# Killing surface gravity for a Kerr black hole

1. Oct 27, 2009

### stevebd1

From wikipedia 'http://en.wikipedia.org/wiki/Surface_gravity" [Broken]'-
For a static black hole, this is-

$$\kappa=\frac{c^2}{4M}=\frac{c^2}{2r_s}$$

where $M=Gm/c^2$ and rs is the Schwarzschild radius (2M)

this is synonymous to Newtons equation for gravity $g=Gm/r^2$ at the event horizon where r=2M (the Schwarzschild radius), the results being the same.

When looking at Kerr black holes, the Killing surface gravity is-

$$\kappa=c^2\frac{(r_+-r_-)}{2(r_+^2+a^2)}$$

where r+ is the outer event horizon $r_+=M+\sqrt(m^2-a^2)$, r- is the inner event horizon $r_+=M+\sqrt(m^2-a^2)$ and a is the spin parameter $a=J/mc$.

which reduces to-

$$\kappa=c^2\frac{r_s}{2r_s^2}$$

when a=0 which is equivalent to the first equation for a Schwarzschild black hole.

The Killing surface gravity works out less for the Kerr black hole than that for the static solution. Obviously this has something to do with spin and the Newtonian equation for gravity is 'modified' but what is this based on? I would have thought the reduction of gravity would be a consequence of the vacuum solution (Kerr metric) as apposed to something that seems related to the equivalent of Newtonian gravity. I understand that κ is abstract but it seems odd that it should reduce the higher the spin when normally it's considered that rotational kinetic energy may contribute to the stress energy tensor.

Last edited by a moderator: May 4, 2017
2. Oct 28, 2009

### Orion1

surface gravitational acceleration...

According to my understanding, increasing angular_momentum and angular velocity have the effect of decreasing surface gravitational acceleration.

Classical Newtonian surface gravitational acceleration:
$$F_a = F_g + F_c$$

$$F_a$$ - force due to acceleration
$$F_g$$ - gravitational force
$$F_c$$ - centripetal force

Integration via substitution:
$$m \cdot \frac{d^2 s}{dt^2} = - \frac{G m^2}{r^2} + mr \omega^2$$

Factoring out mass factor m, the resulting surface gravitational acceleration is:
$$\boxed{\frac{d^2 s}{dt^2} = - \frac{G m}{r^2} + r \omega^2}$$

If $$F_g = -F_c$$ then:
$$\boxed{\frac{d^2 s}{dt^2} = 0}$$

Angular_momentum is related to angular velocity through moment of inertia:
$$J = I \cdot \omega$$

Therefore, the greater the angular velocity, the greater the angular_momentum and the less surface gravitational acceleration.

The equation for the dimensionless spin parameter is:
$$a = \frac{c J}{Gm^2}$$

Moment of inertia of a sphere:
$$I = \frac{2 m r^2}{5}$$

Integration via substitution:
$$J = \frac{2 m r^2 \omega}{5} = \frac{Gm^2 a}{c}$$

Extremal angular velocity of a sphere:
$$\boxed{\omega = \frac{5Gma}{2cr^2}}$$

Integration via substitution:
$$\frac{d^2 s}{dt^2} = - \frac{G m}{r^2} + r \left( \frac{5Gma}{2cr^2} \right)^2 = - \frac{G m}{r^2} + \left( \frac{5Gma}{2c} \right)^2 \frac{}{r^3}$$

Extremal surface gravitational acceleration of a sphere:
$$\boxed{\frac{d^2 s}{dt^2} = - \frac{G m}{r^2} + \left( \frac{5Gma}{2c} \right)^2 \frac{}{r^3}}$$

Maximum angular_momentum of a Kerr black_hole corresponds to a dimensionless spin parameter $$a = 1$$.

Relativistic killing surface gravitational acceleration:
$$\kappa = \frac{}{4} \frac{d^2 s}{dt^2} = \frac{}{4} \left( - \frac{c^2}{r} + r \omega^2 \right)$$

$$\boxed{\kappa = \frac{}{4} \frac{d^2 s}{dt^2} = \frac{}{4} \left[- \frac{G m}{r^2} + \left( \frac{5Gma}{2c} \right)^2 \frac{}{r^3} \right]}$$

$$\boxed{r = \frac{}{2} \frac{(r_+^2 + r_a^2)}{(r_+ - r_-)}}$$

Relativistic Kerr killing surface gravitational acceleration:
$$\boxed{\kappa = \frac{}{4} \frac{d^2 s}{dt^2} = \frac{}{4} \left[- G m \left( \frac{2(r_+ - r_-)}{(r_+^2 + r_a^2)} \right)^2 + \left( \frac{5Gma}{2c} \right)^2 \left( \frac{2 (r_+ - r_-)}{(r_+^2 + r_a^2)} \right)^3 \right]}$$

Reference:
http://en.wikipedia.org/wiki/Centripetal_force" [Broken]
http://en.wikipedia.org/wiki/Angular_momentum" [Broken]
http://en.wikipedia.org/wiki/Moment_of_inertia" [Broken]
http://en.wikipedia.org/wiki/Surface_gravity" [Broken]

Last edited by a moderator: May 4, 2017
3. Oct 29, 2009

### stevebd1

Thanks for the reply Orion1. It appears the key word is surface gravity, as centripetal acceleration (ac) can affect the Earth's gravity at the equator (i.e. reducing it) it's also taken into account for the Kerr black hole, even though the centripetal acceleration for the Kerr BH is a result of frame dragging which is a relativistic effect, the resulting ac is still Newtonian in its roots and therefore included when calculating κ. I also had a look at how the 'short hand' equation for the κ in post #1 was derived-

The total gravitational acceleration is gravity minus the centripetal acceleration-

$$a_T=a_g-a_c$$

where aT is the total gravitational acceleration (κ), ag is gravity and ac is centripetal acceleration (all quantities as observed from infinity and not a representation of the local effect of gravity)

At the outer event horizon (r+)-

$$a_T=\frac{Gm}{r_+^2}-\frac{v^2}{r_+}$$

v (as observed from infinity) can be calculated through two different ways. The first is to multiply the angular velocity by the radius, v=ωr but in the case of a rotating black hole, r needs to be replaced with the reduced circumference, R-

$$R=\frac{\Sigma}{\rho}sin \theta$$

where

$$\] \Sigma^2=(r^2+a^2)^2-a^2\Delta sin^2\theta\\ \\ \Delta= r^{2}+a^{2}-2Mr\\ \\ \rho^2=r^2+a^2 cos^2\theta\\ \[$$

