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Relativity and Cosmology

This is a Blog on relativity and cosmology for engineers and the like. You are welcome to comment upon or question anything said on my website (http://www.relativity-4-engineers.com), in the eBook or in the snippets I post here.

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Regards, Jorrie

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18 comments

Extreme Frame Dragging

Posted April 10, 2007 11:00 PM by Jorrie

To wrap-up this mini-series on rotating black holes, I analyze the strong frame dragging near a rotating black hole. The rather formidable math is simplified somewhat by considering the equatorial plane only, thereby sacrificing generality, but not losing accuracy. The emphasis is on using the equations for calculations in order to get a feel for the magnitudes.

Figure 1 below shows, for ease of reference, a slightly modified diagram of the Kerr black hole picture from the original Blog article of this mini-series. Extreme frame dragging occurs near and inside the dotted circle representing the Schwarzschild radius of an equivalent non-rotating black hole.

Figure 1:

On the equatorial plane of a spinning black hole, the frame dragging can be expressed by the angular velocity (ω) of the spacetime as a function of the radial distance parameter r, the mass M and the spin parameter a of the hole. Put differently, ω is the angular rate, as observed from afar, by which a local observer, stationary relative to the spacetime in the vicinity of the hole, is being dragged around the hole.

Summary of the meanings and units of the parameters used:

Radial parameter2 (Δ) converts local distances to coordinate distances. Also see end-notes(1,3).

Figure 2 below shows how the spacetime is dragged around a spinning black hole in its equatorial plane. The spiraling radial lines are the paths of photons on the equatorial plane, aimed directly at the hole from afar. Photons move at the same local velocity in all directions (i.e., isotropic), hence the spiraling paths indicate that the local space is being dragged around the hole.

Figure 2:Credit:(1)

In order to make the equations more palatable, lets put in values and get some tangible answers. For our Sun, with M ≈ 1.99 x 1030 kg and J ≈ 1.63 x 1041 kg m2 s-1, lets first check how large the frame dragging will be on Earth's orbit, with r ≈ 1.5 x 1011 m. This gives:

a = J/(Mc) ≈ 273 m, Mbar ≈ 1470 m, Δ ≈ 2.32 x 1022 m2, giving

ω ≈ 7.16 x 10-20 radians per second, or ≈ 10-5 arcseconds per century.

As expected, this is negligible since we are still in the weak gravitational field environment and frame dragging falls off with roughly the cube of distance. To find out what will happen in a strong field, lets visit a relatively small black hole of 1.5 solar masses with the same angular momentum than our Sun. We will first check the frame dragging rate at rtest = 4Mbar ≈ 8800 m. This gives a ≈ 182 m, Δ ≈ 3.92 x 107 m2 and ω ≈ 350 radians per second, or a frequency of 56 Hz.

That's one whopping angular rate! Anything that hasn't got some means of propulsion will be dragged around the hole at this rate. However, this is not very fast as far as black holes are concerned. The fastest a black hole can spin is when a = Mbar (the extreme Kerr black hole), or a ≈ 2200 m for this case. At rtest = 8800 m in the equatorial plane, the frame dragging will be: ω ≈ 3876 radians per second, or 617 Hz.

It gets more severe as one goes down to the static limit, rstatic = 2Mbar ≈ 4400 m at the equator, giving a frame drag rate of ω ≈ 22600 radians per second, or 3.6 KHz. Inside rstatic, i.e., in the ergosphere, no amount of propulsion can prevent a spacecraft from being dragged around the black hole, because to do so would require a local speed through spacetime exceeding c.

It doesn't stop there. Frame dragging can theoretically be observed down to the outer horizon of this black hole, at router= Mbar ≈ 2200 m, yielding ω ≈ 67800 radians per second, or 10.8 KHz. Figure 3 shows the main frame dragging rates for a 1.5 solar mass black hole, spinning at the maximum allowable rate, a =Mbar.

Figure 3:

Interestingly enough, due to the gravitational redshift,(2) a distant observer would not observe an unpowered spacecraft at the static limit to whiz around the hole at c, but rather at about 0.4c. You can check this roughly by calculating ωr, but see end-note(3).

