Ganesh Ujwal Posted December 20, 2014 Share Posted December 20, 2014 I'm going through Kerr metric, and following the 'Relativist's toolkit' derivation of the surface gravity, I've come to a part that I don't understand. Firstly, the metric is given by [latex]\mathrm{d}s^2=\left(\frac{\Sigma}{\rho^2}\sin^2\theta\omega^2-\frac{\rho^2\Delta}{\Sigma}\right)\mathrm{d}t^2-2\frac{\Sigma}{\rho^2}\sin^2\theta\omega \mathrm{d}\phi \mathrm{d}t+\frac{\Sigma}{\rho^2}\sin^2\theta \mathrm{d}\phi^2+\frac{\rho^2}{\Delta}\mathrm{d}r^2+\rho^2 \mathrm{d}\theta^2[/latex] With [latex]\rho^2=r^2+a^2\cos^2\theta,\quad \Delta=r^2-2Mr+a^2[/latex], [latex]\Sigma=(r^2+a^2)^2-a^2\Delta\sin^2\theta,\quad \omega=\frac{2Mar}{\Sigma}[/latex] The Killing vector that is null at the event horizon is [latex]\chi^\mu=\partial_t+\Omega_H\partial_\phi[/latex] where[latex] \Omega_H[/latex] is angular velocity at the horizon. Now I got the same norm of the Killing vector [latex]\chi^\mu\chi_\mu=g_{\mu\nu}\chi^\mu\chi^\nu=\frac{\Sigma}{\rho^2}\sin^2\theta(\Omega_H-\omega)^2-\frac{\rho^2\Delta}{\Sigma} [/latex] And now I should use this equation [latex]\nabla_\nu(-\chi^\mu\chi_\mu)=2\kappa\chi_\nu[/latex] And I need to look at the horizon. Now, on the horizon [latex]\omega=\Omega_H[/latex] so my first term in the norm is zero, but, on the horizon [latex]\Delta=0[/latex] too, so how are they deriving that side, and how did they get [latex]\nabla_\nu(-\chi^\mu\chi_\mu)=\frac{\rho^2}{\Sigma}\nabla_\nu\Delta[/latex] if the [latex]\Delta=0[/latex] on the horizon? Since [latex]\rho[/latex]and [latex]\Sigma[/latex] both depend on [latex]r[/latex], and even if I evaluate them at [latex]r_+=M+\sqrt{M^2-a^2}[/latex] they don't cancel each other. How do they get to the end result of [latex]\kappa[/latex]? Link to comment Share on other sites More sharing options...
xyzt Posted December 23, 2014 Share Posted December 23, 2014 I'm going through Kerr metric, and following the 'Relativist's toolkit' derivation of the surface gravity, I've come to a part that I don't understand. Firstly, the metric is given by [latex]\mathrm{d}s^2=\left(\frac{\Sigma}{\rho^2}\sin^2\theta\omega^2-\frac{\rho^2\Delta}{\Sigma}\right)\mathrm{d}t^2-2\frac{\Sigma}{\rho^2}\sin^2\theta\omega \mathrm{d}\phi \mathrm{d}t+\frac{\Sigma}{\rho^2}\sin^2\theta \mathrm{d}\phi^2+\frac{\rho^2}{\Delta}\mathrm{d}r^2+\rho^2 \mathrm{d}\theta^2[/latex] With [latex]\rho^2=r^2+a^2\cos^2\theta,\quad \Delta=r^2-2Mr+a^2[/latex], [latex]\Sigma=(r^2+a^2)^2-a^2\Delta\sin^2\theta,\quad \omega=\frac{2Mar}{\Sigma}[/latex] The Killing vector that is null at the event horizon is [latex]\chi^\mu=\partial_t+\Omega_H\partial_\phi[/latex] where[latex] \Omega_H[/latex] is angular velocity at the horizon. Now I got the same norm of the Killing vector [latex]\chi^\mu\chi_\mu=g_{\mu\nu}\chi^\mu\chi^\nu=\frac{\Sigma}{\rho^2}\sin^2\theta(\Omega_H-\omega)^2-\frac{\rho^2\Delta}{\Sigma}[/latex] And now I should use this equation [latex]\nabla_\nu(-\chi^\mu\chi_\mu)=2\kappa\chi_\nu[/latex] And I need to look at the horizon. Now, on the horizon [latex]\omega=\Omega_H[/latex] so my first term in the norm is zero, but, on the horizon [latex]\Delta=0[/latex] too, so how are they deriving that side, and how did they get [latex]\nabla_\nu(-\chi^\mu\chi_\mu)=\frac{\rho^2}{\Sigma}\nabla_\nu\Delta[/latex] if the [latex]\Delta=0[/latex] on the horizon? Since [latex]\rho[/latex]and [latex]\Sigma[/latex] both depend on [latex]r[/latex], and even if I evaluate them at [latex]r_+=M+\sqrt{M^2-a^2}[/latex] they don't cancel each other. How do they get to the end result of [latex]\kappa[/latex]? What you need is to FIRST calculate the derivative and THEN to apply the boundary conditions. You are doing the steps in reverse order. 1 Link to comment Share on other sites More sharing options...
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