While this seems relativistic, it does provide the correct tangential velocity as observed at infinity, it's also backed up by the following simpler solution for v-

The angular momentum of a rotating black hole is based on J=vmr which is based on a spinning ring. This is also supported by the equation for the spin parameter which is a variation of this- a=J/mc where a is the radius and c is the velocity. Rearrange the equation to find the tangential velocity at r+-

$$v=\frac{J}{m r_+}$$

where J=aGm2/c, a being the unitless spin parameter between 0 and 1.

substituting for J in the equation for v-

$$v=\frac{a_{\ast}Gm}{r_+ c}[/itex] the results are equivalent to $v=\omega R$ Frame dragging rate (or angular velocity) in rads/sec- $$\omega=\frac{2Mrac}{\Sigma^2}$$ The equation for the surface gravity- $$a_T=\frac{Gm}{r_+^2}-\frac{(\omega R)^2}{r_+}$$ Using a 3 sol mass rotating black hole with a spin parameter of 0.95, the results for the original Killing surface gravity equation in post #1 are 2.4136e12 m/s2 while the results for the above equation are 3.678e12 m/s2, while not perfect, it does provide an answer fairly close to the original equation. _________________________________________________ An exact solution (in the equatorial plane at least) to the Killing surface gravity equation in post #1 is- $$a_T=\frac{Gm}{(r_+^2 +a^2)}-\omega_+^2 R_+$$ where ω+ is the frame dragging rate at r+ as observed from infinity and R+ is the reduced circumference at r+ (which always equals 2M or rs at the equator, regardless of the quantity of spin) The above also reduces to zero at a/M=1 which is analogous with κ. Last edited: Oct 29, 2009 4. Oct 30, 2009 ### stevebd1 A couple of equations from the following paper http://www.lsw.uni-heidelberg.de/users/mcamenzi/CObjects_06.pdf" [Broken] by Max Camenzind, page 253- $$\kappa_\pm=c^2\frac{r_\pm-r_\mp}{2(r_\pm^2+a^2)}$$ where r+ and r- are the outer and inner horizons respectively which implies that from infinity the Killing surface gravity at the inner horizon is negative. This is supported by the results for aT in post #3 where the frame dragging becomes so extreme that the resulting centripetal acceleration becomes greater than the gravitational pull. another equation for the Killing horizon is- $$\Omega_\pm=c\ \frac{a}{\left(r_{\pm}^2+a^2\right)}$$ which is equivalent to the frame dragging rate as observed from infinity at r+ and r-, $\omega_\pm$. The equation for surface gravity in post #3 seems to work analogously with κ until you move out of the equatorial plane (i.e. theta begins to vary from 90). According to the zeroth law of BH thermodynamics, surface gravity (κ) is constant over a black holes event horizon. When considering different latitudes using the post #3 equation, while the frame dragging rate remains constant, the reduced circumference begins to drop which means the centripetal acceleration drops and aT begins to increase on the surface of the event horizon towards the poles. The equation for κ in post #1 implies that the surface gravity is constant so there should be a way to modify the equations for ag or ac to reflect this. Last edited by a moderator: May 4, 2017 5. Oct 31, 2009 ### Orion1 killing surface gravitational acceleration... Is this the solution for the relativistic killing surface gravitational acceleration with respect to the Kerr metric? The total gravitational acceleration is gravitational acceleration minus the centripetal acceleration: $$a_T = \frac{d^2 s}{dt^2} = a_g - a_c$$ Spin parameter: $$\alpha = \frac{J}{mc} = \frac{Gma_{\ast}}{c^2} = \frac{r_s a_{\ast}}{2}$$ $$\boxed{\alpha = \frac{r_s a_{\ast}}{2}}$$ $$a_{\ast}$$ being the dimensionless spin parameter between 0 and 1. An exact solution in the equatorial plane to the killing surface gravitational acceleration equation: $$a_T = \frac{Gm}{(r_+^2 + \alpha^2)} - \omega_+^2 R_+$$ Frame dragging rate (or angular velocity) in rads/sec at $$r_+$$: $$\omega_+ = \frac{2Mr_ + \alpha c}{\Sigma_+^2}$$ Reduced circumference at $$r_+$$: $$R_+ = \frac{\Sigma_+}{\rho_+} \sin \theta$$ Radial metric criterion: $$r_s = 2M$$ $$\frac{Gm}{r_s^2} = \frac{c^2}{2 r_s}$$ $$\boxed{\frac{Gm}{(r_+^2 + \alpha^2)} = \frac{c^2}{2 \sqrt{r_+^2 + \alpha^2}}}$$ Integration via substitution: $$a_T = \frac{Gm}{(r_+^2 + \alpha^2)} - \left( \frac{2M r_+ \alpha c}{\Sigma_+^2} \right)^2 \frac{\Sigma_+}{\rho_+} \sin \theta = \frac{c^2}{2 \sqrt{r_+^2 + \alpha^2}} - \left( r_s r_+ \alpha c \right)^2 \frac{\sin \theta}{\Sigma_+^3 \rho_+} = c^2 \left( \frac{1}{2 \sqrt{r_+^2 + \alpha^2}} - \frac{\left( r_s r_+ \alpha \right)^2 \sin \theta}{\Sigma_+^3 \rho_+} \right)$$ Relativistic killing surface gravitational acceleration: $$\boxed{a_T = c^2 \left( \frac{1}{2 \sqrt{r_+^2 + \alpha^2}} - \frac{\left( r_s r_+ \alpha \right)^2 \sin \theta}{\Sigma_+^3 \rho_+} \right)}$$ Reference: http://en.wikipedia.org/wiki/Kerr_metric" [Broken] Last edited by a moderator: May 4, 2017 6. Nov 1, 2009 ### stevebd1 The above provides the Killing surface gravity as observed from infinity (i.e. taking into account only the Newtonian quantities of gravity and centripetal acceleration), the equation for tangential velocity (v) is as observed from infinity (v=ωR), if you want the local v (i.e. incorporating GR), then you have to divide by the redshift (α)- $$\alpha=\frac{\rho}{\Sigma}\sqrt{\Delta}$$ this is confirmed when considering the local v at the ergosphere- $$v_e=\frac{\omega_e R_e}{\alpha}=c$$ where the coordinate radius for the ergosphere is $r_e=M+\sqrt{(M^2-a^2cos^2\theta)}$ Obtaining the local quantity for gravity isn't as clean cut. As with Schwarzschild metric, you could simply multiply the Newtonian quantity of gravity by the radial curvature $\sqrt{(g_{rr})}$ where in Kerr metric $g_{rr}=\rho^2/\Delta$ An