History and geometry of flat universe

[Ranku]

1. The present universe is observed to be flat. Was it always flat, before it started its accelerated expansion?

2. Is the Riemann tensor zero for this flat universe? Is its geometry that of special relativity?

I'd appreciate if Marcus or Ich or any other science advisor weighed in on this. Thanks.

 

[nicksauce]

If the universe is flat, is has always been flat and always will be. That is because the geometry of the universe is determined by a **constant**, k.

The Riemann tensor is non-zero. According to my calculations...
$$R^t_{xtx} = a\dot{\dot{a}}$$
$$R^x_{yxy} = (\dot{\dot{a}})^2$$
and all the other values come from symmetry between x<->y<->z.
The geometry is not that of special relativity. In special relativity the metric is diag(-1,1,1,1). In a flat expanding universe the metric is diag(-1,a^2(t),a^2(t),a^2(t)), where a(t) is the scale factor of the universe.

 

[George Jones]

For Friedmann-Walker-Robertson universes, space is flat but spacetime is curved.

The Riemann tensor for Friedmann-Walker-Robertson universes is not zero because FRW spactimes are not flat. Three-dimensional spatial hypersurfaces orthogonal to cosmic time (i.e., space) are intrinsically flat, i.e., the Riemann tensor constructed from the spatial metric induced on these hypersurfaces by the spacetime metric is zero.

 

[Chalnoth]

1. The present universe is observed to be flat. Was it always flat, before it started its accelerated expansion?

This is not expected to be the case. Rather, it is expected that our region of the universe started off with very significant curvature, but as it was dominated by an inflaton field that drove a very rapidly-accelerated expansion, it was driven to be almost perfectly flat in virtually no time.

This happens because the effect of the curvature scales with the expansion as $$1/a^2$$, but during inflation, the dominant energy density was almost independent of expansion. As the scale factor increased by a factor of $$10^{30}$$ or more, the spatial curvature was driven to very near zero.

2. Is the Riemann tensor zero for this flat universe? Is its geometry that of special relativity?

I'd appreciate if Marcus or Ich or any other science advisor weighed in on this. Thanks.

Nicksauce has responded to this point well.

 

[Ranku]

Thank you all for the responses. Two more questions.

1. Can the presently accelerating universe be flat without inflation?

2. In a flat universe, omega = 1.
Omega = matter(regular matter + dark matter) + dark energy.
Can we have Omega = 1 without dark matter?

 

[Chalnoth]

Thank you all for the responses. Two more questions.

1. Can the presently accelerating universe be flat without inflation?

If there happens to be an alternative explanation for our current observations that point to inflation, I suppose it's possible.

2. In a flat universe, omega = 1.
Omega = matter(regular matter + dark matter) + dark energy.
Can we have Omega = 1 without dark matter?

In principle it's a different issue. The flatness is related to the relationship between the average expansion rate and the total energy density. If the expansion rate is too fast compared to the energy density, then it's open. If it's too slow, then it's closed. If it's "just right", then it's flat.

 

[Ranku]

In principle it's a different issue. The flatness is related to the relationship between the average expansion rate and the total energy density. If the expansion rate is too fast compared to the energy density, then it's open. If it's too slow, then it's closed. If it's "just right", then it's flat.

Is not the average expansion rate controlled by the total energy density, since dark energy that is driving the accelerated expansion is part of it?

 

[Chalnoth]

Is not the average expansion rate controlled by the total energy density, since dark energy that is driving the accelerated expansion is part of it?

Right, so, the first of the Friedmann equations is:

$$H^2 = \frac{8\pi G}{3} \rho - \frac{k c^2}{a^2}$$

So basically if $$H^2 > \frac{8\pi G}{3} \rho$$, $$k < 0$$. Likewise, if $$H^2 < \frac{8\pi G}{3} \rho$$, then $$k > 0$$.

 

[Ranku]

So basically if $$H^2 > \frac{8\pi G}{3} \rho$$, $$k < 0$$. Likewise, if $$H^2 > \frac{8\pi G}{3} \rho$$, then $$k > 0$$.

Both the inequalities are same, yet k changes. I'm confused.

 

[Chalnoth]

Both the inequalities are same, yet k changes. I'm confused.

Ack, sorry, typo. Fixed.

 

[Ranku]

Ack, sorry, typo. Fixed.

Thanks