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?

  • $\begingroup$ 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? $\endgroup$
    – Qmechanic
    Commented May 20, 2015 at 17:20
  • $\begingroup$ 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. $\endgroup$
    – user81619
    Commented May 20, 2015 at 17:28
  • $\begingroup$ 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. $\endgroup$ Commented May 20, 2015 at 17:37
  • $\begingroup$ @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. $\endgroup$
    – user81619
    Commented May 20, 2015 at 18:55

1 Answer 1


The theorem does not apply, as we do not have spherical symmetry. All we have is rotational symmetry about a preferred axis.

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.

Not only are real black holes expected to have some rotation (and therefore not be Schwarzschild), it would not be surprising to find most of them coming close to the upper bound on how fast they can spin.

  • $\begingroup$ Thanks very much Chris, my next question is/was going to be (and this might be a balancing act between 2 opposing effects), say for a neutron star just below the B.H. mass category, could sufficient rotation act to stop further collapse or would this addditional K.E. act as effective mass, through mc^2, to actually promote collapse? Before asking this question, and trying to answer it myself, I will study the Kerr textbook material to check if it is a valid physical argument. $\endgroup$
    – user81619
    Commented May 21, 2015 at 8:07

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