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For the Kerr metric, with line element $$ ds^2 = -\frac{\Delta-a^2\sin^2\theta}{\rho^2}dt^2 - \frac{4Mar\sin^2\theta}{\rho^2}dtd\phi +\frac{(r^2+a^2)^2-a^2\Delta\sin^2\theta}{\rho^2}\sin^2\theta d\phi^2 +\frac{\rho^2}{\Delta}dr^2+\rho^2d\theta^2 $$ in the Boyer-Linquist coordinates $$ \Delta = r^2 - 2Mr +a^2\\ \rho^2 = r^2 + a^2\cos^2\theta $$ where $M$ and $a$ are constants and rotation axis is $\theta =0$. How can I show that the world-line $$ \vec{R} = (t, r_+,\theta_0,\phi_0+\Omega t) $$ is null for $$ \Omega\equiv\Omega_+ = \frac{a}{r^2_++a^2} $$ ? Note that $r_+$ and $r_-$ are the solutions for $\Delta = 0$.

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  • $\begingroup$ Is this homework? $\endgroup$ – mmeent Jun 3 '20 at 12:13
  • $\begingroup$ Past exam question $\endgroup$ – user3613025 Jun 3 '20 at 12:20
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I think your metric is wrong. The $dtd{\phi}$ component in particular. I just tried it with the metric from my notes and it's quite simple. Just find $\frac{d\vec{R}}{d\lambda}$ and use the metric to find its magnitude

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  • $\begingroup$ I just checked again and I don't think I've copied it down wrong. $\endgroup$ – user3613025 Jun 3 '20 at 12:42
  • $\begingroup$ Ah yes, sorry. In the form I have it's written with a delta in it, but simplifies to what you have once subbing in delta. So where are you getting stuck? Have you worked out $\frac{d\vec{R}}{d\lambda}$ and subbed it in? Remember that since the worldline has $r=r_+$ you can sub that in to the metric (and so get rid of all deltas). $\endgroup$ – baker_man Jun 3 '20 at 13:59
  • $\begingroup$ Yea I've solved it but too lazy to write my own solution. It didn't occur to me that deltas would be zero. I did it by direct substitution without doing $\frac{d\vec{R}}{d\lambda}$(How would u do that?).Anyway I'll just accept your answer. You can edit to add stuff in if u want. $\endgroup$ – user3613025 Jun 3 '20 at 14:02
  • $\begingroup$ Thanks, laziness is the same reason I didn't type out my solution on here. That and I think it's frowned upon here to write out solutions to homework style questions (even though this is a past paper). $\endgroup$ – baker_man Jun 3 '20 at 14:04
  • $\begingroup$ We don't get solutions for our past papers. So we wouldn't know if we've calculated something right or wrong. $\endgroup$ – user3613025 Jun 3 '20 at 14:48

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