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George Jones
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#26
Nov21-06, 10:01 AM
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Quote Quote by cesiumfrog View Post
Could you elaborate a little on exactly where those closed timelike curves are in the Kerr solution?
Let me elaborate a bit on what Chris said.

O'Neill, in his book The Geometry of Kerr Black Holes, proves:

there is a closed timelike curve through any event inside the inner (Cauchy) horizon, i.e., through any event for which r < r-.

Carroll gives the following simple example. Consider a curve for which [itex]\phi[/itex] varies, and for which [itex]t[/itex], [itex]r[/itex], [itex]\theta[/itex] are held constant. Because of periodicity with respect to [itex]\phi[/itex], any such curve is closed.

Now, the timelike part.

Take [itex]r < 0[/itex] with [itex]|r|[/itex] small, and [itex]\theta = \pi/2[/itex]. Note [itex]r[/itex] is a coordinate, not a radial distance, and negative [itex]r[/itex] is part of (extended) Kerr. Because [itex]0 = dt = dr = d \theta[/itex], the line element along the curve is

[tex]
ds^2 = \left( r^2 + a^2 + \frac{2Mr a^2}{r^2} \right) d\phi^2
[/tex]

For [itex]r[/itex] negative and small. the last term, whcih is negative, dominates, and thus [itex]ds^2[/itex] is the line element for a timilike curve.