Is there an aether to determine if an object rotates?

[WannabeNewton]

Using this notion of local (While I"m sure it's standard, I dont' alas have a reference for it) the partial derivatives of the metric at a point give you both the readings of "linear" accelerometers and of "rotational" accelerometers. So there really isn't any problem determining when something is rotating - we have instruments that can measure it, directly, and additionally, given a metric, you can compute rotation mathematically from the Christoffel symbols - the same Christoffel symbols that you need mathematically to compute the acceleration from the metric.

There's no need for a reference pervect. While we only need a single event to write down quantities that can be completely described by the basis vectors of a frame, the test for Fermi transport requires covariant derivatives of the frame basis vectors and as such we need neighborhoods of events to perform the Fermi transport test. The same goes for accelerometer tests since such tests require covariant derivatives of the time-like basis vector.

 

[Fantasist]

Look carefully at your picture of the gravitational hammer thrower - it's not correctly describing the rotational behavior of an object in gravitational orbit. The way you've drawn it, the side that is facing the top of the picture now will be facing the bottom of the picture when the object has gone 180 degrees around the orbit as the object keeps the same side facing inwards. That's an object rotating on its own axis once per orbit, and that rotation will be detectable with a local accelerometer.
.

You seem to forget that we are not dealing with a homogeneous gravitational field here. Its direction changes through 360 deg over one orbit. So a detector co-rotating with this should be in an inertial frame and not detect anything.

 

[PeterDonis]

Ignoring tidal effects means ignoring the gravitational force change dF from front to back of the mass compared to the force F itself.

Agreed; but that's not the only thing determining how the object's individual parts move. See below.

It is not a tidal lock, on the contrary: if you ignore tidal effects (in the sense as defined above), then you can as well assume that all parts of the mass orbit independently of each other.

No, you can't, because the object's parts are held together by non-gravitational forces: in your model, the springs. Forcing the object to rotate as you drew it will cause the springs to stretch, because the outside pieces of the object are moving faster than the inside pieces. It doesn't matter how small you make the object: it's geometrically impossible for the object to rotate as you drew it without inducing forces in the springs.

In reality, as I noted previously, gravity will determine the motion of the object's center of mass; the motion of the other parts of the object will be determined by the springs. If the springs stay in equilibrium, which is what I understood you to be stipulating, then, as I noted, the object's motion will look like the right side of pervect's diagram.

 

[Dale]

The point is that this rotation is also inertial like the linear orbital motion, so it would not be detected by an accelerometer. It is simply a consequence of the spherical gravitational potential (in contrast, in the case of the non-rotating orbiting planet in your own diagram, the accelerometer would actually measure a rotation, as that rotation is not an inertial rotation).

As PeterDonis and Nugatory mentioned, you have this exactly backwards. In fact, gravity probe B was designed to detect small deviations from the inertial non-rotating satellite due to frame dragging.

So it is obvious that a purely local measurement can not only provide no answer to the question whether one is actually accelerating or not, but neither to the question whether one is actually rotating or not. The interpretation of any accelerometer measurement can only be unambiguous if the global situation is taken into account.

The phrases "actually rotating" and "actually accelerating" are not scientifically well defined. There is coordinate acceleration and rotation or there is proper acceleration and rotation. Purely local measurements give you proper acceleration and rotation, and they are invariant. The coordinate acceleration and rotation can be made to take any value simply through a coordinate transform, so it is something that is defined rather than measured.

 

[PeterDonis]

You seem to forget that we are not dealing with a homogeneous gravitational field here. Its direction changes through 360 deg over one orbit.

First of all, you were the one who said to neglect tidal effects, which means treating the field as homogeneous; so we're following directions.

Also, as I noted in my last post, the gravity of the central object only determines the motion of the object's center of mass. The motion of the other pieces of the object is determined by the requirement that the springs stay unstressed.

Look at it this way: the motion of the object's center of mass describes a circle. Since we're neglecting tidal effects, then *every* piece of the object should move in an *identical* circle, just shifted slightly in position from the circle described by the center of mass. In other words, each circle must be of the *same* radius. (Of course only one such circle, the one described by the motion of the CoM, will be exactly centered on the object; but neglecting tidal effects means neglecting the small correction due to the other circles being off-center.)

