Why the Moon always shows the same face

[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.

 

[PeterDonis]

Music to my ears!

Don't rejoice too soon; the Moon is not like a pendulum in some key respects. See my post #61.

 

[sophiecentaur]

The equilibrium point is not a potential energy minimum, as it is with a pendulum.

Of course, because there is motion in any orbit too (unlike with a stationary pendulum) - but the total Energy reduces due to losses until it reaches a situation where the Energy will not be dissipated any more. That is 'a' minimum and I have already made this point. The pendulum analogy is simpler but you can't just kick the Energy approach because it's hard to include. There are certain principles which run through all of Physics; it's only a matter of finding where they are lurking.
Let's face it, no one here has demonstrated in detail how the 'torque' argument produces stable locking. Wiki has a very nice animation but does that constitute a proof?

 

[PeterDonis]

the total Energy reduces due to losses until it reaches a situation where the Energy will not be dissipated any more.

No, because, strictly speaking, there is no such situation. At least, not as long as we want the Earth and the Moon to be intact. If you wait long enough and dissipation reduces the total energy enough, the Moon will end up inside the Roche limit and will break apart and the Earth will have a system of rings. (This is considering the Earth-Moon system in isolation only; in our actual case, I believe the Sun will become a red giant long, long before the Earth-Moon system gives up enough energy from dissipation to bring the Moon inside the Roche limit.)

no one here has demonstrated in detail how the 'torque' argument produces stable locking.

I have already said, several times now, that the equilibrium of zero torque is not stable. The zero torque equilibrium is not a potential energy minimum. So the reason no one has demonstrated how torques produce stable locking is that they don't.

 

[PeterDonis]

If you wait long enough and dissipation reduces the total energy enough, the Moon will end up inside the Roche limit

I should probably clarify this. This is a very, very long-term effect of dissipation. Perhaps it will help to describe how we would expect the Earth-Moon system to evolve in the future, assuming it to be isolated (i.e., ignoring things like the Sun becoming a red giant).

Right now, the Moon's rotation and revolution periods match, but the Moon's orbit is elliptical and is inclined to the Earth's equatorial plane. So there are unbalanced torques in the system that tend to make the Moon's orbit circular and to bring its orbital plane into Earth's equatorial plane. Eventually those things will happen.

However, there are also unbalanced torques in the system because the Earth is rotating much faster than the Moon's rate of revolution about the Earth. These torques tend to slow the Earth's rotation and move the Moon's orbit further out. Eventually, the Earth's rotation will have slowed to match the Moon's rate of revolution, at which point there will indeed be a zero torque equilibrium (assuming that the Moon's orbit has also become circular and in the Earth's equatorial plane by that time). If we ignore dissipation, the system will oscillate with some small amplitude about this equilibrium forever.

Dissipation, while the above equilibrium has not yet been reached, will mainly be friction inside the Earth and Moon due to them not being perfectly elastic. This will act to reduce the total energy of the system, which I think will make the zero torque equilibrium described above happen at a slightly smaller Earth-Moon separation than it would have in the absence of dissipation (but still a significantly larger separation than exists today).

Once the system has reached the point where it is oscillating about the zero torque equilibrium, dissipation will be, I think, much smaller than it is today, because the difference at any point in time between the actual rotation/revolution rates and their equilibrium ones will be much smaller than any differences today, so there will be much less frictional damping. Dissipation will act to reduce the amplitude of the oscillations, but it will also still be reducing the total energy of the system, which will act to reduce the Earth-Moon separation, making the system more tightly bound.

If a point is reached at which the oscillations have been damped away, there will still be dissipation within the system due to the emission of gravitational waves, but this will be much, much smaller than any frictional damping was. Even so, it will still act to (very, very slowly) reduce the Earth-Moon separation by reducing the total energy of the system. And at some point, the separation will be small enough that the Moon will be inside the Roche limit. That is what I was referring to in the statement of mine quoted above.

 

[sophiecentaur]

I should probably clarify this. This is a very, very long-term effect of dissipation. Perhaps it will help to describe how we would expect the Earth-Moon system to evolve in the future, assuming it to be isolated (i.e., ignoring things like the Sun becoming a red giant).

Right now, the Moon's rotation and revolution periods match, but the Moon's orbit is elliptical and is inclined to the Earth's equatorial plane. So there are unbalanced torques in the system that tend to make the Moon's orbit circular and to bring its orbital plane into Earth's equatorial plane. Eventually those things will happen.

However, there are also unbalanced torques in the system because the Earth is rotating much faster than the Moon's rate of revolution about the Earth. These torques tend to slow the Earth's rotation and move the Moon's orbit further out. Eventually, the Earth's rotation will have slowed to match the Moon's rate of revolution, at which point there will indeed be a zero torque equilibrium (assuming that the Moon's orbit has also become circular and in the Earth's equatorial plane by that time). If we ignore dissipation, the system will oscillate with some small amplitude about this equilibrium forever.

