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).