Gravitation has an effect analogous
to magnetism in electromagnetics. You can derive the analog of the electromagnetic
"Biot-Savart" law (in the weak gravity limit) by calculating the 4-Force
on a moving object from a stationary gravitating object, then transforming
into multiple Lorentz frames, then separating the results into "electric"
and "magnetic" components. Once you have your gravitational Biot-Savart
law, you can then integrate over the mass/velocity of a rotating body to
get the frame dragging effect. I emphasize that this procedure will only
be *quantitatively* accurate for "weak"
fields such as from a planet. It will not be quantitatively accurate for
strong fields such as near the event horizon of a rotating black hole,
but qualitatively, the effect is the same.

One of the immediate differences one sees in this exercise is that with gravitational "magnetism", like currents repel and opposite currents attract, as opposed to the reverse in electromagnetism. This is why a moving object with otherwise sufficient kinetic energy does not self gravitate into a black hole (however, two moving objects with sufficient kinetic energy colliding head on CAN become a black hole.)

There are a lot of fascinating effects in regard to a rotating black hole. On cursory observation, the Kerr metric for a rotating black hole is very hard to understand, especially the interpretation of "cross-product" terms such as (dr)(dt). I think Thorne's book, and a lot of others are very misleading in describing the frame dragging effect. They give one the impression that space is somehow "viscous" and one is pulled along by the swirling current. This is not really what is happening. If you suspend a plumb bob over the equator of a rotating black hole, it will hang vertically. It will not list either spinward or anti-spinward, even if you are only an arbitrarily small distance above the so-called "static limit".

Now if one is stationary (with respect to the fixed stars) above a rotating black hole and starts to fall or move inward, he will immediately be pulled spinward. If he moves outward he will be pulled anti-spinward. This effect is qualitatively exactly analogous to a negative charge moving about a positively charged rotating sphere. This is the interpretation of the cross product terms in the metric. Movement in one direction causes forces in another direction (in this case right angles).

If we have a figure skater who is skating over the equator of a rotating black hole and she is parallel to the hole's axis of rotation (I know, she's skating on the walls instead of the floor, but this is irrelevant to the argument) and she is initially stationary and she extends her arms, she will be torqued into a counter rotation to that of the hole's, if she brings her arms in she will be torqued into the other direction. If she is initially rotating at some speed which is a function of the hole's rotation and her distance and she extends her arms, nothing will happen

Now we come to a critical observation! Due to the principle of equivalence, gravitationally caused forces are (locally) indistinguishable from accelerations. The rotational speed at which, when the skater extends her arms, nothing happens, is her local reference for non rotation. This local reference is rotating with respect to the fixed stars. If she is stationary and she holds a gyroscope, she will see it precess at this rate and counter rotating to the hole's rotation (it will precess in the same direction over the hole's pole). Picture our skater as a planetary gear between an inner ring and an outer ring, with the inner ring rotating faster than the outer. She rotates in the opposite direction to the inner ring (the hole) (the outer ring being the outside universe) (If this doesn't tell you anything about Mach's principle, nothing will).

On to orbits. Moving in and out causes spinward and anti-spinward forces respectively, likewise, moving spinward and anti-spinward causes outward an inward forces respectively (moving north or south has no effect.) There is a bias to the orbital rate at which point one does not "feel" like one is in orbit. This bias point introduces an anisotropy or preferred direction of motion to an outside observer.

Now with a rotating black hole there is a boundary, distinct from the event horizon, called the "static limit". Below the static limit one cannot remain at rest with respect to the fixed stars. One must orbit the hole in the spinward direction. As mentioned in the above paragraph, there is an orbital bias rate. At the static limit, this bias rate is circling the hole (locally) at speed "c". One way to view this is if someone is orbiting at the bias rate, he sees stationary coordinate points whip by him at speed "c". Also, remember, "c" is decreasing to an outside observer using flat space Boyer/Lindquist coordinates (rotating extension to Schwartzchild coords) since we are going lower into a gravity well.

Between the static limit and the event horizon we have a region called the "ergosphere". As mentioned above, one cannot remain at rest with respect to the fixed stars while in the ergosphere. One can however, remain outside the event horizon and even return back into space if one desires. Inside the ergosphere, one sees stationary coordinate points whiz by him at FASTER than the speed of light.

I mentioned awhile back that a stationary plumb bob suspended over the equator does not list. The astute reader may now ask what happens if we try to SLOWLY lower the plumb bob into the ergosphere. Since a stationary object does not experience anything other than a radial force, what causes it to move spinward in the ergosphere? What happens is that, as the plumb bob approaches the static limit, the RADIAL force on the cable diverges to infinity. It therefore must fall. As soon as it starts to fall it gets an induced gravomagnetic force spinward. As soon as it starts to move spinward Lorentz transformations and gravomagnetically induced outward radial forces, negate the infinite inward radial force. A stationary observer near the static limit will also experience diverging time dilation, with respect to the outside universe, as he approaches the static limit.

Let me bring up another metaphor. Assume we surround our rotating black hole with a stationary spherical ice shell, slightly outside the static limit. We again bring our skater on stage (this time she is now skating on the floor). If she skates in the anti-spinward direction she will weigh more than if she skates in the spinward direction. As the radius of our shell slowly approaches the static limit, if she she doesn't start skating spinward she will eventually weigh so much (even without porking out on Big Macs) that she will break and fall thru the ice.

Finally, as one approaches the event horizon, the speed of light, in external flat space Boyer/Lindquist coordinates, goes to zero, but, and this is critical, the orbital bias rate remains, and continues to have a finite non-zero value in boyer/Lindquist coordinates. At the event horizon an observer is synchronized at the rotation rate of the black hole. He can only move at a delta velocity from the bias rate limited by the speed of light, but since c goes to zero at the event horizon, so does the delta v, and our observer crosses the event horizon orbiting exactly at the bias rate (in the view of an outside observer.)

This is one of the most interesting things
about a rotating black hole. The orbital bias rate, or so called "swirl
of space" is *not limited by the local speed of light*. While most
other features of a black hole are modified by the gravity well and gravitational
redshift, the rotation rate is not! That is why the static limit exists.
Space is "swirling" around locally faster than the local speed of light,
so if a local observer shines a light anti-spinward, the photons will still
have a spinward rotation, and since no one can outrun a photon, one cannot
stand still inside the static limit.