# Spin-flip

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A black hole spin-flip occurs when the spin axis of a rotating black hole undergoes a sudden change in orientation due to absorption of a second (smaller) black hole. Spin-flips are believed to be a consequence of galaxy mergers, when two supermassive black holes form a bound pair at the center of the merged galaxy and coalesce after emitting gravitational waves. Spin-flips are significant astrophysically since a number of physical processes are associated with black hole spins; for instance, jets in active galaxies are believed to be launched parallel to the spin axes of supermassive black holes. A change in the rotation axis of a black hole due to a spin-flip would therefore result in a change in the direction of the jet.

Schematic diagram of a black hole spin-flip.

## Physics of spin-flips

A spin-flip is a late stage in the evolution of a binary black hole. The binary consists of two black holes, with masses ${\displaystyle M_{1}}$  and ${\displaystyle M_{2}}$ , that revolve around their common center of mass. The total angular momentum ${\displaystyle J}$  of the binary system is the sum of the angular momentum of the orbit, ${\displaystyle {L}}$ , plus the spin angular momenta ${\displaystyle {S}_{1,2}={S}_{1}+{S}_{2}}$  of the two holes. If we write ${\displaystyle \mathbf {M_{1}} ,\mathbf {M_{2}} }$  as the masses of each hole and ${\displaystyle \mathbf {a_{1}} ,\mathbf {a_{2}} }$  as their Kerr parameters,[1] then use the angle from north of their spin axes as given by ${\displaystyle \theta }$ , we can write,

${\displaystyle \mathbf {S} _{1}=\{\mathbf {a} _{1}\mathbf {M} _{1}\cos(\pi /2-\theta ),\mathbf {a} _{1}\mathbf {M} _{1}\sin(\pi /2-\theta )\}}$

${\displaystyle \mathbf {S} _{2}=\{\mathbf {a} _{2}\mathbf {M} _{2}\cos(\pi /2-\theta ),\mathbf {a} _{2}\mathbf {M} _{2}\sin(\pi /2-\theta )\}}$

${\displaystyle \mathbf {J} _{\rm {init}}=\mathbf {L} _{\rm {orb}}+\mathbf {S} _{1}+\mathbf {S} _{2}.}$

If the orbital separation is sufficiently small, emission of energy and angular momentum in the form of gravitational radiation will cause the orbital separation to drop. Eventually, the smaller hole ${\displaystyle M_{2}}$  reaches the innermost stable circular orbit, or ISCO, around the larger hole. Once the ISCO is reached, there no longer exists a stable orbit, and the smaller hole plunges into the larger hole, coalescing with it. The final angular momentum after coalescence is just

${\displaystyle \mathbf {J} _{\rm {final}}=\mathbf {S} ,}$

the spin angular momentum of the single, coalesced hole. Neglecting the angular momentum that is carried away by gravitational waves during the final plunge—which is small[2]—conservation of angular momentum implies

${\displaystyle \mathbf {S} \approx \mathbf {L} _{\rm {ISCO}}+\mathbf {S} _{1}+\mathbf {S} _{2}.}$

${\displaystyle S_{2}}$  is of order ${\displaystyle (M_{2}/M_{1})^{2}}$  times ${\displaystyle S_{1}}$  and can be ignored if ${\displaystyle M_{2}}$  is much smaller than ${\displaystyle M_{1}}$ . Making this approximation,

${\displaystyle \mathbf {S} \approx \mathbf {L} _{\rm {ISCO}}+\mathbf {S} _{1}.}$

This equation states that the final spin of the hole is the sum of the larger hole's initial spin plus the orbital angular momentum of the smaller hole at the last stable orbit. Since the vectors ${\displaystyle S_{1}}$  and ${\displaystyle L}$  are generically oriented in different directions, ${\displaystyle S}$  will point in a different direction than ${\displaystyle S_{1}}$ —a spin-flip.[3]

The angle by which the black hole's spin re-orients itself depends on the relative size of ${\displaystyle L_{\rm {ISCO}}}$  and ${\displaystyle S_{1}}$ , and on the angle between them. At one extreme, if ${\displaystyle S_{1}}$  is very small, the final spin will be dominated by ${\displaystyle L_{\rm {ISCO}}}$  and the flip angle can be large. At the other extreme, the larger black hole might be a maximally-rotating Kerr black hole initially. Its spin angular momentum is then of order

${\displaystyle S_{1}\approx GM_{1}^{2}/c.}$

The orbital angular momentum of the smaller hole at the ISCO depends on the direction of its orbit, but is of order

${\displaystyle L_{\rm {ISCO}}\approx GM_{1}M_{2}/c.}$

Comparing these two expressions, it follows that even a fairly small hole, with mass about one-fifth that of the larger hole, can reorient the larger hole by 90 degrees or more.[3]