Is there an aether to determine if an object rotates?

[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.