General relativity and tidal forces

[altergnostic]

Tides on Earth are described with Newton's theory of gravitation. Relativistic effects on tides theoretically become measurable on very strong gravitational fields, possibly becoming twice as strong as tides predicted by Newtonian gravity: http://adsabs.harvard.edu/abs/1983ApJ...264..620N

Tides are presumably outcomes of gravitational forces. Einstein ditched forces and the concept of inertia in GR (http://archive.org/stream/TheBornEinsteinLetters/Born-TheBornEinsteinLetters_djvu.txt). So how is GR used to calculate tidal forces? If different parts of the body travel different geodesics, this would cause the body to tear apart over time. How can tides be described with the geometry of space-time?

 

[Nugatory]

So how is GR used to calculate tidal forces? If different parts of the body travel different geodesics, this would cause the body to tear apart over time. How can tides be described with the geometry of space-time?

If the forces holding the body together are strong enough, they will accelerate the different parts of the body off their geodesics and onto non-geodesic worldlines that stay close enough that the body doesn't tear apart. The center of mass of the object follows a geodesic and the other parts of the body experience fictitious forces that tend to pull the body apart and are resisted by whatever forces hold the body together.

These fictitious forces are tides.

 

[pervect]

Tides on Earth are described with Newton's theory of gravitation. Relativistic effects on tides theoretically become measurable on very strong gravitational fields, possibly becoming twice as strong as tides predicted by Newtonian gravity: http://adsabs.harvard.edu/abs/1983ApJ...264..620N

Tides are presumably outcomes of gravitational forces. Einstein ditched forces and the concept of inertia in GR (http://archive.org/stream/TheBornEinsteinLetters/Born-TheBornEinsteinLetters_djvu.txt). So how is GR used to calculate tidal forces? If different parts of the body travel different geodesics, this would cause the body to tear apart over time. How can tides be described with the geometry of space-time?

Tidal forces, when suitably defined, can be identified as being components of the Riemann curvature tensor.

Under most circumstances, taking the tidal force as one would measure it via Newtonian means ( a couple of accelerometers separated by a rigid rod) is an excellent approximation to (one of the) geometric definitions, which is related to the apparent relative acceleration of nearby geodesics which are initially parallel.

In fact, you write earlier (this is a very good insight)

If different parts of the body travel different geodesics, this would cause the body to tear apart over time.

The point is that when you measure the forces needed to hold a rigid body together, to keep it rigid, you are indirectly measuring "how fast" the geodesics would expand (accelerate away from each other) if said restoring forces did not exist.

MTW's textbook "Gravitation", and a number of other textbooks, take this approach, though MTW is perhaps the textbook which invites the reader to take it most seriously.

The full Riemann curvature can, given a local description of time (a frame of reference, for instance, more formally a timelike congruence of worldlines) be decomposed by the Bel Decomposition http://en.wikipedia.org/w/index.php?title=Bel_decomposition&oldid=512613685 into three parts. One part, called the electrogravitic tensor, describes static gravity, and givenby the above "tidal forces", so the Newtonian tidal tensor can be pretty much directly be linked to the electrogravitic part of the Riemann tensor.

Another part, called the magnetogravitic tensor, described frame-dragging effects (which affect moving bodies, but don't directly affect static bodies). A third part, the topogravitic tensor, describes spatial curvature.

The Bel decomposition, unfortunately, is usually given short shrift in textbooks, so it may be hard to find a formal treatment I was introduced to it rather informally here on PF, for instance.

 

[altergnostic]

All right, so I think I understand the basis of how GR creates tides, even though I have a minor issue with fictitious forces counteracted by real forces in this particular case, but that's a subject for another thread, I don't want to get into this here.
My problem with this analysis is that different parts of the body are not only trying to travel different geodesics, but also, in order to stay rigid, the outer parts of the rotating orbiter must have a faster tangential component than the inner parts. This is a bigger problem when you consider a body in tidal lock, such as the moon. Different geodesics would tend to rip the body apart radially, but different velocities would tend to shear te body in the line of orbit. In all gravitational theories, the tangential component of the velocity can't be caused by the gravitational field and is a constant. Newton called it the body's innate velocity.

Think of it this way: a body is at a constant linear velocity, so all parts of the body travel at the same linear speed. Then it is captured by a gravitational field. It starts to orbit that planet and the linear velocity is not changed, but the body is accelerated into a curved motion (or continues to have a constant velocity in curved spacetime). The tangential component must still be the same for all parts of the body since gravity imparts no (fictitious) forces tangentially. This would tend to cause shearing, especially in a body in tidal lock. If it doesn't start to shear, presumably the tangential velocities are different on different parts of the body, with the far side faster then the near side (this would be true for all bodies, not only the ones in tidal lock), but how can that be? What forces act on the body to change the tangential (innate) velocities if gravity has no way to do so neither with Newton nor with GR? Are tides capable of changing tangential velocities in either GR or Newton's theory?