But the motion you drew does not meet this requirement: it has the outside pieces of the object describing a circle of larger radius than the circle described by the CoM, and the inside pieces describing a circle of smaller radius. That will cause the springs to show stress.

 

[WannabeNewton]

Seriously, this is such a trivial issue that you're trying to augment into something non-trivial. Nugatory gave the resolution already and it complements pervect's original drawing.

If we have an observer in free fall in some space-time geometry, an accelerometer carried by the observer will obviously measure zero acceleration: $\xi^{\gamma}\nabla_{\gamma} \xi^{\mu} = 0$ where $\xi^{\mu}$ is the observer's 4-velocity. An isotropic mass-spring system carried by the observer (of ideal size) would not be displaced from natural equilibrium during the free fall trajectory.

The observer can also carry with him an apparatus consisting of a sphere with beaded prongs attached isotropically. Clearly if the observer had non-zero spin then the beads would be thrown outwards due to centrifugal forces; if the observer had vanishing spin then the beads would remain in place. For observers in free fall this is equivalent to $\xi^{\gamma}\nabla_{\gamma} \eta_{i}^{\mu} = 0$ where $\eta_{i}^{\mu}$ are the spatial axes of the observer's frame.

These are all obviously local measurements-no global measurements are necessary.

Finally and most importantly, there are both mathematical and operational ways of differentiating spinning from orbital rotation. See pp. 221-224 of the notes I linked to yuiop earlier in this thread: http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf

In a nutshell, an observer can always Fermi transport the spatial axes of his frame along his worldline. This immediately gives us the second of pervect's drawings i.e. the one in which the black marker on the particle "always faces the same direction"; we have thus determined a non-spinning frame for our observer before even talking about orbital rotation. If we now consider an observer who is possibly in a circular orbit in a stationary axisymmetric space-time, such that for simplicity the circular orbit has as its tangent field the axial killing field $\psi^{\mu}$ (not to be confused with the 4-veocity of the observer), then the observer can test to see if he is actually in a circular orbit simply by seeing if $\psi^{\mu}$ rotates relative to the spatial axes of his non-spinning frame i.e. he just has to see if $\psi^{\mu}$ changes angle continuously relative to said spatial axes along the worldline of the observer. If it does then the observer is in the circular orbit described above and if it doesn't then the observer is not in such a circular orbit. This is proven mathematically in the notes I linked above.

 

[WannabeNewton]

I was looking through a couple of my other favorite GR texts and came across additional lucid discussions of the relevant subject matter: see exercise 6.8 in MTW and section 3.4.3. of "Gravitation and Inertia"-Ciufolini and Wheeler.

 

[Fantasist]

In reality, as I noted previously, gravity will determine the motion of the object's center of mass; the motion of the other parts of the object will be determined by the springs. If the springs stay in equilibrium, which is what I understood you to be stipulating, then, as I noted, the object's motion will look like the right side of pervect's diagram.

If the springs stay in equilibrium, then you might as well forget about them and consider the free-fall orbit of each part of the object separately. Now, how should in this case a part circling the central mass on an inner orbit go to an outer orbit (and vice versa) after half a revolution (which is what the right hand side of pervects diagram implies)?

 

[PeterDonis]

If the springs stay in equilibrium, then you might as well forget about them and consider the free-fall orbit of each part of the object separately.

Only if the "free-fall orbit" you consider is consistent with all your other assumptions. See below.

Now, how should in this case a part circling the central mass on an inner orbit go to an outer orbit after half a revolution (which is what the right hand side of pervects diagram implies)?

That's not what is happening. Read my previous post again. We are ignoring tidal effects, so *all* of the parts of the object are traveling on the *same* orbit, i.e., on circles with the *same* radius. Their orbits are just centered on slightly different points. So there isn't a "inner orbit" and an "outer orbit".

Here's another way of seeing it: if we ignore tidal effects, then all of the pieces of the object see the same "acceleration due to gravity"--i.e., with respect to an observer who is static at the altitude of the object's center of mass, all the pieces of the object "fall" vertically the same distance in the same amount of time. That means the spatial curvature of all of their paths must be the same. But what you are calling the "inner orbit" and "outer orbit" have different spatial curvatures from the orbit of the object's CoM, so those are not permissible paths for the object's parts to follow. The only permissible paths are the ones I described, circles with the same radius but slightly offset centers.

