Is there an aether to determine if an object rotates?

[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.

 

[pervect]

I agree with Peter Donis' more careful answer, if you try to ask "what happens if a mass suddenly disappears" in GR, and work through the mathematics, you find that it can't happen because of local conservation laws.

I would suggest that doing some research on tidal locking would be helpful to the OP. You don't need GR for this, either. For instance, Wikki http://en.wikipedia.org/wiki/Tidal_locking has some of the basics:

Tidal locking results in the Moon rotating about its axis in about the same time it takes to orbit Earth. Except for libration effects, this results in the Moon keeping the same face turned towards Earth, as seen in the figure on the left. (The Moon is shown in polar view, and is not drawn to scale.) If the Moon were not spinning at all, it would alternately show its near and far sides to Earth, while moving around Earth in orbit, as shown in the figure on the right.

and

The angular momentum of the whole A–B system is conserved in this process, so that when B slows down and loses rotational angular momentum, its orbital angular momentum is boosted by a similar amount (there are also some smaller effects on A's rotation). This results in a raising of B's orbit about A in tandem with its rotational slowdown. For the other case where B starts off rotating too slowly, tidal locking both speeds up its rotation, and lowers its orbit.

Wikki has some basic info on the well known behavior of angular momentum, including the breakdown of total angular momentum into spin and orbital components

http://en.wikipedia.org/wiki/Angular_momentum

The section below would be of particular interest:

In orbits, the angular momentum is distributed between the spin of the planet itself and the angular momentum of its orbit:

$\mathbf{L}_{\mathrm{total}} = \mathbf{L}_{\mathrm{spin}} + \mathbf{L}_{\mathrm{orbit}}$

 

[PeterDonis]

Wikki http://en.wikipedia.org/wiki/Tidal_locking has some of the basics:

One interesting item that I note from this is that, for the case where the orbiting body is rotating more slowly than it revolves in its orbit (which would be the case, of course, for the "non-rotating" case we've been discussing here), tidal locking *lowers* its orbit as it increases its spin. (This makes sense, of course, in terms of total angular momentum conservation.) Tidal locking only raises the orbit if the orbiting body starts off rotating more rapidly than it revolves.

 

[yuiop]

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.

Unfortunately, your choice of initial conditions (if I understand them correctly) does not seem to yield the desired result of no spin relative to the distant stars. I assume you mean that initially A and B are in perfectly circular orbits. The non relativistic equation for tangential velocity of a circular orbit with orbital radius R is:

$v =\sqrt{\frac{GM}{R}}$

For simplicity, when A and B are on the same radial vector they will be connected by a rigid rod of length 2*r, rather a spring. For reference purposes, imagine a third observer C that has zero spin relative to the distant stars at the COM of A and B, when they are connected. If R is the orbital radius of C, then when A and B are connected by a rod, the average angular spin velocity of A and B in the rest frame of C, is (if I have calculated correctly):

$\omega = \sqrt{\frac{GM}{2(R^2r-r^3)}}-\sqrt{\frac{GM}{R^3}}$

where r is the distance from C to A or B. The desired result is that a line connecting A and B constantly points at a distant star and for this to happen $\omega$ has to be zero. The solution to the above equation for $\omega=0$ has two complex solutions and one real solution of r = -1.19149*R which is physically unrealistic.

If a spring is used instead of a rigid rod, then the spring will tend to stretch and to conserve angular momentum the spin rate of A and B will tend to slow down. The amount of stretch depends on the material used and there is no reason for the spin rate to naturally go to zero (relative to the stars) except by chance.

There is no reason to say the left or right hand side of Pervect's diagram represents the natural condition of the 'orbiting hammer' as it depends entirely on the choice of initial conditions. All we can say is that the zero angular spin momentum condition of the hammer is represented by the left hand diagram, where the hammer does not rotate relative to the stars.

Your initial condition is that A and B both have identical tangential velocities, *not* correct for their altitudes.

More precisely, my initial conditions were that A and B have identical orbital angular velocities.

 

[PeterDonis]

I assume you mean that initially A and B are in perfectly circular orbits.

At slightly different radii, yes. However, the spring was also part of the intended initial condition, not a rigid rod; see below.

when A and B are on the same radial vector they will be connected by a rigid rod of length 2*r, rather a spring.

But if the rod is rigid the distance between A and B can't vary, and if the distance between A and B can't vary, they can't possibly be in circular orbits at different radii to start with; their initial velocities must already have components other than the tangential velocities required to keep them in orbit at their respective radii.

If a spring is used instead of a rigid rod, then the spring will tend to stretch and to conserve angular momentum the spin rate of A and B will tend to slow down.

No, if a spring is used instead of a rigid rod, we can specify initial velocities that are equal to the orbital tangential velocities at each radius, because we aren't constrained by the distance between A and B having to remain constant. Only after the spring has started to stretch will other forces come into play.

There is no reason to say the left or right hand side of Pervect's diagram represents the natural condition of the 'orbiting hammer' as it depends entirely on the choice of initial conditions.