 

[pervect]

While "centrifugal" forces, i.e. forces due to rotation, do contribute to the strain on a rigid bar, they do not contribute to the Riemann curvature tensor, which is ultimately based on how fast geodesics separate (or converge). The "force-on-a-bar" idea is very useful, but it can only be used if/when the bar isn't rotating.

So in your orbiter example, either you'd need to imagine that your spacecraft was not rotating (in which case in 1/2 an orbit the outer side would be the inner side, assuming no frame dragging effects), or if your space-craft is tide-locked, you'd have to manually subtract the forces due to its absolute rotation (once per orbit, again assuming no frame dragging) from the measured strain on the bar to get the tensor components.

The Electrogravitic component of the Riemann tensor must be traceless. The centrifugal forces on a rotating sphere are not traceless, this is one way you can tell if a system is rotating.

Accurate measurements of the gravity tensor, typically using rather exotic means such as superconductors and SQUID's for the detectors, are an expensive, but semi-routine, part of modern prospecting. Some interesting references are http://www.physics.umd.edu, http://www.bellgeo.com/tech/technology_theory_of_FTG.html , and http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA496707/GRE/NASA_SGG.pdf . These papers describe some of the modern techniques that are actually used to measure the gravity tensor. The last has some discussion of the physics as well, though it's oriented mostly towards Newtonian gravity.

The Wiki article is also mildly helpful, http://en.wikipedia.org/w/index.php?title=Gravity_gradiometry&oldid=508813691, giving a list of some of the basic systems that have been implemented.

 

[altergnostic]

While "centrifugal" forces, i.e. forces due to rotation, do contribute to the strain on a rigid bar, they do not contribute to the Riemann curvature tensor, which is ultimately based on how fast geodesics separate (or converge). The "force-on-a-bar" idea is very useful, but it can only be used if/when the bar isn't rotating.

I'm not sure i follow. Forces due to rotation cause strain, but can only be used if the bar is not rotating?

So in your orbiter example, either you'd need to imagine that your spacecraft was not rotating (in which case in 1/2 an orbit the outer side would be the inner side, assuming no frame dragging effects), or if your space-craft is tide-locked, you'd have to manually subtract the forces due to its absolute rotation (once per orbit, again assuming no frame dragging) from the measured strain on the bar to get the tensor components.

If the orbiter does not rotate on its axis, as in your first example, everything is fine, because there's no variation in tangential speeds. But if the body is in tidal lock, then how does the tangential velocities adjust to maintain the body in position? Or, if they don't adjust, how come there's no shearing?

 

[russ_watters]

... different velocities would tend to shear te body in the line of orbit.

Why do you say that? Sounds like an issue with geometry to me, like you think that different velocities mean different parts of the object are moving apart. What you're missing is that the object is rotating. Spin a pencil on your desk and you'll see that different parts travel at different speeds, but there is no shear.

But if the body is in tidal lock, then how does the tangential velocities adjust to maintain the body in position? Or, if they don't adjust, how come there's no shearing?

A torque is produced if a body's tidal bulge is not aligned with the source of the bulge: http://en.wikipedia.org/wiki/Tidal_locking

 

[altergnostic]

Why do you say that? Sounds like an issue with geometry to me, like you think that different velocities mean different parts of the object are moving apart. What you're missing is that the object is rotating. Spin a pencil on your desk and you'll see that different parts travel at different speeds, but there is no shear. A torque is produced if a body's tidal bulge is not aligned with the source of the bulge: http://en.wikipedia.org/wiki/Tidal_locking

Sorry, I meant to say that equal tangential velocities would tend to cause shearing, not different velocities. I'm making a diagram to clarify.

 

[altergnostic]

Ok, I'm having a hard time putting the problem into words so I drew a diagram that may help to clarify the whole thing:
http://www.pictureshoster.com/files/aix43ezq7zp99daihgzu.jpg
http://www.pictureshoster.com/files/aix43ezq7zp99daihgzu.jpg

If different parts of the body have different velocities, there's no shearing. But if all parts of the body have the same linear velocity, then a body in tidal lock should exhibit shearing.

If there's no shearing, it follows that different parts of the body have different tangential components. Suppose that a body is traveling in a straight line at a constant velocity. It passes near a second body and starts to orbit it. If nothing else happens, the body wouldn't start to rotate on it's own axis and in "1/2 an orbit the outer side would be the inner side" just like pervect said. For a body in tidal lock, something seems to affect linear velocities on the near and far side so that the far side orbits faster than the near side.

My question is what causes the changes in these tangential velocities. Or alternatively, why is there no shearing. Since gravity has no tangential component neither in Newton nor in GR, I'm lost.

At first I thought maybe it had something to do with the rigidity of the body and tidal forces. I thought that gravity would accelerate the near and far sides differently and force the body in tidal lock, accelerating the far side more than the near side, but this turned out to be a dead end.