 

[pervect]

My suggestion with respect to tidal effects - keep them in the analysis, but use a simple, circular shape, then demonstrate there is no net tidal torque due to tidal forces.

I hope it's already obvious that there is no net force due to tidal forces - that by definition the center of mass moves in such a manner that the tidal forces sum to zero.

I wasn't originally going to do this here, but the argument for zero tidal torque is simple enough that I'll give it a shot. Using the approach suggested by Goldstein in his textbook "Classical mechanics", IIRC called potential theory, one adds up the total potential -GmM/r of the shape to the central mass (which we will assume is a point). Goldstein is an advanced undergraduate/beginning graduate level text IIRC, but the argument is simple enough to present here.

If the summed potential changes when the orientation of the shape changes, there are tidal torques. It's apparent from the circular symmetry that this potential will not change for a disk when it rotates. Therefore there can be no net torque on a disk due to tidal forces.

If one insists on using a "hammer" shape, one will find that there are tidal torques. But there is no compelling reason to use such a complex shape. It obscures the arguments rather than illuminates them.

Using the simper circular shape, because there is no net tidal torque, the angular velocity of a rotating disk will not change. The rotation of a circular disk around its axis of symmetry is completely independent of its orbital motion.

One sees numerous examples of this in astronomy, planets all rotate at different and mostly constant rates. There are some small effects that make the rate of rotation non-constant in the real world, these are due to the lack of perfect spherical symmetry.

In astronomy, the rate of rotation of a celestial body relative to the fixed stars is called the "sidereal day". This is the absolute rate of rotation. The sidereal day can have any value.

Lets consider now a planet orbiting the sun. When the sidereal day is equal to the solar day, the planet will have one side always facing the sun, as per one of my earlier diagrams. The other example on my diagram had no rotation relative to the fixed stars. Not shown on the diagram are the infinite number of other possibilites corresponding to other rotational rates.

 

[WannabeNewton]

The question of whether or not a given extended body experiences tidal torques can be directly analyzed using the formalism of GR. Assume the extended body is small enough so that the Riemann curvature is constant across its volume and attach to the center of mass of the body a spin vector $S^{\mu}$ with $S^{\mu}u_{\mu} = 0$ where $u^{\mu}$ is the 4-velocity of the center of mass. The center of mass is then allowed to go into free fall; it can be shown that the body experiences tidal torques given by $u^{\gamma}\nabla_{\gamma}S^{\rho} = \epsilon^{\rho \beta \alpha \mu}u_{\mu}u^{\sigma}u^{\lambda}t_{\beta\nu}R^{\nu}_{\sigma\alpha \lambda}$ where $t_{\beta\nu}$ is the reduced mass quadrupole moment of the body such that $t_{\beta\nu}u^{\nu} = 0$ (see e.g. exercise 2.23 in Straumann). So whether or not $u^{\gamma}\nabla_{\gamma}S^{\rho} = 0$ depends on whether or not the body has a vanishing mass quadrupole moment (the canonical example of a body with vanishing quadrupole moment is of course a sphere).

 

[Fantasist]

Here's another way of seeing it: if we ignore tidal effects, then all of the pieces of the object see the same "acceleration due to gravity"--i.e., with respect to an observer who is static at the altitude of the object's center of mass, all the pieces of the object "fall" vertically the same distance in the same amount of time.

But the direction of the vertical changes over the orbit.

That means the spatial curvature of all of their paths must be the same. But what you are calling the "inner orbit" and "outer orbit" have different spatial curvatures from the orbit of the object's CoM, so those are not permissible paths for the object's parts to follow. The only permissible paths are the ones I described, circles with the same radius but slightly offset centers.

Consider two astronauts on a spacewalk, with astronaut A 2 m inside astronaut B along the local radius vector of the earth. Are you suggesting that half a revolution later the situation has reversed and B is 2 m inside A?

P.S.: I am away now for week, and won't be able to reply to any posts during that time.
Merry Christmas to everbody

 

[WannabeNewton]

Merry Christmas to everbody

You too bud! Have a good one.

 

[PeterDonis]

But the direction of the vertical changes over the orbit.

Yes, but what does that mean? It means that the direction of the "acceleration due to gravity" that each piece of the object is subjected to changes; but the *magnitude* of that acceleration is still the same for all pieces of the object if we ignore tidal effects. And the magnitude being the same is what makes the curvature of all the circular paths the same (the changing direction is what makes the paths circular).

