Super-Radiance homework

  • #1
1,444
0
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 [itex]j^\mu[/itex] flux 4 vector.
(i) Where does this come from? Should I just accept it?

(ii) He says it's future directed since [itex]-k \cdot j>0[/itex]. 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
[itex]\int_M d^nx \sqrt{-g} \nabla_a X^a = \int_{\partial M} d^{n-1}x \sqrt{|h|} n_a X^a[/itex]
So I'm assuming that his measure [itex]dS_\mu = d^{n-1}x \sqrt{|h|} n_\mu[/itex], 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 [itex]E_1-E_2[/itex]? Is it basically saying that (looking at the diagram) we have some field below [itex]\Sigma_1[/itex] which enters the enclosed region, interacts and loses some energy such that when it comes out of [itex]\Sigma_2[/itex] it has energy [itex]E_1-E_2[/itex] that can be transfered through the horizon?

(vi)In 4.64, is he missing a [itex]dv[/itex] in the measure?

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

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

[itex]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)}[/itex]
But don't know how to proceed?

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

Thanks.
 
Last edited:

Answers and Replies

Related Threads on Super-Radiance homework

  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
9K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
0
Views
927
  • Last Post
Replies
0
Views
671
  • Last Post
Replies
1
Views
2K
Top