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Stable orbits in Kerr metric

  1. Jun 15, 2010 #1
    The equation for the surface gravity of a black hole in Kerr metric is-


    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" [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].
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 15, 2010 #2
  4. Jun 16, 2010 #3
    On a slight side note, what exactly is the significance of putting a dot above a coordinate as appears to be the case in most calculations that relate to E/m and L/m, as shown on page 19 in http://www.tat.physik.uni-tuebingen.de/~kokkotas/Teaching/Relativistic_Astrophysics_files/GTR2009_4.pdf" [Broken]?
    Last edited by a moderator: May 4, 2017
  5. Jun 16, 2010 #4


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    [tex]\dot x [/tex] usually denotes [itex]dx/d\tau[/itex], where [itex]\tau[/itex] is proper time along a worldline, although in some contexts it may denote differentiation with respect to some other parameterisation of a worldline.

    The slight complication is that, in the context of Lagrangians, you consider the dotted variables to be completely independent of the same undotted variables (e.g. in this case [tex]\mathcal L(t, r, \theta, \phi, \dot t, \dot r, \dot \theta, \dot \phi)[/tex] should be considered a function of 8 independent variables), but when you evaluate its value along a worldline, by setting [tex]\dot t = dt/d\tau[/tex], etc, the function becomes constant.
    Last edited by a moderator: May 4, 2017
  6. Jun 20, 2010 #5
    Thanks for the responses. http://arxiv.org/abs/gr-qc/0407004" [Broken] (page 10) provides a relativistic equation for gravity at the equator in a Kerr field-


    If we remove the reduction factor (i.e. multiply by the redshift [itex]\alpha[/itex]), the equation reduces to-


    where [itex]\Sigma_{\phi}^2=(r^3+a^2r+2Ma^2)r,\ \omega_{\phi}=2Mra/\Sigma_{\phi}^2[/itex] and [itex]R_{\phi}=\Sigma_{\phi}/r[/itex].

    which is the gravitational field as observed from infinity in the equatorial plane and reduces to-


    at the event horizon which is equivalent to the Killing surface gravity equation in post #1. There is still the issue of [itex]a\neq0[/itex] when [itex]\omega=\Omega_s[/itex].
    Last edited by a moderator: May 4, 2017
  7. Jun 21, 2010 #6
    According to the paper in post #5, page 12, the local gravitational acceleration for an object in orbit at the equatorial plane is-


    where [itex]\theta_{\hat{\phi}}[/itex] is described as the shear vector and [itex]k_{(\text{lie})}[/itex] is the ZAMO Lie relative curvature vector.

    [tex]\vec{g}=-\vec{a}(n)[/tex] (from post #5)



    where for a stable orbit-

    [tex]\nu_\pm=\frac{r^2+a^2\pm2a\sqrt{Mr}}{\sqrt{\Delta}\left[a\pm r\sqrt{r/M}\right]}[/tex]

    which is equivalent to another equation for [itex]\nu_\pm[/itex] where [itex]\nu_\pm=(\Omega_s-\omega)R/\alpha[/itex] where [itex]\Omega_s[/itex] is the angular velocity for a stable orbit as the equation in post #1, [itex]\omega[/itex] is the frame dragging rate, R is the reduced circumference and [itex]\alpha[/itex] is the redshift, though for a(U)=0 to be exact, the equation from the paper should be used.

    There appears to be a slight conflict of signs as to obtain a(U)=0 for a stable orbit, g should simply equal a(n) or v should be negative in order to obtain either [itex]0=-a_g+a_c[/itex] or [itex]0=a_g-a_c[/itex]. It also appears in the paper that [itex]\gamma[/itex] is represented by two equations, [itex]\sqrt{|1-\nu^2|}[/itex] on page 7, and [itex]1/\sqrt{(1-\nu^2)}[/itex] on page 8, though it's likely the latter that applies to the above equation.

    The equations also reduce to the Schwarzschild solution when the spin parameter a=0.


