Big bang in Robertson - Walker space - time

  • #1
WannabeNewton
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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!)?
 

Answers and Replies

  • #2
fzero
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Since the scale factor evolves with time and curvature invariants (the Ricci scalar would be a more obvious choice) diverge at [tex]R(t_0)=0[/tex], it makes sense to treat [tex]t_0[/tex] as the origin of the time coordinate. This can always be achieved by a constant shift in time. The singularity at [tex]r=0[/tex] is just a coordinate singularity, analogous to the same one in polar coordinates for Cartesian space. Curvature invariants don't do anything special there.
 
  • #3
WannabeNewton
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I'm sorry when you say r = 0 do you mean the scale factor or just the radial coordinate?
 
  • #4
fzero
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I'm sorry when you say r = 0 do you mean the scale factor or just the radial coordinate?
The radial coordinate.
 
  • #5
WannabeNewton
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Ok thank you very much.
 

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