Time-averaged stationary flux of a Kerr horizon mode

Since $\chi$ is normal to the null hypersurface $H^{+}$ and both $T$ and $\mathrm{\Phi }$ are tangent to it by [F3], $g(\chi ,T)=g(\chi ,\mathrm{\Phi })=0$ on $H^{+}$. Hence the metric-trace term in [F2] vanishes when contracted with ${\chi }^a{T}^b$, so $Q(\chi ,T)=(\chi \psi )(T\psi )$.

Fix $p\in P_{\omega ,T}$ and put $A:=u(p)$. Along its stationary integral-curve segment, $\frac{d}{d\tau }u({\gamma }_p(\tau ))=(Tu)({\gamma }_p(\tau ))=-i\omega u({\gamma }_p(\tau ))$ by [F4], so $u({\gamma }_p(\tau ))=A{e}^{-i\omega \tau }$. The mode equations [F1] give $T\psi =\mathrm{Re}(-i\omega u)$ and $\chi \psi =\mathrm{Re}(-i(\omega -m{\mathrm{\Omega }}_H)u)$.

If $A=0$, both differentiated real fields vanish along this orbit and the displayed average identity is $0=0$. Otherwise put $\vartheta =\arg A-\omega \tau$. Then $T\psi =\omega \mathrm{\mid }A\mathrm{\mid }\sin \vartheta$ and $\chi \psi =(\omega -m{\mathrm{\Omega }}_H)\mathrm{\mid }A\mathrm{\mid }\sin \vartheta$. Averaging their product over $\tau \in [0,2\pi \mathrm{/}\omega ]$ uses $\frac{\omega }{2\pi }{\int }_0^{2\pi \mathrm{/}\omega }{\sin }^2(\arg A-\omega \tau )d\tau =\frac{1}{2}$, and gives the displayed identity.

If $m{\mathrm{\Omega }}_H>\omega >0$, the coefficient $\omega (\omega -m{\mathrm{\Omega }}_H)\mathrm{/}2$ is strictly negative. Since $\mathrm{\mid }u{\mathrm{\mid }}^2\ge 0$, step 3.1 gives a nonpositive averaged density, with equality exactly at $u=0$ and strict negativity exactly at $u\mathrm{\ne }0$. A mode that is not identically zero need not be nonvanishing at every point.

On the compact patch $D$, smoothness bounds $\mathrm{\mid }u{\mathrm{\mid }}^2$ and the induced area density, so both displayed integrals are finite. Integrating step 3.1 gives the formula for ${\mathcal{F}}_D$. If $u{\mathrm{\mid }}_D$ is not identically zero, continuity and density of the interior give an interior point where $\mathrm{\mid }u{\mathrm{\mid }}^2>0$ and a smaller coordinate disk where $\mathrm{\mid }u{\mathrm{\mid }}^2\ge b>0$. The positive smooth area density has a positive lower bound $a>0$ on a still smaller closed disk; its positive Euclidean area makes ${\int }_D\mathrm{\mid }u{\mathrm{\mid }}^{2}d{\mu }_D\ge ab\mathrm{area}(\text{disk})>0$. Step 4.1 then makes ${\mathcal{F}}_D<0$. To find a patch near a point $p\in P_{\omega ,T}$ with $u(p)\mathrm{\ne }0$, cover the compact integral-curve segment ${\gamma }_p([0,2\pi \mathrm{/}\omega ])$ by finitely many neighborhoods supplied by [F6] and subdivide the parameter interval so each piece lies in one such neighborhood. Composing the corresponding local solution maps gives smooth dependence on initial data through the whole interval. Since ${\gamma }_p([0,2\pi \mathrm{/}\omega ])\subset P$ and $P$ is open, shrink the initial neighborhood of $p$ so all these curves remain in $P$; nearby points then also lie in $P_{\omega ,T}$. By [F3], $T_p{H}^{+}={\chi }_p^{\perp }$ for a nonzero null normal ${\chi }_p$. Nondegeneracy gives $v$ with $g({\chi }_p,v)=-1$; then $k=v+\frac{1}{2}g(v,v){\chi }_p$ is null and $g({\chi }_p,k)=-1$. The plane spanned by ${\chi }_p,k$ has one negative and one positive direction, so its orthogonal complement is a positive two-plane in $T_p{H}^{+}$. Choose a smooth hypersurface chart whose first two coordinate vectors span that plane at $p$, fix its third coordinate, and shrink a closed coordinate disk about $p$. Positivity persists by continuity of its metric matrix, and the disk can be chosen inside the neighborhood and the nonzero set. No integral over a full noncompact horizon is used.