I agree that the initial conditions you proposed make a difference as compared to mine. See further comments below.

the zero angular spin momentum condition of the hammer is represented by the left hand diagram, where the hammer does not rotate relative to the stars.

I think it was actually the right-hand diagram in pervect's post a while back. I agree that the "non-rotating" diagram he posted is the zero angular momentum one, and this is really the main point.

More precisely, my initial conditions were that A and B have identical orbital angular velocities.

But that's *not* the same as having identical tangential velocities, because A and B are at different radii. (It's also not the same condition I was assuming; see below.) Identical angular velocities means B has a larger tangential velocity, because it's higher up; and that already predetermines that the A-B system is spinning--essentially you've picked the "hammer thrower" state of motion as the initial condition.

Identical tangential velocities at least does not start the A-B system out spinning, but it does mean the tangential velocities of A and B don't match the orbital velocities for their altitudes. The initial condition I was assuming was that A and B each have the correct tangential velocity for their altitude; that means B has a slightly *smaller* tangential velocity than A to start with.

 

[Akashks001]

Rotation with respect to "itself".;)

 

[Akashks001]

Rotation is nothing but motion with respect to itself... Isn't it guys? This may seem absurd, please think over it and correct me

 

[WannabeNewton]

Rotation is nothing but motion with respect to itself... Isn't it guys?

As has been stated multiple times in this thread, local rotation (spin!) is with respect to a complete set of local torque-free gyroscopes.

 

[Fantasist]

In orbits, the angular momentum is distributed between the spin of the planet itself and the angular momentum of its orbit:

L_total = L_spin + L_orbit

So in a reference frame where L_orbit vanishes you should be left with L_spin. But for a bound rotation, L_spin vanishes as well (in a reference frame co-rotating with the orbital motion, you have a non-rotating object fixed in the same place).

The crucial point here is that L_spin has to represent an additional degree of freedom. But if you have the orientation of the object locked to the orbital path, then there is (by definition) no additional degree of freedom i.e. there is no spin angular momentum. I have illustrated this in the below graphic: both for the linear and orbital motion, the orientation of the object is locked to the orbital path, and in both cases there is no spin angular momentum; the total angular is in both cases the same and given by the orbital angular momentum only.

In contrast, what others described here as the "non-rotating" case would correspond to a spin as it represents an additional degree of freedom (the orientation of the object would change with regard to the orbital path).

 

[PeterDonis]

So in a reference frame where L_orbit vanishes

But it doesn't. $L_{orbit}$ is still nonzero after the object starts moving in a straight line (because the outgoing shell of radiation has passed it). $L_{orbit}$ is, in Newtonian terms, $\vec{r} \times \vec{p}$; that doesn't magically go to zero just because the body's trajectory has changed.

[Edit: I may be misunderstanding you here: if by "a reference frame where $L_{orbit}$ vanishes" you mean the "rotating frame", see below.]

for a bound rotation, L_spin vanishes as well

No, it doesn't. In an inertial frame, this is easy to see; $L_{spin}$ is just $I \omega$, moment of inertia times angular velocity. But in a non-inertial frame, things are not so simple:

(in a reference frame co-rotating with the orbital motion, you have a non-rotating object fixed in the same place).

But in that frame, when the outgoing shell passes the object and it starts moving in a straight line, it's no longer in the same place; it's now moving in this frame. [Edit: Of course this applies to $L_{orbit}$ as well; even if that is zero in this frame to start with, which I'm not sure it is--see below--it won't stay zero once the object starts moving in a straight line.]

(I'm also not sure that the formula for angular momentum necessarily gives zero for an object at rest in a non-inertial frame, but that will take some more thought to work out.)

The crucial point here is that L_spin has to represent an additional degree of freedom. But if you have the orientation of the object locked to the orbital path, then there is (by definition) no additional degree of freedom

No, the additional degree of freedom is (temporarily) constrained if you assume tidal locking. When the outgoing shell passes the object and it starts moving in a straight line, the constraint goes away (because it was being enforced by the gravity of the central body, which is no longer there), and spin is no longer coupled to "orbital" motion.

 

[pervect]

So in a reference frame where L_orbit vanishes

There isn't any inertial reference frame where L_orbit vanishes. And there seems to be enough of a difficulty understanding things in inertial frames that considering non-inertial frames would be counterproductive.

The crucial point here is that L_spin has to represent an additional degree of freedom.

It does represent an additional degree of freedom. One can imagine and draw diagrams of spinning hammers, for instance.

But if you have the orientation of the object locked to the orbital path, then there is (by definition) no additional degree of freedom i.e. there is no spin angular momentum.

It is true that when you constrain the system you remove a degree of freedom via the constraint. It is not true that "there is no spin angular momentum". What is true is that if you constrain the system according to your diagram, the spin period becomes constrained to be equal to the orbital period. Because the spin angular momentum is I * omega = 2*pi*I / T, I being the moment of inertia, and T being the orbital period, the spin angular momentum of the system constrained according to your diagram is NOT zero.

 

[ZapperZ]

This thread no longer has any resemblance to the original question, which has been answered. It is now done.

Zz.