Rotation of astrophysical black holes

Question

It's usually stated that "astrophysical black holes are expected to have non-zero angular momentum, due to their formation via collapse of rotating stellar objects". In other words: rotating stellar objects carry orbital angular momentum, which is expected to be in the final black hole configuration.

However, the Kerr solution doesn't carry an orbital angular momentum, but the computation of the ADM angular momentum only provides a Pauli-Lubanski contribution, which is supposed to represent the intrinsic angular momentum of a system in General Relativity:

$W_\mu=\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}J^{\nu\rho}P^{\sigma}$

Where is the orbital angular momentum in the astrophysical black hole created after the collapse? If the astrophysical object only has orbital angular momentum in the collapse, where does the intrinsic angular momentum of the Kerr black hole come from? Or is the usual interpretation of the Pauli-Lubanski contribution in General Relativity wrong?

Answer 1

Albert_RD asked: "Where is the orbital angular momentum in the astrophysical black hole created after the collapse?"

If an orbiting body plunges into a black hole its orbital angular momentum $L$ and intrinsic angular momentum $J$ get converted into the intrinsic angular momentum $J$ of the BH, so it isn't lost.

If a body with a prograde orbit (positive $L$) gets swallowed by the BH that increases the BH's $J$, and if it had a retrograde orbit (negative $L$) it will decrease the BH's $J$.

If the orbiting body collapses on its own without plunging into the BH, its intrinsic angular momentum $J$ will also be conserved, as is the orbital angular momentum $L$.

If the earth became a BH on it's own, its path around the sun (which gives you the $L$) would not change because of it, and neither would it if the sun became a BH.

If two BHs orbit each other and then plunge into each other to form a bigger black hole, their orbital angular momentum $L$ will be converted into the new BH's intrinsic angular momentum $J$.

Some parts of $L$ and $J$ may radiate away via gravitational waves in the process if you leave the test particle regime, but to first order the sum $J+L$ is conserved and they all go into the new BH's $J$.

Answer 2

The decomposition of angular momentum into orbital and instrinsic depends on the point about which we compute the angular momentum. The Earth has orbital angular momentum about the sun and "instrinsic" angular momentum about its center of mass. In a relativistic theory $$ M^{\mu\nu}= x^\mu p^\nu - x^\nu p^\mu+ S^{\mu\nu} $$ The skew-symmetric tensor $M^{\mu\nu}$ is the total angular momentum, $x^\mu p^\nu - x^\nu p^\mu=L^{\mu\nu}$ is the orbital part, and $S^{\mu\nu}$ is the intrinsic angular momentum. The vector $x^\mu$ is the position (usually, but not always, the center of mass in the body's rest frame) of the object with respect to the point about which we are computing $M$, and $p$ is the linear 4-momentum.

The Pauli-Lubanski vector (which was, I believe, discovered by Myron Mathisson) is derived by dualizing the totally antisymmetric P-L tensor $$ P^{\lambda\mu\nu}= p^\lambda M^{\mu\nu}+p^\mu M^{\nu\lambda}+p^\nu M^{\lambda\mu}. $$ The advantage of the P-L tensor (or vector) is that arbitrary-chosen $x$'s cancel and $$ P^{\lambda\mu\nu}= p^\lambda S^{\mu\nu}+p^\mu S^{\nu\lambda}+p^\nu S^{\lambda\mu}. $$ The $S$ tensor is still not unique. One must impose conditions such as $p_\mu S^{\mu\nu}=0$ decide "where" in the body are we considering to be its position.

For a massless object there is no frame-independent notion of "center of mass" and the decomposition is necessily frame dependent. Even for spinning Black holes there are effects such as bobbing due the ambiguity of the decomposition.