I think the make up of the problem is clearer now with my diagram, and I believe my logic is consistent. A body in this situation must either shear or achieve different tangential velocities on the near and on the far sides. If this is the case, what is the cause of these changes?

 

[D H]

I think the make up of the problem is clearer now with my diagram, and I believe my logic is consistent.

Your logic is incorrect.

You are looking in the wrong directions. First look at a spec of mass on the orbiting body at the point furthest from the central mass. The velocity of that spec of mass is a function of the radius of the orbiting body and the velocity of the orbiting body's center of mass. Now imagine what would happen if that orbiting body wasn't there; all you have is the spec of mass as an orbiting body. That free particle will follow a different path than would our spec of mass. In particular, it would move outward. The tidal force at this point is radially outward away from the central mass, not tangential. Similarly, the tidal force on the point closest to the central mass is radially inward, toward the central mass. In both cases, the tidal force is away from the center of mass of the orbiting body. There is also a lesser effect (about half as much) for a particle at the leading and trailing points on the orbiting body. A free particle at those leading and trailing points would follow different paths than will a particle fixed to the orbiting body, but now the tidal force is directed toward the center of mass of the orbiting body. The end result is that the tidal forces act to pull the object apart radially, squeeze it together tangentially.

Newtonian mechanics and general relativity agree on the above description so long as the central mass isn't particular massive or the distance to the central mass is sufficiently large. The reason for this agreement is that space-time is locally flat. "Locally" is a fairly large volume in these weak field circumstances, at least as far as physicists are concerned. (Mathematicians will disagree; local means infinitesimally small to them.) Make the central body massive enough or close enough and those weak field approximations become invalid. You'll start seeing effects that result from the curvature of space-time. Newtonian mechanics and relativity diverge at this point. Newtonian mechanics does not properly describe the extreme spaghettification that results from close proximity to extremely massive objects.

 

[pervect]

If you connect the "red x"s on the same point on a non-rotating body, you'll get an elliptical orbit more like the one I'll attach, the picture you drew isn't right.

If you go through the math (using just Newtonian theory), you'll get the results DW quoted.

In particular, if the radial Newtonian force is -GM/r^2, differentiating this with respect to r gives the well-known result for the radial tidal force 2GM/r^3. See also the wiki article, http://en.wikipedia.org/w/index.php?title=Tidal_tensor&oldid=332450104.

It takes more work to go through the math to get the compressive tidal forces, the results dW quotes are correct however.

The results for GR are formally similar to the Newtonian results in a local frame-field, if you replace the radial distance "r" with the radial coordinate "r" for the Schwarzschild metric. Local frame fields seem to confuse more readers than they should, the math to compute them is somewhat involved, but the end result is just the forces/fields/tensors that a local observer would measure with local clocks and local rulers.

 

[altergnostic]

First look at a spec of mass on the orbiting body at the point furthest from the central mass. The velocity of that spec of mass is a function of the radius of the orbiting body and the velocity of the orbiting body's center of mass. Now imagine what would happen if that orbiting body wasn't there; all you have is the spec of mass as an orbiting body. That free particle will follow a different path than would our spec of mass. In particular, it would move outward. The tidal force at this point is radially outward away from the central mass, not tangential. Similarly, the tidal force on the point closest to the central mass is radially inward, toward the central mass. In both cases, the tidal force is away from the center of mass of the orbiting body.

I completely agree. But this does not address the problem. In the diagram, I hid the effects of tides, I want to focus on the tangential components of the velocities. I am aware that the specs would move apart, I actually said that explicitly in my first post.

What keeps the near and far side from moving apart, which was answered in the beginning of the post, are the forces holding the body together, so it is clear that tidal forces tend to elongate the body radially, and it is clear that neither tides nor gravity have mechanisms to act tangentially on the near and far sides to change these velocities.

Please be patient since my question has not been answered. Leave tidal forces out for a moment and consider this:

- A body Y moves in a straight line at a constant velocity V.
- All parts of the body have the same linear velocity V.
- This body starts to orbit a central mass M.
- M imparts a centripetal acceleration due to gravity on Y.
- Y feels no tangential forces from M, and no drag.
- Y achieves tidal lock: the far side has a faster tangential velocity than the inner side.

Now, if Y did not achieve tidal lock, and no forces other then gravity from M act on it, we may assume that there is no change in the tangential velocities and there would be no fixed near and far sides, the tides would travel, the body would appear to rotate as seen from M, the body would not be rotating in it's own axis locally (relative to it's own orbit) and everything is fine.

But if Y orbits in tidal lock, the far side orbits faster than the near side, and we must assume a change in the tangential velocities of each sides, namely, a positive acceleration on the far side and a negative one on the near side.