Consider two astronauts on a spacewalk, with astronaut A 2 m inside astronaut B along the local radius vector of the earth. Are you suggesting that half a revolution later the situation has reversed and B is 2 m inside A?

If we ignore tidal effects, and if the astronauts are connected by a spring, then yes, that's what would happen.

You appear to have a mistaken notion of what "ignoring tidal effects" means, so let's put them back into see exactly what it is that we're ignoring. Consider the two astronauts, and suppose that they are *not* connected by a spring, so there is nothing constraining their relative motion. And suppose we can make measurements accurate enough to detect tidal effects over a 2 meter range. What will we observe?

Since astronaut B is in an orbit with 2 meters higher altitude, his orbital velocity is slightly slower; so the effect of tidal gravity will be to make B slowly fall *behind* A. In other words, when A has completed exactly one orbit, B will have completed slightly *less* than one orbit. So a line drawn between them will no longer be exactly radial (and its length will be a bit longer than 2 meters)--but the *direction* in which it is turned is *opposite* to the direction in which you drew the masses turning in your "gravitational hammer thrower" drawing.

Now suppose we put a spring between the two astronauts, with an equilibrium length of 2 meters. What will it do? It will pull astronaut B towards astronaut A, along a line which is slightly tilted "backwards" from radial (because B falls slightly behind A). But as this is happening, the "radial" direction is also changing, so the direction in which A and B are falling due to the Earth's gravity is changing. The net effect is that B and A remain 2 meters apart along the *original* "radial" direction--so with respect to the *new* radial direction, B is at an altitude slightly *less* than 2 meters above A, and a small distance "behind" A.

(How do we know this is the net effect? Because we are taking tidal effects into account, so B falls a slightly *smaller* distance than A; the extra motion due to the spring just makes up the difference.)

Continuing this around the orbit, after a quarter orbit, with respect to the "radial" direction there, B and A will be at the same altitude, but B will be 2 meters behind A in the orbit. After a half orbit, B will be 2 meters *lower* than A, with respect to the current "radial" direction. And so on.

All this was taking tidal effects into account: but the only way tidal effects appeared in the analysis was in the small force that the spring had to exert to keep A and B's separation constant at 2 meters. Ignoring tidal effects therefore simply means ignoring that small force. It does *not* mean changing the motion of A and B; otherwise ignoring tidal effects would not be a good approximation.

 

[Nugatory]

Consider two astronauts on a spacewalk, with astronaut A 2 m inside astronaut B along the local radius vector of the earth. Are you suggesting that half a revolution later the situation has reversed and B is 2 m inside A?

Yes, if you ignore tidal forces.

If the two astronauts are not connected together at all, or if they are connected with a tether more than 2 meters long, then no matter how small the tidal forces are they'll be more than the zero force holding the astronauts together, so you can't ignore them.

But for any moderately rigid connection between them (and a hammer is way more than just "moderately" rigid) then, yes, you will get exactly what PeterDonis says.

 

[Dale]

Are you suggesting that half a revolution later the situation has reversed and B is 2 m inside A?

Yes, provided A and B experience 0 proper rotation (meaning ring interferometers detect no rotation and local gyroscopes maintain their orientation wrt A and B).

This is experimentally well proven, most dramatically in gravity probe B. There simply isn't any ambiguity on this point.

 

[yuiop]

.. You appear to have a mistaken notion of what "ignoring tidal effects" means, so let's put them back into see exactly what it is that we're ignoring. ...

Now suppose we put a spring between the two astronauts, with an equilibrium length of 2 meters. What will it do? It will pull astronaut B towards astronaut A, along a line which is slightly tilted "backwards" from radial (because B falls slightly behind A). But as this is happening, the "radial" direction is also changing, so the direction in which A and B are falling due to the Earth's gravity is changing. The net effect is that B and A remain 2 meters apart along the *original* "radial" direction--so with respect to the *new* radial direction, B is at an altitude slightly *less* than 2 meters above A, and a small distance "behind" A.

(How do we know this is the net effect? Because we are taking tidal effects into account, so B falls a slightly *smaller* distance than A; the extra motion due to the spring just makes up the difference.)

