Undergrad Components of *J in Kerr geometry

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The discussion centers on the integral of the dual current ##\int_{\mathcal{N}} \star J## over a null hypersurface in Kerr geometry, where ##J_a = -T_{ab} k^b##. Participants express confusion regarding the presence of the energy-momentum tensor ##T_{ab}## in a vacuum spacetime, as it is typically zero in Kerr geometry. The integral's formulation and its relation to the Penrose diagram are explored, suggesting that the problem may involve analyzing a "test field" on a Kerr background, which does not create curvature. Additionally, references to superradiant scattering indicate a potential approximation approach to the matter fields involved. The conversation highlights the complexities of integrating in a vacuum spacetime and the nuances of theoretical physics in this context.
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
 
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etotheipi said:
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?
 
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).
 
etotheipi said:
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 said:
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.
 
A good one to everyone. My previous post on this subject here on the forum was a fiasco. I’d like to apologize to everyone who did their best to comment and got ignored by me. In defence, I could tell you I had really little time to spend on discussion, and just overlooked the explanations that seemed irrelevant (why they seemed irrelevant, I will tell you at the end of this). Before we get to the point, I will kindly ask you to comment having considered this text carefully, because...

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