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Let's say we can observe expansion in a supercluster.
We define origin of our frame of reference at the center of the supercluster.
We observe an object/atom at point A at time T. The object is motionless relative to the origin.
We wait for expansion until T+ΔT and again observe the object.
Is the object at A or somewhere else?

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The question is an interesting one, but I think it needs to be further defined. In particular, how do you define point A? The wording of your question indicates that you intend it to be a position relative to the center of the supercluster, but the "center" is also poorly defined. Why not define the position relative to the observer, which would have to be dealt with eventually anyway? Even more difficult, are the inherent problems of measuring position in quantum mechanics, i.e. the uncertainty principle. –  AdamRedwine Sep 7 '11 at 18:41
@AdamRedwine: If we say the object is of size 1 cm^3, can we forget about quantum mechanics to not further complicate the matter? –  Mitja Sep 7 '11 at 18:56
Not really... if you say that the 1 cm^3 object is centered at A, how well defined is you point A? If your point is defined to a volume of 1 cm^3, then after time has elapsed, the point has expanded and the object could still be located at the same "point" even though it could have "moved" in the traditional sense. If you define your point to be very small, the same is the case though the bounds are tighter. By the time you get to an exact location (or even an infinitesimally small one), you run smack dab back into quantum territory. –  AdamRedwine Sep 7 '11 at 19:26
Seems to me it is related to the classic "can one cross the same river" . The river is never the same, as the water flows, so the answer is no; on the other hand if one defines the bed of the river and call that the river, the answer is yes. Paradoxes arise when one mixes up meta-levels. In the case of this cluster, if one defines the x,y,z position with respect to another cluster, the answer is no. If one defines it with respect to the center of mass of the cluster, the answer is yes. –  anna v Sep 8 '11 at 5:04

1 Answer 1

In the classical description of general relativity, the spacetime points constitute a smooth manifold with local coordinates $x^{\mu}$. In order to compute distances and intervals between points, an extra piece of information is needed, namely the metric tensor field $g_{\mu\nu}$(x). Space expansion can be thought of as not moving the points around, but rather as simply a change of the spatial part of the metric tensor with time.

For example, if we write the spacetime coordinates as $x^{\mu}$ = (t, $x^i$) where i=1,2,3 then we can write the metric of a spatially expanding universe as $$ds^2 = c^2dt^2-a(t)^2g_{ij}dx^idx^j$$ where the spatial metric components $g_{ij}$ do not depend on time.

So spatial expansion is a change in the metric tensor field, rather than any "motion" of the spacetime points themselves.

An alternative way to think of this is that points of a spacetime manifold in isolation (i.e. with no other physical fields defined) do not have any physical significance. This is connected with Einstein's hole argument.

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