Gravity can't cause these tangential accelerations, so there's no way M causes it. The body Y also cannot impart forces on itself, so again, it cannot be the cause of these changes.

My question is: what causes the changes in the tangential velocities of the near and far sides?

 

[Nugatory]

The body Y also cannot impart forces on itself, so again, it cannot be the cause of these changes.

The body Y cannot apply forces to itself, but one part of the body can apply forces to another part. Replace the body with a cloud of dust, such that each dust particle is connected to its neighbors with a spring. Now if two of the particles are on diverging geodesics, the spring between them will be stretched, applying a force to both particles that will accelerate them off their inertial path and onto other geodesics. Obviously this process can change the shape of the body/cloud; but it can also change the velocity of one part of the cloud relative to another, as in tidal locking.

 

[Nugatory]

Gravity can't cause these tangential accelerations

Even though it acts only in a radial direction, gravity can produce tangential accelerations. Consider, for example, dropping a horizontally oriented bar towards the surface of the earth. Each end of the bar will experience a gravitational force pointing towards the center of the earth. These two vectors are not parallel; if I extend them far enough they will intersect at the center of the earth, so they are ever so slightly converging. Thus, both vectors have a small tangential component that is pushing the ends of the rod towards the middle and is resisted by the rigidity of the rod.

It's true that these forces are balanced so they cannot affect the center-of-mass movement of the rod - but that's true of tidal locking of a rotating body as well.

 

[altergnostic]

That's a good point, but isn't that one of the reasons the body should resist changes on tangential velocities on the near and far sides? For example, if I try to push only the far side, these internal bonds (rigidity) would tend to transfer forces to the near side (and everywhere else). But the force is only transferred in this case, isn't it? That means that there must be an external force acting on the tangent in this case, and this force would be resisted by rigidity.

You mention tangential accelerations caused by gravity in a rigid rod falling horizontally towards the central mass. But in this case, the body is not in orbit, and if it was (even if slowly falling), the forces would be on the trailing and leading sides, wouldn't they? And this would only help the radial bulging anyway. These tangential components are not capable of altering tangential velocities on the near and far sides. A good analogy would be to hold the same rod vertically wrt Earth and throw it straight towards the horizon (ignore drag). Would the far side of the rod start to speed up relative to the near side? Of course not. But that would be a rod in tidal lock.

Since there's no tangential forces in gravity, I don't see where these changes in tangential velocities come from. The stretching you mentioned with the springs, caused by gravity, is only radial, it would cause radial stretching (tidal bulges). Exactly how can this radial force change the tangential velocities of the near and far sides?

 

[Mentz114]

It might be constructive to look at some actual figures. For a test particle in the Schwarzschild vacuum the proper acceleration felt by an observer in a circular path u is

<br /> \frac{du}{d\tau} = \frac{m\,{\cos\left( \theta\right) }^{2}}{{r}^{2}-2\,m\,r}\ dr -\frac{m\,\cos\left( \theta\right) \,\sin\left( \theta\right) }{r-3\,m}\ d\theta<br />
If the particle is entirely in the equatorial plane (EP, θ = π/2) both components are zero, and the orbit is a geodesic. Considering a small sphere in orbit, with its centre following the geodesic we can make some deductions about the forces required to keep the whole body together.

The acceleration in the θ-direction ( the one pointing up and down from the equatorial plane) will change sign between the two halves so it always acts towards the EP causing compressive stress. In the radial direction, leaving the EP in either direction will cause an increase in the acceleration in the +ve r-direction. I don't know what deformation this causes. Both components get smaller as the radius increases so it looks like there is also some stretching in the r and θ-directions.

The acceleration above is calculated from the Schwarzschild circular orbit geodesic in the coordinate basis with 4-velocity
<br /> u=\frac{\sqrt{r}\,\sqrt{m\,{\sin\left( \theta\right) }^{2}+r-3\,m}}{\sqrt{r-3\,m}\,\sqrt{r-2\,m}}\ \partial_t -\frac{\sqrt{m}}{r\,\sqrt{r-3\,m}}\ \partial_\phi<br />
This is only a geodesic if sin(θ) = 1. So any circular path off the EP must apply forces to offset the accelerations engendered.

 

[zonde]

But if Y orbits in tidal lock, the far side orbits faster than the near side, and we must assume a change in the tangential velocities of each sides, namely, a positive acceleration on the far side and a negative one on the near side.

Gravity can't cause these tangential accelerations, so there's no way M causes it. The body Y also cannot impart forces on itself, so again, it cannot be the cause of these changes.

My question is: what causes the changes in the tangential velocities of the near and far sides?

Let's say that this body in tidal lock does not have perfect spherical symmetry. Then as it rotates it's gravity affects large mass M differently. Now let's say that body M is not entirely rigid so that there are slightly different responses to small body in different rotation angle positions.
Another possibility. Body in orbit itself is not entirely rigid and then radial stretch elongates the body in different directions as it rotates. If this response to stretch in different directions is different then shouldn't it be possible to dump angular momentum using that radial force?

