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Based on http://casa.colorado.edu/~ajsh/phys5770_08/bh.pdf" (page 37), closed timelike curves exist within a boundary dictated by [itex]g_{\phi \phi}=0[/itex] which occurs where
[tex]\frac{R^4}{\Delta}=a^2 \sin^2 \theta[/tex]
where [itex]R=\sqrt(r^2+a^2)[/itex] and [itex]\Delta=R^2-2Mr+Q^2[/itex]
I'd appreciate it if someone could give some indication of how to solve for r in the following equation-
[tex]\frac{(r^2+a^2)^2}{(r^2+a^2-2Mr+Q^2)}=a^2\sin^2\theta[/tex]
[tex]\frac{R^4}{\Delta}=a^2 \sin^2 \theta[/tex]
where [itex]R=\sqrt(r^2+a^2)[/itex] and [itex]\Delta=R^2-2Mr+Q^2[/itex]
I'd appreciate it if someone could give some indication of how to solve for r in the following equation-
[tex]\frac{(r^2+a^2)^2}{(r^2+a^2-2Mr+Q^2)}=a^2\sin^2\theta[/tex]
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