to make the mass asymmetry radial (whereas with a Moon that was perfectly symmetric in the absence of tidal gravity, there would be no preference for any particular orientation once it was tidally locked--but that would not stop some orientation from being tidally locked). However, there are two local equilibria for this: more mass towards the Earth, and more mass away from the Earth. The latter equilibrium is only local; the former is the global one. But if the Moon were stuck in the latter local equilibrium, it's quite possible that it could stay there, since the forces acting on it might be insufficient to kick it out of the local equilibrium hard enough to overcome the potential barrier between that local equilibrium and the global one.

 

[sophiecentaur]

@Orodruin gave a good explanation of tidal gravity in post #20.

Yes. That post describes well how the force / acceleration on each part of the object varies over the object and with its orientation. However, the term "tidal gravity" is confusing because it implies more than that simple description. It can cause 'tidal' movement when the object is not rigid so why use the term "tidal" when what is really meant is "tide-causing" or even "anisotropic", which would be precise? I guess my problem here is that 'tidal' can refer to both a cause and an effect. Many of the red herrings that have appeared in the thread seem to be based on just that confusion.
The rest of your post is fine by me - even if the 'details' of the Earth - Moon motions are more involved than necessary for a general discussion.
I still think that these things are better dealt with in terms of Energy, rather than forces and torque. Every such system will eventually reach the condition of least potential energy but the forces approach doesn't make that as clear as it should, imo.
On the whole, this thread has been pretty useful at getting to the right conclusion; I got a lot from it.

 

[PeterDonis]

I guess my problem here is that 'tidal' can refer to both a cause and an effect.

In both Newtonian gravity and General Relativity, the term "tidal gravity" is pretty standardly used to refer to the cause: in Newtonian terms, the difference in the "acceleration due to gravity" from place to place; in GR terms, spacetime curvature (or sometimes specifically Weyl curvature). One of the advantages of GR is that it gives a separate term, "spacetime curvature", which unambiguously refers to the cause and doesn't give rise to any confusion with various effects, as you say the word "tidal" does.

 

[PeterDonis]

Every such system will eventually reach the condition of least potential energy

I'm not sure I would specify potential energy here. Consider the Earth-Moon system: as tidal gravity slows the Earth's rotation, it moves the Moon further away (to conserve angular momentum), so the potential energy of the system increases. (This is balanced by a decrease in kinetic energy due to slowing the Earth's rotation and the Moon's orbital speed.)

 

[sophiecentaur]

I'm not sure I would specify potential energy here.

You are right, of course - the total orbital Energy needs to be considered and it will go to a minimum when locking is reached.

 

[snorkack]

The causes I gave certainly do. See below.

Yes, there is: tidal gravity. See my explanation of tidal locking in post #30. That explanation applies perfectly well to a Moon that would be spherically symmetric in the absence of tidal gravity, because in the presence of tidal gravity, it is not spherically symmetric: there are tidal bulges on it, and if the Moon's period of rotation is different from its period of revolution around the Earth, those tidal bulges will move, which will create a restoring force tending to make the two periods the same.

Does not follow.
If Moon were perfectly elastic, the tidal bulges would move - but without creating any force.
Even if a force existed, it need not be a restoring force.

It is true that, with a Moon that would be slightly asymmetric in the absence of tidal gravity, there will be an additional force tending to make the mass asymmetry radial (whereas with a Moon that was perfectly symmetric in the absence of tidal gravity, there would be no preference for any particular orientation once it was tidally locked--but that would not stop some orientation from being tidally locked).

It would.
A rubber tire is not axisymmetric when supported on road. Because the weight on the tire presses into the point of support.
But if you turn it then it rolls without bumps. Any part of the tire will be equally compressed when loaded. Therefore there will be no restoring force.
A rubber tire does exert a force against movement. But it is a retarding force, not restoring one. A rolling wheel comes to a stop, but at a random new position - it will not return to its previous position.
So, a wheel without a parking brake on will move slightly but permanently on any small nudge. Over time and many small nudges, it adds up.
Whereas a wheel with a parking brake on does have a restoring force and returns to a fixed position when displaced by a small nudge.

Moon shows one side because Moon possesses a permanent asymmetry - Moon acts like a wheel with parking brake on, not like a wheel without a parking brake on.

 

[sophiecentaur]

- but without creating any force.

No net force at equilibrium? That always has to be true but surely what counts is not Force but Work done on the system, with or without Hysteresis.

no preference for any particular orientation once it was tidally locked--but that would not stop some orientation from being tidally locked)

It would.

