Rotation in the Kerr metric?

This means that the large mass rotates with respect to the local inertial frame free falling from spatial infinity.
  • #1
exmarine
241
11
I understand the Kerr metric has an off-diagonal term between the rotation and the time degrees-of-freedom? That a test mass falling straight down toward a large rotating mass from infinity will begin to pick up angular momentum? Is that what’s called “frame dragging”? Did the Gravity Probe B verify that effect?

Finally my main question: what is the large mass rotating with respect to? Before someone says “the distant stars”, remember that the Kerr metric is a MODEL of an otherwise empty universe. I don’t think there are any “distant stars” in the MODEL. I am not that familiar with this metric yet, but I assume that like the Schwarzschild it asymptotically goes to Minkowski at infinity?
 
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  • #2
exmarine said:
what is the large mass rotating with respect to?
Rotation is non inertial. It is not relative like velocity. You don't need a "with respect to" for rotation.
 
  • #3
exmarine said:
I understand the Kerr metric has an off-diagonal term between the rotation and the time degrees-of-freedom?

"Degrees of freedom" is not really the right term, but yes, there is an off-diagonal ##dt d\phi## term in the Kerr metric. More precisely, there is one off-diagonal term in the Kerr metric in Boyer-Lindquist coordinates (which are basically the analogue in Kerr spacetime of Schwarzschild coordinates in Schwarzschild spacetime); in other charts there may be more than one. A coordinate-independent way of putting it is to say that the "time translation" Killing vector field in Kerr spacetime is not orthogonal to the spacelike hypersurfaces in which the orbits of the "rotation" Killing vector field lie.

exmarine said:
a test mass falling straight down toward a large rotating mass

Strictly speaking, the Kerr metric does not describe a large rotating "mass" such as a planet or star. It describes a rotating black hole. It is believed that the Kerr metric (or at least a portion of it) also describes spacetime, at least approximately, around a rotating mass like a planet or star, but this has not been proven.

exmarine said:
from infinity will begin to pick up angular momentum?

No, it will begin to pick up angular velocity. One of the counterintuitive things about Kerr spacetime is that nonzero angular velocity does not always mean nonzero angular momentum. A test object that starts with zero angular momentum at infinity and falls into a Kerr black hole will have zero angular momentum throughout its fall--angular momentum is a constant of the motion for freely falling objects in Kerr spacetime, just as it is in Schwarzschild spacetime. But because of the non-orthogonality described above, this zero angular momentum object will pick up angular velocity as it falls.

exmarine said:
Is that what’s called “frame dragging”?

It's one manifestation of frame dragging, yes. There are others as well.

exmarine said:
Did the Gravity Probe B verify that effect?

It verified a different manifestation of frame dragging, its effect on a gyroscope in a free-fall nearly circular orbit about a rotating mass. (Note, again, that the presence of frame dragging in the spacetime around the Earth does not prove that the Kerr metric exactly describes that spacetime. It only proves that there is one particular term in the metric around the Earth that corresponds to what you would get if you did a weak-field approximation based on the Kerr metric--more precisely, based on expressing the Kerr metric as the Schwarzschild metric plus small perturbations.)
 
  • #4
exmarine said:
I am not that familiar with this metric yet, but I assume that like the Schwarzschild it asymptotically goes to Minkowski at infinity?

Yes.
 
  • #5
exmarine said:
Is that what’s called “frame dragging”?

"Frame dragging" is a very general term when applied to general stationary axisymmetric space-times that takes on varied meanings based on the context. There are three rather canonical examples of frame dragging: a test particle is dropped from spatial infinity with zero angular momentum and free falls toward the central rotating mass and in doing so, it gains an angular velocity (not angular momentum, due to conservation of said quantity) around the central mass because the space-time itself rotates relative to spatial infinity; a gyroscope at rest relative to spatial infinity whose axis is fixed to the distant stars will start precessing relative to local Fermi-Walker transported gyroscope axes; a gyroscope in orbit around the central mass with the zero angular momentum parameters whose axis is fixed to the local space-time symmetries will precess relative to comoving Fermi-Walker transported gyroscope axes. The second effect is called the Lense-Thirring precession and the latter effect is a combination of Lense-Thirring, geodetic, and Thomas precessions.

exmarine said:
Did the Gravity Probe B verify that effect?

It verified Lense-Thirring and geodetic precessions to different accuracy.

exmarine said:
Finally my main question: what is the large mass rotating with respect to?

It rotates with respect to the asymptotic Minkowskian (flat space-time) Lorentz frame at spatial infinity.
 

1. How does rotation affect the Kerr metric?

In the Kerr metric, rotation has a significant effect on the spacetime geometry. It causes the spacetime to become curved and twisted, resulting in the appearance of a black hole. The rotation also introduces a property known as frame-dragging, where nearby objects will be dragged along the rotation of the black hole.

2. Can the Kerr metric be applied to any rotating object?

No, the Kerr metric is specifically designed to describe the spacetime of a rotating black hole. It is not applicable to other rotating objects, such as planets or stars.

3. How is the Kerr metric different from the Schwarzschild metric?

The Kerr metric is a more complex and general solution to Einstein's field equations than the simpler Schwarzschild metric. The Kerr metric takes into account the effects of rotation, while the Schwarzschild metric only describes non-rotating black holes.

4. What is the significance of the Kerr metric in astrophysics?

The Kerr metric is a crucial tool in understanding the properties and behavior of rotating black holes. It has been used to make predictions about the behavior of matter and radiation near black holes and has been confirmed by observations of astrophysical phenomena, such as accretion disks and jets.

5. Are there any limitations to the Kerr metric?

Yes, the Kerr metric is limited in its applicability to situations where the black hole is rotating at a constant rate. It does not account for changes in the rotation of the black hole over time or when the black hole is not in a steady state. Additionally, the Kerr metric assumes that the black hole has no charge, which may not be the case in all astrophysical scenarios.

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