General relativity and tidal forces

[Nugatory]

The problem is that if tidal lock occur due to torque, there should be skewing, meaning the far side would lag behind the near side.

Why so? When the far side lags, the torque acts to increase the rotation of the body, reducing the lag; when the far side leads the torque acts to reduce the rotation of the body and hence to reduce the lead. It's only when the body is rotating once per orbit so that the far side is neither lagging nor leading, the situation that we call tidal lock, that there is no net torque to further change the rotation.

 

[Nugatory]

Are there not internal shearing forces at work even when the body is in tidal lock (unless it's a ideal rod, in which case it it will be oriented radially and only under tension)? The body is being stretched in the radial direction and compressed tangentially.

However, these are internal forces trying to change the shape of the body and resisted by the rigidity of the body - they produce neither net force at the center of mass nor torque around the center of mass, so do not disturb the one revolution per orbit equilibrium of tidal lock.

Perfect.

Hmmm... I hope I have not inadvertently contributed to the confusion here. To be clear:

These forces (radial stretching and tangential compression) are there no matter what the rotation of the body is; they arise from gravity acting in slightly different directions and with slightly different strength on the various parts of the body.

Only when the body is rotating exactly once per orbit (so that the same side always faces in, the situation that we call gravitational lock) do these forces balance and produce no net torque at the center of mass.

 

[D H]

You point out one thing I figured out yesterday before I read your post just now: this is closer to an engineering problem that happens to be inside celestial mechanics, so it's no surprise we haven't given it a good amount of thought.

That is utter nonsense.

Any thoughts?

Yes. You need to stop posting nonsense and you need to start using math. You need to start doing that right now.

You posted nonsense in your other thread, which you apparently are trying to reinvoke. You repeatedly posted that there are no forces in general relativity, that scientists don't know / don't study the shapes of objects. Regarding the former, physicists talks about proper acceleration, model the interiors of stars with relativistic hydrodynamics. They wouldn't do either if there are no forces in general relativity. Regarding the latter bit of nonsense, which you have now repeated in this thread (see above), I suggest you google the phrase "lunar k2 love number" as a start.

Regarding your use (misuse) of logic: Start using math. If you don't know the math, you need to learn it before you start saying what physicists and astronomers know / don't know.

 

[russ_watters]

Are there not internal shearing forces at work even when the body is in tidal lock (unless it's a ideal rod, in which case it it will be oriented radially and only under tension)?

You are correct, but the OP's diagram appears to me to be analyzing just three points on that circular body in his diagram, so I was indeed treating it as a rod.

 

[russ_watters]

...but I urge you to notice that my problem is not with tidal forces, it is with the result shape of the body.

Ok...you haven't really explained why that would be though.

What I intended to point out in that diagram is the difference in tangential velocities only, later we entered the discussion regarding torque which is the main cause of tidal lock, so it is clear that torque must speed up the far side so it stays aligned with the near side towards the central mass. My problem is that this torque should cause shearing stress and strain, and the far side should be skewed behind the near side, but we don't see that in any of the bodies in tidal lock (at least I haven't found any evidence of this).

The torque exists until the body reaches tidal lock. Once in tidal lock, the bulges are directly aligned with the object being orbited, so there can't be any more torque.

On Earth, for example, our oceans move due to tidal forces, but since there is friction with the land, they aren't exactly aligned with the moon. Hence, a torque exists which is slowing our rate of rotation. Once the rate of rotation slows to equal the moon's rate of revolution, two sides of Earth will have permanent high tides.

And yes, I read the post, and it makes sense, I just think the forces would necessarily cause skewing, just like any torque causes stress, strain, shearing, distortion, yada yada.

What is caused, internally, by the tidal force is the tidal bulges.

 

[altergnostic]

Yes, apparently the OP's intuition is leading him astray somehow. I'm not quite sure how, exactly. Currently, all I"m hoping for is that he'll realize from the number of SA's and other posters that are disagreeing with his conclusions (about Newtonian gravity, nevermind general relativity) that he probably made a mistake somewhere.

It's hard to tell what the OP's background is in order to help him figure out where he's gone astray. From the lack of calculus in any of his arguments, I' suspect the OP's background probably doesn't include calculus :-(.

