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zaarn | 3 years ago

Spin and Angular momentum are two very different things. Angular momentum measures the velocity of a black hole in, eg, an orbit. Spin measures the rotation of a black hole against it's rest frame.

These are independent quantities.

discuss

order

raattgift|3 years ago

In a Kerr black hole spacetime there is no orbital angular momentum (OAM, L), only the intrinsic spin angular momentum (SAM, S). One can get OAM in a relativistic n-body black hole problem. Each of those black holes will have its own SAM.

See §5.11. ANGULAR MOMENTUM in Misner, Thorne & Wheeler (MTW) and in particular Box 5.6 D (Intrinsic Angular Momentum) and E (Decomposition of Angular Momentum into Intrinsic and Orbital Parts). The latter gives J = L + S. Admittedly, in MTW §33 the authors prefer to use S (e.g. eqn. 33.4) but that raises the important caveat for binary black hole (BBH) mergers at the end of Box 33.4 (I)(A)(4), which refers back to Box 5.6). Newer textbooks and other sources (including Wikipedia [1]) prefer J, although commonly it gets called spin angular momentum (as in [1] and my earlier comment). Carroll's textbook calls it the Komar angular momentum (near eqn. 6.73, referring to eqn. 6.48) and "spin (angular momentum)" (above eqn. 6.47). This is the sort of thing that annoys mathematicians and non-relativist physicists about relativists; confusion is completely understandable.

A binary black hole (BBH) is not a Kerr solution. No exact analytical solutions to the Einstein Field Equations for a BBH have been found, only approximations and numerical solutions (see the comprehensive review by Baker et al. at <https://arxiv.org/abs/1010.5260> and \vec{L} therein, notably at section D(2) in the second column on PDF p. 26, "In a related phenomenon, the direction of the total angular momentum (\vec{L} + \vec{S}_1 + \vec{S}_2) may change.").

No change of coordinates can turn a BBH into a Kerr solution; the former radiates gravitational waves (if there is no incoming gravitational radiation), the latter doesn't.

(Another way of distinguishing is in the algebraic symmetries of the Weyl curvature tensor. Kerr is a Petrov type D spacetime, BBH spacetimes are generically type I up to some degeneracy measure.)

Finally, I can tie this in to neutron stars: the (exterior) Hartle-Thorne metric is an approximation of the Kerr metric useful for relativistic stars without horizons (neutron stars, white dwarfs) and without regard to interior differentiation. Its usual write-down uses J, but sometimes S, and sometimes both (e.g. <https://arxiv.org/abs/1507.04264>, where at the top of p. 2 the authors give J = GS/c^3).

[1] <https://en.wikipedia.org/wiki/Kerr_metric#Overview>, "J represents its spin angular momentum" and see eqn (6) further down, "a = J/Mc".

zaarn|3 years ago

You're still referring to two distinct properties with Angular Moment being one and Spin being the other. Because a black hole can in fact orbit things and that would give it angular momentum. The spin value is merely how fast it rotates around an axis (which you can define but that's just an observational data point) and is unrelated to the external movement in spacetime.