 

[Nugatory]

A good analogy would be to hold the same rod vertically wrt Earth and throw it straight towards the horizon (ignore drag). Would the far side of the rod start to speed up relative to the near side? Of course not.

When you're holding the rod vertically, it's exactly lined up with the radial gravitational force; both ends of the rod and the center of the Earth lie in the same line, and the gravitational force acts along that line. But when you throw the rod it moves off that vertical line. Now the forces acting on the two ends are no longer exactly parallel (extend the vectors and they meet at the center of the earth, therefore are very slightly converging). Furthermore, the forces have very slightly different magnitudes, because the distance to the center of the Earth is different for the two ends.

So we have forces of different magnitude acting in different directions on the two ends of the rod. Why shouldn't one end speed up relative to the other?

 

[altergnostic]

Mentz114, Thanks for this. Correct me if I'm wrong, i take this to mean that anything above the equatorial plane suffers an accel towards the equatorial plane. That would be the flattening at the poles of tidal theory. As you also pointed out, there should also be stretching in the equatorial plane (i assume only on the near and far sides) which would be the tidal bulges. Still, i see no component that would cause different tangential forces on the near and far sides that could adjust velocities to cause a body to achieve tidal lock, is this correct?

 

[altergnostic]

Let's say that this body in tidal lock does not have perfect spherical symmetry. Then as it rotates it's gravity affects large mass M differently. Now let's say that body M is not entirely rigid so that there are slightly different responses to small body in different rotation angle positions.
Another possibility. Body in orbit itself is not entirely rigid and then radial stretch elongates the body in different directions as it rotates. If this response to stretch in different directions is different then shouldn't it be possible to dump angular momentum using that radial force?

I have considered this, and i believe that would tend to cause shearing. If the tangential velocities on the near and far sides remain unchenged, these forces that you mention would work against this stretching, but not overcome it. If a body has almost no rigidity, it could not hold itself together and it would shear (near side ahead of the far side), stretch radially and tear apart.

 

[altergnostic]

When you're holding the rod vertically, it's exactly lined up with the radial gravitational force; both ends of the rod and the center of the Earth lie in the same line, and the gravitational force acts along that line. But when you throw the rod it moves off that vertical line. Now the forces acting on the two ends are no longer exactly parallel (extend the vectors and they meet at the center of the earth, therefore are very slightly converging). Furthermore, the forces have very slightly different magnitudes, because the distance to the center of the Earth is different for the two ends.

So we have forces of different magnitude acting in different directions on the two ends of the rod. Why shouldn't one end speed up relative to the other?

This is a more thorough analysis. But wouldn't this cause the near side to accelerate more than the far side? Garvity is stronger nearer the central mass than farther away. The rod would actually start to rotate slightly backwards, the near side being faster than the far side, the far side getting behind. Also, these forces still have no tangential component, they are only radial. I don't see these forces accelerating the far side more than the near side in any way, especially tangentially. Agreed?

 

[Nugatory]

But wouldn't this cause the near side to accelerate more than the far side? Garvity is stronger nearer the central mass than farther away. The rod would actually start to rotate slightly backwards, the near side being faster than the far side, the far side getting behind.

I threw the rod, so it's moving in some direction. The tiny horizontal components of the gravitational forces acting on the ends of the rod are acting against the direction of movement so are slowing the ends. The inner end, subject to the stronger horizontal force, is slowed more so ends up behind.

But that's a detail... The more important point is

Also, these forces still have no tangential component, they are only radial. I don't see these forces accelerating the far side more than the near side in any way, especially tangentially. Agreed?

No. If one end is getting behind the other, then the speed of one or both has necessarily changed since the moment they were thrown. That's acceleration, by definition.

Also important is that you have to be very careful with the word "radial" and "tangential" when you're dealing with non-point objects, because the directions identified by those words change from point to point, and therefore from one place to another on a non-point object like a planet or a rod or a cloud of dust particles connected by springs. A purely radial force at the center of gravity of an object has a tangential component at another part of the object.

At least in classical mechanics, the net force on the body must sum to being a radial force through the center of mass, but other parts of the body can feel forces that try to move them relative to that center of mass. These can rotate the object (thrown vertical rod), compress it horizontally (stationary horizontal rod), as well as stretch it vertically or change its shape.

 

[altergnostic]

I threw the rod, so it's moving in some direction. The tiny horizontal components of the gravitational forces acting on the ends of the rod are acting against the direction of movement so are slowing the ends. The inner end, subject to the stronger horizontal force, is slowed more so ends up behind.