alternative, also analogues to the Schwarszchild metric, might be to multiply the Newtonian quantity of gravity by the inverse of the redshift, which in Kerr metric is also sometimes referred to as the reduction factor. Both quantites are similar with α-1 being marginally higher. So a solution to gravity in GR around a Kerr black hole in the equatorial plane might be- $$a_T=\frac{Gm}{(r^2+a^2)}\sqrt{g_{rr}}-\frac{\omega^2 R}{\alpha^2}$$ where $\sqrt{(g_{rr})}$ might be replaced with $\alpha^{-1}$ The results using either grr or α are fine outside the ergosphere where the local gravity is stronger than the local ca but inside the ergoregion, ca begins to increase over gravity which would imply that infalling matter might hold stable orbit inside the ergoregion which doesn't seem right. Maybe the equations need to be modified once inside the ergoregion. As expected, both the local ag and ac become infinite at the event horizon. ____________________________________________ On the subject of the zeroth law of BH thermodynamics and κ being constant over the surface of r+, another equation for the Killing surface gravity is- $$\kappa_+=c^2\frac{\sqrt{M^2-a^2}}{2Mr_+}$$ which can be rewritten- $$\kappa_+=\frac{c^2}{2r_+}\frac{\sqrt{M^2-a^2}}{M}$$ where $c^2/2r_+=Gm/(r_+^2+a^2)$ which means there is a reduction factor of $\sqrt{(M^2-a^2)}/M$ at the surface of the black hole. This is taken care of by the centripetal acceleration induced by frame dragging at the equator but the question is, what causes the reduction at the poles? This extract from a NASA page regarding Saturn seems to shed some light on the subject (http://saturn.jpl.nasa.gov/faq/FAQSaturn/#q14" [Broken])- so it seems acceptable to say that the long range gravity of a spinning object might be stronger at the equator than at the poles, removing the reductive effects of frame dragging, the reduction factor might be something like $\sqrt{(M^2-a^2cos^2\theta)}/M$. Using this equation on regular objects such as main sequences stars, the difference is slight but for relativistic objects such as neutron stars and black holes, the difference is significant. incorporating this into aT now provides a near-correct value for κ over the entire horizon regardless of theta (there appears to be a slight blip at theta=45 where the surface gravity reduces by approx. a factor of 0.986 which tells me that the introduction of $cos^2\theta$ isn't entirely perfect). Last edited by a moderator: May 4, 2017 7. Nov 5, 2009 ### stevebd1 While $\sqrt(M^2-a^2cos^2\theta)/M$ provides an exact solution for the gravity field at the poles and at the equator, it didn't provide a constant surface gravity over the event horizon from $\theta=0 \rightarrow 90$, the following provides an exact solution as observed from infinity- $$\kappa_+=\frac{r_+-r_-}{2(r_+^2+a^2)}=\frac{M}{(r_+^2+a^2)}\left(\frac{\sqrt{M^2-a^2}}{M}+\frac{a^2 sin\theta}{M\rho_+}\right)-\left(\frac{2Mr_+a}{\Sigma_+^2}\right)^2\frac{\Sigma_+}{\rho_+}sin\theta$$ which is κ = Killing surface gravity = (gravity x gravity distribution relative to spin) - (centripetal acceleration due to frame dragging). This provides a constant surface gravity over the surface of the event horizon. The asymmetrical gravitational field is reduced at the equator by frame dragging to match the poles exactly at the event horizon while the long range field will remain stronger than the poles. For a star with a spin parameter of a/M=0.2, the poles would be x0.98 compared to the equator, for a neutron star with a spin parameter of a/M=0.4, this would be x0.92 at the poles and for a black hole with a/M=0.95, this would be x0.31. It appears this reduction is uncannily matched in the equatorial plane at the event horizon when taking into account frame dragging regardless of the spin parameter. The weaker gravity at the poles may go some way to explaining GRB's. The above reduce to $\kappa=M/r_s^2=1/4M$ for a static black hole. The equation changes slightly for the inner horizon but still provides an exact result- $$\kappa_-=\frac{r_--r_+}{2(r_-^2+a^2)}=\frac{M}{(r_-^2+a^2)}\left(-\frac{\sqrt{M^2-a^2}}{M}+\frac{a^2 sin\theta}{M\rho_-}\right)-\left(\frac{2Mr_-a}{\Sigma_-^2}\right)^2\frac{\Sigma_-}{\rho_-}sin\theta$$ The main difference being that $\sqrt{(M^2-a^2)}$ becomes negative, which appears to be analogues with M and a changing places- $-\sqrt{(|a^2-M^2|)}$. While the negative gravity of the inner event horizon at the equator can be easily explained with the extreme frame dragging, the negative gravity at the poles isn't so easy to explain, maybe the nature of the poles of the event horizon is dictated by the nature of the equator. It's worth noting that while r is variable in most cases to get an idea of the gravitational field, $\rho_\pm$ in gravity distribution relative to spin is not variable and is relative to event horizon specifically, $\rho_+$ for outside r+ and $\rho_-$ for inside r-. Both equations can be rewritten as- $$\kappa_\pm=\frac{r_\pm-r_\mp}{2(r_\pm^2+a^2)}=\frac{M}{(r_\pm^2+a^2)}\left(\pm \frac{\sqrt{M^2-a^2}}{M}+\frac{a^2 sin\theta}{M P_\pm}\right)-\left(\frac{2Mr_\pm a}{\Sigma_\pm^2}\right)^2\frac{\Sigma_\pm}{\rho_\pm}sin\theta$$ where the r in $P_\pm$ (rho) varies only between r+ and r-, the above can be reduced to- $$\kappa_\pm=\frac{\pm\sqrt{M^2-a^2}\ P_\pm + a^2 sin\theta}{(r_\pm^2+a^2)P_\pm}\ -\ \frac{4(M r_\pm a)^2}{\Sigma_\pm^3\rho_\pm}sin\theta$$ though the separate components are not as clear. A better short hand version might be- $$\kappa_\pm=\frac{M}{(r_\pm^2+a^2)}D_\pm-\Omega_\pm^2R_\pm$$ where $D_\pm$$ is the gravity distribution factor (1 at the equator) and [itex]\Omega_\pm$ is the Killing horizon which is equivalent to the frame dragging rate as observed from infinity at r+ and r-.

