# Black holes, frame dragging and the effect on gravity

1. Mar 20, 2008

### stevebd1

In regard of frame-dragging within the ergosphere of a rotating black hole, what effect would this have on gravity?

It seems accepted that with frame dragging (or lense-thirring), the fabric of space is dragged around with the black hole but that light within the ergosphere still travels at c (though it could be said that in the direction of rotation, it's actually travelling faster relative to light outside the ergosphere due to the fact that the space it's travelling in is also moving). In relation to the attached picture, I'm under the impression that light has to travel further in line B than say line A (for a static black hole) as it is still travelling at c. My question is that based on the notion that gravity propagates at the at the speed of light, and that gravity (along with light) travels along geodesics, would gravity increase if it was to travel through line B compared to A or would the gravity acceleration be equal for both lines but stretched in line B? Would gravity still relate to the mass in a direct radial manner or would the gravity in effect be travelling the new distances created by the frame-dragging?

Any feedback welcome.

Steve

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2. Mar 21, 2008

### stevebd1

I'm reaching the conclusion that while light and matter follow geodesics, gravity propogates unaffected in a radial manner and remains unchanged.

Regarding the ergosphere of a rotating black hole, most literature I find states that the ergosphere is rotating at the speed of light. Surely this depends on the rotation of the black hole itself (which in itself depends on the angular momentum of the star which collapsed, forming the black hole). Based on the factor a ranging from 0 to 1, 0 = no rotation and 1 = maximum rotation, or unity (rotation = speed of light), surely this would also apply to the ergosphere. For example, if a black hole was rotating at half the speed of light, then the rotation of the fabric of space within the ergosphere would be equal to or less than this at the event horizon, dissipating as it reached the static limit (though it also seems to be accepted knowledge that there is some frame dragging outside the static limit but to much less degree). Any points of view welcome.

Steve

Last edited: Mar 22, 2008
3. Mar 22, 2008

### stevebd1

how velocity of rotation of a Kerr black hole might be estimated

Based on

$$a=\frac{cJ}{GM^{2}}$$

which converts to

$$J=\frac{aGM^{2}}{c}$$

and

$$J = vMR$$

which converts to

$$v=\frac{J}{MR}$$

figures for angular momentum and velocity of rotation can be estimated, using the outer event horizon radius for each case to caclulate velocity*. Even though the event horizon is a boundary and not a surface as such, and doesn't technically have a velocity to establish rotation, a figure could still be estimated by making a measurement based on the moment the neutron star/core collapses. If we have a factor of a, and a mass, we can estimate the velocity based on the radius being fractions of a millimetre outside the event horizon while technically it was still a star. Any frame dragging and rotation established at this stage would probably remain as the star collapsed further within the event horizon and the surface of the star became a boundary.

*once inside the ergosphere, space itself begins to warp, moving with the black hole, and the velocities/frequencies provided for just outside the event horizon are relative to inside the ergosphere. They will, in fact, be travelling slightly faster compared to static space outside the ergosphere (while light within the ergosphere still travels at 299,792.5 km/s, due to the space it's travelling in being dragged around by the black hole, it's actually travelling faster than light outside the ergosphere). The velocities and frequencies provided would be relative only to the black hole itself.

Based on the above, regarding the velocity of rotation at the outer event horizon, regardless of the size of the black hole, if a = 1, the velocity at the outer event horizon will always equal c (hence the term ‘unity’ when a = 1) as the following demonstrates-

$$J=\frac{aGM^{2}}{c}$$

if a=1, it can be discarded) so

$$J=\frac{GM^{2}}{c}$$

if angular momentum is

$$J = vMR$$

then velocity is

$$v=\frac{J}{MR}$$

outer event horizon radius when a is 1

$$R=\frac{GM}{c^{2}}$$

when a is 1, velocity at outer event horizon

$$v=\frac{J}{MR}$$

$$=\frac{GM^{2}/c}{MGM/c^{2}}$$

$$=\frac{GM^{2}/c}{GM^{2}/c^{2}}$$

$$=c$$

Likewise, regardless of the size of the black hole, at a = 0.2, the speed of rotation at the event horizon would be ~30,285 km/s, at a = 0.6, the speed would be ~99,931 km/s and at a = 0.8, the speed would be ~149,896 km/s. The quantity of rotation that would change would be the frequency (Hz) which would reduce as the mass increased.

