Friedmann-Lemaître-Robertson-Walker metric...

[Orion1]

Friedmann-Lemaître-Robertson-Walker metric:

[math]c^{2} d\tau^{2} = -c^{2} dt^2 + a(t)^2 \left(\frac{dr^2}{1 - k r^2} + r^2 d\theta^2 + r^2 \sin^2 \theta \; d\phi^2 \right)[/math]

 

[math]g_{\mu \nu} = \left( \begin{array}{ccccc} \; & dt & dr & d \theta & d\phi \\ dt & -1 & 0 & 0 & 0 \\ dr & 0 & a(t)^2 g_{rr} & 0 & 0 \\ d\theta & 0 & 0 & a(t)^2 g_{\theta \theta} & 0 \\ d\phi & 0 & 0 & 0 & a(t)^2 g_{\phi \phi} \end{array} \right) \; \; \; \; \; \; J = 0[/math]

 

I noticed that the Friedmann-Lemaître-Robertson-Walker metric scale factor function [math]a(t)[/math] operates on the [math]g_{rr}, g_{\theta \theta}, g_{\phi \phi}[/math] metric tensors as part of these tensors and that this metric does not have rotation. If the metric had both rotation and scale factor, would the scale factor function also operate on the [math]g_{t \phi}[/math] metric tensors?

 

[math]g_{\mu \nu} = \left( \begin{array}{ccccc} \; & dt & dr & d \theta & d\phi \\ dt & -1 & 0 & 0 & a(t)^2 g_{t \phi} \\ dr & 0 & a(t)^2 g_{rr} & 0 & 0 \\ d\theta & 0 & 0 & a(t)^2 g_{\theta \theta} & 0 \\ d\phi & a(t)^2 g_{t \phi} & 0 & 0 & a(t)^2 g_{\phi \phi} \end{array} \right) \; \; \; \; \; \; J \neq 0[/math]

 

The solution that I have calculated for a metric with both rotation and scale factor is:

[math]c^{2} d\tau^{2} = -c^2 \; dt^2 + a(t)^2 \left(\frac{dr^2}{1 - kr^2} + \rho^2 d\theta^2 + ((r^2 + \alpha^2)(r^2 + \alpha^2 - \alpha^2 \sin^2 \theta) + r_s r \alpha^2 \sin^2 \theta) \frac{\sin^2 \theta \; d\phi^2}{\rho^2} + \frac{4r_s r \alpha \sin^2 \theta \; c \; dt \; d\phi}{2r^2 + \alpha^2 + \alpha^2 \cos 2 \theta} \right)[/math]

 

[math]\rho^{2} = r^{2} + \alpha^{2} \cos^{2} \theta[/math]

 

This metric reduces into the Friedmann-Lemaître-Robertson-Walker metric in the limiting case where there is no rotation.

Reference:

Metric tensor - General Relativity - Wikipedia

Friedmann–Lemaître–Robertson–Walker metric - Wikipedia

Kerr metric - Wikipedia

Edited October 13, 2012 by Orion1

[ajb]

Question: Now your metric contains a term that mixes time and an angular coordinate, does it admit CTCs?

[Orion1]

Question: Now your metric contains a term that mixes time and an angular coordinate, does it admit CTCs?

I cannot answer that question without a metric tensor based theorem that gives definite positive result. There is a pattern for metric solutions that contain closed timelike curves (CTC) in General Relativity, one is the existence of poly-horizons in General Relativity, in which the inner horizon is a Cauchy horizon.

 

The Cauchy horizon is obtained by setting [math](g_{rr})^{-1} = 0[/math] then normalizing the equation, then solving for the zeros using the quadratic equation, which gives two radial results, with one of the results being the Cauchy horizon.

 

An example of a metric with a inner Cauchy horizon is the Reissner–Nordström metric:

[math](g_{rr})^{-1}= 1 - \frac{r_{s}}{r} + \frac{r_{Q}^{2}}{r^{2}} = \frac{1}{r^2}(r^2 - r_sr + r_Q^2) = 0[/math]

[math]r_\pm = \frac{r_{s} \pm \sqrt{r_{s}^2 - 4r_{Q}^2}}{2}[/math]

 

The Kerr metric inner horizon:

[math](g_{rr})^{-1} = \frac{\Delta}{\rho^2} = 0[/math]

[math]\Delta = r^{2} - r_{s} r + \alpha^{2} = 0[/math]

[math]r_{-} = \frac{r_{s} + \sqrt{r_{s}^{2} - 4 \alpha^{2}}}{2}[/math]

 

The Kerr metric outer horizon:

[math]g_{tt} = \left(1 - \frac{r_s r}{r^2 + \alpha^2 \cos^2 \theta} \right) = 0[/math]

[math]r_{+} = \frac{r_{s} + \sqrt{r_{s}^{2} - 4\alpha^{2} \cos^{2}\theta}}{2}[/math]

 

For a BTZ black hole the poly-horizons are defined within the metric tensor:

[math]g_{rr} = \frac{l^2 r^2}{(r^2 - r_{+}^2)(r^2 - r_{-}^2)}[/math]

 

Although the existence of poly-horizons in General Relativity models is not proof of the existence of closed timelike curves in such models, it does present a weak indication.

 

For the rotational scale factor metric listed above, it may contain a closed timelike curve at [math]g_{t \phi}[/math], because its relative the Kerr metric probably contains one at that specific metric tensor near the singularity.

Another important note is that the Universe has not been observed to be rotating, however it seems improbable that rotation in any given model in General Relativity is absolutely zero.

Reference:

Reissner–Nordström metric - charged black holes - Wikipedia

Kerr metric = important surfaces - Wikipedia

BTZ black hole metric - Wikipedia

Closed time-like curves - General Relativity - Wikipedia

Godel metric - cosmological interpretation - Wikipedia

Edited October 14, 2012 by Orion1