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I think to answer this question we would have to fully understand the nature of black holes. It seems to me that the smaller black hole could have a lower density than the larger one. In this case the smaller black hole might be effected by the tidal forces of the larger black hole. But if the densities of the two black holes were the same, wouldn't the smaller black hole be immune to spaghettification?

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  • $\begingroup$ If we trust GR, then both black holes are already ring-singularities. What does it mean to spaghettify spaghetti? $\endgroup$ – CuriousOne Jul 6 '15 at 5:58
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The black hole event horizon is not a thing i.e. not a physical object. It is just a surface in spacetime from which light can never escape to infinity.

Also, if we take the Schwarzschild description of a (non-rotating) black hole then it is a point mass hidden away behind the event horizon. You can't spaghettify a point mass.

When two black holes merge, the surface in spacetime from which light can never escape evolves with time. It starts out as two separate spheres, evolves through a complex geometry as the black holes merge then ends up as a single (bigger) sphere. There is nothing in this process that can reasonably be described as spaghettification.

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  • $\begingroup$ There is a serious question in here, but I don't think the OP is aware of it. How do two ring singularities merge? I am kind of curious myself now. $\endgroup$ – CuriousOne Jul 6 '15 at 8:23
  • $\begingroup$ @CuriousOne How do they merge? Very quickly. ;-P $\endgroup$ – John Apr 28 '20 at 15:55

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