Continuing this around the orbit, after a quarter orbit, with respect to the "radial" direction there, B and A will be at the same altitude, but B will be 2 meters behind A in the orbit. After a half orbit, B will be 2 meters *lower* than A, with respect to the current "radial" direction. And so on.

All this was taking tidal effects into account: ...

I do not think this is quite right. The lack of circular symmetry of the 'two astronauts and a spring assembly' will cause tidal locking, such that astronaut A will always be *lower* than astronaut B.

The COM of the assembly will have the correct orbital velocity which means astronaut A will have a tangential velocity lower than the required orbital velocity at his altitude, meaning that A will be pulled towards the gravitational body. Conversely, B will have a tangential velocity greater than the required orbital velocity at his altitude, causing B to experience an outward force. The combined tidal effect keeps the connecting spring parallel to the instantaneous radial direction and causes a force that pulls A and B apart. For a natural orbiting body, the radius at which these force tend to tear the orbiting body apart, is known as the Roche limit.

To minimise this effect, the orbiting body should be small and ideally have spherical symmetry.

 

[Nugatory]

I do not think this is quite right. The lack of circular symmetry of the 'two astronauts and a spring assembly' will cause tidal locking, such that astronaut A will always be *lower* than astronaut B.

Eventually, yes, but it will take many orbits for the lock to develop, and this branch of the thread is discussing what happens halfway around the first orbit after the initial conditions have been set up. It's reasonable to consider tidal forces but not the cumulative effect of the locking torque under those circumstances.

 

[PeterDonis]

The COM of the assembly will have the correct orbital velocity which means astronaut A will have a tangential velocity lower than the required orbital velocity at his altitude, meaning that A will be pulled towards the gravitational body.

This is a different initial condition from the one I was considering. I was considering an initial condition where both objects, A and B, have the correct (different) tangential velocities for their altitudes, but then we put a spring between them. Your initial condition is that A and B both have identical tangential velocities, *not* correct for their altitudes. Changing the initial condition can change the result.

 

[PeterDonis]

Eventually, yes, but it will take many orbits for the lock to develop, and this branch of the thread is discussing what happens halfway around the first orbit after the initial conditions have been set up.

Correct.

 

[PeterDonis]

The lack of circular symmetry of the 'two astronauts and a spring assembly' will cause tidal locking, such that astronaut A will always be *lower* than astronaut B.

I should also add that, as I understand it, tidal locking is a dissipative process--for example, the Moon became tidally locked to the Earth because when it wasn't locked, there was friction within its interior as the tidal bulge caused by the Earth moved. In order to include this effect in our scenario, we would have to add dissipation to the spring; with an idealized perfectly elastic spring, there wouldn't be any tidal locking.

 

[Fantasist]

Since astronaut B is in an orbit with 2 meters higher altitude, his orbital velocity is slightly slower; so the effect of tidal gravity will be to make B slowly fall *behind* A. In other words, when A has completed exactly one orbit, B will have completed slightly *less* than one orbit. So a line drawn between them will no longer be exactly radial (and its length will be a bit longer than 2 meters)--but the *direction* in which it is turned is *opposite* to the direction in which you drew the masses turning in your "gravitational hammer thrower" drawing.

Yes, an object in the outer orbit will fall behind, and an object in a lower orbit will advance in its orbit relatively to the center of mass. That's why you have to accelerate the former and decelerate the latter to preserve the spatial configuration, i.e. you have to rotate the whole system forward (as indicated in my diagram).
Of course, for a rigid body, its constituents would be kept together anyway due to the electrostatic force, but the tendency for the parts to drift apart would create a stress force in the body. You can eliminate this stress force by rotating the body forward synchronously in its orbit. So the bound rotation of an object should correspond to an inertial state (or something close to it anyway, as I don't think the stress force can be completely neutralized through simple rotations or translations other than in first order of r/R, as the orbital velocity is ~1/sqrt(R+r)).

All this was taking tidal effects into account: but the only way tidal effects appeared in the analysis was in the small force that the spring had to exert to keep A and B's separation constant at 2 meters. Ignoring tidal effects therefore simply means ignoring that small force. It does *not* mean changing the motion of A and B; otherwise ignoring tidal effects would not be a good approximation.