If there was perfect symmetry, the orientation at which it would lock would be indeterminate. It would just go slower and slower, as with any exponential damping. Any slight Energy Minimum / dip would eventually be the orientation where locking would occur. (See where Energy considerations can be useful?)

A rubber tire does exert a force against movement. But it is a retarding force, not restoring one.

Why make the distinction? Both forces are present; if you cut the drive, the tyre will spring back to another position (restoring force) which will change the slip angle. The tyre analogy doesn't really add to this argument, imo; we are past analogies at this point.

 

[PeterDonis]

If Moon were perfectly elastic, the tidal bulges would move - but without creating any force.

So what? The actual Moon is not perfectly elastic. But a perfectly symmetric Moon (i.e., symmetric in the absence of tidal gravity) would still experience tidal locking, which is what I've been saying.

Even if a force existed, it need not be a restoring force.

If you don't like the term "restoring force", then just say "force". The point is that even if the Moon were perfectly spherically symmetric in the absence of tidal gravity, there will still be a force tending to make the Moon's orbit circular and in the Earth's equatorial plane, and to make the Earth's period of rotation the same as the Moon's period of revolution around the Earth. You don't need a asymmetric Moon for that.

Moon shows one side because Moon possesses a permanent asymmetry

No, it shows one side because that part of tidal locking in the Earth-Moon system has had enough time to run to completion. Whereas the parts where the Moon's orbit is circular in the equatorial plane, and the Earth's rotation has the same period as the Moon's revolution, have not.

What is true is that, if the Moon did not have an asymmetry (meaning an asymmetry that would be there in the absence of tidal gravity), which side of the Moon permanently faced the Earth as a result of tidal locking would have been a random result, since no side would have been preferred. But since the Moon does have such an asymmetry, if some other side had ended up originally facing the Earth as a result of tidal locking, i.e., if the asymmetry were not radial, there would have been an additional force that gradually changed things until the asymmetry was radial.

 

[PeterDonis]

If Moon were perfectly elastic, the tidal bulges would move - but without creating any force.

Actually, even this is not true. There would still be torques on the tidal bulge even with a perfectly elastic Moon.

 

[PeterDonis]

the total orbital Energy needs to be considered and it will go to a minimum when locking is reached.

Actually, the main mechanism of tidal locking is torques on the tidal bulges, for which no dissipation is required, so the total energy of the system actually remains constant.

In actual objects there is also dissipation going on, but its effects are very small compared to the effects of the torques. Even in the absence of dissipation, the tidally locked configuration is an equilibrium because any small perturbation from it will set up torques that tend to restore it.

The Wikipedia page gives a good overview:

https://en.wikipedia.org/wiki/Tidal_locking

 

[sophiecentaur]

for which no dissipation is required, so the total energy of the system actually remains constant.

No problem there but. without losses there will always be oscillatory motion about some Energy minimum. The Moon would go past its maximum orbit size and then return ad infinitum.
Tell me why there is such bad reception for the Energy - based argument. Somewhere along the line, the Force approach will need to consider Work in and Work out and a 'hysteresis' style of diagram. A good answer is available, based on Energy which needs no complicated calculations; it's based on fundamentals and needs no details.

 

[PeterDonis]

without losses there will always be oscillatory motion about some Energy minimum
The Moon would go past its maximum orbit size and then return ad infinitum.

That's not quite true either as you state it. There will indeed be oscillations about an equilibrium, but these are not oscillations about an energy minimum, because there is no energy minimum; all the relevant states of the system have the same total energy. The equilibrium is an equilibrium of zero torque, but it's only meta-stable (if that's the right word); it's not an energy minimum.

The effect of dissipation is to slowly reduce the total energy of the system; what that does is slowly move the equilibrium of zero torque to a smaller average distance between the two bodies (i.e., make the system more tightly bound). This will not damp oscillations about that equilibrium, because the oscillations are not due to the system having more energy than some minimum at equilibrium.

Tell me why there is such bad reception for the Energy - based argument.

I don't think it's a matter of "bad reception". I just think that for this particular problem, since the primary process of tidal locking conserves total energy, thinking in terms of energy doesn't help to understand that primary process.

 

[sophiecentaur]

Isn’t a stable equilibrium situation at an energy minimum? I realize my assumptions are not based on orbits but where is the difference in principle? How can friction forces not lose energy and bring the system to a limiting low level of energy?
An alternative view can be valid also even if it’s not conventional.

