I Components of *J in Kerr geometry

[etotheipi]

I am in the middle of a problem for the Kerr geometry, I need to do the integral $\int_{\mathcal{N}} \star J$ over a null hypersurface $\mathcal{N}$ which is a subset of $\mathcal{H}^+$, where $J_a = -T_{ab} k^b$ and the orientation on $\mathcal{N}$ is $dv \wedge d\theta \wedge d\chi$ so that $\int_{\mathcal{N}} \star J = \int_{\phi{(\mathcal{N})}} dv d\theta d\chi (\star J)_{v\theta \chi}$. It's supposed to be that $(\star J)_{v\theta \chi} = (r_+^2 + a^2)\sin\theta \xi^a J_a$, but how do you get this? I tried to work backward from this to $(\star J)_{v\theta \chi} = \dfrac{1}{3!} g^{ba} \epsilon_{v\theta \chi b} J_a$ but not successfully. I had thought that maybe from the Rayachudri equation with $\hat{\sigma} = \hat{\omega} = 0$ that \begin{align*}
0 = R_{ab} \xi^a \xi^b \vert_{\mathcal{H}+} = 8\pi T_{ab} \xi^a \xi^b \vert_{\mathcal{H}+} &= 8\pi T_{ab} \xi^a \left(k^b + \dfrac{a}{r_+^2 + a^2} m^b \right) \vert_{\mathcal{H}+} \\

0 &= \left( -8\pi \xi^a J_a + \dfrac{a}{r_+^2 + a^2} 8\pi T_{ab} m^b \right) \vert_{\mathcal{H}+}
\end{align*}so that $(r_+^2 + a^2) \sin{\theta} \xi^a J_a \vert_{\mathcal{H}+} = a \sin{\theta} T_{ab} m^b \vert_{\mathcal{H}+}$. But now I don't know what to do with $T_{ab} m^b$? Thanks

 

[PeterDonis]

I need to do the integral $\int_{\mathcal{N}} \star J$ over a null hypersurface $\mathcal{N}$ which is a subset of $\mathcal{H}^+$, where $J_a = -T_{ab} k^b$

Kerr spacetime is a vacuum spacetime, so $T_{ab} = 0$ everywhere. So this doesn't make sense.

Where is this problem coming from?

 

[etotheipi]

It is question 6: https://www.damtp.cam.ac.uk/user/examples/3R3c.pdf. For the first part I already wrote that since Penrose diagram would show two lines representing $\Sigma$ and $\Sigma'$ starting at $i_0$ and meeting $\mathcal{H}^+$ in the 2-spheres $H$ and $H'$, and because on the diagram the subset of $\mathcal{H}^+$ connecting $H$ and $H'$ represents $\mathcal{N}$, the hypersurfaces $\Sigma$, $\Sigma'$ and $\mathcal{N}$ bound a spacetime region $R$, so\begin{align*}E(\Sigma) - E(\Sigma') + E(\mathcal{N}) = - \int_{\partial R} \star J = - \int_R d \star J = 0 \\\end{align*}and so $E(\Sigma) - E(\Sigma') = -E(\mathcal{N}) = \int_{\mathcal{N}} \star J$. I'm not completely sure that's right, but it seems reasonable. And for (b) the orientation is fixed by Stokes. But I am totally stuck on (c).

 

[PeterDonis]

It is question 6: https://www.damtp.cam.ac.uk/user/examples/3R3c.pdf.

Hm. The question still doesn't make sense to me, since, as I said, Kerr spacetime is a vacuum spacetime, so $T_{ab} = 0$ everywhere, but the question is talking about "matter fields". Perhaps it is talking about some kind of approximation where the behavior of a matter field is being analyzed on a background Kerr spacetime, where the matter field is considered a "test field" which doesn't produce any spacetime curvature on its own.

 

[PeterDonis]

Perhaps it is talking about some kind of approximation where the behavior of a matter field is being analyzed on a background Kerr spacetime, where the matter field is considered a "test field" which doesn't produce any spacetime curvature on its own.

The reference in part (e) to superradiant scattering seems to bear this out, since other treatments of superradiance, such as the one in MTW, take a similar approach.