= Solution
For the <scalar wave separation in Kerr spacetime>, continue to use $A=r^2+a^2$ and $s=\sin\theta$. First verify the determinant in the hint. Direct multiplication of the covariant $t,\phi$ components gives
$$
\begin{aligned}
\Sigma^2(g_{tt}g_{\phi\phi}-g_{t\phi}^2)&=-s^2(\Delta-a^2s^2)(A^2-\Delta a^2s^2)-a^2s^4(A-\Delta)^2\\
&=-\Delta s^2(A-a^2s^2)^2=-\Delta s^2\Sigma^2.
\end{aligned}
$$
Hence the block determinant is $-\Delta\sin^2\theta$. Inverting this block gives
$$
g^{tt}=-\frac{A^2-\Delta a^2s^2}{\Sigma\Delta},\qquad g^{t\phi}=-\frac{a(A-\Delta)}{\Sigma\Delta},\qquad g^{\phi\phi}=\frac{\Delta-a^2s^2}{\Sigma\Delta s^2}.
$$
The other inverse components are $g^{rr}=\Delta/\Sigma$, $g^{\theta\theta}=1/\Sigma$, and $\sqrt{-g}=\Sigma s$. The <covariant wave operator> on a scalar consequently has the divergence form
$$
\Box\Psi=\frac1{\Sigma s}\partial_\mu(\Sigma s\,g^{\mu\nu}\partial_\nu\Psi).
$$
Let $m_\phi$ denote the azimuthal mode number, to distinguish it from the axial vector $m$. Insert the mode $\Psi=e^{-i\omega t+i m_\phi\phi}R(r)\Theta(\theta)$ in the massless <Klein-Gordon equation>. Single-valuedness makes $m_\phi$ an integer. The $t,\phi$ derivatives give
$$
\Sigma\left(-\omega^2g^{tt}+2\omega m_\phi g^{t\phi}-m_\phi^2g^{\phi\phi}\right)=\frac{(A\omega-a m_\phi)^2}{\Delta}+2a\omega m_\phi-a^2\omega^2\sin^2\theta-\frac{m_\phi^2}{\sin^2\theta}.
$$
With $K(r)=A\omega-a m_\phi$, division by the mode factor gives, on patches where $R\Theta\ne0$,
$$
\frac{(\Delta R')'}R+\frac{K^2}\Delta+2a\omega m_\phi-a^2\omega^2+\frac{(\sin\theta\,\Theta')'}{\sin\theta\,\Theta}+a^2\omega^2\cos^2\theta-\frac{m_\phi^2}{\sin^2\theta}=0.
$$
The radial and angular expressions must be opposite constants. Defining the <separation constant> as $\Lambda$, we obtain the two <ordinary differential equations>
$$
\boxed{\frac1{\sin\theta}\frac d{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\left(a^2\omega^2\cos^2\theta-\frac{m_\phi^2}{\sin^2\theta}+\Lambda\right)\Theta=0,}
$$
$$
\boxed{\frac d{dr}\left(\Delta\frac{dR}{dr}\right)+\left[\frac{((r^2+a^2)\omega-a m_\phi)^2}{\Delta}-a^2\omega^2+2a\omega m_\phi-\Lambda\right]R=0.}
$$
These equations also hold at zeros of a mode by continuity, without dividing there. Regular angular solutions are <scalar spheroidal harmonics>, with discrete $\Lambda=\Lambda_{\ell m_\phi}(a\omega)$. For $a\omega=0$ the angular equation becomes the <associated Legendre function> equation, with $\Lambda=\ell(\ell+1)$ and $\ell\geq|m_\phi|$, providing a useful check of the signs and normalization. The radial function here is exactly $R$ in the chosen ansatz, without an additional factor of $1/r$.
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