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

1. Dec 15, 2013

### 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?

2. Dec 15, 2013

### Staff: Mentor

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.

3. Dec 15, 2013

### Meir Achuz

But rotation with respect to what?

4. Dec 15, 2013

### Staff: Mentor

The same thing that proper acceleration is with respect to.

5. Dec 15, 2013

### WannabeNewton

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.

6. Dec 15, 2013

### greypilgrim

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

7. Dec 15, 2013

### WannabeNewton

I explained that in post #5.

8. Dec 15, 2013

### Bill_K

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.

9. Dec 15, 2013

### pervect

Staff Emeritus
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.

10. Dec 15, 2013

### WannabeNewton

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.

11. Dec 15, 2013

### WannabeNewton

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 [Broken]).

Last edited by a moderator: May 6, 2017
12. Dec 16, 2013

### bcrowell

Staff Emeritus
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.

13. Dec 16, 2013

### Fantasist

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)?

14. Dec 16, 2013

### PAllen

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.

15. Dec 16, 2013

### bcrowell

Staff Emeritus
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.

16. Dec 16, 2013

### Staff: Mentor

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

17. Dec 16, 2013

### Fantasist

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.

18. Dec 16, 2013

### PAllen

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.

Last edited: Dec 16, 2013
19. Dec 16, 2013

### Staff: Mentor

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.