I haven't seen calculus in any posts, so I don't know why you are taking me to task for this. But regardless, I was trying some calculus solutions with my father yesterday (who is a mechanical engineer). One thing you will notice if you try it is that this is a huge, complex problem. Really, you can't calculate it with high precision overnight unless you are very obstinate, there are so many things to consider. But there's a way to solve it very roughly with algebra:

- Position the moon vertically with the axis aligned towards the earth.
- Divide the mass of the moon equatorially in two halves so that each half has the length of one moon radius and diagram each center of mass m1 and m2. Consider each half as a cube with half the volume of the moon - this will give you a block shape)
- Find the lengths of the sides of the cubes.
- Put the center of this system Mm at a distance D from a point Me, with D=moon-earth distance.
- Move the system (the moon) a bit in the direction of the constant innate tangential velocity as if there was no gravity - I moved it half the moon's radius). – You have to so this because if you don't, there's no orbit, no linear velocity and the moon would fall straight to earth, I hope you can see that.
- Turn on Earth's gravity at point Me and draw both gravity vectors from Me to point m1 and m2 and give them the right magnitudes (the magnitudes are not the same, you have to consider the distance between them - the correct value would be one moon radius, but since we're looking for the forces on the near and far sides I use the magnitudes of gravity on the near and far sides).
- Project the gravity vectors in the opposite direction of the linear innate velocity of the moon at right angles from the vertical axis of the moon (the line connecting m1, Mm and M2) and find their magnitudes.
- Subtract them and find the net force F.
- You can ditch the Earth now, we won't need it anymore. Apply F on either point m1 or m2 (it doesn't matter because all we have to do now is find out how much a force on one point skews the whole system).
- Consider a moon of iron (or choose the material you find closer to what's the main composition of the moon)
- Find the elasticity modulus of the material E (you can find it easily on the internet).
- The displacement of the point where F is applied is Δb.
- Calculate Δb:

Δb = FL³//3EI

where:
I = bh³//12

h=b= the length of the sides of the cubes = 1 moon radius
L= the diameter of the moon

If you do this you will see there's no possible way whatsoever that the moon could ever maintain it's shape with no shearing. The forces are too great and the moon is not nearly strong enough to overcome the shear stress.
The value I found for Δb was so large I can't believe it, either I've done something wrong, or... I don't even want to think about it. I'll take it back to my father to analyze, but meanwhile, why don't you guys try to calculate it and then we can compare the numbers?

The numbers I used are:
3474km - Diameter of the moon
3.84x10⁹ - Distance from the center of the Earth to the near side of the moon
3.85x10⁹ - Distance from the center of the Earth to the far side of the moon
211 GPa (Giga Pascal) - elasticity module of iron - I took the moon as a ball of iron for simplicity

Notice this gives us a very rough estimate and it's already a bit dense. If you want to do the complete analysis, you have to take the moon as a sphere and do a lot of calculus indeed, but it seems pointless to try and get a more precise number since just for this simple estimate the work is already a bit tiring, and to consider it fully we'd have to know the full composition of the moon, where each element is located, know their densities on the moon, test our numbers against each element (and do it in a laboratory since you won't find these on the net), know the strength of the bonds between the various elements, consider pressure from the inside out (mantle pressure, etc) and outside in (moon's own gravity), plug in the eccentricity of the moon and of it's orbit and on and on and on. Since all our numbers would be rough estimates from the start, I can live with the above analysis. Also see that final number is probably rounded up, since the moon has mainly 60% the density of the Earth (and so much less pressure and much less rigidity).

I conclude that a body in tidal lock should always be skewed, with the inner side ahead of the far side by a very clear amount, or even breaking up entirely. Either in GR or Newton. If you have forces of torsion, torques and anything like that, the body must suffer distortion. Always. Engineers know this very well and take very much care to consider the forces involved and how much of it a material can take without breaking up.

The point is that, if tidal lock is the result of a torque, the near side of the body should travel ahead of the far side in the direction of orbit, dragging the far side with it, skewing the shape of the body by a large amount. Since there's no absolute rigidity and instantaneous transmission of forces and gravity drops with the square of the distance, I'm starting to think that the only plausible scenario shearing wouldn't happen without reconsidering current theory is if the rotation is innate and just happens to match the orbit period by coincidence, but that can't be, there are too many bodies in tidal lock, too much to be just an accident. There must be a mechanism... but if it is due to torque, the shape of the bodies in tidal lock don't show any sign of it (namely, shearing).

I wasn't expecting this. I thought there should be shearing, but I was open to consider that it was too small to notice, that's why I went and did the calculations. The results just blew me away. I'm starting to believe that no one has ever thought of applying "engineering" equations to check if a body like the moon could hold itself together with the forces predicted/assumed by current theory. We see that it does, so we just assumed the forces were not strong enough to cause any significant stress, but I urge you to try the math yourselves.

