7
$\begingroup$

I've read on Newton's Bucket and the pseudo-special frame of reference bound to CMB but I didn't find anything that would bind the two well - so, is the CMB frame "rest frame" for a spinning body not locally gravitationally bound? If not, is any other?

Let's perform a thought experiment: we have a space probe with a powered Control Moment Gyroscope which allows us to put the probe in arbitrary spin, and accelerometers that measure local acceleration in every direction (without care what's the source of that acceleration: gravity, engines, centripetal force etc) placed at extended locations on the probe (way off center of mass).

We place the probe squat in the middle of the Giant Void so it gets as little local disturbances as possible from nearby celestial bodies (since there are none; all very distant.) Then we drive the CMG in such a way as to minimize readouts of all the accelerometers.

Will we reach a flat zero readout? And with the readout so minimized that any operation only increases it, will be the frame of reference bound to the probe non-rotating relative to the CMB-based pseudo-master frame, or will it be something else - bound to what (besides the probe)?

Improve this question
$\endgroup$
1
  • $\begingroup$ It may be interesting to bring in the effects of quantum mechanics into the answer to this question. To some extent, a superfluid can give you an absolute rotation-free reference frame (or at least you could tell of you aren't in such a frame). $\endgroup$
    – KF Gauss
    Commented Feb 27, 2017 at 0:51

1 Answer 1

Reset to default
4
+50
$\begingroup$

Yes, there exists a frame in which all accelerometers can in principle reach zero. If fact, there are infinitely many such frames. In some of those frames, the CMB would appear isotropic; in others, it wouldn't. (I'm assuming that you're neglecting the radiation pressure of the CMB on the probe, since that can be made arbitrarily small by making the probe small enough.)

The statement that accelerometers register no acceleration in an inertial frame (i.e. a frame in which constant-spatial-coordinate curves obey the geodesic equation), which follows from the equivalence principle, is a statement about the fundamental nature of spacetime - it would hold true in any universe described by general relativity, regardless of that universe's initial or boundary conditions, and in particular it has nothing at all to do with the existence of the CMB.

The fact that there exists a frame in which the CMB appears isotropic is a statement about the initial conditions of our universe - it certain doesn't follow from general relativity. And the existence of the CMB does not in any way affect the readings of locally moving accelerometers (except through a truly tiny amount of radiation pressure, as mentioned above). Put another way, general relativity just says that the universe appears locally like Minkowski spacetime in small enough regions. The existence of a CMB frame asserts that there exists a global frame in which the universe is (approximately) space-translationally and rotationally invariant (though not time-translationally- or boost-invariant) - a completely logically independent statement.

Also, note that "rotationally invariant" does not mean "looks the same no matter how you're rotating" - it means "looks the same both before and after you begin and end rotating, so that you are no longer rotating but are now pointing in a different direction from before."

Improve this answer
$\endgroup$
5
  • $\begingroup$ In some of those frames, the CMB would appear isotropic; in others, it wouldn't. - this is not enough for me to accept the answer. If I put a probe in a frame where CMB appears isotropic, then send it into a wild spin, CMB will remain isotropic. But CMB may appears anisotropic and the anisotropy may be invariable in relation to the probe (probe moves in a straight line relative to CMB) or rotate (translation + rotation) around the probe - are there frames of reference where the latter would still display 0 on the accelerometers? Will the first?) $\endgroup$
    – SF.
    Commented Feb 27, 2017 at 2:28
  • $\begingroup$ @SF. In the first situation, where the probe sees the CMB as anisotropic but constant in time, its accelerometers will read zero, because it is not rotating with respect to the spacetime metric. In the second situation, where the probe sees the anisotropy to be rotating, its accelerometers will measure acceleration, because it is rotating w.r.t. to the metric. There's also a third possible situation in which the CMB appears isotropic and the accelerometers read zero, and a fourth possible situation in which the CMB again appears isotropic but the accelerometers detect an acceleration. $\endgroup$
    – tparker
    Commented Feb 27, 2017 at 5:50
  • $\begingroup$ @SF. The fourth situation shouldn't be considered "rotation-neutral" because it's not locally inertial - the probe won't see nearby free objects as traveling in straight lines. I suppose you could say that the third situation is the one where the probe sees "the universe [as] most rotation-neutral." But the key point is that the first situation is almost as good, because the laws of physics themselves (though not the CMB) are perfectly rotationally-invariant in this frame, and everything except for the CMB will be just as well-behaved as in the third situation. $\endgroup$
    – tparker
    Commented Feb 27, 2017 at 5:58
  • $\begingroup$ due to spacetime being curved, we can't really take 'nearby free objects trajectory' for granted. (I purposely chose Giant Void to avoid this; imagine finding the inertial frame in low orbit of a neutron star!). As for 'everything except CMB, it's all in constant flux, filaments stretching, clusters moving, colliding or pulling apart, that makes it a poor reference for a frame of reference. $\endgroup$
    – SF.
    Commented Feb 27, 2017 at 7:23
  • $\begingroup$ @SF. I just depends how far away you look. The equivalence principle says that locally, any spacetime looks flat - and in particular, free particles move in perfectly straight lines. Specifically, "locally" means "over distances much less than the length scale set by the curvature." It's true that near a neutron star, spacetime is quite strongly curved (small radius of curvature), so you don't out have to look out very far before gravitational effects become noticeable. But even there, gravity seems to "vanish" as long as your restrict your attention to a small enough region of spacetime $\endgroup$
    – tparker
    Commented Feb 27, 2017 at 8:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.