- #1

WannabeNewton

Science Advisor

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## Main Question or Discussion Point

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

[tex]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}}][/tex]

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!)?

[tex]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}}][/tex]

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!)?