Last edited: Nov 6, 2009
8. Nov 5, 2009

### Orion1

Killing surface gravity for a Kerr black hole...

$$\boxed{M = r_g = \frac{r_s}{2} = \frac{Gm}{c^2}}$$

Kerr metric spin parameter:
$$\alpha = \frac{J}{mc} = \frac{Gma_{\ast}}{c^2} = \frac{r_s a_{\ast}}{2} = r_g a_{\ast}$$

$$\boxed{\alpha = r_g a_{\ast}}$$
$a_{\ast}$ being the dimensionless spin parameter between 0 and 1.

The $r$ in $P_{\pm}$ (rho) varies only between $r_+$ and $r_-$.

Kerr metric gravitational distribution factor:
$$D_{\pm} = \pm \frac{\sqrt{r_g^2 - \alpha^2}}{r_g} + \frac{\alpha^2 \sin \theta}{r_g P_\pm}$$

Kerr metric killing horizon $\Omega_{\pm}$, is equivalent to the frame_dragging rate $\omega_{\pm}$, (or angular velocity) in radians per second, as observed from infinity at $r_+$ and $r_-$:
$$\Omega_{\pm} = \omega_{\pm} = \frac{2 r_g r_{\pm} \alpha c}{\Sigma_{\pm}^2}$$

Reduced circumference at $r_+$ and $r_-$:
$$R_{\pm} = \frac{\Sigma_{\pm}}{\rho_{\pm}} \sin \theta$$

Killing surface gravitational acceleration equation:
$$\kappa_{\pm} = \frac{c^2 r_g}{2(r_{\pm}^2 + \alpha^2)} D_{\pm} - \Omega_{\pm}^2 R_{\pm}$$

Integration via substitution:
$$\kappa_{\pm} = \frac{c^2 r_g}{2(r_{\pm}^2 + \alpha^2)} \left( \pm \frac{\sqrt{r_g^2-\alpha^2}}{r_g} + \frac{\alpha^2 \sin \theta}{r_g P_\pm} \right) - \left( \frac{2 r_g r_{\pm} \alpha c}{\Sigma_{\pm}^2} \right)^2 \frac{\Sigma_{\pm}}{\rho_{\pm}} \sin \theta = c^2 \left( \frac{\pm \sqrt{r_g^2 - \alpha^2} P_{\pm} + \alpha^2 \sin \theta}{2(r_{\pm}^2 + \alpha^2) P_{\pm}} - \frac{4(r_g r_{\pm} \alpha)^2 \sin \theta}{\Sigma_{\pm}^3 \rho_{\pm}} \right)$$

Killing surface gravitational acceleration:
$$\boxed{\kappa_{\pm} = c^2 \left( \frac{\pm \sqrt{r_g^2 - \alpha^2} P_{\pm} + \alpha^2 \sin \theta}{2(r_{\pm}^2 + \alpha^2) P_{\pm}} - \frac{4(r_g r_{\pm} \alpha)^2 \sin \theta}{\Sigma_{\pm}^3 \rho_{\pm}} \right)}$$

Reference:
http://en.wikipedia.org/wiki/Kerr_metric" [Broken]
http://en.wikipedia.org/wiki/Surface_gravity" [Broken]

Last edited by a moderator: May 4, 2017
9. Nov 8, 2009

### stevebd1

Some other equations that shed some light on the gravitational nature of Kerr black holes-

Relativistic stable orbit in the equatorial plane-

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

Stable angular velocity in the equatorial plane as observed from infinity-

$$\tag{2}\Omega_s=\frac{\pm\sqrt{M}}{r^{3/2}\pm a\sqrt{M}}$$

which appears to based on a slightly different expression for gravity as observed from infinity compared with the one used in previous posts $a_g=(M/(r^2+a^2))$

The relativistic stable orbit can be rewritten-

$$v_s=(\Omega_s-\omega)\frac{R}{\alpha}$$

$$v_s=(\Omega_s-\omega)\frac{\Sigma^2 sin\theta}{\rho^2 \sqrt{\Delta}}$$

which matches the first equation.

Some notion of radial coordinate acceleration $(x)$ can be derived from rearranging the above-

$$\frac{v_s^2}{R}=a_{g(\infty)}x-a_{c(local)}$$

$$\frac{\left(v_s^2R^{-1}+a_{c(local)}\right)}{a_{g(\infty)}}=x$$

where $a_{c(local)}=\omega^2R/\alpha^2$, the relativistic centripetal acceleration contributed by frame dragging, and $a_{g(\infty)}=M/(r^2+a^2)$, gravity as observed from infinity.

The results outside the ergosphere are close to $\alpha^{-1}$ but inside the ergoregion, $x$ increases significantly over $\alpha^{-1}$ which means radial coordinate acceleration ensures that gravity is always greater than the increasing local centripetal acceleration as they both approach infinity.

http://www.iop.org/EJ/article/0067-...uest-id=99398a5d-17c3-4c96-a4c7-516cf1ef178b" for equations (1) and (2).

Similar (and other) equations-
http://www.icra.it/MG/mg12/talks/apt1_slany.pdf

Last edited by a moderator: Apr 24, 2017
10. Nov 9, 2009

### Orion1

solving for the angular velocity killing horizon...