Finally, we will check the frame dragging magnitude experienced by that "GIGO-city"(4) with the clean energy. For this, we must find the radial distance of the city from the hole for the required "1g" of local gravity. When talking a low 1g gravity, we can easily find the distance with Newton's dynamics: r = √[GM/9.81] ≈ 4.5 x 109 m. This is less than one tenth of Mercury's distance from the Sun.

Even for an extreme Kerr hole, with all the other variables remaining the same, the frame dragging is ≈ 3 x 10-14 radians/sec, or some 20 arc-seconds per century. Not too bad at all - Earth "wobbles" more than that over the course of a century!

This concludes this mini-series on spinning black holes - and some heads are probably spinning too (including mine)! Now it's time for rants, raves or rotten tomatoes. I'll be grateful if someone would check my calculations - it's all too easy to goof at these sums!

Regards, Jorrie

(1) Equations are adapted for "engineering use" (i.e., straight SI units) from http://www.astro.ku.dk/~cramer/RelViz/text/exhib4/exhib4.html. The referenced website tends to use SI units and geometric units (where c=G=1) interchangeably and this can be very confusing, but it has great computer-generated graphics of frame dragging.

(2) The gravitational redshift factor (the ratio: frequency observed at r1 -infinity, to frequency emitted at r) in the plane of the equator of a Kerr black hole is given by:

The other parameters are as defined in the text above. For an extreme Kerr black hole, the redshift factor approaches zero as r approaches Mbar. It's hard to see in the equation, but when spin parameter a = 0, it reduces to the Schwarzschild redshift factor: dτ/dt = √[1-2GM/(rc2)]. (Edit: Changed equation to be compatible with MTW, i.e., using dτ/dt in place of α (alpha), which has a different meaning there.)

(3) The radial parameter r is not quite a circumferential radius, where the circumference is 2Πr. Parameter r includes a spin or rotational component and is used to generalize the Schwarzschild radial distance for rotating space-times. It is defined in Boyd-Lindquist coordinates.

(4) The "GIGO-city" is discussed in this Blog entry.

-J


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#1

Re: Extreme Frame Dragging

04/11/2007 4:47 AM

Hi Jorrie, great stuff again, but yes, it does make my head spin!

One question: I see in your reference (1), they use the spin parameter as a dimensionless quantity with value ranging from 0 to 1. Why is yours different?

SL

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#2
In reply to #1

Re: Extreme Frame Dragging

04/11/2007 7:28 AM

Hi SL, yep, they seem to call a the specific angular momentum, but then they go and add a2 to r2 in many places, meaning that you must realise that r must also be dimensionless by dividing it by Mbar. Otherwise, you must remember to multiply a by Mbar to convert it into meters everywhere it is used. All unnecessarily confusing.

I like my scheme better - more engineering-like, if you like...

Regards, Jorrie

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#3
In reply to #2

Re: Extreme Frame Dragging

04/12/2007 6:12 AM

I sometimes wonder why academics have to make things so confusing - is it to frighten the ordinary citizen away from their work?

SL

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#4
In reply to #3

Re: Extreme Frame Dragging

04/12/2007 6:52 AM

Hi SL, I share your frustration - and to make matters worse, relativists like to work with geometric units all in cm, apparently to remain compatible with a large volume of legacy work of the past.

I guess the academic institutions require their students to learn to live with this complexity (or confusion, if you like), because the real world is not plain sailing!

Regards, Jorrie

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#5

Re: Extreme Frame Dragging

04/13/2007 10:57 AM

"For our Sun, with M ≈ 1.99 x 1030 kg and J ≈ 1.63 x 1041 kg m2 s-1, lets first check how large the frame dragging will be on Earth's orbit, with r ≈ 1.5 x 1011 m. This gives: ..."

Question: you state that for the sun r ≈ 1.5 x 1011 m. But you also say that r is not quite the radial distance. How does this tie up?