Something is wrong.
Or I am wrong and desperately in need a careful consideration of the variables and calculations and a good explanation on why there's no shearing at all in tidal locked bodies.

 

[zonde]

I conclude that a body in tidal lock should always be skewed, with the inner side ahead of the far side by a very clear amount, or even breaking up entirely.

There are no forces in tangential direction when body is in tidal lock. All parts of body are moving inertially in tangential direction. The only forces are radial.

So please explain why do you think there should be forces that cause shear?

 

[altergnostic]

There are no forces in tangential direction when body is in tidal lock. All parts of body are moving inertially in tangential direction. The only forces are radial.

So please explain why do you think there should be forces that cause shear?

The problem is that you are considering the forces on a body after it has achieved tidal lock. But the forces that work to put a body in tidal lock are the same forces that cause shearing. Torque causes shearing every day in the real world. And if the force is constant, so is the shearing. Take a long plastic stick and spin it around, if you keep the force constant, you keep the stick bent by the same amount all the way around.

 

[russ_watters]

The problem is that you are considering the forces on a body after it has achieved tidal lock. But the forces that work to put a body in tidal lock are the same forces that cause shearing. Torque causes shearing every day in the real world. And if the force is constant, so is the shearing. Take a long plastic stick and spin it around, if you keep the force constant, you keep the stick bent by the same amount all the way around.

That's all fine. So what's the problem?

I conclude that a body in tidal lock should always be skewed, with the inner side ahead of the far side by a very clear amount...

Why do you believe that a body can be "skewed" and in tidal lock at the same time? Why do you not believe that if the body is "skewed", there will be a torque on it that is trying to "unskew" it? That's the entire point of tidal locking in a nutshell.

 

[altergnostic]

That is utter nonsense.

If it is nonsense please show me where these calculations have been done, where has shear strain have ever been calculated for a body in tidal lock? I looked for it and found none, I'd be glad if you post it here, though.

Yes. You need to stop posting nonsense and you need to start using math. You need to start doing that right now.

Done - previous to your "warning" (I've been preparing my last post since yesterday).

You posted nonsense in your other thread, which you apparently are trying to reinvoke. You repeatedly posted that there are no forces in general relativity, that scientists don't know / don't study the shapes of objects. Regarding the former, physicists talks about proper acceleration, model the interiors of stars with relativistic hydrodynamics. They wouldn't do either if there are no forces in general relativity. Regarding the latter bit of nonsense, which you have now repeated in this thread (see above), I suggest you google the phrase "lunar k2 love number" as a start.

Regarding the first bit of nonsense, gravity is not a force in GR, and I'm quoting Einstein, what's the problem? Other forces are still forces, just gravity isn't, if that wasn't clear, i hope it's cleared up now.
And I think love numbers are related to tidal compression and stretching, not to how the body holds itself against shear stress and strain, and this was already on your first wiki link on tidal locking where I clearly responded further that I've seen no consideration of shearing. Using your own words, I suggest you google "elasticity modulus" as a start.

Regarding your use (misuse) of logic: Start using math. If you don't know the math, you need to learn it before you start saying what physicists and astronomers know / don't know.

Regarding your use (misuse) of courtesy: start using math. If you don't know the math, you need to learn it before you start saying what I know / don't know. I'm not judging what you know/don't know, we don't even know each other, so drop it.

I think you're tone is starting to be a little disrespectful here. Have I been rude to you or anyone? I have been asking questions nicely and I just did post the math. If you are convinced I'm mistaken and don't feel like addressing my questions, that's fine, let others try to explain what am I missing. This has been a nice discussion so far. Maybe you can post some math or draw some diagrams as well instead of shushing and scoffing me off to wikipedia before you even consider my questions respectfully.

I don't want to take this unscientific debate any further, I will continue to stay focused on my questions and on physics (if you allow me). And I'd be happy if you did the same. The only "content" of your post was "google love number".

 

[altergnostic]

That's all fine. So what's the problem? Why do you believe that a body can be "skewed" and in tidal lock at the same time? Why do you not believe that if the body is "skewed", there will be a torque on it that is trying to "unskew" it? That's the entire point of tidal locking in a nutshell.

The torque is what causes skewing. If the body wasn't feeling any torques, there would be no reason for skewing! And I don't "believe" it, it is both logical and I have calculated it. The plastic stick is in tidal lock: the period of one revolution matches the period of rotation. If you are at the center turning around holding this stick, you will always be facing the near side and the far side will always be facing away. That's the entire point of tidal locking in a nutshell.

