Can Empty Space Curve? FRW Cosmology Equations Explored

[copernicus1]

If I start with the standard FRW cosmology equations,
$${\eqalign{
3\dot a^2/a^2&=8\pi\rho-3k/a^2\cr
3\ddot a/a&=-4\pi\left(\rho+3P\right),}}$$
and set [/tex]\rho=P=0[/itex] (or $T^{\mu\nu}=0$), I have
$${\eqalign{
3\dot a^2/a^2&=-3k/a^2\cr
3\ddot a/a&=0.}}$$
The second equation gives $$\ddot a=0,$$ but $$\dot a$$ seems to depend on the value of k.

Namely, if I set k=0, then $$\dot a=0$$ and this leads to an ordinary Minkowski space metric. If I choose k=+1, then a is complex and that doesn't seem physical, but if I set k=-1, then I can get $$3\dot a^2/a^2=3/a^2~~~~\Longrightarrow~~~~\dot a=1~~~~\Longrightarrow~~~~a=t,$$ which I suppose describes a spatially hyperbolic universe (k=-1) with no energy/matter content, where spatial distances increase linearly in time.

Do we just ignore this solution based on the assumption that nonzero k implies the presence of mass/energy by definition, or have I gone wrong in my reasoning somewhere?

 

[George Jones]

if I set k=-1, then I can get $$3\dot a^2/a^2=3/a^2~~~~\Longrightarrow~~~~\dot a=1~~~~\Longrightarrow~~~~a=t,$$ which I suppose describes a spatially hyperbolic universe (k=-1) with no energy/matter content, where spatial distances increase linearly in time.

Do we just ignore this solution based on the assumption that nonzero k implies the presence of mass/energy by definition, or have I gone wrong in my reasoning somewhere?

This is the Milne universe, which has curved 3-dimensional spatial sections, and which also has zero spacetime curvature.

Just a short, redundant comment for clarity.

For example, an empty FRW model is characterized by hyperbolic spatial curvature.

Here spacetime is flat and space is curved.

For a more plausible kinematic analysis, one might want to compare the supernova data to an FRW Omega=1 spatially flat dust model using light travel distance from time of emission.

Here spacetime is curved and space is flat.

Mass/energy (the stress-energy tensor) is the source of spacetime curvature. Spatial curvature is somewhat arbitrary, i.e., it depends on how the particular 3-dimensional hypersurfaces are chosen.

You're talking about the Milne universe, which is a a patch of Minkowski spacetime in somewhat unusual coordinates.

Start with Minkowski spacetime in spherical coordinates,

<br /> ds^2 = dt&#039;^2 - dr^2 - r^2 d \Omega^2 ,<br />

and make the coordinate transformation

<br /> t&#039; = t \cosh \chi<br />

<br /> r = t \sinh \chi.<br />

 

[copernicus1]

Awesome thanks. How can I tell what the spacetime curvature is, as opposed to just the spatial curvature?

 

[George Jones]

Awesome thanks. How can I tell what the spacetime curvature is, as opposed to just the spatial curvature?

Calculate the spacetime curvature tensor (from the metric). Note that if all the components of a tensor are zero in one coordinate system, then they are all zero in all coordinate systems.

In this case, because the components of the Minkowski metric are constant in an inertial coordinate system, the components of the curvature tensor are all zero in inertial coordinates, are zero in spherical coordinates, and are all zero in cosmological coordinates for the Milne universe.

 

[copernicus1]

Thanks this is very helpful.

 

[stevmg]

Without going through the mathematics as posted above, it would appear that a theoretical totally empty universe with nothing in it would have no mass to "bend" it. Once light is introduced to observe this, a theoretical "mass equivalent" m=E/c^2 is brought in, which would change that and curvature would be effected.