Steve

4. Mar 24, 2008

### stevebd1

I found a solid set of equations that helped provide a set of co-ordinates for frame-dragging around a black hole and produced the following results for a 3 sol mass black hole with an angular momentum factor of a = 0.95 (figures shown are the frequencies and velocities of spacetime itself rotating)-

http://cr4.globalspec.com/blogentry/1670/Extreme-Frame-Dragging

(Rg = GM/c^2)

at marginally stable orbit (~6Rg)- 91.67 Hz (angular v = 0.053c)

at photon sphere (~3Rg)- 649.31 Hz (angular v = 0.19c)

at static limit (ergosphere edge)- 1,762.40 Hz (angular v = 0.35c)

at outer event horizon (R+)- 3,898.17 Hz (angular v = 0.53c)

While it's considered that space and time swap properties at the event horizon and that it is still ambiguous what happens beyond the event horizon, in the Kerr rotating black hole, there is a hypothetical second event horizon (the inner event horizon or cauchy horizon) where time and space swap back to their original properties while the space within the Cauchy horizon would be very curved considering it's proximity to the ring singularity. Some figures for frame-dragging within the Cauchy horizon are-

at Cauchy (inner event) horizon (R-)- 7,437.856 Hz (angular v = 0.58c)

at 500 mm from ring singularity- 10,722.147 Hz (angular v = 0.263c)

Incidentally, frame dragging velocity seems to peak at the gravitational radius (Rg or GM/c^2) which resides directly between the inner and outer event horizons-

at Rg- 5,794.005 Hz (angular v = 0.62c)

While this more or less answers a lot of questions, it seems to raise some others-

1.) Is the centripetal acceleration (v^2/R), which is supposed to cancel out some of the gravitational acceleration, based on the velocity of the frame dragging? i.e. at the outer event horizon, centripetal acceleration (based on the velocity of the frame dragging) would equal 3.893569x10^12 m/s^2. Are objects (including light) expected to rotate at the same velocity as the frame-dragging?

2.) Frame dragging seems to be just as extreme within the marginally stable orbit as it is within the static limit, yet the ergosphere within the static limit is treat as a significant zone of frame dragging. Is there something else that's suppose to occur within this zone that sets it apart from the other areas?

3.) I also found the below quote on wiki under 'Light-dragging effects'

"Rotation-dragging effects-
Under general relativity, the rotation of a body gives it an additional gravitational attraction due to its kinetic energy, and light is also pulled around (to some degree) by the rotation (Lense-Thirring effect)."

http://en.wikipedia.org/wiki/Light-dragging_effects

I'd be interested to know how to calculate the kinetic energy of a rotating black hole and how it might contribute to the overall gravity.

4.) If an object was moving with an angular velocity of 0.3c with the orbit of the frame-dragging at the marginally stable orbit (angular v = 0.053c), would the object have an overall comoving velocity of 0.353c?

cheers
Steve

Last edited: Mar 25, 2008
5. Mar 28, 2008

### stevebd1

As I could find nothing on the web regarding the effects of frame-dragging on gravity, I decided to give it a bash myself. The following is my own opinion which I’ve tried to keep as tangible as possible. For the sake of this exercise, I've assumed that gravity and light might take the same path.

Source for equating frame dragging (which appear to be based on Kerr metric)
http://cr4.globalspec.com/blogentry/1670/Extreme-Frame-Dragging

Image 1 shows the frame dragging for a 3 sol mass rotating black hole with an angular momentum factor of 0.95. The image shows a snapshot of rotation, a change of 8.416x10^-4 seconds (or based on the black hole having a frequency of 5941 Hz, half a turn) light would have travelled 25,230 m towards the black hole (approx. from the marginally stable orbit to the singularity).

Image 2 shows the construct of a photon path (or path of gravity) into the black hole. This is based on the opinion that light and gravity might be fast enough to propagate through the frame dragging and not be pulled in concentric circles as would be the case for non-relativistic matter, though the frame-dragging would still have a detrimental affect on the direct path of the light. Based on a photon travelling directly towards the centre of the rotating black hole, depending on the velocity of the frame dragging, the photon would be 'knocked' off course by the frame dragging (a frame dragging velocity of zero would result in the photon being diverted by zero degrees, a maximum frame dragging velocity of c would result in the photon being diverted by 90 degrees), based on this, for a frame dragging velocity of 0.530c, the angle would be 47.7 degrees. This produced the path shown in image 2.

In trying to find an equation that takes this into account and could be applied to gravity, the closest I could get was- effect of frame-dragging = c/(c-v) (or using the results directly from the extreme frame-dragging equations- c/(c-(rad/s x r)). This is based on the idea that as the path of gravity is elongated, it is still accelerating as it approaches divergence (or infinite) at the event horizon. Basically, the gravity well would become significantly deeper as the frame-dragging increased (namely within the marginally stable orbit) so for a frame-dragging velocity of 0.530c, the actual gravitational acceleration at this point would be g x 2.128. The equation also derives from c being the speed of gravity and c-v comes from zero frame dragging = zero change and maximum frame-dragging (~c) = infinite increase in gravity.