 

[PeterDonis]

Isn’t a stable equilibrium situation at an energy minimum?

It depends on what you mean by "stable equilibrium". In the case under discussion, the equilibrium of zero torque for a system subject to tidal locking, like the Earth-Moon system, is not an energy minimum, since the forces involved are conservative and so all states of the system accessible with those forces have the same energy. So if "stable equilibrium" requires an energy minimum, then this equilibrium is not one. That's a matter of terminology, not physics.

How can friction forces not lose energy and bring the system to a limiting low level of energy?

Friction forces do. But friction forces aren't what make tidal locking happen. Please read the Wikipedia article I linked to in post #41. Friction forces do what I said the "effects of dissipation" are in post #43; that's not the same as tidal locking. Briefly, friction forces change the equilibrium of zero torque; but they are not what drives the system towards whatever that equilibrium is.

 

[sophiecentaur]

But friction forces aren't what make tidal locking happen.

Have I misunderstood the actual meaning of Tidal Locking then? Is it correct to say that he Moon faces the Earth due to Tidal Locking? Is it not true that it is losses that caused that basically stable situation? Without losses (purely elastic deformations) why would the situation be as it is?

 

[PeterDonis]

Have I misunderstood the actual meaning of Tidal Locking then?

I don't know. What do you think tidal locking is? Did you read the Wikipedia article?

Is it correct to say that he Moon faces the Earth due to Tidal Locking?

Yes, with a correct understanding of what "tidal locking" means.

Is it not true that it is losses that caused that basically stable situation?

No, it's not true. Tidal locking would happen even with a perfectly elastic Earth and Moon.

Without losses (purely elastic deformations) why would the situation be as it is?

Because tidal locking is due to torques exerted by the gravity of each body on the tidal bulges in the other. Please read the Wikipedia article I linked to; it gives a good explanation of the mechanism. Feel free to ask questions if something there isn't clear to you.

 

[sophiecentaur]

The Wikipedia page gives a good overview:

It's full of different instances of the effect and it has some good graphics but it doesn't actually include the words 'energy' or 'loss' and the word 'friction' only occurs in the bibliography. That, to my mind does not constitute a full explanation of what happens between Earth and Moon. The article makes mention of the rapid drop-off of the effect (inverse cube law), which explains why the effect is only seen in bodies with close separation.
If you are using the Wiki article as your source of information for me then there is no wonder you are not including the Energetic principles used. Wiki should always be read with care and using it as a slope source can be risky. Physics (and Science in general) advances by starting with simple systems and trying to understand the actual mechanism at work.

with a correct understanding of what "tidal locking" means.

Wiki does not supply that. It would be nice if you were to address my issue, rather than to ignore it and refer me only to that article. If you don't understand what I am getting at then, fair enough, say so but don't just dismiss it.

 

[snorkack]

Actually, even this is not true. There would still be torques on the tidal bulge even with a perfectly elastic Moon.

No. If Moon were perfectly elastic, there would be tidal bulges, but there would be no torques whatsoever on the tidal bulges, because the tidal bulges would move freely with rotation of Moon and be always exactly aligned with the tidal forces.

 

[sophiecentaur]

No. If Moon were perfectly elastic, there would be tidal bulges, but there would be no torques whatsoever on the tidal bulges, because the tidal bulges would move freely with rotation of Moon and be always exactly aligned with the tidal forces.

+1
No loss element would mean no phase shift and, hence, no net torque.

 

[PeterDonis]

It's full of different instances of the effect and it has some good graphics but it doesn't actually include the words 'energy' or 'loss' and the word 'friction' only occurs in the bibliography.

Yes. That is because you do not need the concepts of energy or friction to explain tidal locking. All you need are tidal bulges, torques acting on them, and the finite response time of elastic objects to distorting forces. The process conserves energy and it works on perfectly elastic objects so friction is not required.

That, to my mind does not constitute a full explanation of what happens between Earth and Moon.

I'm sorry, but this is not a valid argument. You need to respond to what the article does say, not what it doesn't say.

If you are using the Wiki article as your source of information

I asked you to read it because it seems to me to give a good simple presentation of the mechanism. What about that explanation do you think is wrong? Please be specific.

Wiki does not supply that. It would be nice if you were to address my issue, rather than to ignore it and refer me only to that article.

What about the mechanism described in the article don't you understand?