I realized now that it was wrong to assume the absence of tidal effects here, as the whole issue in fact is about tidal effects.
Even for perfectly spherical bodies, the inclusion of tidal effects (i.e. the variation of the gravitational force inside the body) is actually already implied by the 'Shell Theorem' (i.e. Gauss's law), so it would be a contradiction in terms to neglect them.

 

[PeterDonis]

Yes, an object in the outer orbit will fall behind, and an object in a lower orbit will advance in its orbit relatively to the center of mass.

*If* they are both moving freely. But if they're connected by springs, they're not moving freely.

That's why you have to accelerate the former and decelerate the latter to preserve the spatial configuration

Yes, that's what the spring does--but that just maintains the system's orientation in the *original* radial direction. If the spring were not there, a line connecting A and B would rotate *backwards* relative to the original radial direction. So this...

you have to rotate the whole system forward (as indicated in my diagram).

...is not correct. The spring just counteracts the *backward* rotation that would be present if A and B were moving freely. It does not add any forward rotation.

Of course, for a rigid body, its constituents would be kept together anyway due to the electrostatic force, but the tendency for the parts to drift apart would create a stress force in the body.

Which the parts of the body would move, if possible, to eliminate, just as A and B move to eliminate the stress in the spring. The equilibrium configuration of the body is still unstressed; you can't keep it stressed in equilibrium if the body as a whole is in free fall (in the absence of tidal effects--see below).

You can eliminate this stress force by rotating the body forward synchronously in its orbit.

No, you eliminate the stress force by *not* rotating the body at all, instead of letting it rotate backwards due to tidal effects. See above.

So the bound rotation of an object should correspond to an inertial state

Yes, in the absence of tidal effects. See below.

(or something close to it anyway, as I don't think the stress force can be completely neutralized through simple rotations or translations other than in first order of r/R, as the orbital velocity is ~1/sqrt(R+r)).

In the absence of tidal effects, there is zero stress force if the body as a whole is in free fall. If tidal effects are included, then yes, there will be small internal stresses present due to tidal effects. However, those by themselves won't cause the object to rotate; you need dissipation as well. See further comments below.

I realized now that it was wrong to assume the absence of tidal effects here, as the whole issue in fact is about tidal effects.

No, there are two separate issues, as the discussion between yuiop, Nugatory, and myself in this thread shows. Tidal effects will *eventually* cause the orbiting body to rotate once per revolution, as per your "hammer thrower" drawing; but this happens on a much longer timescale than a single orbit. And the mechanism of tidal locking involves dissipation, so it requires damping in the springs in your simplified model; idealized perfectly elastic springs would *not* cause tidal locking, and so would *not* cause any forward rotation.

Even for perfectly spherical bodies, the inclusion of tidal effects (i.e. the variation of the gravitational force inside the body) is actually already implied by the 'Shell Theorem' (i.e. Gauss's law), so it would be a contradiction in terms to neglect them.

Huh? The "Shell Theorem" talks about variation due to the object's own self-gravity; it has nothing to do with variation of the gravity seen by a small object orbiting a much larger one.

 

[Fantasist]

No, you eliminate the stress force by *not* rotating the body at all, instead of letting it rotate backwards due to tidal effects. See above.

Sorry, but I disagree. If you pull on an object, you can release the associated stress by accelerating the object forward (in the direction of the pull), not be leaving it stationary or even pulling it backward. So if the center of mass pulls on the orbitally lagging part of the object, the stress is released (or minimized) by moving the latter in the direction of the pull (i.e. by rotating it forward).

No, there are two separate issues, as the discussion between yuiop, Nugatory, and myself in this thread shows. Tidal effects will *eventually* cause the orbiting body to rotate once per revolution, as per your "hammer thrower" drawing; but this happens on a much longer timescale than a single orbit. And the mechanism of tidal locking involves dissipation, so it requires damping in the springs in your simplified model; idealized perfectly elastic springs would *not* cause tidal locking, and so would *not* cause any forward rotation.

The cause of the tidal locking wasn't the issue here. Only whether the tidally locked (i.e. bound) rotation is the inertial situation or not. Intuitively, one should expect that all systems should eventually settle into an inertial state, and most moons in the solar systems for instance seem to confirm this as they have a bound rotation.

 

[Dale]

I do not think this is quite right. The lack of circular symmetry of the 'two astronauts and a spring assembly' will cause tidal locking, such that astronaut A will always be *lower* than astronaut B.

