[zaarn]

The spin axis isn't a valuable reference since it depends on your frame of reference, which is a concept that gets rather ambiguous as you get close to a black hole anyway.

Plus you can just "rotate" a black hole to get it to have the same spin axis as another black hole. You can't "rotate" or "translate" a black hole in space to make the other three numbers change. Those require ingesting matter or emitting hawking radiation and that is the only thing that changes those properties.

[bell-cot]

I'm figuring that the low-rent planets, stars, etc. in the vague vicinity of an actual black hole would provide a fine frame of reference.

> Plus you can just "rotate" a black hole to get it to have the same spin axis as...

Quip: If you have the tech & budget to meaningfully rotate a spinning black hole, then you've got the tech & budget to change the other parameters, too.

FWIW - Wikipedia's answer is that 11 numbers (or 2 scalars and 3 vectors) are needed to fully spec. a stable black hole - https://en.wikipedia.org/wiki/Rotating_black_hole#Types_of_b...

[zaarn]

There is a cheap and even for current humanity achievable way to rotate a black hole; change your frame of reference. That is not only true for angular momentum but linear momentum and position. Those are entirely dependent on the observer and their frame of reference. Spin, magnetic charge and mass are not.

Two black holes who differ only by their position, linear and/or angular momentum but are equal in all other parameters are not distinguishable from simply seeing the same black hole twice from a different perspective.

Two black holes who differ in any of the three properties of mass, spin or magnetic charge are distinguishable by those properties (but even that is arguable to some extend).

edit: The rent prices of a planet don't matter since frame of reference is an actual term here, there is no frame of reference more valid than any other for determining the linear or angular velocity or the position of a black hole.

[raattgift]

The quantity a = cJ/GM^2, the spin parameter of the Kerr metric, is not observer-dependent. No observer (or system of coordinates) can turn an axisymmetric system into a spherically symmetric one.

Also,

> Two black holes who differ only by their ... angular momentum ... are not distinguishable

is contradicted by

> Two black holes who differ in ... spin ... are distinguishable

(Spin) Angular momentum is J in my first line above.

[zaarn]

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.

[raattgift]

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]

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.