On 2/13/16 2/13/16 12:34 AM, RichD wrote:
> On February 10, tjrob137 wrote:
>> The usual standard for "no rotation" is a locally inertial frame,
>> and any observer in freefall is at rest in one. How can such an
>> observer observe the black hole "itself"? (to check if it is rotating
>> wrt her locally inertial frame). The answer, of course, is that she
>> cannot, because the black hole "itself" has no substance to observe.
> But if no such substance, it raises the question of what
> exactly is rotating, besides space itself, which is a
> 'thing', in GR, and so should be observable.
No. You need to distinguish between world and model. GR is a MODEL of the world.
Space is part of the MODEL, not the world, and is not at all a 'thing", and is
not ITSELF "observable". Ditto for spacetime.
But the structure of spacetime can be observed indirectly via
its effect on the motions of objects (e.g. the geodesic paths
followed by small freefalling objects).
> Perhaps
> 'frame dragging' comes ino play (although I barely
> understand that bit), which was recently onserved by
> a Stanford team.
Yes. A rotating black hole "drags nearby frames" vastly more than the earth.
> As another point, there appears to be a pardox here.
Only due to your misunderstandings.
> Free fall is supposedly an inertial frame, hence not
> rotating, according to SR.
And GR, locally.
> But the infalling obserever
> must be rotating with the black hole, as he falls toward
> the center. Therefore this should be observable, as
> rotation is absolute.
But in GR rotation is not "absolute" in the sense you mean.
For any LOCAL observation, a locally inertial frame is the standard relative to
which rotation can be measured. But for NON-local observations, there is no such
standard. But one can, in some circumstances including this one, observe signals
sent from a distant inertial frame and measure one's own inertial frame's
rotation relative to the distant one.
> That is, the observer must 'spiral' as he falls, if
> that makes sense, given the weird geometry inside.
For a Kerr black hole this just so happens to be true, but your approach to it
is wrong. Near the black hole's horizon, locally inertial frames rotate relative
to distant (locally out there) inertial frames.
> We know the universe isn't infinite, though it may
> expand forever.
One must be more precise. We know the universe is not infinite in the past
temporal direction. But we have no knowledge whether it is infinite or finite
spatially, or in the future temporal direction.
> However, if it is rotating, there must be an
> axis of rotation, implying a preferred reference frame.
> Which could be seen by an observer in some higher dimension.
Perhaps. But WE are not such observers, so this is useless speculation outside
of physics.
Moreover, there can be regions in which the rotation is different from that of
other regions, so this "preferred frame" need not be universal or global (e.g. a
universe containing multiple Kerr black holes orientated differently).
>> For a compact universe (i.e. one with no boundary and thus
>> no need for boundary conditions), I don't think there
>> can be an overall rotation that is consistent with any possible
>> topology.
>> I'm not sure about a universe that is compact spatially but
>> not temporally.
>
> ?
>
>> The Kerr manifold is a universe with an overall rotation. It is
>> quite complex.... It is not compact.
>
> In math, compact means finite domain, so I'm not sure of your usage.
I use it in the usual topological sense: a compact manifold is closed (i.e.
contains all its limit points), and thus has no boundary. In the geometry of
physics this implies that it has finite volume (though there are unphysical
mathematical manifolds for which this is not true).
Tom Roberts