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The equation for the surface gravity of a black hole in Kerr metric is-
[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 at is total acceleration, ag is gravitational acceleration and ac 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- (at 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], ag is divided by [itex]\alpha[/itex] and ac 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 at, I would have anticipated at to be zero but this doesn't seem to be the case. For prograde orbits, ag is marginally greater than ac and in the case of retrograde orbits, ag is marginally smaller than ac. 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 at=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" (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].
[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 at is total acceleration, ag is gravitational acceleration and ac 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- (at 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], ag is divided by [itex]\alpha[/itex] and ac 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 at, I would have anticipated at to be zero but this doesn't seem to be the case. For prograde orbits, ag is marginally greater than ac and in the case of retrograde orbits, ag is marginally smaller than ac. 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 at=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" (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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