Big bang in Robertson - Walker space - time

[WannabeNewton]

Hey guys. I was looking at the Robertson - Walker metric and went about calculating $$R^{\mu \nu }R_{\mu \nu }$$ as a way of looking at coordinate independent curvature properties through the Ricci Tensor. Considering I did it correctly I ended up with:

$$R^{\mu \nu }R_{\mu \nu } = \frac{1}{R(t)^{2}}[9\ddot{R} + 3\frac{(\ddot{R}R + 2\dot{R}^{2} + 2k)^{2}}{R(t)^{2}}]$$

Where R(t) is the scale factor, k is the spatial curvature index, and dots denote time derivatives. So if R(0) = 0 is taken to be the origin then can it be interpreted as being that moment right at the big bang? I ask this because the big bang is said to be a real, geometric singularity and in this coordinate - invariant expression R(0) = 0 induces infinite curvature in the manifold. Also, what would r = 0 mean in the manifold (r being the radial coordinate as defined by the metric) since it doesn't really show up in this expression (if I did it right of course!)?

 

[fzero]

Since the scale factor evolves with time and curvature invariants (the Ricci scalar would be a more obvious choice) diverge at $$R(t_0)=0$$, it makes sense to treat $$t_0$$ as the origin of the time coordinate. This can always be achieved by a constant shift in time. The singularity at $$r=0$$ is just a coordinate singularity, analogous to the same one in polar coordinates for Cartesian space. Curvature invariants don't do anything special there.

 

[WannabeNewton]

I'm sorry when you say r = 0 do you mean the scale factor or just the radial coordinate?

 

[fzero]

I'm sorry when you say r = 0 do you mean the scale factor or just the radial coordinate?

The radial coordinate.

 

[WannabeNewton]

Ok thank you very much.