Is there an aether to determine if an object rotates?

[greypilgrim]

Is there an "aether" to determine if an object rotates?

Hi,

In modern physics aether theories are obsolete, all inertial systems are equivalent. For rotational motion, the situation is different: In a rotating system, it is possible to measure the rotation without any outside reference (e.g. by measuring the centripetal force needed to keep a body on its orbit). So there is an objective rest state where a body does not rotate. Does this somehow imply that there must be a "directional aether" throughout the universe that determines when an object does not rotate?

 

[Dale]

Rotation involves acceleration, so it shouldn't be surprising that it is locally detectable in the same manner that acceleration is. In fact, I cannot see how it could be otherwise with rotation given the detectability of linear acceleration.

 

[Meir Achuz]

But rotation with respect to what?

 

[Dale]

The same thing that proper acceleration is with respect to.

 

[WannabeNewton]

But rotation with respect to what?

Lorentz frames rotate relative to local gyroscopes. More precisely, we can define a Fermi-transported Lorentz frame (or just Fermi frame for short) as a Lorentz frame $\{e_{\alpha}\}$, where $e_0 = u$ is the 4-velocity of the observer whose measuring apparatus corresponds to this frame, such that $\nabla_u e_{\alpha} = g(a,e_{\alpha})u - g(u,e_{\alpha})a$ where $a = \nabla_u u$ is the 4-acceleration. The spatial axes of the Fermi frame can be thought of as mutually perpendicular gyroscopes. An arbitrary Lorentz frame is then deemed to be rotating at a given event if it does so relative to a Fermi frame coincident at that event.

 

[greypilgrim]

Lorentz frames rotate relative to local gyroscopes.

And what determines the non-rotating state of a gyroscope?

 

[WannabeNewton]

I explained that in post #5.

 

[Bill_K]

And what determines the non-rotating state of a gyroscope?

WbN gave you a full mathematical definition of a nonrotating frame, and said it can be thought of as a gyroscope. Here's another way, maybe more intuitive.

A nonrotating frame can be defined entirely in terms of the local paths of light rays, a "photon gyroscope" if you will. Stand at the origin O and fire a laser. The light pulse goes out, hits a mirror and returns to O. By definition, it returns in the same (i.e. nonrotated) direction that it was fired in. Repeatedly fire lasers in three mutually perpendicular directions, the x, y, and z axes, and their returning directions define what we mean by a nonrotating frame.

 

[pervect]

Note that a nonrotating frame, according to a mechanical gyroscope or the laser variant that Bill K describes, may or may not be rotating relative to distant "fixed" stars, due to an effect called frame dragging.

Near a massive rotating body, a gyroscope (whether mechanical or optical) will rotate relative to a body that uses distant "fixed" stars as a reference, due to the frame-dragging caused by the massive rotating body.

I don't have a lot of interest in aether theories, but this is not what you'd expect out of such a theory. You can always start adding patches to your aether theories, such as the attempts to explain the MM experimental data by "aether drag". Ultimately, though, relativity has proven to be simpler (not having any free parameters), has explained all known observations, and have suggested new, previously unexpected phenomenon. Aether theory is generally regarded (with good reason) as a dead end.

 

[WannabeNewton]

Does this somehow imply that there must be a "directional aether" throughout the universe that determines when an object does not rotate?

See the other surface level problem with this is that even if we consider a fluid with 4-velocity field $\xi^{\mu}$ that is locally non-rotating, which by definition means that $\xi^{[\gamma}\nabla^{\mu}\xi^{\nu]} = 0$, $\xi^{\mu}$ and hence the fluid are not unique. There can obviously be multiple fluids whose 4-velocity fields are twist-free. What natural choice is there for a fluid that could be a candidate for a "directional aether"?

Furthermore, the condition $\xi^{[\gamma}\nabla^{\mu}\xi^{\nu]} = 0$ depends on the space-time because of $\nabla^{\mu}$ so the class of locally non-rotating fluids associated with one space-time do not even have to be the same as those associated with a different space-time. In fact, if we take a stationary space-time and consider the fluid described by the time-like killing field $\xi^{\mu}$ then $\xi^{[\gamma}\nabla^{\mu}\xi^{\nu]} = 0$ if and only if the observers following orbits of $\xi^{\mu}$ are non-rotating in the sense defined in post #5. In particular, the observers following orbits of $\xi^{\mu}$ would be non-rotating outside of a non-rotating star but would be rotating outside of a rotating star so clearly there is no universal prescription for the kind of fluid you seek.