    Gravitational field in the equatorial plane as observed from infinity-



    where for a stable orbit-

    Last edited: Jun 21, 2010
  8. Jun 23, 2010 #7
    Gravity at the equator is reduced by the centripetal acceleration brought on by frame dragging which reduces to, when observed from infinity, the equation for the Killing surface gravity at the event horizon but this only appears to apply when [itex]\theta=\pi/2[/itex]. As [itex]\theta[/itex] approaches zero, while frame dragging itself is constant over the event horizon regardless of [itex]\theta[/itex], centripetal acceleration begins to reduce in accordance with the reduced circumference, [itex]R=(\Sigma\sin \theta)/\rho[/itex] where [itex]a_c=\omega^2R[/itex], which would imply that the surface gravity of the BH would vary from equator to poles which would violate the zeroth law of black hole thermodynamics. Spin redistributes mass and hence gravity can be stronger at the equator while centripetal acceleration brought on by the spin will in some way reduce this increase. Is there an equation that demonstrates the redistribution of mass in an object due to spin and it's effect on the gravitational field?
    Last edited: Jun 23, 2010
  9. Jun 30, 2010 #8
    In order for the gravitational field to apply for [itex]\theta\neq\pi/2[/itex] in Kerr metric, two things should to be taken into account- 1). [itex]\kappa[/itex] has to be constant over the event horizons when the solution reduces to a black hole (in accordance with the zeroth law in BH thermodynamics), 2). [itex]a_t\approx M/r^2[/itex] should apply at infinity regardless of [itex]\theta[/itex] (which is backed up to some degree when considering the gravitational field of a torus). To maintain a constant surface gravity over the event horizons, [itex]M(r^2+a^2)^2/\Sigma^3}[/itex] (or [itex]M/(r^2+a^2)[/itex] at the event horizons) should be multiplied by-

    [tex]d_\kappa=\frac{(r_\pm-M)\rho_\pm+a^2\sin \theta}{M\rho_\pm}[/tex]

    which is simply 1 at the equator and where [itex]\rho_\pm[/itex] is [itex]\rho[/itex] relative to either the outer or inner horizon. This could be related to mass distribution and with the centripetal acceleration brought on by frame dragging, provides a smooth surface gravity over the EH's irrespective of [itex]\theta[/itex] in conjunction with the Killing surface gravity equation in post #1, positive (attractive) at r+ and negative (repulsive) at r-. To apply to the field, [itex]d_\kappa[/itex] needs to reduce to 1 at infinity when [itex]\theta\neq\pi/2[/itex]-

    [tex]D_{g\pm}=\frac{d_\kappa r}{r_\pm+d_\kappa r-d_\kappa r_\pm}[/tex]

    where Dg+ applies for r>r+ and Dg- theoretically applies for r<r-. The above provides a smooth surface gravity over [itex]r_\pm[/itex] regardless of a/M and reduces to simply 1, irrespective of [itex]\theta[/itex], for the Schwarzschild solution. The equation for gravity as observed from infinity that applies for a variable [itex]\theta[/itex] for an object in free fall with zero angular momentum might be written-


    which matches exactly with the Killing surface gravity equation at the top of post #1 at the event horizons.
    Last edited: Jun 30, 2010
  10. Dec 13, 2010 #9
    Relativistic equation for gravity around a Kerr black hole for an object in free fall from rest at infinity for various [itex]\theta[/itex]-

    [tex]\vec{a}(n)=\frac{M}{\Sigma^2 \rho^3}\left( \left[(r^2-a^2\cos^2\theta)(r^2+a^2)^2-4Mr^3a^2\sin^2\theta\right] \Delta^{-1/2},-2ra^2\sin\theta\cos\theta(r^2+a^2)\right)[/tex]

    when [itex]r=r_\pm[/itex] and [itex]\vec{a}(n)[/itex] is multiplied by the reduction factor [itex]\alpha[/itex]-


    meaning [itex]\kappa[/itex] is constant over both the inner and outer event horizons irrespective of [itex]\theta[/itex].

    A 'shorthand' version for as observed from infinity would be-

    [tex]\vec{a}(n)=\frac{M}{\Sigma^3 \rho^2}\left(\left[(r^2-a^2\cos^2\theta)(r^2+a^2)^2-4Mr^3a^2\sin^2\theta\right],-2ra^2\sin\theta\cos\theta(r^2+a^2)\sqrt{\Delta} \right)[/tex]

    It's a bit patchy but the equation can also be rewritten as [itex]a(n)_\infty=a_{g,\infty}-a_{c,\infty}[/itex] with centripetal acceleration due to frame dragging, [itex]\omega^2R[/itex], expressed (boxed)-

    [tex]\vec{a}(n)_\infty=\frac{M}{\Sigma^3 \rho^2}\left((r^2-a^2\cos^2\theta)(r^2+a^2)^2-2ra^2\sin\theta\cos\theta(r^2+a^2)\sqrt{\Delta}\right)- \boxed{\frac{4M^2r^2a^2\sin\theta}{\Sigma^3 \rho}}\frac{r\sin\theta}{\rho}[/tex]

    while it's no suprise that [itex]a_{g,\infty}[/itex] is complex (and does indeed reduce at the poles locally), the introduction of [itex]r\sin\theta/\rho[/itex] to [itex]\omega^2R[/itex] was unexpected unless this can somehow be incorporated into [itex]a_{g,\infty}[/itex].

    source- http://arxiv.org/abs/gr-qc/0407004 page 10
    Last edited: Dec 13, 2010
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