 

[PeterDonis]

If Moon were perfectly elastic, there would be tidal bulges, but there would be no torques whatsoever on the tidal bulges, because the tidal bulges would move freely with rotation of Moon and be always exactly aligned with the tidal forces.

This is not correct, because the response time of an elastic object to distorting forces is not instantaneous. It takes a finite time for the tidal bulges on the Moon to move in response to the changing tidal gravity of the Earth because of its orbital motion. That means the tidal bulges are never precisely aligned with the radial/tangential axes of the Moon, so there is a nonzero torque on them. The Wikipedia article makes precisely this point.

 

[sophiecentaur]

What about the mechanism described in the article don't you understand?

Why are you defending the article as a matter of principle? What it says is the equivalent of a pendulum with no loss will end up pointing vertically. It does not explain (doesn't even seem to consider) how the process of locking actually stops. The torques that it describes will pull one body round but why will it stop being pulled round? Why would the torques, once they have produced angular acceleration (which initially reduces the Moon's rotation), then produce a counter acceleration to stop it? If you can't answer with a technical reason, there is no point in referring me to the Wiki article again. Someone who 'knows' the right answer will have a satisfactory answer to the Energy question; do you?

 

[PeterDonis]

Why are you defending the article as a matter of principle?

I'm not "defending" the article "as a matter of principle". I'm saying the mechanism it describes looks valid to me.

It does not explain (doesn't even seem to consider) how the process of locking actually stops.

Sure it does:

"Tidal locking (also called gravitational locking or captured rotation) occurs when the long-term interaction between a pair of co-orbiting astronomical bodies drives the rotation rate of at least one of them into the state where there is no more net transfer of angular momentum between this body (e.g. a planet) and its orbit around the second body (e.g. a star); this condition of "no net transfer" must be satisfied over the course of one orbit around the second body."

There's the stopping condition.

Why would the torques, once they have produced angular acceleration (which initially reduces the Moon's rotation), then produce a counter acceleration to stop it?

We've already discussed this. You said:

The Moon would go past its maximum orbit size and then return ad infinitum.

And I agreed:

There will indeed be oscillations about an equilibrium

What I disagreed with was your claim that the equilibrium in question is an "energy minimum". It isn't: the torques acting to produce the oscillation about this equilibrium are conservative: total energy and angular momentum are conserved. That's why this equilibrium is not stable.

There have also been two additional factors discussed in this thread:

(1) The Moon's mass asymmetry, which produces an additional torque that acts to align the mass asymmetry radially. This torque is also conservative, so there will be oscillations about this equilibrium as well.

(2) Dissipation due to friction in both the Earth and the Moon. This acts to reduce the total energy of the system, which in turn means that the equilibrium point (the point where all torques are zero) shifts to a smaller average separation between the Earth and the Moon (i.e., the system becomes more tightly bound).

The only other point that hasn't yet been discussed is whether dissipation will also decrease the amplitude of oscillations about the equilibrium point. It seems to me that it would. So if that's the point you're trying to make, it's a valid point: but it certainly isn't the same as saying dissipation is what causes tidal locking. If the torques on the tidal bulges of the Earth and Moon (and on the Moon's mass asymmetry) were not present, the Moon would not be facing the same side to the Earth at all, and dissipation wouldn't change that; it would just slowly decrease the Moon's average distance from Earth, without driving its period of rotation toward its period of revolution.

 

[PeterDonis]

Dissipation due to friction in both the Earth and the Moon. This acts to reduce the total energy of the system, which in turn means that the equilibrium point (the point where all torques are zero) shifts to a smaller average separation between the Earth and the Moon (i.e., the system becomes more tightly bound).

Actually, this is incomplete. Dissipation due to friction in the Earth acts to slow the Earth's rotation, which in turn transfers angular momentum to the Moon and acts to increase its average distance from the Earth. This effect will continue until the Earth's rotation period has slowed to match the Moon's orbital period, i.e., until the Earth is tidally locked with the Moon. Once both bodies are tidally locked, I think dissipation will act as I described in the above quote.