 

[WannabeNewton]

A nonrotating frame can be defined entirely in terms of the local paths of light rays, a "photon gyroscope" if you will. Stand at the origin O and fire a laser. The light pulse goes out, hits a mirror and returns to O. By definition, it returns in the same (i.e. nonrotated) direction that it was fired in. Repeatedly fire lasers in three mutually perpendicular directions, the x, y, and z axes, and their returning directions define what we mean by a nonrotating frame.

That is unequivocally a more intuitive definition!

For anyone interested, the very last page of Geroch's GR notes has a proof of the equivalence of Bill's definition to that of Fermi transport (second link in the following: http://home.uchicago.edu/~geroch/Links_to_Notes.html ).

 

[bcrowell]

There is a discussion of this sort of thing in the very readable intro of Einstein's 1916 paper on GR, "The foundation of the general theory of relativity" (section A.2). (You can find English translations in various places, including the back of my GR book, http://www.lightandmatter.com/genrel/ .) It later turned out that Einstein's interpretation of his own theory was wrong, and it was less Machian than he thought. A useful alternative test theory, in which rotation really is relative rather than absolute, is Brans-Dicke gravity. The original paper on B-D gravity has a readable intro that discusses this point. The paper is available on on Brans's web page: http://loyno.edu/~brans/ST-history/ . Solar system tests show that B-D gravity is not viable, so in this sense rotation is absolute and non-Machian. Re the "aether" idea, an aether is basically a preferred vector field, but in B-D gravity the extra equipment that comes attached to spacetime is a scalar field.

 

[Fantasist]

Rotation involves acceleration, so it shouldn't be surprising that it is locally detectable in the same manner that acceleration is. In fact, I cannot see how it could be otherwise with rotation given the detectability of linear acceleration.

How would you locally detect then whether a planet is going around its sun in a circular orbit or whether it sits just at a stationary location relatively to some hollow (and thus effectively massless) sphere (assuming the latter to be featureless and the universe otherwise empty, and ignoring tidal effects on the planet)?

 

[PAllen]

How would you locally detect then whether a planet is going around its sun in a circular orbit or whether it sits just at a stationary location relatively to some hollow (and thus effectively massless) sphere (assuming the latter to be featureless and the universe otherwise empty, and ignoring tidal effects on the planet)?

Orbit is not rotation. Do you see a difference between a merry go round spinning versus a merry go round orbiting a planet? Two completely different scenarios. The rotation is trivially locally detectable, the orbit is not.

 

[bcrowell]

How would you locally detect then whether a planet is going around its sun in a circular orbit

You can't, if "locally" refers to a region of space that's small compared to the orbit. The planet's motion is inertial, and there is nothing detectable about the fact that it's orbiting the sun.

In general, the effects of rotation are proportional to the enclosed area. E.g., the Sagnac effect is proportional to the enclosed area.

 

[Nugatory]

How would you locally detect then whether a planet is going around its sun in a circular orbit or whether it sits just at a stationary location relatively to some hollow (and thus effectively massless) sphere (assuming the latter to be featureless and the universe otherwise empty, and ignoring tidal effects on the planet)?

You don't. They're both in free fall, following a geodesic locally straight line through spacetime.
However, neither is a rotation.

 

[Fantasist]

Do you see a difference between a merry go round spinning versus a merry go round orbiting a planet?

Yes, I see one: in the latter case the force that holds the merry go round together is of the action-at-a-distance type, in the former it isn't. Otherwise I don't see any difference.

 

[PAllen]

Yes, I see one: in the latter case the force that holds the merry go round together is of the action-at-a-distance type, in the former it isn't. Otherwise I don't see any difference.

1) Gravity has nothing to with holding the merry go round together.

