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Understanding Flux Vectors and Super-Radiance in General Relativity
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[QUOTE="latentcorpse, post: 3310010, member: 132470"] I'm working through p90/91 of these notes: [url]http://arxiv.org/PS_cache/gr-qc/pdf/9707/9707012v1.pdf[/url] 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 transferred 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. [/QUOTE]
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Understanding Flux Vectors and Super-Radiance in General Relativity
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