The method for solving for the angular velocity killing horizon $\Omega$, regarding the Kerr metric:
$$c^{2} d\tau^{2} = \left( 1 - \frac{r_{s} r}{\rho^{2}} \right) c^{2} dt^{2} - \frac{\rho^{2}}{\Delta} dr^{2} - \rho^{2} d\theta^{2} - \left( r^{2} + \alpha^{2} + \frac{r_{s} r \alpha^{2}}{\rho^{2}} \sin^{2} \theta \right) \sin^{2} \theta \ d\phi^{2} + \frac{2r_{s} r \alpha c \sin^{2} \theta }{\rho^{2}} dt d\phi$$

The Kerr metric is rewritten in the following spherical form:
$$\boxed{c^{2} d\tau^{2} = g_{tt} dt^{2} + g_{rr} dr^{2} + g_{\theta \theta} d\theta^{2} + g_{\phi \phi} d\phi^2 + 2 g_{t \phi} dt d\phi}$$

$$g_{t\phi} = \frac{r_{s} r \alpha c \sin^{2} \theta }{\rho^{2}}$$
$$g_{\phi\phi} = - \left( r^{2} + \alpha^{2} + \frac{r_{s} r \alpha^{2}}{\rho^{2}} \sin^{2} \theta \right) \sin^{2} \theta$$

Integration via substitution:
$$\Omega = \frac{d \phi}{dt} = - \frac{g_{t\phi}}{g_{\phi\phi}} = - \left( \frac{r_{s} r \alpha c \sin^{2} \theta }{\rho^{2}} \right) \left( - \frac{1}{\left( r^{2} + \alpha^{2} + \frac{r_{s} r \alpha^{2}}{\rho^{2}} \sin^{2} \theta \right) \sin^{2} \theta} \right) = \frac{r_{s} r \alpha c}{\rho^{2} \left( r^{2} + \alpha^{2} \right) + r_{s} r \alpha^{2} \sin^{2}\theta}$$

Angular velocity killing horizon:
$$\boxed{\Omega = \frac{r_{s} r \alpha c}{\rho^{2} \left( r^{2} + \alpha^{2} \right) + r_{s} r \alpha^{2} \sin^{2}\theta}}$$

Another interpretation of the Kerr metric in Boyer-Lindquist coordinates:
$$c^2d \tau^2 = \frac{\Delta - \alpha^2 \sin^2 \theta}{\rho^2} c^2 dt^2 - \frac{\rho^2}{\Delta}dr^2 - \rho^2 d\theta^2\ - \frac{\Sigma^2}{\rho^2} \sin^2 \theta d\phi^2\ + \frac{2 r_s r \alpha c \sin^2 \theta}{\rho^2} dt d\phi$$

$$g_{t\phi} = \frac{r_s r \alpha c \sin^2 \theta}{\rho^2}$$
$$g_{\phi\phi} = - \frac{\Sigma^2}{\rho^2} \sin^2 \theta$$

Integration via substitution:
$$\Omega = \frac{d \phi}{dt} = - \frac{g_{t\phi}}{g_{\phi\phi}} = - \left( \frac{r_s r \alpha c \sin^2\theta}{\rho^2} \right) \left( - \frac{\rho^2}{\Sigma^2 \sin^2 \theta} \right) = \frac{r_s r \alpha c}{\Sigma^2}$$

Angular velocity killing horizon:
$$\boxed{\Omega = \frac{r_s r \alpha c}{\Sigma^2}}$$

$$\Omega = \frac{r_s r \alpha c}{\Sigma^2} = \frac{r_{s} r \alpha c}{\rho^{2} \left( r^{2} + \alpha^{2} \right) + r_{s} r \alpha^{2} \sin^{2}\theta}$$

Kerr metric interpretation identity:
$$\boxed{\Sigma^2 = \rho^{2} \left( r^{2} + \alpha^{2} \right) + r_{s} r \alpha^{2} \sin^{2} \theta}$$

$$\boxed{\Omega = \frac{d \phi}{dt} = - \frac{g_{t\phi}}{g_{\phi\phi}}}$$

Reference:
http://en.wikipedia.org/wiki/Kerr_metric#Frame_dragging"
http://albert51.tripod.com/bhole.htm" [Broken]
https://www.physicsforums.com/blog.php?b=569" [Broken]

Last edited by a moderator: May 4, 2017
11. Nov 9, 2009

### stevebd1

At the Killing horizons (r+ and r-), exact solutions are-

$$\Omega_\pm=\frac{a}{\left(r_{\pm}^2+a^2\right)}= \frac{2Mr_\pm a}{\Sigma_\pm^2}=-\frac{g_{t\phi\pm}}{g_{\phi\phi\pm}}}$$

$$\omega=\frac{2Mra}{\Sigma^2}=-\frac{g_{t\phi}}{g_{\phi\phi}}}$$

where $\omega$ is the frame dragging rate as observed from infinity.

There seems to be a discrepancy of 2 in various expressions of Kerr metric regarding the equivalent of $g_{t\phi}$ but in these two papers, http://www.math-inst.hu/pub/algebraic-logic/AndrekaNemetiWuthrich07.pdf" [Broken] page 2, it's clearly expressed as-

$$g_{t\phi}=\frac{2Mra}{\rho^2}sin^2\theta$$

As apposed to the 2 being a 4 in some formal expressions of the metric, the 2 seeming to work better with the frame dragging rates and Killing horizons.

Last edited by a moderator: May 4, 2017
12. Nov 9, 2009

### Orion1

formal interpretation for the kerr metric...

With respect to formal solution derivations, the formal interpretation for the Kerr metric elements in spherical form must be:
$$c^{2} d\tau^{2} = g_{tt} dt^{2} + g_{rr} dr^{2} + g_{\theta \theta} d\theta^{2} + g_{\phi \phi} d\phi^2 + 2 g_{t \phi} dt d\phi$$

Where the formal $dt d\phi$ element is interpreted as:
$$\boxed{2 g_{t \phi} dt d\phi}$$

I noticed that extracting the $d \phi$ and $dt$ elements from this particular Kerr metric that is equivalent to a co-rotating reference frame that is rotating with angular velocity $\omega$ relevant to both $r$ and $\theta$:
$$c^2 d \tau^2 = \left(g_{tt} - \frac{g_{t \phi}}{g_{\phi \phi }} c \right)dt^2 + g_{rr}dr^2 + g_{\theta \theta}d\theta^2 + g_{\phi \phi} \left( d\phi + \frac{g_{t\phi}}{g_{\phi \phi}} dt \right)^2$$

Extract elements and expand terms:
$$g_{\phi \phi} \left( d\phi + \frac{g_{t\phi}}{g_{\phi \phi}} dt \right)^2 = g_{\phi \phi} d\phi^2 + 2 g_{t \phi} dt d\phi + \frac{g_{t \phi}^2}{g_{\phi \phi}} dt^2$$

This results in an expansion in which there is a $2$ in the $dt d\phi$ element that is independent of $g_{t \phi}$ as a result of identity.

$$g_{t\phi} = \frac{r_s r \alpha c \sin^2 \theta}{\rho^2}$$
$$g_{\phi\phi} = - \frac{\Sigma^2}{\rho^2} \sin^2 \theta$$

However, this produces another $dt^2$ element that I have not examined before in a formal Kerr metric.

