Timeline for What frame of reference in the universe is (most) rotation-neutral?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 2, 2017 at 14:50 | vote | accept | SF. | ||
| Mar 2, 2017 at 14:50 | history | bounty ended | SF. | ||
| Feb 27, 2017 at 8:54 | comment | added | tparker | @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 | |
| Feb 27, 2017 at 7:23 | comment | added | SF. | 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. | |
| Feb 27, 2017 at 5:58 | comment | added | tparker | @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. | |
| Feb 27, 2017 at 5:50 | comment | added | tparker | @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. | |
| Feb 27, 2017 at 2:28 | comment | added | SF. | 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?) | |
| Feb 27, 2017 at 1:38 | history | answered | tparker | CC BY-SA 3.0 |