# Does Birkhoff's theorem apply to rotating collapsing stars?

Birkhoff's theorem states that every spherically symmetric vacuum solution to $R_{\alpha\beta} = 0$ is static, which greatly assists in the solution to the Schwarzschild solution by eliminating time derivatives.

For collapsing stars with angular momentum, (i.e. all real ones, as far as I know), is this theorem applicable?

I ask this because I am not sure if real stars collapse in a sufficiently spherical manner.

I am presuming, and I am open to correction on this, that a rotating star's angular momentum would result in an equatorial bulge, with a non spherically symmetrical collapse. In particular, would this have consequences for the formation of black holes?

• Comment to the question (v3): Are you aware that rotating black holes are described by the Kerr metric (or Kerr-Newman metric) not the Schwarzschild metric? – Qmechanic May 20 '15 at 17:20
• Yes, vaguely by reading ahead a bit, but I only covered Schwartzchild solution today and still have to go through the consequences of that before it sinks in. If my question is dealt with in the Kerr related material, that's great, and probably an answer to my question, and I will wait till then. – user81619 May 20 '15 at 17:28
• Even if we had a perfectly rigid sphere that was rotating about an axis, that wouldn't count as spherically symmetric for the purposes of Birkhoff's Theorem. Roughly, the spin axis gives us a "preferred direction" on the sphere, breaking its symmetry. – Michael Seifert May 20 '15 at 17:37
• @MichaelSeifert thank you and now I do follow the idea of an axis causing symmetry breaking. It's difficult, for me at least, to get the idea,"it's a star, it must spin" out of my head. But your point really pins down the notion of absolutely no rotation involved, just in/out motion only. – user81619 May 20 '15 at 18:55

In fact, the gravitational field outside a rotating object will be Kerr, which only reduces to Schwarzschild in the case of no rotation. Otherwise, there will be time-space terms in the metric, making it not static. Still, Kerr is stationary, meaning the metric does not depend on time ($\partial_0 g_{\mu\nu} = 0$).
As a matter of fact, rotation is far from negligible in most contexts of astrophysical collapse. Things generally have too much angular momentum to collapse further, and this is true even of stars becoming black holes. For some rough numbers, the Sun has a mass $M = 2\times10^{30}\ \mathrm{kg}$, a radius $R = 7\times10^8\ \mathrm{m}$, and a period $P = 25\ \mathrm{days}$. Approximating it as a uniform sphere undergoing uniform rotation, it has an angular momentum $$L = \frac{4\pi MR^2}{3P} = 2\times10^{42}\ \mathrm{kg\cdot m^2/s}.$$ A Kerr black hole cannot have $cL/GM^2$ exceed $1$, or else there would be a naked singularity. But $cL/GM^2 \approx 2$ for the Sun.