 

[sophiecentaur]

There have also been two additional factors discussed in this thread:

There are a lot of separate factors but to understand more about a topic does not require more factors - it needs the prime factors to be identified in a very simple model. We were getting somewhere when the topic of a spherically symmetrical (when unstressed) planet was considered (the simplest situation, I think). The question is then about what will happen in the absence of losses and then when losses are present. For some reason, the Energy aspect seems to be ignored (including in the whole of the Wiki article). I find this quite amazing when pretty much every other topic in Physics can be analysed usefully in Energy terms. There is no surprise that I am complaining about that and I have been interpreting that in terms of lack of depth in the analysis.
The sort of explanation which says "It works because it works and here is an example" is not helpful and there has been a lot of that (Actually, you have been more helpful than most in that respect and I appreciate it).
You have picked me up about using the term Potential Minimum and then the term Energy Minimum but there has to be a valid term somewhere to describe the condition reached when the locking process is taken to its limit and two bodies have no more rotational energy to be exchanged between them. I am perplexed that you seemed to dismiss this idea initially and I'm really not impressed with what Wiki does not say about that.
I have come to terms with what is generally meant by the term 'locking' but there is a serious need for some sort of term to describe the final state which is reached due to 'locking'. I can't think why the term Drag can't be used for the cause and Locking for the final result. Too late now, as with many other terminologies used in Science. No doubt it was historical and the term came, not from Physicists but more practical 'observers'.
I guess this has run its course for me. Thanks for responding at great length. Fewer than 60 posts in one thread is no big deal on PF!

 

[snorkack]

We were getting somewhere when the topic of a spherically symmetrical (when unstressed) planet was considered (the simplest situation, I think). The question is then about what will happen in the absence of losses and then when losses are present. For some reason, the Energy aspect seems to be ignored (including in the whole of the Wiki article). I find this quite amazing when pretty much every other topic in Physics can be analysed usefully in Energy terms. There is no surprise that I am complaining about that and I have been interpreting that in terms of lack of depth in the analysis.
The sort of explanation which says "It works because it works and here is an example" is not helpful and there has been a lot of that (Actually, you have been more helpful than most in that respect and I appreciate it).
You have picked me up about using the term Potential Minimum and then the term Energy Minimum but there has to be a valid term somewhere to describe the condition reached when the locking process is taken to its limit and two bodies have no more rotational energy to be exchanged between them. I am perplexed that you seemed to dismiss this idea initially and I'm really not impressed with what Wiki does not say about that.
I have come to terms with what is generally meant by the term 'locking' but there is a serious need for some sort of term to describe the final state which is reached due to 'locking'. I can't think why the term Drag can't be used for the cause and Locking for the final result.

"Drag" is a necessary but not sufficient cause.
In order to lock a pendulum to a specific position, you will need two very different forces.
A restoring force and a retarding force.
If you don't have both, then there will be no locking.
If there is only a restoring force but no retarding force then the effect of a small nudge on a pendulum is to set it in oscillation - and the oscillation would last forever without damping. The average result of cumulation of many small nudges would be oscillating with increased amplitude, and eventually full swing.
If there is only a retarding drag force but no restoring force then the effect of a small nudge would be to move the object slightly and the drag would stop it in a new position after a finite and small distance covered (though not in finite time) - but there would be no return to starting position. The cumulative result of many small nudges would be drifting randomly a long way from starting position.

Moon is tidally locked because, and only because, Moon has both restoring and retarding forces.

 

[sophiecentaur]

Moon is tidally locked because, and only because, Moon has both restoring and retarding forces.

Music to my ears! That sums it up.
Now all I want is some sort of way of describing the Energy Situation that allows my idea that Stability means some sort of Energy Minimum, which would give an 'explanation' that's along the same lines as for any other ' locking' situation in life, rather than a description of what we see.

 

[snorkack]

Now all I want is some sort of way of describing the Energy Situation that allows my idea that Stability means some sort of Energy Minimum, which would give an 'explanation' that's along the same lines as for any other ' locking' situation in life, rather than a description of what we see.

Yes. An "energy minimum" is called "stable equilibrium", because it causes restoring force towards the minimum of potential energy.
You will also need a retarding force for tidal locking - otherwise the energy is conserved and the potential energy minimum is reached at kinetic energy such that total energy is not diminished and cannot reach the minimum of potential energy.

 

[PeterDonis]

An "energy minimum" is called "stable equilibrium", because it causes restoring force towards the minimum of potential energy.

And this is not the case for the restoring force on the Moon (the torque on the tidal bulge if the rotation and revolution periods do not match). The equilibrium point is not a potential energy minimum, as it is with a pendulum.

Moon is tidally locked because, and only because, Moon has both restoring and retarding forces.

But the "retarding force" (by which I assume you mean friction in the Moon's structure because it's not perfectly elastic) changes the equilibrium point by reducing the total energy. It does not simply damp oscillations about a fixed equilibrium, as it would with a pendulum.