I see what you mean, it is tricky indeed. Let's look at it with GR: the far side follows a larger geodesic than the near side of the rod. There are no forces or accelerations. The tangential velocity of the rod is the same all over, so it would travel the outer geodesic at the same speed as the inner geodesic, and since the inner geodesic is smaller, the near side gets ahead of the far side. Just like two race cars with the same velocity in their speedometers, making a turn in an inner and an outer track lane – the car in the outer lane has to travel a longer distance, and ends up behind the car in the inner lane.

Considering Newton, it's a bit trickier (who would imagine!): in visualizing the rod oriented vertically to the central mass, traveling linearly (trying to describe no curves - curves would be the result of gravity alone) with a constant linear velocity, I see your point. I can also visualize it differently, though: the inner side has a greater pull than the far side, and drags the far side down with it (more than gravity alone would if acted on a particle at the same distance as the far side). That would cause the far side to experience basically the same acceleration down as the near side for a rigid rod. The vector addition of the tangential velocity and the centripetal acceleration would result in vectors with the direction and magnitude for both sides, and since the far side has a longer distance to travel, we end up with the same situation as the analysis above.

Also important is that you have to be very careful with the word "radial" and "tangential" when you're dealing with non-point objects, because the directions identified by those words change from point to point, and therefore from one place to another on a non-point object like a planet or a rod or a cloud of dust particles connected by springs. A purely radial force at the center of gravity of an object has a tangential component at another part of the object.

I agree we have to be very careful. In the diagram I posted, you can see that a change the directions of the tangents precisely the way you propose. But it's very tricky without diagrams, I admit.
But I disagree with your analysis of the tangential components... The tangential component you specified is only apparent, meaning, it is relative to other parts of the body, not relative to the central mass. The centripetal acceleration have no tangential components of its own. You have to consider that all resulting motions are vector additions of a centripetal acceleration and a linear tangential velocity, nothing else. If one side has a different tangential velocity and describes a different curve than the other, it must be due to this vector addition, and the only tangential component is the body's innate linear velocity. Remember that Einstein also claimed that gravity can't work on the tangents in GR.

At least in classical mechanics, the net force on the body must sum to being a radial force through the center of mass, but other parts of the body can feel forces that try to move them relative to that center of mass. These can rotate the object (thrown vertical rod), compress it horizontally (stationary horizontal rod), as well as stretch it vertically or change its shape.

Agreed. Now, none of that can make the far side orbit faster than the near side so a body achieves tidal lock, can it?

The best way to visualize all this, I think, is to consider 2 small bodies with the same linear velocity, one starting to orbit closer to the central mass than the other. The one closer to the central mass would complete a revolution before the one farther away. These forces should be equivalent to the forces experienced by different parts of a large orbiting body, the only difference is the reaction to these forces due to rigidity. But these reactions can't move the far side faster than the near side on a body in tidal lock, so I still can't figure out what creates the tidal lock.

 

[altergnostic]

One more thing: both the centripetal acceleration and the forces due to rigidity take time to travel from the near side to the far side in Newton's theory, so the far side would lag behind the near side. In GR, the same is true for the forces due to rigidity.

 

[Nugatory]

The best way to visualize all this, I think, is to consider 2 small bodies with the same linear velocity, one starting to orbit closer to the central mass than the other. The one closer to the central mass would complete a revolution before the one farther away.

That's not possible, at least not if the orbits are circular and I'm understanding "linear velocity" correctly as "the speed of the object, which is also its velocity in its tangential direction at a particular moment". The speed needed to keep an object in a circular orbit in an inverse-square field like gravity is a function of the radius of the orbit (compare the acceleration \frac{v^2}{r} with the force producing that acceleration \frac{K}{r^2} to see this) so they can't both have the same linear velocity.

The best way to visualize all this, I think, is to consider 2 small bodies with the same linear velocity, one starting to orbit closer to the central mass than the other...These forces should be equivalent to the forces experienced by different parts of a large orbiting body, the only difference is the reaction to these forces due to rigidity. But these reactions can't move the far side faster than the near side on a body in tidal lock, so I still can't figure out what creates the tidal lock.

OK, back to tidal lock...

Consider your two bodies orbiting at slightly different radii and joined by a spring. The tidal lock problem comes down to explaining how this system ends up oriented vertically so that both objects complete an orbit in the same time - which implies that the ensemble of two objects and a spring is rotating about its own center of mass once per orbital period.

Assume for definiteness that both the orbit and the positive sense of rotation of the body-spring system is counter-clockwise. If the body-spring system is rotating too slowly, less than one rotation per orbit (this also covers the case of zero rotation and clockwise rotation), then the outer object will lag behind the inner object. Conversely if the body-spring system is rotating at more than one rotation per orbit, then the outer object will move ahead of the inner object.

Now look at the torque around the center of mass of the body-spring system when the outer object is leading and when it is lagging; we know that there will be some torque because both the direction and the strength of the forces on the two bodies is different. When the outer body is leading, that torque acts against the rotation, and when the inner body is leading that torque tends to act with the rotation. That is, when the body-spring system is rotating around its center of mass at less than one rotation per orbit, the torque from the different forces on the two ends acts to increase the rotation rate; and when the body-spring system is rotating at more than that rate, the torque acts to reduce it.

