Local Kerr Hamilton-Jacobi separation

Put $A=r^2+a^2$ and $q={\sin }^2\theta$. The determinant of the $t,\varphi$ block from [F1] is $-\mathrm{\Delta }q$. Here $r>r_{+}>0$ and $r_{-}<r_{+}$, so [F1] gives $\mathrm{\Delta }=(r-r_{+})(r-r_{-})>0$ and $\mathrm{\Sigma }\ge r^2>0$. On the stated smooth chart $q>0$, so this block is invertible. By [F2] the full metric and time orientation meet the hypotheses of [F3]; the smooth Lorentzian chart convention uses [A1].

Inverting the block from step 1.1 and using the diagonal $r,\theta$ entries gives $\mathrm{\Sigma }{g}^{tt}=-A^2\mathrm{/}\mathrm{\Delta }+a^{2}q$, $\mathrm{\Sigma }{g}^{t\varphi }=-a(A\mathrm{/}\mathrm{\Delta }-1)$, $\mathrm{\Sigma }{g}^{\varphi \varphi }=1\mathrm{/}q-a^2\mathrm{/}\mathrm{\Delta }$, $g^{rr}=\mathrm{\Delta }\mathrm{/}\mathrm{\Sigma }$, and $g^{\theta \theta }=1\mathrm{/}\mathrm{\Sigma }$. Direct multiplication with the covariant block verifies these entries as required by [F3].

For the separated ansatz, ${\mathrm{\partial }}_{t}S=-E$, ${\mathrm{\partial }}_{\varphi }S=L$, ${\mathrm{\partial }}_{r}S=S_r^{\mathrm{\prime }}$, and ${\mathrm{\partial }}_{\theta }S=S_{\theta }^{\mathrm{\prime }}$. Substituting these derivatives and the inverse entries of step 2.1 into the equation in the Statement, then multiplying by $\mathrm{\Sigma }$, yields $\mathrm{\Delta }(S_r^{\mathrm{\prime }}{)}^2-\frac{[E(r^2+a^2)-aL{]}^2}{\mathrm{\Delta }}+{\mu }^2{r}^2+(S_{\theta }^{\mathrm{\prime }}{)}^2+(L\mathrm{/}\sin \theta -aE\sin \theta {)}^2+{\mu }^2{a}^2{\cos }^2\theta =0$.

Put $C(\theta ):={\cos }^2\theta [a^2({\mu }^2-E^2)+L^2\mathrm{/}{\sin }^2\theta ]$. Expanding the square and using ${\sin }^2\theta +{\cos }^2\theta =1$ gives $(S_{\theta }^{\mathrm{\prime }}{)}^2+(L\mathrm{/}\sin \theta -aE\sin \theta {)}^2+{\mu }^2{a}^2{\cos }^2\theta =(S_{\theta }^{\mathrm{\prime }}{)}^2+C(\theta )+(L-aE{)}^2$. Thus the equation of step 3.1 is the sum of a radial expression and $(S_{\theta }^{\mathrm{\prime }}{)}^2+C(\theta )$. Set the angular expression equal to the separation constant $Q$; this gives $(S_{\theta }^{\mathrm{\prime }}{)}^2=Q-C(\theta )=\mathrm{\Theta }(\theta )$. The remaining radial equation is $\mathrm{\Delta }(S_r^{\mathrm{\prime }}{)}^2-[E(r^2+a^2)-aL{]}^2\mathrm{/}\mathrm{\Delta }+{\mu }^2{r}^2+(L-aE{)}^2+Q=0$, equivalently ${\mathrm{\Delta }}^2(S_r^{\mathrm{\prime }}{)}^2=\mathcal{R}(r)$ with $\mathcal{R}$ as in the Statement.

On the stated intervals the nonnegative square roots are continuous. Choose either sign on each interval and take primitives of $\sqrt{\mathcal{R}}\mathrm{/}\mathrm{\Delta }$ and $\sqrt{\mathrm{\Theta }}$ using [F4]; since step 1.1 gives $\mathrm{\Delta }>0$, the radial integrand is continuous. The resulting $S$ is $C^1$ and satisfies the two separated equations of step 4.1, hence satisfies the Hamilton-Jacobi equation by reversing that algebra.

Where the radicands are strictly positive, repeated use of the derivative formula in [F5] shows the square-root compositions have derivatives of every order; [F6] then gives smooth integrands, and their primitives are smooth because their first derivatives are smooth.

Thus the separated form is a local solution under the stated radicand conditions. The argument makes no claim across a zero of either radicand or about global geodesic completeness.