# Rotation in the Kerr metric?

## Main Question or Discussion Point

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?

Related Special and General Relativity News on Phys.org
Dale
Mentor
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.

PeterDonis
Mentor
2019 Award
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.

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.

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.

Is that what’s called “frame dragging”?
It's one manifestation of frame dragging, yes. There are others as well.

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.)

PeterDonis
Mentor
2019 Award
I am not that familiar with this metric yet, but I assume that like the Schwarzschild it asymptotically goes to Minkowski at infinity?
Yes.

WannabeNewton