hellfire said:
pervect, I think you are right if your answer is related only to material objects. The question whether rods expand or not can be analized taking a look to the stress and therefore to the Riemann tensor. However, one may ask also whether two points in space separated by a "small" distance increase its distance or not. As far as I know, this is indeed measurable. This has nothing to do with stresses on rods and the Riemann tensor, but only with the increase of the scale factor: in a linearly expanding model with homogeneity at arbitrary scales and with zero Riemann tensor, the scale factor increases and two arbitrary points do always increase its distance. So, in my opinion, the question here reduces to know what metric is relevant at what scales.
The problem with using a metric to define the "expansion of space" is that we have the option of using many different metrics.
If we insist that the metric be homogeneous, isotropic, and spatially flat, and if we ignore "lumps", we pin the form of the metric down very closely, but one issue remains -defining the time scale for the 't' coordinate.
If we furthermore define the time scale 't' to be that of the proper time of an observer who observes the universe to be isotropic, we wind up with a unique metric. However, operationally, I believe that this definition is equivalent to my more physical defintion, as long as the universe is not a vacuum universe (more on this later). In a vacuum universe, there's nothing to look at to define the "preferred frame", so my proposal fails.
I agree that if a(t) is linear, there are no tidal forces on a meter stick. However, the Riemann is not zero everywhere in a FRW universe where a(t) is linear. Though the tidal force components of the Riemann are zero, other components, such as R^{x}_{yxy} are not zero. The Einstein tensor G is also not zero for a(t) linear, it reqires the presence of matter to have this solution.
[add]\rho is non-zero, P is non-zero, but \rho+3P is zero for a(t) linear - the solution requries negative pressures.
\rho = 3a_t/a^2 if you set the problem up with an orthonormal basis of one-forms.
The required presence of matter (a non-zero Einstein tensor which implies a non-zero stress-energy tensor) implies that a hypothetical observer has something to "look at" to define the "rest frame" of the linearly expanding universe, which is why I say my more physical definition is operationally equivalent to yours for this case.
Note that the observers at both ends of a meter-stick cannot both simultaneously be in frames which are "at rest" with respect to the universe, i.e. cannot be in the priveleged frame from which the universe appears to be isotropic. At most one obserer on a meter-stick can be in this special frame.
Thus there is no incosistency in saying that meter sticks do not expand and that space does expand. Usually there is a tidal force on the meter stick, but even when there is no such force, both ends of the meter stick are not and cannot be in the preferred "rest" frame which is how we are defining our distances to expand.
This is always true, even in the very special case (linear expansion) where there is no tidal force on a meter-stick.
I'll note that for the vacuum universe (where my defintion fails) we have an ambiguity as to what metric to use, the static, spatially flat, non-expanding universe, or the expanding non-spatially flat Milne universe. Thus, while my observational method fails to define an expansion factor when there is nothing to look at, your metric approach also has some ambiguities to deal with too. The Milne solution is the only expanding isotropic vacuum solution. Other solutions have the da/dt = 0.
The remaining issue is how to deal with "lumps". I'm not sure how best to do that, anyway this post is already too long.