# Can empty space curve?

1. Jan 17, 2014

### copernicus1

{\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?

2. Jan 17, 2014

### George Jones

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

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.

Last edited: Jan 17, 2014
3. Jan 17, 2014

### copernicus1

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

4. Jan 17, 2014

### George Jones

Staff Emeritus
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.

5. Jan 17, 2014