Integration via substitution:
$$\frac{g_{t \phi}^2}{g_{\phi \phi}} dt^2 = \left( \frac{r_s r \alpha c \sin^2 \theta}{\rho^2} \right)^2 \left(- \frac{\rho^2}{\Sigma^2 \sin^2 \theta} \right) dt^2 = - \frac{(r_s r \alpha c)^2 \sin^2 \theta }{ \Sigma^2 \rho^2} dt^2 = - \left( \frac{r_s r \alpha c \sin \theta}{\Sigma \rho} \right)^2 dt^2$$

$$\boxed{\frac{g_{t \phi}^2}{g_{\phi \phi}} dt^2 = - \left( \frac{r_s r \alpha c \sin \theta}{\Sigma \rho} \right)^2 dt^2}$$

$$\boxed{g_{\phi \phi} \left( d\phi + \frac{g_{t\phi}}{g_{\phi \phi}} dt \right)^2 = g_{\phi \phi} d\phi^2 + 2 g_{t \phi} dt d\phi + \frac{g_{t \phi}^2}{g_{\phi \phi}} dt^2 = - \frac{\Sigma^2}{\rho^2} \sin^2 \theta d\phi^2 + \frac{2r_s r \alpha c \sin^2 \theta}{\rho^2} dt d\phi - \left( \frac{r_s r \alpha c \sin \theta}{\Sigma \rho} \right)^2 dt^2}$$

Last edited: Nov 10, 2009
13. Nov 11, 2009

### stevebd1

Based loosely on the Kepler orbit equations from post #9, the following is an approximate solution to the local gravitational field of a Kerr black hole-

$$a_T=\left[\frac{M}{(r^2+a^2)}D\,\alpha^{-1}\left(1+\frac{\left|v_\omega \pm v_o|}{c}\right)\right]-\frac{|\omega \pm \omega_o|^2}{\alpha^2}R$$

where D is the gravity distribution factor from post #7 which keeps the Killing surface gravity constant, alpha is the reduction factor, vω is the tangential velocity due to frame dragging, vo is the tangential velocity of the object falling into the black hole which can be positive (corotating), negative (retrograde) or zero (free fall), ωo is the angular velocity of the object falling into the BH and R is the reduced circumference.

$\alpha^{-1}=\sqrt{(\rho^2/\Delta)}$ at the poles and the above reduces to the Schwarzschild solution when a/M=0. Gravity now increases steadily over the increasing centripetal acceleration regardless of the spin parameter.

It's apparent that gravity increases significantly once inside the ergosphere and this couldn't be accounted for by just the reduction factor alpha or $\sqrt{(g_{tt})}$. It might be suggested that as an object rotates rapidly about an object of mass, the circumference of the orbit will want to reduce due to length contraction (SR). We know that this isn't literal otherwise the circumference at the ergosphere would be zero (tangential velocity = c due to frame dragging) but maybe the want to contract manifests as an increase in gravity, which is taken into account in the equation above. It would seem that as an object orbits closer to the EH of a Kerr black hole, the velocity associated with the extreme centripetal acceleration begins to work against the object, just as pressure begins to work against a collapsing neutron star, increasing gravity towards the object being orbited. The above, while not a perfect match, does produce results at least analogues to those of the Kepler equations.

Last edited: Nov 12, 2009
14. Nov 14, 2009

### stevebd1

Re: formal interpretation for the kerr metric...

For the record, there's a power of 2 missing from the first $g_{t \phi}$ which would also result in c relative to the first dt2 being squared. The complete form would read-

$$c^2d\tau^2=\left(g_{tt}-\frac{g_{t\phi}^2}{g_{\phi \phi }}c^2\right)dt^2\ +\ g_{rr}dr^2\ +\ g_{\theta\theta}d\theta^2\ +\ g_{\phi\phi}\left(d\phi+\frac{g_{t\phi}}{g_{\phi\phi}}cdt\right)^2$$

source- http://en.wikipedia.org/wiki/Kerr_metric#Frame_dragging, the slight difference being that the version on the wiki page includes c in the $-g_{t \phi}/g_{\phi \phi }$ equation and hence isn't expressed specifically in the form.

Last edited: Nov 14, 2009
15. Nov 14, 2009

### Orion1

Kerr metric that is equivalent to a co-rotating reference frame that is rotating with angular velocity $\omega$ relevant to both $r$ and $\theta$ in factored form:
$$c^2 d\tau^2 = \left(g_{tt} - \frac{g_{t\phi}^2}{g_{\phi \phi}} \right)dt^2 + g_{rr}dr^2 + g_{\theta \theta}d\theta^2 + g_{\phi\phi} \left(d\phi+\frac{g_{t\phi}}{g_{\phi \phi}} dt \right)^2$$

The expanded form of this co-rotating reference frame metric must be:
$$\boxed{c^2 d\tau^2 = g_{tt} dt^2 - \frac{g_{t\phi}^2}{g_{\phi \phi }} dt^2 + g_{rr}dr^2 + g_{\theta\theta}d\theta^2 + g_{\phi \phi} d\phi^2 + 2 g_{t \phi} dt d\phi + \frac{g_{t \phi}^2}{g_{\phi \phi}} dt^2}$$

Which reduces to:
$$\boxed{c^2 d\tau^2 = g_{tt} dt^2 + g_{rr}dr^2 + g_{\theta\theta}d\theta^2 + g_{\phi \phi} d\phi^2 + 2 g_{t \phi} dt d\phi}$$

$$g_{tt} = \frac{(\Delta - \alpha^2 \sin^2 \theta)c^2}{\rho^2}$$
$$g_{t\phi} = \frac{r_s r \alpha c \sin^2 \theta}{\rho^2}$$
$$\frac{g_{t \phi}^2}{g_{\phi \phi}} dt^2 = - \left( \frac{r_s r \alpha c \sin \theta}{\Sigma \rho} \right)^2 dt^2$$

