The equation for the surface gravity of a black hole in Kerr metric is-(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\kappa_\pm=\frac{r_\pm-r_\mp}{2(r_\pm^2+a^2)}[/tex]

where r_{+}is the outer event horizon- [itex]r_+=M+\sqrt(M^2-a^2)[/itex], r_{-}is the inner event horizon- [itex]r_-=M-\sqrt(M^2-a^2)[/itex] and [itex]a[/itex] is the spin parameter in metres- [itex]a=J/mc[/itex].

An exact solution in the equatorial plane is-

[tex]\kappa_\pm \equiv \frac{M}{(r_\pm^2+a^2)}-\Omega_\pm^2\,R_\pm[/tex]

where [itex]\Omega_\pm[/itex] is the frame dragging at r_{+}and r_{-}as observed from infinity- [itex]\Omega_\pm=a/(r_{\pm}^2+a^2)}[/itex] and [itex]R_\pm[/itex] is the reduced circumference at the outer and inner event horizons.

This equation also suits [itex]a_t=a_g-a_c[/itex] where a_{t}is total acceleration, a_{g}is gravitational acceleration and a_{c}is centripetal acceleration, brought on by frame dragging in this case [itex](a_c=v^2/r \equiv (\Omega\,R)^2/R=\Omega^2\,R)[/itex].

The solution is exact and gives equivalent results for both r_{+}and r_{-}(a_{t}and [itex]\kappa_-[/itex] being negative at r_{-}due to extreme frame dragging).

The above works at the inner and outer event horizons and may be extrapolated to get an idea of the gravitational field at the equator around a rotating black hole-

[tex]a_t= \frac{M}{(r^2+a^2)}-\Omega^2\,R[/tex]

where [itex]\Omega[/itex] (or [itex]\omega[/itex]) is the frame dragging rate- [itex]\omega=2Mrac/\Sigma^2[/itex] and R is the reduced circumference- [itex]R=(\Sigma/\rho) \sin \theta[/itex] where-

[tex]\Sigma^2=(r^2+a^2)^2-a^2\Delta \sin^2\theta[/tex]

[tex]\Delta= r^{2}+a^{2}-2Mr[/tex]

[tex]\rho^2=r^2+a^2 \cos^2\theta[/tex]

The above would be described as as observed from infinity (coordinate). The relativistic (local) solution might be-

[tex]a_{t(rel)}= \frac{M}{(r^2+a^2)\,\alpha}-\frac{\Omega^2\,R}{\alpha^2}\alpha[/tex]

where proper tangential velocity is [itex]v_{rel}=\Omega\,R/\alpha[/itex] where [itex]\alpha[/itex] is the reduction factor or redshift-[itex]\alpha=(\sqrt{(\Delta)}\rho/\Sigma)[/itex], a_{g}is divided by [itex]\alpha[/itex] and a_{c}is multiplied by the [itex]\alpha[/itex] (as discussed in https://www.physicsforums.com/showthread.php?t=407909"). The equation can be written-

[tex]a_{t(rel)}= \frac{M}{(r^2+a^2)\,\alpha}-\frac{\Omega^2\,R}{\alpha}\equiv\frac{a_t}{\alpha}[/tex]

The above would apply to an object in free fall with no angular momentum (ZAMO- zero angular momentum observer).

The Kepler equation for an object in stable orbit in Kerr metric is-

[tex]\Omega_s=\frac{\pm\sqrt{M}}{r^{3/2}\pm a\sqrt{M}}[/tex]

where [itex]\pm[/itex] denotes prograde and retrograde orbits. When using [itex]\Omega_s[/itex] in the equation for a_{t}, I would have anticipated a_{t}to be zero but this doesn't seem to be the case. For prograde orbits, a_{g}is marginally greater than a_{c}and in the case of retrograde orbits, a_{g}is marginally smaller than a_{c}. While Kepler orbits in Kerr metric are complex, I'm sure there is a solution that can apply to both the coordinate and local quantities to obtain a_{t}=0 when [itex]\Omega=\Omega_s[/itex]. I've tried to find derivation for the Kepler stable orbit equation in order to see exactly what it's based on but the only source I could find was http://www.roma1.infn.it/teongrav/VALERIA/TEACHING/ONDE_GRAV_STELLE_BUCHINERI/AA2006_2007/DISPENSE/ch20.pdf" [Broken] (page 321) which links [itex]\Omega_s[/itex] to the equation for A, the redshift for an object in orbit- [itex]A=\sqrt{(g_{tt} + 2\Omega g_{\phi t}-\Omega^2 g_{\phi \phi})}[/itex].

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# Stable orbits in Kerr metric

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