Thus, we expect the system to stabilize at exactly one revolution per orbit, and that's tidal lock.
----

There are some complexities that I've glossed over. Among them:
- As long as we aren't tidally locked, there will be some points in the orbit when the two objects are at the same distance from the center and the spring is compressed, and other points where they are at different distances and the spring is stretched. Energy is dissipated in the stretching and relaxing of the spring as the body-spring system orbits (or in a real stone-and-rock satellite, as the tidal bulge moves through its crust with the rotation of the satellite) until the stable tidal lock situation is reached. Thus, tidal locking may result in some loss of orbital kinetic energy, and we may end up with a lower orbit post-lock than pre-lock.
- Angular momentum has to be conserved. When locking changes the angular momentum of the body-spring system about its center of mass, something else has to change as well. This will either be the orbital angular momentum of the body-spring system (see above for how it can change) or the rotation of the parent body (so far we've assumed the parent body to be a fixed point source, but of course it's not, so these tidal effects are at work on it as well).

 

[altergnostic]

That's not possible, at least not if the orbits are circular

You are correct, maybe I oversimplified the problem. When I said "started to orbit" I wasn't implying the bodies would complete the orbit, but then I negligently stated that the inner body would complete one revolution before the outer body, which deserves your criticism. What I should have said is that, since the outer body has to travel a longer distance in order to stay aligned with the inner body towards the central mass (conjunction configuration), it would get behind the inner body because they have the same linear velocities.

Now look at the torque around the center of mass of the body-spring system when the outer object is leading and when it is lagging; we know that there will be some torque because both the direction and the strength of the forces on the two bodies is different.

When the outer body is leading, that torque acts against the rotation, and when the inner body is leading that torque tends to act with the rotation. That is, when the body-spring system is rotating around its center of mass at less than one rotation per orbit, the torque from the different forces on the two ends acts to increase the rotation rate; and when the body-spring system is rotating at more than that rate, the torque acts to reduce it.

Following your logic, when the bodies are supposedly stabilized in tidal lock, we will always have a greater force from gravity on the inner body than on the outer body, so there's a tendency to increase the distance from one to the other. Also, the force that would cause the torque has to travel through the spring until it reaches the outer body, at a later time. As a result, the outer body will always be behind, and this is precisely the shearing I was talking about in the beginning of this thread. When you have torque in the real world, you can be certain that you'll get some shearing.

Thus, we expect the system to stabilize at exactly one revolution per orbit, and that's tidal lock.

Ok, plus shearing.

 

[altergnostic]

I believe we have struck the heart of the problem. If tidal lock was really achieved by these mechanisms, there should be measurable shearing, but we have no evidence of that on the moon. I'd also like to point out that we've been mostly using Newton, but if you take GR, where there are no forces (only geodesics), you get to shearing much faster and with a much simpler analysis. The outer body has to travel a larger geodesic – a longer distance – than the inner body. Both have the same constant linear velocity and cross the same distance over the same time, so logically the outer body gets behind. Connect them with a spring and you get shearing, just like Newton.

I think we are facing a real problem here. It is clear that the rigidity of the body (the spring we are using to simplify the problem) has no possible mechanism to produce enough torque to keep the far side aligned towards the central mass without shearing – that would mean infinite rigidity, and transmission of forces at infinite speed! Something else must be accelerating the far side so it stays aligned. In other words, tidal locked bodies have some mechanism that adjusts the tangential components of the orbital velocities on the near and far sides so they match precisely the magnitudes needed to maintain alignment towards the central mass, and it can't be the torque due to rigidity, and it can't be gravity directly.

Any thoughts?

 

[Nugatory]

Following your logic, when the bodies are supposedly stabilized in tidal lock, we will always have a greater force from gravity on the inner body than on the outer body, so there's a tendency to increase the distance from one to the other

Yes, that is exactly what happens. As long as that outward pull doesn't exceed the strength of the forces holding the orbiting body together, the resulting situation is stable. The Earth's tidally locked moon is an example - the near and the far side are being pulled apart, but the lunar rock is strong enough that the moon doesn't fall apart. If the tidal forces were stronger or the forces holding the moon together were weaker the Earth wouldn't have a moon, it would have a ring like Saturn. But as long as the moon doesn't disintegrate the tidal locked situation is stable (Try googling for "Roche limit").

Also, the force that would cause the torque has to travel through the spring until it reaches the outer body, at a later time.

Also quite true, but for most of the weak-field situations that we're considering here not all that important. For example, a force applied to one side of the moon will propagate to the other side in a thousand seconds or so while the moon rotates about its axis once every million seconds or so... If we treat the propagation as instantaneous, we'll be close enough.