2) Do you think someone on the rim of the merry go round would feel the same thing in the two cases? If you admit they would not, this is the difference between rotation and orbit. If you don't, you reject reality, and it is not clear how to discuss anything.

 

[Dale]

How would you locally detect then whether a planet is going around its sun in a circular orbit or whether it sits just at a stationary location relatively to some hollow (and thus effectively massless) sphere (assuming the latter to be featureless and the universe otherwise empty, and ignoring tidal effects on the planet)?

How would you detect the colors of a rainbow if you closed your eyes and didn't use any instruments to detect light? You can always make some perfectly measurable thing undetectable by forbidding enough measurements.

If you want to ask an honest question I will be glad to answer.

 

[greypilgrim]

What if the universe was completely empty apart from a single object. What would determine the frame in which the object does not rotate? There are no reference points whatsoever, the situation is completely isotropic.

 

[WannabeNewton]

Otherwise I don't see any difference.

PeterDonis and I had a really long discussion about the various differences, both locally and globally, between orbital rotation and spin (as well as discussions of other things of course) in this thread: https://www.physicsforums.com/showthread.php?t=702423

Hope you find it useful!

 

[PeterDonis]

Yes, I see one: in the latter case the force that holds the merry go round together is of the action-at-a-distance type, in the former it isn't. Otherwise I don't see any difference.

An observer orbiting a planet is weightless; an observer on the rim of an ordinary merry go round is not. That is a significant difference.

 

[Nugatory]

Do you see a difference between a merry go round spinning versus a merry go round orbiting a planet?

Yes, I see one: in the latter case the force that holds the merry go round together is of the action-at-a-distance type, in the former it isn't.

There's another difference, more important because it is completely local:

In the orbiting case there exists a reference frame in which all points of the merry-go-round are at rest (if we can ignore tidal effects) AND Newton's first and second laws are obeyed. In the rotating case, there does not exist such a frame. Thus, I can distinguish the two cases by performing local experiments.

I cannot, however, distinguish the other two cases you asked about:

whether a planet is going around its sun in a circular orbit or whether it sits just at a stationary location relatively to some hollow (and thus effectively massless) sphere

by performing completely local experiments. Newton's laws are obeyed in both cases in a frame in which all points of the merry-go-round are at rest.

 

[bcrowell]

What if the universe was completely empty apart from a single object. What would determine the frame in which the object does not rotate? There are no reference points whatsoever, the situation is completely isotropic.

You would, for example, observe that gyroscopes precessed relative to the object. For more on this see the references in #12.

 

[PAllen]

Aether puts me to sleep.

 

[pervect]

I think it would be useful to illustrate the difference between "rotating" and "orbiting", the two seem to be conflated in this thread.

I will do this by exhibiting a diagram of two orbiting bodies. Several "snapshots" are taken throughout the orbit to show the position and orientation of the body at different times. There is a "black mark" on the body to allow its orientation to be determined. While both bodies are orbiting, only one body is rotating, which clearly (to my mind at least) illustrates that the concept of "orbiting" is distinct and different from the concept of "rotating".

I haven't included examples of non-orbiting bodies in a state of rotation and non-rotation, but I hope such can be easily imagined without a diagram. Thus we see that orbiting and rotation are different concepts, a body can have any of the four possible combinations of rotation/non-rotation and orbiting / non-orbiting.

In the attached diagram, the body on the left is orbiting and rotating (with respect to the fixed stars). The body on the right is orbiting and not rotating (with respect to the fixed stars).

I hope this will be helpful and will prompt a more coherent explanation of what the question(s) are. At least some of the posters in this thread seem to be interested more in Newtonian ideas ('instantaneous action at a distance") than GR ideas ("following a geodesic").

The diagram itself is basically Newtonian as it uses the Newtonian idea of "rotation relataive to the fixed stars".

I wouldn't mind talking some about the difference between the Newtonian and GR views, but I'm not actually sensing any interest in the GR point of view ("follwing a geodesic") at this point in time. Or perhaps the interest is there, and there's a language barrier.

 

[yuiop]

How would you locally detect then whether a planet is going around its sun in a circular orbit or whether it sits just at a stationary location relatively to some hollow (and thus effectively massless) sphere (assuming the latter to be featureless and the universe otherwise empty, and ignoring tidal effects on the planet)?

