How to derive the Kerr metric by cheating quite a bit. Part 3

( … continued from Part 2)

3. Deriving the Kerr metric

In this section we show how the Kerr metric may be (in a large part) derived by studying Problem 2.23.

The main result that we rely on is a lemma given in Mars’ 1999 paper (If you actually look at the paper, you’d see that there are some factors of 2 differences in a lot of the statements. These are related to the fact that our definitions of anti-self-dual forms differs by a factor of 2, and that our definitions of the Ernst two-form and the Ernst potential also differ by a factor of 2).

Lemma 3.1
We can define the real-valued function $y$ and $z$ by $-\sigma^{-1} = y + i z$. Then there exists a non-negative real number $B$ such that $B \geq z^2$, and
$\displaystyle (\nabla y)^2 = \frac{y^2 - 2y + B}{M^2(y^2+z^2)}$
$\displaystyle (\nabla z)^2 = \frac{B-z^2}{M^2(y^2+z^2)}$

The proof is omitted here. I’ll give a basic sketch of the idea. Write $k$ and $l$ for the two mutually-normalized vector fields corresponding to the principal null directions. Consider the integral curves of $k$ (or $l$ respectively). By the Goldberg-Sachs theorem the congruences defined by the family of integral curves are geodesic and shear-free. Observe also that the only components of $\mathcal{C}_{abcd}$ comes from the $\mathcal{C}(k,l,k,l)$ component and others that can be related to it using the algebraic symmetries of Weyl fields. Now consider the second Bianchi identity applied to $\mathcal{C}_{abcd}$ (it is here we use the Ricci-flat condition: that the Weyl field obeys the second Bianchi identity), one sees that this
implies

$\epsilon_{abcd}l^ak^b\nabla^cy = 0 ~,\quad l^a\nabla_az = k^a\nabla_az = 0$

which, in particular, shows that $\nabla z$ is space-like. A rather complicated calculation then shows that

$M^2(y^2+z^2)(\nabla z)^2 + z^2$

is constant, using the equation induced on $\sigma$ from the second Bianchi identity. Using that $\nabla z$ is space-like, we have that the constant $B$ must be greater than or equal to $z^2$. The statement for $(\nabla y)^2$ follows from the observation

$\displaystyle (\nabla\sigma)^2 = \mathcal{F}_{ab}\tau^a\mathcal{F}^{cb}\tau_c = \frac14\mathcal{F}^2\tau^2 = - \frac{1}{M^2}\sigma^4\tau^2$

So

$\displaystyle (\nabla \frac{1}{\sigma})^2 = - \frac{\tau^2}{M^2} = \frac{1+2\Re\sigma}{M^2}$

By simple algebraic manipulations from the definition of $y$ and $z$, we obtain the statement on $(\nabla y)^2$.

By the above lemma and its proof, we see that at points where $B\neq z^2$ and $y^2-y + B\neq 0$, $y$ and $z$ are independent, non-degenerate scalar functions. Furthermore, as they are geometric quantities defined from objects that are symmetric under the $\tau$ and $\eta$ actions, we must have $\tau(y) = \tau(z) = \eta(y) = \eta(z) = 0$. So we can take $y$ and $z$ to be coördinate functions on (subsets of) the surface orthogonal to $\tau,\eta$.

We will make also the following guess: by the form of $(\nabla z)^2$, we can reasonably expect the change of variables $z = \sqrt{B}\cos\theta$ may patch together where $z^2 = B$ and resolve the coördinate singularity. Furthermore we will take $a = M\sqrt{B}$. In addition, let $r = M y$. Then we write