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#6
In reply to #5

Re: Extreme Frame Dragging

04/14/2007 2:04 AM

Hi Guest,

At large distances there is negligible difference between the normal radial distance rn and the Boyer-Lindquist r, where r = √[rn2+a2], because rn >> a , hence r ≈ rn.

It is obviously not so when close to the rotating hole, say at rn = 2Mbar, where for an extreme Kerr hole, a = Mbar and rn = 2.236Mbar.

Jorrie

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#7
In reply to #6

Re: Extreme Frame Dragging

04/14/2007 3:54 AM

I spotted a typo just too late to edit in my: "It is obviously not so when close to the rotating hole, say at rn = 2Mbar, where for an extreme Kerr hole, a = Mbar and rn = 2.236Mbar."

It should read: It is obviously not so when close to the rotating hole, say at rn = 2Mbar, where for an extreme Kerr hole, a = Mbar and r = 2.236Mbar.

-J

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#8

Re: Extreme Frame Dragging

03/25/2008 7:36 AM

First of all, great post. Secondly, I'd appreciate if you could consider the questions below regarding extreme frame-dragging-

1.) Are objects (including light) expected to rotate at the same velocity as the frame-dragging? 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?

2.) Frame dragging seems to be just as extreme within the marginally stable orbit and photon sphere 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 some other factor that applies to this zone that sets it apart from the other areas?

3.) If an object was moving with an angular velocity of 0.3c with the orbit of the frame-dragging at the marginally stable orbit, would the object have an overall comoving velocity of 0.3c + velocity of frame-dragging rotation at this radius?

Any feedback would be welcome.

Steve D

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#9
In reply to #8

Re: Extreme Frame Dragging

03/28/2008 7:54 AM

Hi Steve, thanks.

1.) The frame dragging moves the space-time around the rotating black hole and things, including light, seem to have that speed 'added', when viewed from a large distance. Orbits are faster for their radii when with the rotation than when against it.

2.) Inside the static limit, the frame dragging is so severe that not even light can move against the spin. This means that irrespective of its thrust level, no craft can be held static there - hence the name. The static limit is the radial distance* where an object can just barely be held static by near infinite thrust.

3.) It's a bit difficult to decide what the coordinate speed of an orbit is in Kerr coordinates - speed relative to what and how would it be measured? At some distances, you cannot have a static observer to measure the orbital period. When viewed from a large distance, gravitational and velocity time dilation influence the measurement of orbital speed. Loosely speaking I think one can say that the "speed" of the frame dragging adds to the normal Schwarzschild orbital speed.

* Radial distance is a 'slippery' concept around black holes and is usually named "circumferential radius".

Hope it helps.

Jorrie

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#10
In reply to #9

Re: Extreme Frame Dragging

05/09/2008 8:46 AM

Hello Jorrie and thanks for the response.

I've looked a little more into the ergoregion and the static limit- *The Killing vector field for a Kerr black hole ranges from 0 to -1 and is time-like (c^2 t^2 > r^2) to infinite outside the black hole (from what I can gather, in very basic terms, the Killing vector field ensures smoothness of the metric).

Divergence of the Killing field is at the event horizon for a static (Schwarzschild) black hole (where it become positive) and at the ergosphere for a Kerr rotating black hole; co-ordinates become light-like (c^2 t^2 = r^2) at the point of divergence and then space-like beyond (in the case of the rotating black hole, within the ergoregion).

It might be reasonable to say only one vector in the ergoregion becomes space-like (i.e. the vector parallel to the direction of rotation which increases over t (c^2 t^2 < r^2) while the other two vectors remain time-like. While there are no static observers relative to the black hole within the ergoregion, you can (with enough power) move away from the black hole in the direction of one of the other vectors which still remain time-like. It might be said that the ergosphere is where spacetime begins to become space-like and the event horizon (where space itself diverges) is where it becomes completely space-like in all three vectors.

But when it comes to calculating the Lense-Thirring effect, the angular velocity of frame-dragging at the ergosphere edge (oddly enough, regardless of sol mass) ranges from 0.1c at a/M=0.2 to 0.333c at a/M=1 which implies the spacetime should remain time-like well into the ergosphere.

