An intriguing comment below this question links to this Quora question and points to the answer by Robert Walker.

The sentence that particularly intrigued me is:

It turns out that there are two possible solutions, as the spin rate increases. You can get an oblate spheroid, or a triaxial ellipsoid - the solution "bifurcates". But the triaxial ellipsoid is the most stable of the two as Jacobi found out in his paper published in 1834. Figures of Equilibrium - Historical Account* - Chandrasekar

It seems to refer both to an 1834 paper by Jacobi and it's reference in the 1964 paper Ellipsoidal figures of equilibrium—an historical account by S. Chandrasekhar, Communications on Pure and Applied Mathematics, 20 (2), May 1967, 251–265. While the latter is paywalled, there seems to be a readable version here in google in the book A Quest for Perspectives: Selected Works of S. Chandrasekhar : with Commentary By Subrahmanyan Chandrasekhar, Kameshwar C. Wali, Volume 1. Imperial College Press, 2001.

Mathematical bifurcation is a fascinating topic and there is even an entire mathematics journal dedicated to the subject.

In this case, where exactly in the calculation of hydrostatic equillibrium of rotating bodies does this bifurcation (mentioned in the Quora answer) occur? I'm wondering if it simply means that for a given volume and density, below a certain rotation rate the hydrostatic equilibrium shape is an oblate sphere, but above it, the hydrostatic equilibrium shape will be a triaxial ellipsoid? Or is the bifurcation part of the evolution of the shape over time, involving viscocity or external effects?

If the triaxial ellipsoide is always(?) the most stable, then is a oblate spheroid actually an intermediate shape, and not really the shape of hydrostatic equillibrium?

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    $\begingroup$ Doesn't it just mean that for a given mass there are two local minima in the energy density? $\endgroup$ – ProfRob Nov 27 '16 at 19:00
  • $\begingroup$ @RobJeffries I don't know for sure. For example when we talk about bifurcation of orbits in a 3-body problem, two trajectories with almost identical starting conditions will appear to suddenly separate (bifurcate) at a point in time as the orbits evolve. Bifurcation means more than there being two minima. It may mean that there is some rotation rate above-which one minimum changes to two minima, I don't know. $\endgroup$ – uhoh Nov 27 '16 at 19:10
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    $\begingroup$ Oh, it definitely refers to there being more than one possible solution. i.e. there is no unique shape that is stable above some critical rotation threshold. $\endgroup$ – ProfRob Nov 27 '16 at 19:12
  • $\begingroup$ @RobJeffries can you support this with something that shows the mathematical bifurcation? I'm not asking about uniqueness, I am asking specifically about this bifurcation. In what space or along what dimension or axis does a minimum bifurcate? $\endgroup$ – uhoh Nov 27 '16 at 19:25
  • $\begingroup$ "bifurcates" just means "goes one way or another". here it means "the whole situation" goes either one way or another; it heads towards one or the other solution-paradigms. $\endgroup$ – Fattie Nov 28 '16 at 3:39

I think all this means is that up to some critical rotation rate, the equilibrium shape of a rotating, self-gravitating fluid is an oblate spheroid. That is, this shape defines a unique global minimum of energy.

Above this threshold there are two possible equilibrium solutions. Apparently one is an oblate spheroid, whereas the other is a triaxial ellipsoid. These will define local energy minima, but the triaxial ellipsoid is the global minimum.

The gory mathematical details are no doubt contained in Chandrasekhar's textbook on equilibrium ellipsoidal figures. Pages 3-5 of Iurato (2014) appear to summarise the historical development of this bifurcation in the equilibrium solutions.

  • $\begingroup$ OK that makes sense, and this is best followed up by a trip to the library. Iurato (2014) certainly sets the stage from the historical perspective, and it's possible that numerical simulations may show the minimum energy figures are not necessarily pure ellipsoids. Thank you again for another helpful answer! $\endgroup$ – uhoh Nov 28 '16 at 1:03

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