Understanding Flux Vectors and Super-Radiance in General Relativity

[latentcorpse]

I'm working through p90/91 of these notes:
http://arxiv.org/PS_cache/gr-qc/pdf/9707/9707012v1.pdf

In 4.58, he introduces this $j^\mu$ flux 4 vector.
(i) Where does this come from? Should I just accept it?

(ii) He says it's future directed since $-k \cdot j>0$. I can't see why this inequality is true or why it implies future directed?

(iii)He then gets 4.59 using the divergence theorem for manifold integration
$\int_M d^nx \sqrt{-g} \nabla_a X^a = \int_{\partial M} d^{n-1}x \sqrt{|h|} n_a X^a$
So I'm assuming that his measure $dS_\mu = d^{n-1}x \sqrt{|h|} n_\mu$, yes?

(iv) Then in 4.60, why do the second two terms have minus signs? I know it's to do with the way we integrate around that surface and the direction of the normals but I can't make sense of it.

(v)Why in 4.62 is the energy through the horizon $E_1-E_2$? Is it basically saying that (looking at the diagram) we have some field below $\Sigma_1$ which enters the enclosed region, interacts and loses some energy such that when it comes out of $\Sigma_2$ it has energy $E_1-E_2$ that can be transferred through the horizon?

(vi)In 4.64, is he missing a $dv$ in the measure?

(vii) Also in 4.64, how does $\xi_\mu j^\mu = ( \xi \cdot \partial \Phi)( k \cdot D \Phi)$

(viii) I also cannot show 4.67. I find:

$P= \int dA \left( \frac{\Phi_0}{\omega} \sin{( \omega v - \nu \chi)} - \frac{\Omega_H \Phi_0}{\nu} \sin{(\omega v - \nu \chi)} \right) \frac{\phi_0}{\omega} \sin{(\omega v - \nu \chi)}$
But don't know how to proceed?

(ix) Also, does this whole super-radiance thing happen only in the ergoregion?

Thanks.

 

[member 553083]

(i) The j^μ flux 4-vector is derived from the conservation of energy-momentum, which states that the divergence of the stress-energy tensor is equal to zero. This means that there must be some flux of energy-momentum in and out of any volume element of space-time. j^μ is the 4-vector that represents this flux. (ii) -k⋅j>0 implies that the flux is future-directed because k is a future-directed vector. This follows from the fact that the dot product of two vectors is positive when they point in the same direction.(iii) Yes, his measure dSμ=d^n−1x√|h|nμ.(iv) The second two terms have minus signs because they represent outgoing flux through the boundary. The outward normal at a boundary is always pointing away from the enclosed region, so the sign of the flux term has to be negative for it to represent outgoing flux.(v) Yes, the energy through the horizon E1−E2 is the difference between the energy of the field entering the region through Σ1 and the energy of the field leaving the region through Σ2.(vi) No, he is not missing a dv in the measure.(vii) \xi_\mu j^\mu = ( \xi \cdot \partial \Phi)( k \cdot D \Phi) follows from the definition of the flux vector j^μ. (viii) To obtain 4.67, you need to use the identities cos(A+B)=cos(A)cos(B)−sin(A)sin(B) and sin(A+B)=sin(A)cos(B)+cos(A)sin(B).(ix) Super-radiance can occur in any region where there is an ergoregion, i.e. a region of spacetime where the Killing vector field associated with time translation is timelike.