Is there another factor that comes into play that rotates spacetime fast enough that the vector parallel to the direction of rotation becomes space-like within the ergoregion?

Steve

*source- 'Compact Objects in Astrophysics' by Max Camenzind

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#11
In reply to #10

Re: Extreme Frame Dragging

05/09/2008 10:41 AM

Hi Steve; you wrote:

"It might be reasonable to say only one vector in the ergoregion becomes space-like (i.e. the vector parallel to the direction of rotation which increases over t (c^2 t^2 < r^2) while the other two vectors remain time-like."

I think the problem is that once any component of a vector becomes spacelike, the vector is spacelike. This is true for a particle that tries to remain static inside the limit - it must be tachyon! However, when a particle is dragged around with the hole, it can have zero speed in the direction of rotation relative to spacetime and it can hence move outwards under some finite force.

"But when it comes to calculating the Lense-Thirring effect, the angular velocity of frame-dragging at the ergosphere edge (oddly enough, regardless of sol mass) ranges from 0.1c at a/M=0.2 to 0.333c at a/M=1 which implies the spacetime should remain time-like well into the ergosphere."

The mass dependency is not that odd, because the size of the ergosphere scales with the mass. I think the 0.333c is the tangential coordinate speed of light at the static limit- this is what a distant static observer would measure - taking into account all the time dilation factors. I calculated it before as ~0.4c just outside the static limit (I think I mentioned that in the OP).

Jorrie

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#12
In reply to #11

Re: Extreme Frame Dragging

05/09/2008 11:35 AM

Again, thanks for the reply Jorrie.

"It might be reasonable to say only one vector in the ergoregion becomes space-like (i.e. the vector parallel to the direction of rotation which increases over t (c^2 t^2 < r^2) while the other two vectors remain time-like."

I think the problem is that once any component of a vector becomes spacelike, the vector is spacelike. This is true for a particle that tries to remain static inside the limit - it must be tachyon! However, when a particle is dragged around with the hole, it can have zero speed in the direction of rotation relative to spacetime and it can hence move outwards under some finite force.

Just to clarify, regarding the above, might it be correct to say that based on spacetime consisting of 4 dimensions, three vectors (x, y and z) and one of time (t), only one of the vectors would be space-like in the ergoregion, (the one parallel to rotation) and the other two vectors would remain time-like but spacetime would be described as space-like, period; or are you saying if one vector becomes space-like then all the vectors are space-like within the ergoregion?

Steve

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#13
In reply to #12

Re: Extreme Frame Dragging

05/10/2008 12:04 AM

Hi again Steve.

To say "spacetime is timelike", etc. is a bit confusing. It is actually only events in spacetime that can be separated by either timelike, lightlike or spacelike intervals. One can loosely say that a particle's behaviour is timelike, spacelike (tachyons), etc, but it is not strictly true.

The spacetime interval can be broken up into its 4 components, some of which may appear to be spacelike and others timelike, but this does not change to fact that a spacelike interval is spacelike for all observers. In a sense, of you break a spacetime interval up into all its components, you have not two, but four separate spacetime intervals. All may be timelike while the overall interval may be spacelike.

Maybe the easiest way to cope with this is in terms of a photon moving "lightlike" in a given direction. Its velocity components (not vectors!) in the x, y and z dimensions may all be less than c, but the velocity vector has magnitude c for all observers. If one of those components equals c, the others must be zero...

Hope this helps clarifying the issue to some extent.

Jorrie

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#14
In reply to #13

Re: Extreme Frame Dragging

05/10/2008 7:41 AM

Hello again Jorrie, and thanks for your time. If it's no hassle, I have one last question regarding the transition of time-like to space-like intervals at the static limit. Say we have a small 3 sol mass rotating black hole with a low spin parameter of a/M=0.02, the angular velocity of the frame-dragging (fd) at the photon sphere (~13,200 m) would be 1,323,312 m/s (0.00441c); at the static limit (8861.099 m) 2,997,325 m/s (0.00998c) and at the event horizon (8860.213 m) 2,997,924 m/s (0.01c) giving an equatorial distance of 0.886 m between the static limit and the event horizon. How is it that within this small distance, the ergoregion is rotating faster than light when the figures for frame dragging imply the angular velocities of rotation are well below c? I did a quick check and found that while the angular velocity of fd increases by ~600 m/s in the ergoregion over the 0.886 m, the average increase over the same distance between the photon sphere and static limit is only ~333 m/s. Are the space-like intervals created in the ergoregion by the rapid increase in fd rather than the angular velocity of the fd itself?