Integration via substitution:
$$\left(g_{tt} - \frac{g_{t \phi}^2}{g_{\phi \phi}} \right)dt^2 = g_{tt} dt^2 - \frac{g_{t\phi}^2}{g_{\phi \phi }} dt^2 = \frac{(\Delta - \alpha^2 \sin^2 \theta)c^2}{\rho^2}dt^2 + \left( \frac{r_s r \alpha c \sin \theta}{\Sigma \rho} \right)^2 dt^2 = \left[ \frac{(\Delta - \alpha^2 \sin^2 \theta)}{\rho^2} + \left( \frac{r_s r \alpha \sin \theta}{\Sigma \rho} \right)^2 \right] c^2 dt^2$$

$$\boxed{\left(g_{tt} - \frac{g_{t \phi}^2}{g_{\phi \phi}} \right)dt^2 = \left[ \frac{(\Delta - \alpha^2 \sin^2 \theta)}{\rho^2} + \left( \frac{r_s r \alpha \sin \theta}{\Sigma \rho} \right)^2 \right] c^2 dt^2}$$

With respect to the $dt$ element, the constant $c$ must be included in the formal definition of $g_{tt}$ and $g_{t \phi}$ in order to produce formal derivation solutions that have dimensional results that are in a International System of Units (S.I.) form for all solutions.

For example, the angular velocity killing horizon:
$$\Omega = \frac{d \phi}{dt} = - \frac{g_{t\phi}}{g_{\phi\phi}} = \frac{r_s r \alpha c}{\Sigma^2} = \frac{r_{s} r \alpha c}{\rho^{2} \left( r^{2} + \alpha^{2} \right) + r_{s} r \alpha^{2} \sin^{2}\theta}$$

Which has International System of Units (S.I.) of radians per second.

Kerr metric that is equivalent to a co-rotating reference frame that is rotating with angular velocity $\omega$ relevant to both $r$ and $\theta$ in Boyer-Lindquist coordinates:
$$\boxed{c^2 d\tau^2 = \left[ \frac{(\Delta - \alpha^2 \sin^2 \theta)}{\rho^2} + \left( \frac{r_s r \alpha \sin \theta}{\Sigma \rho} \right)^2 \right] c^2 dt^2 - \frac{\rho^2}{\Delta}dr^2 - \rho^2 d\theta^2\ - \frac{\Sigma^2}{\rho^2} \sin^2 \theta d\phi^2\ + \frac{2 r_s r \alpha c \sin^2 \theta}{\rho^2} dt d\phi - \left( \frac{r_s r \alpha c \sin \theta}{\Sigma \rho} \right)^2 dt^2}$$

Reference:
http://en.wikipedia.org/wiki/International_System_of_Units" [Broken]
http://en.wikipedia.org/wiki/Kerr_metric#Frame_dragging"

Last edited by a moderator: May 4, 2017
16. Nov 19, 2009

### stevebd1

A couple more useful equations regarding Kerr orbits-

$$\tag{1}E/m=\tilde{E}=\frac{r^2-2Mr\pm a\sqrt{Mr}}{r\left(r^2-3Mr\pm2a\sqrt{Mr}\right)^{1/2}}$$

$$\tag{2}L/m=\tilde{L}=\pm\frac{\sqrt{Mr}\left(r^2\mp2a\sqrt{Mr}+a^2}{r\left(r^2-3Mr\pm2a\sqrt{Mr}\right)^{1/2}}$$

where the upper sign corresponds to corotating orbits and the lower sign to counterrotating orbits. The above become infinite at the prograde and retrograde photon sphere orbits where vs is c.

Note the above only applies to stable orbits.

Effective potential for Kerr metric (which also applies to L and E for non-stable orbits)-

$$\tag{3}V_{eff}=\frac{kM}{r}\,+\,\frac{\tilde{L}^2}{2r^2}\,+\,\frac{1}{2}(k-\tilde{E}^2)(1+\frac{a^2}{r^2})\,-\,\frac{M}{r^3}(\tilde{L}-a\tilde{E})^2$$

where $$\tilde{L}$$ is the conserved angular momentum per unit mass of the particle orbiting the BH, and $$\tilde{E}$$ is the conserved energy per unit mass of the particle, and where k=0 for light ray geodesicis and k=1 for massive particles. The above also appears to work for the static solution.

On a slightly different note, according to http://www.lsw.uni-heidelberg.de/users/mcamenzi/GR_07.pdf" [Broken] (page 211), $\alpha^{-1}$ does represent coordinate acceleration for gravity (or scalar acceleration as it's described in the paper) where $\alpha$ is the reduction factor or redshift.

(1),(2)-
http://www.tat.physik.uni-tuebingen...Relativistic_Astrophysics_files/GTR2009_4.pdf
page 21

(3)
http://physics.ucsd.edu/students/courses/winter2007/physics161/Lectures/p161.27feb07.pdf [Broken]
page 5

Last edited by a moderator: May 4, 2017
17. Nov 20, 2009

### stevebd1

Based on the idea that gravity is synonymous with blue shift (becoming infinite at the event horizon) and centripetal acceleration is synonymous with redshift (becoming zero at the event horizon and negative beyond the EH) a revised equation for relativistic centripetal acceleration might be established. As in https://www.physicsforums.com/showthread.php?t=354583", ac is multiplied by $\sqrt{g_{tt}}$, this can't be literally translated to Kerr metric as $\sqrt{g_{tt}}$ becomes zero at the ergosphere where tangential light cones tip over. The reduction in ac is relative to radial light cones tipping over so $\sqrt{g_{tt}}$ is replaced with $\alpha$ which is the redshift equation for Kerr metric which reduces to $\sqrt{g_{tt}}$ for the static solution (this is also supported by the fact that stable orbits exist within the ergosphere for BH's with spin higher than 0.94 which requires ac to be positive). Centripetal acceleration can now be expressed as-

$$a_{c}=\frac{v^2}{R}\alpha$$

where $v=\omega R/\alpha$ and R is the reduced circumference. The equation can be rewritten-

$$a_{c}=\frac{\omega^2R}{\alpha}$$

Predicted relativistic gravitational field for an object in free fall-

$$a_{rel}=\frac{M}{(r^2+a^2)}D_\pm\,\alpha^{-1}-\frac{\omega^2R}{\alpha}$$

where $D_\pm$ is 1 at the equator. The equation would expand into-

$$a_{rel}=\frac{M}{(r^2+a^2)} \left(\pm \frac{\sqrt{M^2-a^2}}{M}+\frac{a^2 sin\theta}{M P_\pm}\right) \sqrt{\frac{\Sigma^2}{\rho^2 \Delta}}\ -\ \left(\frac{2Mra}{\Sigma^2}\right)^2 \frac{\Sigma}{\rho}sin\theta \sqrt{\frac{\Sigma^2}{\rho^2 \Delta}}$$

which also produces a repulsive force over the entire inner Cauchy horizon.