 

[russ_watters]

The torque is what causes skewing. If the body wasn't feeling any torques, there would be no reason for skewing!

Maybe it isn't clear what you mean by "skewing", since that isn't a scientific term. It sounded like you meant the moon would be symmetrically oval-shaped (like the theory predicts), but the oval would not aligned with the planet.

Now what bothers me here is that you're saying there is a net torque. But a net torque would cause a continuous acceleration -- the moon would start spinning faster and faster if there was a net torque: it wouldn't stay in your "skewed" alignment.

And I don't "believe" it, it is both logical and I have calculated it.

Er, no. One of the things that's getting on DH's nerves (and mine, frankly) is that you're making statements against established theory, while simultaneously displaying misunderstandings of how the theories work. I'm not sure who you think you are, but these theories have been around for one hundred and several hundred years, respectively and have been examined by millions of scientists in that time -- scientists who actually understand what the theories are saying. They aren't wrong. You are. You need to start dealing with that or this thread will be locked too. We teach established science here, we don't argue with crackpots about whether the science is right or wrong.

Second, your calculation - rather, the description of the setup - was incomprehensible. A diagram would be a big help, for us to know what you are calculating. Odds are, you just caluclated something nonsensical.

The plastic stick is in tidal lock: the period of one revolution matches the period of rotation. If you are at the center turning around holding this stick, you will always be facing the near side and the far side will always be facing away. That's the entire point of tidal locking in a nutshell.

If there is a net torque on it, then it isn't rotating at a constant rate.

...gravity is not a force in GR, and I'm quoting Einstein, what's the problem?

The problem is that you don't understand what that means and won't accept explanations of it.

 

[altergnostic]

Maybe it isn't clear what you mean by "skewing", since that isn't a scientific term. It sounded like you meant the moon would be symmetrically oval-shaped (like the theory predicts), but the oval would not aligned with the planet.

Yes, that's what I meant, it would be elliptical, but the far side would be a little behind in the direction of orbit.

Now what bothers me here is that you're saying there is a net torque. But a net torque would cause a continuous acceleration -- the moon would start spinning faster and faster if there was a net torque: it wouldn't stay in your "skewed" alignment. Er, no. One of the things that's getting on DH's nerves (and mine, frankly) is that you're making statements against established theory, while simultaneously displaying misunderstandings of how the theories work. I'm not sure who you think you are, but these theories have been around for one hundred and several hundred years, respectively and have been examined by millions of scientists in that time -- scientists who actually understand what the theories are saying. They aren't wrong. You are. You need to start dealing with that or this thread will be locked too. We teach established science here, we don't argue with crackpots about whether the science is right or wrong.

The constant torque here is the difference between the centripetal forces on the near and far side. There's a constant force on the near side that is stronger than the one on the far side. If there's a constant acceleration, there's a constant force, and a constant torque. And I am not even saying that current theory is wrong, what I am saying is that following the forces the theory predicts a body suffers to achieve tidal lock, the torque will cause shearing and the far side will not be aligned with the near side (although will be trying to). It isn't, and I am asking why. All I hear is that there should be no shearing, that somehow the torque causes no deformation on the orbiter and the far side reaches the point where it's aligned with the near side towards the central mass, but the only way I see this a possible is if the forces are too small compared to the rigidity of the body to notice, and that's the calculation I am seeking. I did it roughly, if the set up has flaws I would like you to point them out so I can correct them.

Second, your calculation - rather, the description of the setup - was incomprehensible. A diagram would be a big help, for us to know what you are calculating. Odds are, you just caluclated something nonsensical.
If there is a net torque on it, then it isn't rotating at a constant rate. The problem is that you don't understand what that means and won't accept explanations of it.

You are saying there shouldn't be any shearing expected, but I can't see that. The example of the plastic stick is clear, apply a constant torque on the near side, the far side will try to catch up but will never be able to, a constant centripetal force is a constant acceleration, but does not necessarily cause a varying period of rotation, in this problem it doesn't. This is clear from the plastic stick example.
I will try to diagram each step and post it since it is a complex calculation, but it will take some time. You are welcome to ask what is it that you don't understand from the description (I took it straight from a book on torsion, shearing, stress, strain, pressure and all sort of effects different forces have on materials, it's my father's so I don't know the name or the author, I can try to check it for you). I am not asking anyone to believe anything, the problem I need you to address is:

Before a body achieves tidal lock, there is a torque applied on the body. This torque emerges from the differing forces on the near and far sides, the near side experiencing a stronger force. This torque tends to push/pull the far side in line with it towards the central mass, but due to the elasticity/rigidity of the orbiting body, it should suffer shearing – the far side would not catch up to be perfectly aligned with the near side towards the central mass. Depending on the relations between the forces and the rigidity, the shearing will achieve a state of equilibrium at some point or, if the forces are strong enough, the body will tear apart. If the applied force is constant (which it is, gravity is a constant force/acceleration), the far side will always stay behind, it will never catch up. The body will achieve a state of equilibrium where the period of rotation matches the period of orbit, but the far side will not be aligned, the body will be elliptical with the far side behind the near side.

It is just like holding a plastic stick horizontally from one end and start spinning around at a constant rate. You have to apply a constant force, which will result in a constant torque. The period of "orbit" will match the period of rotation and the stick will behave just like if it was in tidal lock, but the far side will be behind, not perfectly aligned with the near side towards you. The plastic stick will be bent. The torque could only make the far side align perfectly with the near side if the stick was perfectly rigid. This is the shearing I am talking about.

 

[zonde]

You are saying there shouldn't be any shearing expected, but I can't see that. The example of the plastic stick is clear, apply a constant torque on the near side, the far side will try to catch up but will never be able to, a constant centripetal force is a constant acceleration, but does not necessarily cause a varying period of rotation, in this problem it doesn't. This is clear from the plastic stick example.

Before a body achieves tidal lock, there is a torque applied on the body. This torque emerges from the differing forces on the near and far sides, the near side experiencing a stronger force. This torque tends to push/pull the far side in line with it towards the central mass, but due to the elasticity/rigidity of the orbiting body, it should suffer shearing – the far side would not catch up to be perfectly aligned with the near side towards the central mass. Depending on the relations between the forces and the rigidity, the shearing will achieve a state of equilibrium at some point or, if the forces are strong enough, the body will tear apart. If the applied force is constant (which it is, gravity is a constant force/acceleration), the far side will always stay behind, it will never catch up. The body will achieve a state of equilibrium where the period of rotation matches the period of orbit, but the far side will not be aligned, the body will be elliptical with the far side behind the near side.

The far side will catch up with the near side and overcome it. Perfectly elastic body would be oscillating around middle point with far side sometimes being ahead and sometimes behind the near side.

The example of the plastic stick is clear, apply a constant torque on the near side, the far side will try to catch up but will never be able to, a constant centripetal force is a constant acceleration, but does not necessarily cause a varying period of rotation, in this problem it doesn't. This is clear from the plastic stick example.

A constant torque on the near side means that you are spinning faster and faster.
And then - yes, the far side will try to catch up but will never be able to.
And then - no, it does cause a varying period of rotation.
And centripetal force is not constant either in that case.

 

[Austin0]

You point out one thing I figured out yesterday before I read your post just now: this is closer to an engineering problem that happens to be inside celestial mechanics, so it's no surprise we haven't given it a good amount of thought. My father is an engineer, so I went to speak with him about this. He is not an astrophysicist, and knows squat about gravitational theories, so I had to give him some background. We talked for over 4 hours about all this, and in the end he said that we probably should expect shearing indeed, he had a book on shearing, skewing, stress, torques and all that regarding different materials, elasticity, etc - he knows a lot about this since he is a mechanical engineer and works with cars, engines, etc. At first he thought the forces wouldn't be great enough to cause shearing, that's why we don't see it, but when we plugged in some (rough) numbers, he started to consider otherwise. We even considered an iron only moon, and we ended up skeptical that there's no shearing. The problem is that if tidal lock occur due to torque, there should be skewing, meaning the far side would lag behind the near side.

It might help if you considered the mechanics of planetary condensation from an orbiting cloud.
If the forces and effects you are proposing as inferred from Keplerian mechanics were operating as you surmise, then how could condensation ever occur?
The differing orbital velocities at the near and far sides of the diffuse mass would prevent anything like spherical symmetry from ever forming wouldn't they? Would keep the matter smeared out around the orbit.
But it seems evident that the combined Ricci and Weyl effects bring it about as there are a bunch of pretty round planets

 

[Bandersnatch]

Hmmm, that stick analogy - isn't the far end lagging behind only due to air resistance?

 

[D H]

The constant torque here is the difference between the centripetal forces on the near and far side. There's a constant force on the near side that is stronger than the one on the far side. If there's a constant acceleration, there's a constant force, and a constant torque.