Steve

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#15
In reply to #13

Re: Extreme Frame Dragging

05/13/2008 1:50 AM

After a bit more reading, I gather the frame-dragging figure of ω 'parameterizes the rotation of space as viewed from infinity'*, which implies that regardless of quantity of spin, space is spinning at c at the ergosphere edge. At what point do the results of the calculations become relativistic (i.e. observed information rather than actual/local information) and does ω still hold as an actual quantity of frame-dragging (as apposed to observed quantity) for objects with less extreme spin such as planets, stars and possibly neutron stars?

*'Experimental Evidence of Black Holes' by Andreas Müller

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#16
In reply to #15

Re: Extreme Frame Dragging

05/23/2008 12:48 AM

Hi Steve.

Yep, ω is a parameter measured in the asymptotically flat coordinate system (observer at 'infinity').

You asked: "At what point do the results of the calculations become relativistic (i.e. observed information rather than actual/local information)..."

I don't quite understand what you are asking. Local info is also observed, I think. The results may just differ from that of 'distant' observed information. A local observer being kept static relative to the distant stars will observe the frame dragging in terms of the path that light and free particles follow.

Weak-field frame dragging follows the same equations, but they can be simplified, as I have shown previously in a discussion on GPB.

Jorrie

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#17
In reply to #16

Re: Extreme Frame Dragging

05/30/2008 8:12 AM

Hi Jorrie

Thanks again for your reply. When I ask when do the results become relativistic, I am under the impression that the information observed globally (calculated using the equations above) differs from what is taking place locally for objects with extreme frame dragging. If you take a look at post 14, I describe a scenario where you have a black hole with a very low rotation but that still has an ergoregion. It seems accepted that space within the ergoregion is rotating faster than the speed of light but when I calculate angular velocity at the static limit using the frame dragging rate, it produces a figure considerably lower than c. The fact that it's been stated that the rate of frame dragging calculated is the information as seen from infinity implies that the rate of rotation local to the ergosphere (which is supposedly travelling at c) differs to what is observed globally (for the black hole with a low rotation, the frame dragging rate calculated produces an angular velocity of only 0.00998c at the static limit). Is there a way of calculating the actual angular velocity at the static limit that would provide evidence that space at the static limit does indeed rotate at c. Hopefully this explains the question a bit better.

Steve

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#18
In reply to #17

Re: Extreme Frame Dragging

05/30/2008 10:36 AM

Hi Steve, you originally wrote: "Say we have a small 3 sol mass rotating black hole with a low spin parameter of a/M=0.02, the angular velocity of the frame-dragging (fd) at the photon sphere (~13,200 m) would be 1,323,312 m/s (0.00441c); at the static limit (8861.099 m) 2,997,325 m/s (0.00998c) and at the event horizon (8860.213 m) 2,997,924 m/s (0.01c) giving an equatorial distance of 0.886 m between the static limit and the event horizon."

Ah, I now see what your issue is. A spin parameter of a/M=0.02 is very small and the static limit is just about equal to the Schwarzschild radius (2M), as your 0.866 m indicates. There the local velocity approaches the speed of light, but coordinate (distantly observed) velocity approaches zero. Recall that, from a previous Blog post,

The redshift factor, as I posted (and edited later) in the notes of the present Blog entry, is the reason for this behaviour. You can use that equation to calculate the local speed of a particle static relative to space, just inside the static limit - presumably measured by a hypothetical observer held static in the distant coordinate system, just outside the static limit. You work out the coordinate speed from ω and then divide it by the redshift factor.

Jorrie

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