An observer on the planet with a gyroscope can determine if the planet is rotating or not. If it is not rotating and the sun appears to move in the sky then we can be sure that the sun and planet are not stationary and must be orbiting about their common centre of mass or barycentre.

If the planet is rotating then it will have an axis of rotation and an observer at the pole can construct a platform that rotates in the opposite direction such that gyroscopes mounted on the observation platform detect no rotation. If the observer sees no motion of the sun in the sky when standing on this non rotating platform, then the sun and planet are not orbiting each other.

Instead of a gyroscope, a Sagnac device or devices based on the Coriolis effect such as a Foucault pendulum, can be used to locally detect the absolute rotation of the planet.

 

[WannabeNewton]

Instead of a gyroscope, a Sagnac device or devices based on the Coriolis effect such as a Foucault pendulum, can be used to locally detect the absolute rotation of the planet.

You have to be careful when talking about rotation of extended bodies like planets in GR (i.e. you have to be careful when talking about global rotation in GR). The mounted gyroscope can yield a positive result for rotation of the planet but at the same exact time the Sagnac effect can yield a negative result for rotation of the planet and vice versa.

 

[yuiop]

You have to be careful when talking about rotation of extended bodies like planets in GR (i.e. you have to be careful when talking about global rotation in GR). The mounted gyroscope can yield a positive result for rotation of the planet but at the same exact time the Sagnac effect can yield a negative result for rotation of the planet and vice versa.

I was assuming devices that could fit in a small lab to comply with the 'local' specification. However, I would be interested in an example of where a global measurement by a Sagnac device (eg a fibre optic that extends all the way around the equator) contradicts gyroscope measurements, as I was not aware of that. Are we talking about frame dragging here?

 

[WannabeNewton]

Are we talking about frame dragging here?

It plays a role yeah. See section 3.2 of the following notes: http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf

 

[Fantasist]

I think it would be useful to illustrate the difference between "rotating" and "orbiting", the two seem to be conflated in this thread.

I will do this by exhibiting a diagram of two orbiting bodies. Several "snapshots" are taken throughout the orbit to show the position and orientation of the body at different times. There is a "black mark" on the body to allow its orientation to be determined. While both bodies are orbiting, only one body is rotating, which clearly (to my mind at least) illustrates that the concept of "orbiting" is distinct and different from the concept of "rotating".

I haven't included examples of non-orbiting bodies in a state of rotation and non-rotation, but I hope such can be easily imagined without a diagram. Thus we see that orbiting and rotation are different concepts, a body can have any of the four possible combinations of rotation/non-rotation and orbiting / non-orbiting.

In the attached diagram, the body on the left is orbiting and rotating (with respect to the fixed stars). The body on the right is orbiting and not rotating (with respect to the fixed stars).

I hope this will be helpful and will prompt a more coherent explanation of what the question(s) are. At least some of the posters in this thread seem to be interested more in Newtonian ideas ('instantaneous action at a distance") than GR ideas ("following a geodesic").

The diagram itself is basically Newtonian as it uses the Newtonian idea of "rotation relataive to the fixed stars".

I wouldn't mind talking some about the difference between the Newtonian and GR views, but I'm not actually sensing any interest in the GR point of view ("follwing a geodesic") at this point in time. Or perhaps the interest is there, and there's a language barrier.

Let me reply with my own diagram

consider a hammer thrower who is rotating around his own axis, keeping a mass attached to a string in a circular orbit around him. In this case, the force that keeps the mass in orbit acts directly only on the part of the mass where the string is attached. The rest of the mass is just passive mass, and thus an internal stress force (i.e. an electrostatic force) must be set up in it to keep it together (this internal stress force is what we feel as 'weight' when we accelerate e.g. in a car).

Now, in contrast to this consider a 'gravitational hammer thrower', who keeps the mass in orbit by means of the gravitational force rather than a mechanical connection. In this case, the force acts directly on all atoms of the mass, so no internal stress force is set up. Ignoring tidal effects, all parts of the mass should orbit the center of force on their own account, maintaining the same orientation with regard to the local radius vector. 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).

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.