While it's ambiguous what happens on the other side of the event horizon, based on the fact that the photon will eventually collide with the ring singularity, I've assumed the frame-dragging to occur within the event horizon which peaks at the gravitational radius and then drops to zero as it approaches the ring singularity.

Image 3 shows the light (gravity) lines for a static Schwarzschild black hole

Image 4 shows the light (gravity) lines for a rotating Kerr black hole, a = 0.95

(While the above is based on the effects of frame dragging applying to both light and gravity, I'm leaning towards the fact that it might only apply to gravity (if it applies at all!) as photons would have more of a tendency to follow geodesicsis and be more effected by frame-dragging while gravity might be more motivated in some way to reach the black hole).

key-
a- Gravitational radius (normalized mass in m)
b- event horizon
c- photon sphere
d- marginally stable orbit

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6. Apr 11, 2008

### stevebd1

In respect of the above post regarding the rotational properties of a black hole and frame-dragging, in theory, would the following be correct-

'Based on a 3 sol mass black hole having a frequency of 5941 Hz, in reality, the black holes frequency would be 9839 Hz. (5941 Hz plus the frame-dragging frequency, 3898 Hz) as it would be a rotating object within a rotating frame. If the black hole has an apparent rotational velocity of 0.72c at the event horizon and the frame-dragging velocity was 0.530c at the EH then the black hole would have a comoving rotational velocity of 1.250c (though the black hole would technically still be rotating at 0.720c). This might be what gives the ergosphere its unique quality. Based on this, a period of 8.416x10^-4 seconds would actually result in 0.87 of a turn (while the black hole itself only produces half a turn).'

The only drawback I can see with this is that technically the poles of the event horizon still have contact with regular space which keeps the black hole relative, though there is the notion that the ergosphere might extend over the poles, keeping the black hole isolated from regular spacetime.

Steve

7. Apr 21, 2008

### stevebd1

In order to calculate the correct gravity acceleration for static black holes, Newtonian gravity needs to incorporate Schwarzschild coordinate acceleration which appears to be based on the same factor used for calculating true radii or incorporating radial stretching due to the steep gravity well. Based on a similar factor I found relating to Kerr black holes and allowing for a potential frame-dragging factor, I pieced together the following-

Newtonian gravity-

$$g = G\frac{m}{r^{2}}$$

Radial stretching due to gravity for a Kerr (rotating ) black hole (Based on equations shown on page F-13, 'exploring black holes' by E. F. Taylor & J. A. Wheeler')-

$$\frac{1}{1-(R_{+}/r)}$$

where

$$R_{+}$$ is the outer event horizon in m-

$$R_{+}=R_{g}+\sqrt{(R_{g}^{2}-a^{2})}$$

$$R_{g}$$ is the gravitational radius in m-

$$R_{g}=\frac{Gm}{c^{2}}$$

$$a$$ is the spin parameter in m-

$$a=\frac{J}{mc}$$

It's likely that the the actual angular momentum of the black hole will not be known and only an angular momentum factor of between 0 and 1 would be assigned. In that case, the following equation would apply-

$$J=\frac{Gm^{2}a_{f}}{c}$$

where

$$a_{f}$$ is the rotation factor between 0 and 1

(The factor is similar to the Schwarzschild coordinates but minus the square root to the bottom half of the fraction which makes for a steeper gravity well. It also allows for the reduced radius of the outer event horizon for a rotating black hole. Source- www.eftaylor.com/pub/SpinNEW.pdf).

The above doesn't appear to allow for radial stretching due to frame dragging as it seems to apply to both equatorial and polar planes of the black hole. The following is an assumption of what a modification allowing for radial stretching due to frame dragging, on the equatorial plane, might look like-

Radial stretching due to frame-dragging for a Kerr (rotating ) black hole (based on Kerr metric. Source- http://cr4.globalspec.com/blogentry/1670/Extreme-Frame-Dragging)-

$$\frac{1}{1-(\omega r/c)}$$

where

$$\omega$$ is the quantity of frame-dragging in rads/s-

$$\omega=\frac{2R_{g}\,r\,a\,c}{(r^{2}\,+\,a^{2})^{2}\,-\,a^{2}\,\Delta}$$

$$\Delta= (r^{2}\,+\,a^{2}\,-\,2\,R_{g}\,r)$$

$$\omega r$$ is angular velocity in m/s

Based on the information above, gravity at a reduced circumference of 7000 m for a 3 sol mass black hole with a spin of 0.95 on the equatorial plane would be-

$$\frac{Gm}{r^{2}} \left(1-\frac{R_{+}}{r}\right)^{-1} \left(1-\frac{\omega r}{c}\right)^{-1}$$

g = 8.1265 x10^12 (5.902)(1.719) $$m/s^{2}$$

g = 8.245 x10^13 $$m/s^{2}$$ at 7000 m