Research over the last 200 years suggests that half or more of visible stars are part of multiple star systems.

I apologise for the number of assumptions in my question but, because of the numbers of binary systems involved, I feel this "what if" type question may be a plausible physical outcome in some binary star systems.

Assume both stars were born at the same time, with approximately the same mass, but spinning in opposite directions. I don't know enough about stellar evolution to say whether this is feasible, but we do know binary star systems exist and they must have some chance, over time, of reaching similar angular momentum values, occasionally in oppositely rotating directions.

My questions are, assuming they are close enough, will this eventually result in a slowdown in their rotations, followed by mutual tidal locking?

If this occurs, and the mass of at least one star is large enough, could we then have a stationary neutron star or black hole?

I realise that the process of creating the dense star will occur after a supernova explosion, which will most likely vaporise the other star in the system.

I note there are related questions such as Binary Star System but I can't immediately see a duplicate of my question.

  • $\begingroup$ You would expect that the the binary stars formed from a single cloud of gas that had some angular momentum, since they are orbiting each other. Therefore, the binary system would have to maintain that momentum. It would be unlikely that they could combine into something that had zero angular momentum. $\endgroup$
    – Peter R
    Jul 24, 2016 at 0:37
  • $\begingroup$ Maybe terminology, but to be clear a black hole with angular momentum is stationary. Without it is static. All after they settle down to their equilibirium states. The stationary case is the Kerr solution, with rotation. The static case is the spherically symmetric nonrotating Schwarzchild black hole. And yes, possible to have zero rotation but highly unlikely because the orbital rotation soon before they collapse is huge, they are accelerating quickly to near c;it requires really fine tuning to make it all come to zero. Simulations do give different specific results for counter to co rotating $\endgroup$
    – Bob Bee
    Jul 24, 2016 at 6:42
  • $\begingroup$ @BobBee thanks Bob, I forgot some basic but important GR terminology. $\endgroup$
    – user108787
    Jul 24, 2016 at 9:29

2 Answers 2


I'm sure that an exactly spinless final black hole is theoretically possible, but it would be a very, very precise fine-tuning, akin to balancing a pencil on it's point.

Also, your configuration is unlikely to produce a spinless black hole, because when the plunge phase of the merger happens, the stars will suddenly fall from the innermost stable orbit, and they will not have time to make a complete orbit. This means that the angular and linear momentum of the radiated gravitational waves will be asymmetric, which means that the final black hole will pick up additional angular momentum and momentum during the plunge.


I'll try to be careful to distinguish between orbital rotation and spin. The requirement for a non-spinning finish is that the sum of the orbital and angular momentums of the material that remains is zero.
Even in clasical physics this is clearly impossible with equal masses and in the absence of loss of mass due to supernovae (which would further impede spin equalisation). This is because the orbital angular momentum increases with orbital radius, and with equal star size this implies that the spin velocity at a given radius would have to exceed the orbital velocity at that radius, so the stars would simply disintegrate. Gravitational waves in the final join would (as stated by Jerry Schirmer be asymmetric), My expectation is that although velocities still increase, gravitational waves reduce angular momentum, but I may have slipped up somewhere; maybe Jerry Shirmer will validate this counter-intuitive result? Low values are of course possible with highly unequal stars, or a collision with an external system could leave you with any level of rotation.
A pedantic point, perhaps, but (other than as a theoretical construct) I'm not familiar with the value zero; however, Heisenberg excepted, it is possible for the final angular momentum of an asymmetric star system to be smaller than any finite value you may specify.


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