There is zero torque on an object that is circularly orbiting some other object and is tidally locked to that other object.

You are saying there shouldn't be any shearing expected, but I can't see that. The example of the plastic stick is clear, apply a constant torque on the near side, the far side will try to catch up but will never be able to, a constant centripetal force is a constant acceleration, but does not necessarily cause a varying period of rotation, in this problem it doesn't. This is clear from the plastic stick example.

That's a bad example, because once again there is no torque once the object does become tidally locked.

Before a body achieves tidal lock, there is a torque applied on the body. This torque emerges from the differing forces on the near and far sides, the near side experiencing a stronger force.

Wrong. The gravitational force at the point closest to the central body, furthest from the central body, or at any point along the line from the central mass to the center of mass of the body in question contribute absolutely nothing to the torque. That's Physics 101. It's the points off that line that result in gravity gradient torque. To first order, the Newtonian gravity gradient torque on some body located a distance $\vec R$ from some central mass and with a moment of inertia tensor $\boldsymbol I$ about the body's center of mass expressed in inertial coordinates is (without derivation; it's long) $3 \mu/R^5\, \vec R \times (\boldsymbol I \vec R)$, where $\mu=GM$ is the standard gravitational coefficient of the central mass.

Note that there is zero torque if the body is aligned such that one of its principal axes points toward the central mass. Now consider an object in a circular orbit whose rotational angular velocity is equal to the orbital angular velocity and that is aligned in this manner. It will always be aligned in this manner. This is tidal lock in its simplest form. (Aside: Note that there is zero torque of the body in question has a spherical mass distribution as any set of axes are principal axes for such a body. Bodies with a spherical mass distribution cannot become tidally locked.)

What if the object has a non-spherical mass distribution and isn't aligned in this manner? Now you will get a non-zero gravity gradient torque, and that torque will be in the direction of tidal lock conditions. If the body truly is a rigid body what you'll get is something akin to harmonic motion. Truly rigid bodies cannot become tidally locked. A non-rigid body will have some plasticity or viscosity to it, and this in turn means that some of that change in rotational energy that would nominally result from that gravity gradient torque instead heats the body (which is eventually radiated away). These dissipative forces, along with a non-spherical mass distribution, are a necessary conditions for tidal lock to occur.

What if the orbit isn't circular, or the central mass doesn't have a spherical mass distribution, or there are other objects around (we're no longer in the two body problem world)? Things become a bit more complicated here. Now there are time-varying tidal forces on the body. The Moon, for example, has small but observable time varying body tides. The observability of these time varying body tides is one of the legacies of the Apollo program. The Apollo astronauts left retroreflectors on the Moon. The McDonald Observatory has collected a large time series of measurements of the distances between the observatory and these retroreflectors. These data yield a very accurate picture of the Moon's orbit and also of the Moon itself, even it's interior.

 

[russ_watters]

altergnostic, are you familiar with the moon's libration? The moon rocks back and forth exactly like a big, solid pendulum would on your desk. When it rocks to the left, the torque pulls to the right. When it rocks to the right, the torque pulls to the left. When centered, there is no torque.

Perhaps if you drew a diagram showing the forces acting on a body in tidal lock, it would help here.

 

[altergnostic]

The far side will catch up with the near side and overcome it. Perfectly elastic body would be oscillating around middle point with far side sometimes being ahead and sometimes behind the near side.

I think the far side would catch up and oscillate if the point from where the torque is applied (the near side) was not accelerating, that's right.

A constant torque on the near side means that you are spinning faster and faster.
And then - yes, the far side will try to catch up but will never be able to.
And then - no, it does cause a varying period of rotation.
And centripetal force is not constant either in that case.

The centripetal force isn't constant?? If it wasn't, the moon would either fly off or fall down. Consider this: the tangential velocity of the moon is constant, if the acceleration from gravitational force gets weaker, the moon would start to orbit farther away, the opposite would be true if the force increases. The force from gravity is constant, the torque is constant, the period of rotation is constant. What am I missing?

 

[altergnostic]

It might help if you considered the mechanics of planetary condensation from an orbiting cloud.
If the forces and effects you are proposing as inferred from Keplerian mechanics were operating as you surmise, then how could condensation ever occur?
The differing orbital velocities at the near and far sides of the diffuse mass would prevent anything like spherical symmetry from ever forming wouldn't they? Would keep the matter smeared out around the orbit.
But it seems evident that the combined Ricci and Weyl effects bring it about as there are a bunch of pretty round planets

But if that is the case, than we wouldn't need torques to create a tidal locked body in the first place, would we? Everything would be naturally in tide lock.