 

[PeterDonis]

Ignoring tidal effects

Which means the mass will *not* move the way you say it will, since tidal effects are what would make it "rotate" as you have drawn it. See below.

The point is that this rotation is also inertial like the linear orbital motion

No, it isn't. Here's how you know it isn't: the way you drew the mass moving in the gravitational case is *not* the way the mass would actually move, if it were moving completely inertially (i.e., zero accelerometer reading everywhere). It would only move the way you drew it if it were tidally locked to the planet it was orbiting, as the Moon is, to a fairly good approximation, to the Earth. But orbiting satellites are *not* tidally locked; their spatial orientation is not in any way tied to the direction the Earth is in, because tidal effects are simply way too weak, over the size scale of a typical satellite.

In other words, a satellite, like, say, the International Space Station, that was really moving completely inertially around the Earth would look like the diagram on the right in pervect's post, *not* the diagram on the left. A satellite that was moving like the diagram on the left would be experiencing tidal effects and so would *not* be moving completely inertially.

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).

No, you have it backwards. See above. The only feature of the motion that is a consequence of the "spherical gravitational potential" is the orbital motion of the center of mass of the satellite.

 

[Nugatory]

Now, in contrast to this consider a 'gravitational hammer thrower', who keeps the mass in orbit by means of the gravitational force rather than a mechanical connection.

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.

If the orbiting object is not rotating about its axis, then it will be inertial and the accelerometer will just tell you that it is in free fall, not accelerating... and the accelerometer will be correct. An object in orbit is not accelerating even though it is following a curved path in three-dimensional space.

 

[pervect]

Let me reply with my own diagram

consider a hammer thrower who is rotating around his own axis, keeping a mass attached to a string in a circular orbit around him. In this case, the force that keeps the mass in orbit acts directly only on the part of the mass where the string is attached. The rest of the mass is just passive mass, and thus an internal stress force (i.e. an electrostatic force) must be set up in it to keep it together (this internal stress force is what we feel as 'weight' when we accelerate e.g. in a car).

Now, in contrast to this consider a 'gravitational hammer thrower', who keeps the mass in orbit by means of the gravitational force rather than a mechanical connection. In this case, the force acts directly on all atoms of the mass, so no internal stress force is set up. Ignoring tidal effects, all parts of the mass should orbit the center of force on their own account, maintaining the same orientation with regard to the local radius vector. 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).

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.

While it is true that rotation of the hammer thrower would not be detected by a linear accelerometer, it would be detected by (for example) a ring laser gyroscope. http://en.wikipedia.org/wiki/Ring_laser_gyroscope. Or an ordinary gyroscope, for that matter.

You don't need any "global" interpretation to interpret the results of a ring laser gyroscope. You do need the ring laser to enclose a finite area, which means that a ring laser gyroscope (for example) can't be pointlike - but the area enclosed can be arbitrarily small.

Similar remarks can be made about an accelerometer, actually. Any real accelerometer will need to have a finite size. If you have a mass on a spring, for instance, the mass has to be able to move to compress or stretch the string. An accelerometer can be very small, but it won't be pointlike.

"Local" does not mean that one restricts measurement to a single point. A "local" measurement of a function includes knowledge of not only the value of the function at a point, but a value of the function "in the neighborhood" of the point. This information is the same information needed to compute the function, and it's derivatives at that point.

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.

 

[Fantasist]

Ignoring tidal effects

Which means the mass will *not* move the way you say it will, since tidal effects are what would make it "rotate" as you have drawn it. See below.

Ignoring tidal effects means ignoring the gravitational force change dF from front to back of the mass compared to the force F itself. Of course, my drawing is not to a realistic scale. For e.g. any planet this should easily be satisfied, but if you want just make the mass arbitrarily small, and the assumption applies exactly.

But orbiting satellites are *not* tidally locked; their spatial orientation is not in any way tied to the direction the Earth is in, because tidal effects are simply way too weak, over the size scale of a typical satellite.

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. And assuming a circular orbit, this means a part orbiting at a larger radius will stay at a larger radius, so the mass overall must rotate once over one orbit. Like I said, this rotation is fully inertial (considering the global gravitational field).

 

[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.