Dissipation, while the above equilibrium has not yet been reached, will mainly be friction inside the Earth and Moon due to them not being perfectly elastic. This will act to reduce the total energy of the system, which I think will make the zero torque equilibrium described above happen at a slightly smaller Earth-Moon separation than it would have in the absence of dissipation (but still a significantly larger separation than exists today).

Once the system has reached the point where it is oscillating about the zero torque equilibrium, dissipation will be, I think, much smaller than it is today, because the difference at any point in time between the actual rotation/revolution rates and their equilibrium ones will be much smaller than any differences today, so there will be much less frictional damping. Dissipation will act to reduce the amplitude of the oscillations, but it will also still be reducing the total energy of the system, which will act to reduce the Earth-Moon separation, making the system more tightly bound.

If a point is reached at which the oscillations have been damped away, there will still be dissipation within the system due to the emission of gravitational waves, but this will be much, much smaller than any frictional damping was. Even so, it will still act to (very, very slowly) reduce the Earth-Moon separation by reducing the total energy of the system. And at some point, the separation will be small enough that the Moon will be inside the Roche limit. That is what I was referring to in the statement of mine quoted above.

I am quite happy to accept that the situation as it is and how it will be is more or less as you describe it. Many of these things have already all been mentioned in the thread. However, I am more interested in what happens in the very simplest system than in a load of instances because the whole thing becomes divergent instead of homing in on some actual understanding. Plenty of time to expand when the basics have been established.
What you are saying does not disprove what I suggest, at all. I realize that you don't want to look at the problem this way but it surprises me that you don't find the Energy analysis approach is of use when you find it elsewhere, all over Physics. You talk about dissipation getting less but you don't want to acknowledge a parallel between this sort of system and other damped oscillating systems. Never mind, I have found out a lot from you and others.

Once the system has reached the point where it is oscillating about the zero torque equilibrium, dissipation will be, I think, much smaller than it is today,

Is it surprising that the rate of Energy Dissipation will follow a basically exponential law? It applies to other systems that we come across.

 

[PeterDonis]

I realize that you don't want to look at the problem this way but it surprises me that you don't find the Energy analysis approach is of use

Didn't I say that dissipation will reduce the total energy of the system? How is that ignoring energy?

What I did say is that the zero torque equilibrium is not a minimum of potential energy, and you agreed with that. That means you can't analyze that equilibrium the way you would analyze an equilibrium that is a minimum of potential energy, like a pendulum.

You talk about dissipation getting less but you don't want to acknowledge a parallel between this sort of system and other damped oscillating systems.

It seems to me that you are the one failing to acknowledge something: the key difference between this system and other damped oscillating systems, which I have just repeated above.

 

[PeterDonis]

it surprises me that you don't find the Energy analysis approach is of use when you find it elsewhere, all over Physics

I've given a pretty complete description of what the system will do. How do you think an energy analysis approach would improve it?

 

[PeterDonis]

Is it surprising that the rate of Energy Dissipation will follow a basically exponential law?

I'm not sure it will in the regime I was referring to in what you quoted (after the Earth and Moon are both tidally locked and the system is oscillating with small amplitude about the zero torque equilibrium). In that regime, as I said, dissipation doesn't just reduce the amplitude of oscillations; it moves the equilibrium, because the equilibrium depends on the total energy of the system (since that determines the Earth-Moon distance, which affects the rotation/revolution rate). So, again, it's not the same as a system like a pendulum, where the equilibrium does not depend on the total energy.

In the regime the actual Earth-Moon system is in now, where the Moon is tidally locked (mostly--its orbit is still not circular in the Earth's equatorial plane) but the Earth isn't, I don't think the rate of energy dissipation is even close to following an exponential law.

 

[PeterDonis]

In actual objects there is also dissipation going on, but its effects are very small compared to the effects of the torques.

Looking at the Wikipedia page on tidal acceleration...

https://en.wikipedia.org/wiki/Tidal_acceleration
...which gives a detailed treatment of the slowing of the Earth's rotation by the Moon (and the corresponding increase in the Moon's orbital distance), it seems that, at least for the case of the Earth's rotation, my intuition here was way off. According to this article (which gives several references to papers describing detailed measurements of tidal dissipation in the Earth), only a small fraction (about 1/30 if I'm reading the article correctly) of the total rotational kinetic energy lost by the Earth as its rotation slows is transferred to the Moon along with angular momentum; the rest is converted to heat.

 

[sophiecentaur]

I've given a pretty complete description of what the system will do. How do you think an energy analysis approach would improve it?

Because it would demonstrate the commonality between such phenomena. Why exclude it from all the other mechanical, electrical, thermal and vibrational studies? The problem here seems to be that you are finding reasons why not to consider what I suggest. I really don't see why parallels between phenomena don't strike you as 'insightful'.