While that certainly is a valid alternative scenario, it is a different scenario from what has been proposed and discussed up until now. In such a scenario the astronauts will experience locally measurable rotation. Ring interferometers will give non-zero results and local gyroscopes will precess wrt the astronauts.

 

[PeterDonis]

if the center of mass pulls on the orbitally lagging part of the object

In the A and B example, the CoM is halfway between A and B. So to the extent that you can view the CoM as "pulling", it pulls B forward and A backward. It doesn't just pull on B. So the object won't "accelerate forward".

(Purely internal forces can't impart a net forward acceleration to an object anyway, by conservation of momentum. They can only change the distribution of the parts around the CoM; they can't change the motion of the CoM itself.)

The cause of the tidal locking wasn't the issue here. Only whether the tidally locked (i.e. bound) rotation is the inertial situation or not.

If tidal effects are present, there is *no* "inertial situation", if by that you mean a state of motion in which all of the individual pieces of the object are in free fall. The best you can do is a state of motion in which the object's CoM is in free fall. But there are many possible states of motion that meet that requirement, including both your "hammer thrower" state (the tidally locked state) and the "non-rotating" state I have been describing. I agree that *eventually*, your "hammer thrower" state will be reached via tidal locking (provided, as I said before, that dissipation is present), but that doesn't mean that's the only possible state, or that that state will be reached within a single orbit; it will take a lot longer than that.

 

[Dale]

The cause of the tidal locking wasn't the issue here. Only whether the tidally locked (i.e. bound) rotation is the inertial situation or not.

It is not. That is already experimentally established.

Intuitively, one should expect that all systems should eventually settle into an inertial state, and most moons in the solar systems for instance seem to confirm this as they have a bound rotation.

When intuition contradicts experiment it is always the intuition which must be discarded.

 

[PeterDonis]

If tidal effects are present, there is *no* "inertial situation"

I should also add that if tidal effects are neglected, the "inertial situation" is the "non-rotating" one I described, *not* the "hammer thrower" situation you described. In the latter state of motion, there will be internal stresses in the object even though its CoM is moving inertially, even if tidal effects are ignored.

 

[Fantasist]

A Happy New Year to everybody!

I should also add that if tidal effects are neglected, the "inertial situation" is the "non-rotating" one I described, *not* the "hammer thrower" situation you described. In the latter state of motion, there will be internal stresses in the object even though its CoM is moving inertially, even if tidal effects are ignored.

You could neglect tidal effects strictly only in case of a homogeneous gravitational field. But in that case you couldn't have any closed orbits at all, so it is an unrealistic assumption here. The whole scenario rests on the presence of tidal effects.

If tidal effects are present, there is *no* "inertial situation", if by that you mean a state of motion in which all of the individual pieces of the object are in free fall. The best you can do is a state of motion in which the object's CoM is in free fall. But there are many possible states of motion that meet that requirement, including both your "hammer thrower" state (the tidally locked state) and the "non-rotating" state I have been describing. I agree that *eventually*, your "hammer thrower" state will be reached via tidal locking (provided, as I said before, that dissipation is present), but that doesn't mean that's the only possible state, or that that state will be reached within a single orbit; it will take a lot longer than that.

I agree largely, but the important point is that the 'non-rotating' state can not be stable in the presence of tidal effects. Unless the body is in a 'locked' rotation, the direction of the gravitational force vector (and thus the internal stress force required to hold the body together) would permanently change inside the body, which requires a permanent drift of charges against each other. This inevitably will lead to an internal friction slowing the rotation down (and I am not sure whether this actually requires (energy)-dissipation; there is after all something like angular momentum transfer (which for instance changes the orbital parameters of the earth-moon system at the expense of the Earth's rotation rate)).

 

[Dale]

the important point is that the 'non-rotating' state can not be stable in the presence of tidal effects.

For an asymmetric object, yes. However, that is irrelevant to the question of which state is inertial.

Forget an orbit and consider only an isolated object with some angular momentum about its center of mass. By conservation of angular momentum the spinning state is stable, but non-inertial. If you are mentally equating "stable" with "inertial" then you are making a mistake.

 

[PeterDonis]

You could neglect tidal effects strictly only in case of a homogeneous gravitational field. But in that case you couldn't have any closed orbits at all

True. But not neglecting tidal effects is not the same as requiring tidal locking, nor does it make the tidally locked state non-rotating. See below.

