altergnostic said: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.
Nugatory said: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.
altergnostic said:Perfect.
That is utter nonsense.altergnostic said: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.
Yes. You need to stop posting nonsense and you need to start using math. You need to start doing that right now.altergnostic said:Any thoughts?
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.Nugatory said: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)?
Ok...you haven't really explained why that would be though.altergnostic said:...but I urge you to notice that my problem is not with tidal forces, it is with the result shape of the body.
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.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).
What is caused, internally, by the tidal force is the tidal bulges.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.
pervect said: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 :-(.
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.altergnostic said: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.
zonde said: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?
That's all fine. So what's the problem?altergnostic said: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.
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.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...
D H said:That is utter nonsense.
Done - previous to your "warning" (I've been preparing my last post since yesterday).D H said:Yes. You need to stop posting nonsense and you need to start using math. You need to start doing that right now.
D H said: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.
D H said: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 said: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.
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.altergnostic said:The torque is what causes skewing. If the body wasn't feeling any torques, there would be no reason for skewing!
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.And I don't "believe" it, it is both logical and I have calculated it.
If there is a net torque on it, then it isn't rotating at a constant rate.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.
The problem is that you don't understand what that means and won't accept explanations of it....gravity is not a force in GR, and I'm quoting Einstein, what's the problem?
russ_watters said: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.
russ_watters said: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.
russ_watters said: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.
altergnostic said: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.
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.altergnostic said: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.
A constant torque on the near side means that you are spinning faster and faster.altergnostic said: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.
altergnostic said: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.
There is zero torque on an object that is circularly orbiting some other object and is tidally locked to that other object.altergnostic said: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.
That's a bad example, because once again there is no torque once the object does become tidally locked.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.
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 [itex]\vec R[/itex] from some central mass and with a moment of inertia tensor [itex]\boldsymbol I[/itex] about the body's center of mass expressed in inertial coordinates is (without derivation; it's long) [itex]3 \mu/R^5\, \vec R \times (\boldsymbol I \vec R)[/itex], where [itex]\mu=GM[/itex] is the standard gravitational coefficient of the central mass.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.
zonde said: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.
zonde said: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 said: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 said:Hmmm, that stick analogy - isn't the far end lagging behind only due to air resistance?
D H said: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.
D H said: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 [itex]\vec R[/itex] from some central mass and with a moment of inertia tensor [itex]\boldsymbol I[/itex] about the body's center of mass expressed in inertial coordinates is (without derivation; it's long) [itex]3 \mu/R^5\, \vec R \times (\boldsymbol I \vec R)[/itex], where [itex]\mu=GM[/itex] is the standard gravitational coefficient of the central mass.
D H said: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.)
D H said: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.
D H said: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 said: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 said:Nope. Imagine a light year long stick in vacuum and start to spin it.
Austin0 said:But it seems evident that the combined Ricci and Weyl effects bring it about as there are a bunch of pretty round planets
Nugatory said: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.
You're argument is wrong, and has gone on far too long.altergnostic said: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.
This is old, old stuff, some going back to George Darwin and A.E.H. Love. More recently, but still old stuff,I haven't seen tide lock explained by heat the way you state, could give me a reference?