 

[altergnostic]

Hmmm, that stick analogy - isn't the far end lagging behind only due to air resistance?

Nope. Imagine a light year long stick in vacuum and start to spin it.

 

[altergnostic]

There is zero torque on an object that is circularly orbiting some other object and is tidally locked to that other object.
That's a bad example, because once again there is no torque once the object does become tidally locked.

I agree, but you are considering the body after it is in tidal lock. The point is that there is a torque for the body to reach tidal lock, and my argument is that the body will reach tidal lock but will suffer shearing.

Wrong. The gravitational force at the point closest to the central body, furthest from the central body, or at any point along the line from the central mass to the center of mass of the body in question contribute absolutely nothing to the torque. That's Physics 101. It's the points off that line that result in gravity gradient torque. To first order, the Newtonian gravity gradient torque on some body located a distance $\vec R$ from some central mass and with a moment of inertia tensor $\boldsymbol I$ about the body's center of mass expressed in inertial coordinates is (without derivation; it's long) $3 \mu/R^5\, \vec R \times (\boldsymbol I \vec R)$, where $\mu=GM$ is the standard gravitational coefficient of the central mass.

I agree, that's why we have proposed tidal bulges to produce the torque needed to lock the body.

Note that there is zero torque if the body is aligned such that one of its principal axes points toward the central mass. Now consider an object in a circular orbit whose rotational angular velocity is equal to the orbital angular velocity and that is aligned in this manner. It will always be aligned in this manner. This is tidal lock in its simplest form. (Aside: Note that there is zero torque of the body in question has a spherical mass distribution as any set of axes are principal axes for such a body. Bodies with a spherical mass distribution cannot become tidally locked.)

I agree as well, torque is produced from traveling tidal bulges. Plus, with no gravity, the linear velocity alone would not cause rotation, so we may assume that the linear velocity would tend to keep the far side from turning in alignment with the near side, but that's not fundamental. I'm considering the situation a bit like Newton solved the orbital speed, he let the body travel straight to the point where it would go without gravity, than he worked on the distance from that point to where the object ends up in an orbit.

What if the object has a non-spherical mass distribution and isn't aligned in this manner? Now you will get a non-zero gravity gradient torque, and that torque will be in the direction of tidal lock conditions. If the body truly is a rigid body what you'll get is something akin to harmonic motion. Truly rigid bodies cannot become tidally locked. A non-rigid body will have some plasticity or viscosity to it, and this in turn means that some of that change in rotational energy that would nominally result from that gravity gradient torque instead heats the body (which is eventually radiated away). These dissipative forces, along with a non-spherical mass distribution, are a necessary conditions for tidal lock to occur.

I haven't seen tide lock explained by heat the way you state, could give me a reference? And the body isn't aligned when forces are acting on the body to achieve tidal lock. And truly rigid bodies can't be tidal locked, I agree.

What if the orbit isn't circular, or the central mass doesn't have a spherical mass distribution, or there are other objects around (we're no longer in the two body problem world)? Things become a bit more complicated here. Now there are time-varying tidal forces on the body. The Moon, for example, has small but observable time varying body tides. The observability of these time varying body tides is one of the legacies of the Apollo program. The Apollo astronauts left retroreflectors on the Moon. The McDonald Observatory has collected a large time series of measurements of the distances between the observatory and these retroreflectors. These data yield a very accurate picture of the Moon's orbit and also of the Moon itself, even it's interior.

You are correct, these would complicate a lot the problem, but I am trying to simplify things to understand the basis of the forces that create tidal lock. Also notice that the semi major axis of the moon is not from the near to the far sides, but from the leading to the trailing sides... actually I found differing statements regarding this online, so I am not completely certain of this, but everywhere you look you will find that the near side is shallower than the rest of the moon. But that will just complicate things up, consider any other body with tidal bulges front and back and the forces that work on it before it is in tidal lock, and notice that these forces, specially the torque, should cause shearing. Everywhere you have, you have to expect shearing, and to keep a body in a constant orbit, you have to apply a constant force, if you turn off the centripetal force, the body flies off at the tangent. So we have a constant centripetal force, a constant linear velocity and a constant torque that must cause shearing, am I wrong? Or is gravity not a constant force/acceleration? If you turn off the tangential velocity, won't the body fall down? Doesn't that imply a constant force/acceleration?