I agree largely, but the important point is that the 'non-rotating' state can not be stable in the presence of tidal effects.

Stable in the long run, yes. But that doesn't mean it's the only possible state of motion for an orbiting body.

Also, your original claim wasn't that the tidally locked state is stable, but that it is non-rotating. That claim is still false. A tidally locked orbiting body is rotating, as can be shown by local measurements (such as gyroscopes); this has been pointed out several times already.

This inevitably will lead to an internal friction slowing the rotation down (and I am not sure whether this actually requires (energy)-dissipation

Of course it does; friction converts rotational energy into heat.

there is after all something like angular momentum transfer (which for instance changes the orbital parameters of the earth-moon system at the expense of the Earth's rotation rate)).

Angular momentum is not energy; angular momentum can be conserved even if mechanical energy is changing (because some mechanical energy is being converted into heat).

 

[Fantasist]

the important point is that the 'non-rotating' state can not be stable in the presence of tidal effects.

For an asymmetric object, yes. However, that is irrelevant to the question of which state is inertial.

An object will be asymmetric in some sense in the presence of tidal effects.

Forget an orbit and consider only an isolated object with some angular momentum about its center of mass. By conservation of angular momentum the spinning state is stable, but non-inertial. If you are mentally equating "stable" with "inertial" then you are making a mistake.

See my reply to PeterDonis below.

 

[Fantasist]

Also, your original claim wasn't that the tidally locked state is stable, but that it is non-rotating. That claim is still false. A tidally locked orbiting body is rotating, as can be shown by local measurements (such as gyroscopes); this has been pointed out several times already.

Well, then consider a simple thought experiment (just taking the hammer-throw analogy a bit further): assume a tidally locked orbiting body and suddenly we switch the gravitational field off. Obviously, the body will now continue in a straight line. The question is, will it be rotating or not?

Angular momentum is not energy; angular momentum can be conserved even if mechanical energy is changing (because some mechanical energy is being converted into heat).

You said 'dissipation is required' for the rotation to become tidally locked. I know the word 'dissipation' only in context of 'energy dissipation' i.e. energy is lost from the system in the shape non-mechanical energy. My point was that the rotation of a body can also change without dissipation in this sense, namely by angular momentum transfer to orbital angular momentum (the orbit would then change to one with a correspondingly higher energy).

 

[PeterDonis]

Well, then consider a simple thought experiment (just taking the hammer-throw analogy a bit further): assume a tidally locked orbiting body and suddenly we switch the gravitational field off.

I'm not sure you can actually consistently formulate such a scenario in GR, because you can't just "switch gravity off". However, here's an alternative formulation that might work: suppose the central object that the body is orbiting is suddenly converted entirely to radiation, which expands outward in a spherically symmetric shell moving at the speed of light. By Birkhoff's theorem, the orbiting body will see no change until the shell passes it on the way out; and after that, spacetime inside the shell will be flat, so the orbiting body will indeed move in a straight line.

The question is, will it be rotating or not?

In the scenario as I described it just now, yes.

You said 'dissipation is required' for the rotation to become tidally locked. I know the word 'dissipation' only in context of 'energy dissipation' i.e. energy is lost from the system in the shape non-mechanical energy.

Yes, mechanical energy is being converted into heat, which is non-mechanical energy.

My point was that the rotation of a body can also change without dissipation in this sense, namely by angular momentum transfer to orbital angular momentum (the orbit would then change to one with a correspondingly higher energy).

Have you done the math to show that both total angular momentum (spin + orbital) and total mechanical energy (kinetic + potential) can be conserved in such a process while still satisfying Kepler's laws for orbits?

Also, I'm a little unclear on what mechanism you think is operating to change the orbiting body's spin angular momentum into orbital angular momentum without any dissipation. To change the object's spin it's not enough just to have stresses inside the object: the stresses need to lead to non-periodic changes in the object's configuration. Without dissipation I'm not sure how that can happen.

 

[Dale]

assume a tidally locked orbiting body and suddenly we switch the gravitational field off. Obviously, the body will now continue in a straight line. The question is, will it be rotating or not?

Yes, it will be rotating. If it was carrying a gyro it would have precessed relative to the body while it was orbiting and unless there is some external torque it will continue to precess.