 

[russ_watters]

I believe we have struck the heart of the problem. If tidal lock was really achieved by these mechanisms, there should be measurable shearing, but we have no evidence of that on the moon.

Any thoughts?

This may be the heart of your misunderstanding, but there is no such "problem" with science's understanding of tidal force.

In the diagram you posted on the first page, the two examples show exactly why shearing occurrs when a body is not in tidal lock (the first diagram) and why there is no shearing when the body is in tidal lock (the second diagram).

It almost looks from your posts like you are saying that both diagrams show the body in tidal lock. Is that the issue? If the body isn't rotating, then it isn't in tidal lock. Tidal lock means the body is rotating at the same rate as it is revolving.

From that post:

My question is what causes the changes in these tangential velocities. Or alternatively, why is there no shearing.

Did you read the wiki link I posted? It has a nice diagram showing the forces involved in causing a tidal lock.

 

[Nugatory]

I think we are facing a real problem here. It is clear that the rigidity of the body (the spring we are using to simplify the problem) has no possible mechanism to produce enough torque to keep the far side aligned towards the central mass without shearing – that would mean infinite rigidity, and transmission of forces at infinite speed! Something else must be accelerating the far side so it stays aligned. In other words, tidal locked bodies have some mechanism that adjusts the tangential components of the orbital velocities on the near and far sides so they match precisely the magnitudes needed to maintain alignment towards the central mass, and it can't be the torque due to rigidity, and it can't be gravity directly.

I'm sorry, but I don't see the problem you see - which probably means that we're differing on some underlying assumption that hasn't come to the surface yet.

Tidal lock develops very slowly over a period of many orbits as the modest torque generated by the imbalanced forces on the leading and lagging parts slowly pushes the system towards one rotation per orbit. There's no need for infinite rigidity or instantaneous transmission of force. All that's necessary is that the forces propagate quickly compared to the orbital and rotational periods, and that the the tidal forces not be so strong that they tear the system apart instead of locking it.

A digression: You've mentioned "shear" several times. We'd probably be better off speaking in terms of tension and compression - shear is a combination of tension and compression working in different directions, and it's much easier to analyze tidal forces in terms of tension and compression than the combination. The compressive forces act tangentially (the horizontal rod is compressed) while the tensile ones act radially (the vertical rod is stretched). Of course the people building bridges and buildings and airplane wings think in terms of shear as a force in its own right... But they're working with a different class of problems.

 

[Nugatory]

the two examples show exactly why shearing occurrs when a body is not in tidal lock (the first diagram) and why there is no shearing when the body is in tidal lock (the second diagram).

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.

 

[pervect]

This may be the heart of your misunderstanding, but there is no such "problem" with science's understanding of tidal force.

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 :-(.

 

[altergnostic]

This may be the heart of your misunderstanding, but there is no such "problem" with science's understanding of tidal force.

Probably science's understanding sees no problem with tidal forces, otherwise I would have found this discussion somewhere, but I urge you to notice that my problem is not with tidal forces, it is with the result shape of the body.

In the diagram you posted on the first page, the two examples show exactly why shearing occurrs when a body is not in tidal lock (the first diagram) and why there is no shearing when the body is in tidal lock (the second diagram).

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).

It almost looks from your posts like you are saying that both diagrams show the body in tidal lock. Is that the issue? If the body isn't rotating, then it isn't in tidal lock. Tidal lock means the body is rotating at the same rate as it is revolving.
From that post:
Did you read the wiki link I posted? It has a nice diagram showing the forces involved in causing a tidal lock.

No, that's not the issue, as I said, I was only trying to point out the difference in tangential velocities. I actually, the diagram has a mistake, now that you mention it: those difference in tangential velocities would never cause shearing as I drew it, it would cause rotation, the shearing I'm talking about here is due to the torque in a body in tidal lock. 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.

 

[altergnostic]

I'm sorry, but I don't see the problem you see - which probably means that we're differing on some underlying assumption that hasn't come to the surface yet.

Tidal lock develops very slowly over a period of many orbits as the modest torque generated by the imbalanced forces on the leading and lagging parts slowly pushes the system towards one rotation per orbit. There's no need for infinite rigidity or instantaneous transmission of force. All that's necessary is that the forces propagate quickly compared to the orbital and rotational periods, and that the the tidal forces not be so strong that they tear the system apart instead of locking it.

A digression: You've mentioned "shear" several times. We'd probably be better off speaking in terms of tension and compression - shear is a combination of tension and compression working in different directions, and it's much easier to analyze tidal forces in terms of tension and compression than the combination. The compressive forces act tangentially (the horizontal rod is compressed) while the tensile ones act radially (the vertical rod is stretched). Of course the people building bridges and buildings and airplane wings think in terms of shear as a force in its own right... But they're working with a different class of problems.

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.

 

[altergnostic]

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.

 

[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