 

[altergnostic]

altergnostic, are you familiar with the moon's libration? The moon rocks back and forth exactly like a big, solid pendulum would on your desk. When it rocks to the left, the torque pulls to the right. When it rocks to the right, the torque pulls to the left. When centered, there is no torque.

Perhaps if you drew a diagram showing the forces acting on a body in tidal lock, it would help here.

Yes, I am. There's a beautiful animation of it on the wikipedia page on the moon. And I don't disagree with you, I am not contesting that bodies are in tidal lock in real life, with no shearing, I am trying to point out that the forces (constant centripetal and torque) that act on the body before it achieves tidal lock should cause shearing, shouldn't they? You are saying that there's no constant torque, because if there was the period of rotation would vary, but there must be a constant torque if the centripetal force is constant - gravity is a constant. Are you saying that gravity is not a constant centripetal force/acceleration?

I'll try to diagram what I am visualizing more accurately and post it, maybe it will really help. I'll try to do it tonight, but I have an astrophysics class today that I don't want to miss... maybe I can talk with the teacher about this and bring something new to this discussion.

 

[Nugatory]

Nope. Imagine a light year long stick in vacuum and start to spin it.

The relativistic effects that matter across a light-year (about 10¹⁶ meters) are completely irrelevant at the diameter of the moon (less than 10⁷ meters. There are simply no relativistic phenomena involved in understanding the tidal lock of these systems (yes, you CAN solve them using the methods of the GR gravitational theory... but the first step in that solution is to apply the weak-field approximation and that gives you the same results as Newtonian gravity).

Bandersnatch is correct. If no torque is applied to the non-relativistic rotating rod it will be straight (although if has been elastically deformed by an applied torque it will take a moment for the deformation to relax and the rod to regain that undeformed straight shape). The only reason for the outer edge to lag would be drag, atmospheric or otherwise.

 

[Nugatory]

But it seems evident that the combined Ricci and Weyl effects bring it about as there are a bunch of pretty round planets

In this context, is "combined Ricci and Weyl effects" anything but a long-winded way of talking about classical Newtonian gravity, which is completely sufficient to explain the phenomenon of near-spherical planets?

It's been said a few times in this thread, but it may need repeating: Relativity, both special and general, is completely irrelevant to the situations we're considering here. The only reason to introduce GR into the discussion would be as a GR learning exercise: Show that weak-field GR yields the same results as the well-understood Newtonian solution. And understanding the Newtonian solution is a prerequisite for taking on that exercise.

 

[altergnostic]

The relativistic effects that matter across a light-year (about 10¹⁶ meters) are completely irrelevant at the diameter of the moon (less than 10⁷ meters. There are simply no relativistic phenomena involved in understanding the tidal lock of these systems (yes, you CAN solve them using the methods of the GR gravitational theory... but the first step in that solution is to apply the weak-field approximation and that gives you the same results as Newtonian gravity).

Bandersnatch is correct. If no torque is applied to the non-relativistic rotating rod it will be straight (although if has been elastically deformed by an applied torque it will take a moment for the deformation to relax and the rod to regain that undeformed straight shape). The only reason for the outer edge to lag would be drag, atmospheric or otherwise.

I was not talking about relativistic effects, i was talking about deformation due to torque. I think the point of our misunderstanding is that you consider that the torque is not constant, but if the force from gravity is constant, so is the torque. And this force is always greater on the near side. Think of it this way: the law of inertia states the if there are no forces on a body, it will remain in its rest state, either traveling at a constant linear velocity or standing still (which are the same thing). To maintain a stable curved path, like an orbit, the centripetal acceleration must be constant. This constant force generates a constant torque, like in a mery go round. If you don't hold on to something, you will fly off at the tangent. This must mean a constant centripetal force, and if the body is not falling towards the central mass, but orbiting, there is a constant force holding it in its curved path. This constant force is gravity, and it generates a constant torque. Why do you believe the torque ceases?

 

[D H]

I agree, but you are considering the body after it is in tidal lock. The point is that there is a torque for the body to reach tidal lock, and my argument is that the body will reach tidal lock but will suffer shearing.

You're argument is wrong, and has gone on far too long.

I haven't seen tide lock explained by heat the way you state, could give me a reference?

This is old, old stuff, some going back to George Darwin and A.E.H. Love. More recently, but still old stuff,

Goldreich P. and Soter S. (1966). Q in the Solar System. Icarus 5, 375-389.
MacDonald, G.J.F (1964). Tidal Friction, Rev. Geophsys 2, 467-541.

You don't even understand the basics, and you are trying to argue? Thread locked.