Last edited by a moderator: Apr 24, 2017
18. Nov 22, 2009

### stevebd1

In respect of a stable orbit, Keplers third law states that 'the square of the orbital period is directly proportional to the cube of the semi-major axis' so the constant T2/r3 should always apply regardless of r. This works for the Schwarzschild solution and for large radii for the Kerr solution but when looking at radii smaller than 20M, even when considering frame dragging, there seems to be something amiss, especially when using the Kepler equation for angular velocity applicable to Kerr metric-

$$\Omega_s=\frac{\pm\sqrt{M}}{r^{3/2}\pm a\sqrt{M}}$$

where T is based on C/v where $v=\Omega_s R$ and $C=2\pi \sqrt{(r^2+a^2)}$. While this is no doubt related to frame-dragging, it would be good to see some confirmation that T2/r3 still applies, even if it's in some abstract sense like the Killing surface gravity.

Also the balance between ag and ac doesn't seem exact when using $\Omega_s$. For a 3 sol mass BH with a spin parameter of a/M=0.95 at just outside the inner photon sphere (6145 m) with a prograde stable orbit in the equatorial plane-

$$a_{rel}=\frac{M}{(r^2+a^2)}\,\alpha^{-1}-\frac{\Omega_s^2R}{\alpha}$$

The answer comes out with ac (RHS of the minus sign) being lower than the gravity by a factor of ~0.86 when they should be equal.

For a retrograde stable orbit just outside the outer photon sphere (17525 m), the ac works out at being just higher than the gravity by a factor of ~1.43

I thought of reconsidering the equation for ag but the above (without the reduction factor $\alpha$ and relative to $\omega$) is an exact solution for the Killing surface gravity regardless of mass and spin.

It appears the Kepler equation might take into account some drag factor which is <1 for prograde orbits and >1 for retrograde orbits. Though the velocities becomes unfeasable, I did notice that as r reduces, the angular velocity for the retrograde orbit increases almost exponentially over the av for the prograde orbit as if more work is being done when rotating against the frame-dragging.

Last edited: Nov 23, 2009
19. Nov 25, 2009

### stevebd1

A few more equations-

Period of a circular orbit in Kerr metric-

$$T=2\pi M\left(\frac{r}{M}\right)^{3/2}\frac{\left[1+\frac{a^2}{r^2}-\frac{2M}{r}\pm\left(\frac{a}{M}\right)\left(\frac{M}{r}\right)^{3/2}\pm\left(\frac{a}{M}\right)^3\left(\frac{M}{r}\right)^{7/2}\mp 2\left(\frac{a}{M}\right) \left(\frac{M}{r}\right)^{5/2}\right]}{\left[1+\frac{a^2}{r^2}-\frac{2M}{r}\right]}$$

which reduces to simply $T=2\pi M(r/M)^{3/2}$ for a static BH.

source- http://books.google.com/books?id=oP...8859-1&output=html&source=gbs_search_r&cad=1" By Derek J. Raine, Edwin George Thomas, page 79, eq. 3.46

Another equation I see in various form is an alternative for the $\alpha$, the redshift-

$$A=\sqrt{g_{tt}+2\Omega g_{t\phi}-\Omega^2g_{\phi \phi}}$$

which produces results identical to $\alpha$ when $\Omega=\omega$, and becomes zero at the inner/outer photon spheres when $\Omega=\Omega_{s+}$ and $\Omega=\Omega_{s-}$ respectively. Reduces to $A=\sqrt{(g_{tt})}$ for a radial free-fall plunger for a static BH.

examples (in various forms)-
http://www.roma1.infn.it/teongrav/VALERIA/TEACHING/ONDE_GRAV_STELLE_BUCHINERI/AA2006_2007/DISPENSE/ch20.pdf" [Broken] page 321, where it seems related to the Kepler stable orbit equation.

http://www.phys.uu.nl/~prokopec/StijnJvanTongeren_bh3.pdf" page 10

http://www.icra.it/MG/mg12/talks/apt1_slany.pdf page 5

It is also referred to in this paper where something referred to as the cumulative drag index is talked about which may (or may not) have something to do with what was mentioned at the end of post #18- http://arxiv.org/PS_cache/gr-qc/pdf/9701/9701008v2.pdf page 4 eq. 5

The equation is also shown in some derivative form representing effective potential in this section from Scholapedia, http://www.scholarpedia.org/article/Accretion_discs/2._Basic_physics_of_accretion_discs/2.1._The_black_hole_gravity" [Broken]

More to do with a free-fall plunger than stable orbits, radial proper velocity of a zero angular momentum observer dropped from at rest at infinity-

$$v=\frac{\sqrt{2Mr(r^2+a^2)}}{\rho^2}$$

which reduces to $v=\sqrt{(2M/r)}$ for a static BH.

Source- http://arxiv.org/PS_cache/arxiv/pdf/0805/0805.0206v2.pdf [Broken] page 2

Last edited by a moderator: May 4, 2017
20. Nov 28, 2009

### Orion1

Keplerian-Newtonian orbital period for the Kerr metric...

According to reference 1, pg. 4 eq. 23 and reference 2, the Keplerian-Newtonian orbital period for the Kerr metric is:
$$\boxed{T = \sqrt{ \frac{r^3}{G(M + m)}} \left( 2 \pi + \frac{3 \pi \alpha^2}{2r^2} \right)}$$

$$M$$ - primary mass
$$m$$ - orbiting mass

Which reduces to the Keplerian orbital period for $m = 0$ and $\alpha = 0$.

Reference:
"arxiv1.library.cornell.edu/pdf/gr-qc/0203035v3"[/URL]
[URL]http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Nonzero_planetary_mass"[/URL]
[URL]http://en.wikipedia.org/wiki/Kerr_metric#Mathematical_form"[/URL]

Last edited by a moderator: Apr 24, 2017