The following is the equation for a Keplerian stable orbit at the equator around a Kerr black hole-(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\tag{1}v_s=\frac{\pm\sqrt{M}(r^2\mp2a\sqrt{Mr}+a^2 )}{\sqrt{\Delta}(r^{3/2}\pm a\sqrt{M})}[/tex]

where [itex]M=Gm/c^2,\ a=J/mc[/itex] and [itex]\Delta=r^2-2Mr+a^2[/itex]

which for a static black hole would reduce to -

[tex]v_s=\frac{\pm\sqrt{M}\,r^2}{\sqrt{r^2-2Mr}\,r^{3/2}}[/tex]

and I would have thought that it was save to assume that for a static black hole-

[tex]\frac{v_s^2}{r}=\frac{M}{r^2}\sqrt{g_{rr}}[/tex]

which is basically [itex]a_c=a_g[/itex] and where [itex]g_{rr}=(1-2M/r)^{-1}[/itex]

rearranging the equation, the following should apply-

[tex]\frac{v_s^2r}{M}=\sqrt{g_{rr}}[/tex]

but for some reason, the answer I get is simply [itex]g_{rr}[/itex] instead of [itex]\sqrt{g_{rr}}[/itex]. Does anyone see why this might be the case?

http://www.iop.org/EJ/article/0067-...uest-id=99398a5d-17c3-4c96-a4c7-516cf1ef178b" for equations (1)

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# Kepler orbits

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