Equation 3.2
$\displaystyle \sigma = -\frac{M}{r + ia\cos\theta} = \frac{-Mr + iMa\cos\theta}{r^2 + a^2\cos^2\theta}$
$\displaystyle \Rightarrow S = 2\Re\sigma + 1 = 1- \frac{2Mr}{r^2+a^2\cos^2\theta}$
$\displaystyle \Rightarrow \Theta = -2\Im\sigma = -\frac{2Ma\cos\theta}{r^2 + a^2\cos^2\theta}$
$\displaystyle (\nabla r)^2 = \frac{r^2 - 2Mr + a^2}{r^2 + a^2\cos^2\theta}$
$\displaystyle (\nabla \theta)^2 = \frac{1}{r^2 + a^2\cos^2\theta}$

so going back to ansatz 1.2, we have that

Equation 3.3
$\displaystyle ds^2 = -(1-\frac{2Mr}{r^2+a^2\cos^2\theta}) dt^2 +2Q dtd\phi + Rd\phi^2 + \frac{r^2+a^2\cos^2\theta}{r^2 - 2Mr + a^2} dr^2 + (r^2+a^2\cos^2\theta)d\theta^2$

Now look at 2.9, we can re-write it as

Equation 3.4
$\displaystyle \frac{\partial_r(Q/S)}{\partial_\theta(Q/S)} = - \frac{V\partial_\theta\Theta}{W\partial_r\Theta} = -\frac{(r^2-a^2\cos^2\theta)\sin\theta}{2r\cos\theta(r^2-2Mr+a^2)}$

Observe now that for $K(r,\theta)$ given by

Equation 3.5
$\displaystyle K(r,\theta) := \frac{2Ma\sin\theta}{(r^2-2Mr+a^2\cos^2\theta)^2}$

we have that

Equation 3.6
$\displaystyle \frac{\partial}{\partial\theta}[(a^2\cos^2\theta - r^2)\sin\theta K(r,\theta)] = \frac{\partial}{\partial r}[2r\cos\theta(r^2-2Mr + a^2)K(r,\theta)]$

which implies that (with a rather non-trivial computation)

Equation 3.7
$\displaystyle \frac{Q}{S} = \frac{2Mar\sin^2\theta}{r^2-2Mr + a^2\cos^2\theta}$

This in turn tells us that

Equation 3.8
$\displaystyle Q = \frac{2Mar\sin^2\theta}{r^2+a^2\cos^2\theta}$

By examining 2.9 again, we can solve for $R$ purely algebraically (a computation I’ll omit here) to arrive at

Equation 3.9
$\displaystyle R = \sin^2\theta\left(r^2 + a^2 + \frac{2Ma^2r\sin^2\theta}{r^2+a^2\cos^2\theta}\right)$

This gives us the Kerr metric.

Notice that, however, the computation of $Q/S$ has the freedom to add a constant. This reflects the fact that our computation, in the end, is completely local. In other words, we know that on a local neighborhood the metric takes a given expression, what we don’t know is whether the Killing vector field represented by $\partial_\phi$ actually has closed orbits! The freedom to add a constant factor to $Q/S$ reflects the fact that any constant coefficient linear combination $\tilde\eta = c_1\tau + c_2\eta$ is again a Killing vector field which commutes with $\tau$. So our local coördinate form may be chosen initially such that $\partial_\phi$ coïncides with $\tilde\eta$, which does not have closed orbits.

A proper argument to get rid of this degree of freedom requires a careful examination of the properties of the bifurcate event horizon. In particular, assuming the space-time has a bifurcate event horizon, one can easily argue that both $\tau$ and $\eta$ are tangent to the bifurcate sphere, and in fact are multiples of each other. This allows us to fix the unknown constant. This argument is similar to the assumption used by Chandrasekhar in Mathematical Theory of Blackholes to analytically obtain the Kerr metric.

Lastly, one may ask about the title of this lecture, in particular about the “cheating quite a bit” part. The algebraic conditions derived in Section 2 are actually not too unreasonable. The primary “cheat” employed here is actually in the integrating factor $K(r,\theta)$ in 3.5. Whereas with our perfect hindsight, we can easily find the correct integrating factor, it is extremely difficult (insofar as my limited computational capabilities is concerned) to find the integrating factor just given 3.4.