# How to show if a world line is null for a particular metric [closed]

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 $$\Delta = 0$$.

• Is this homework? – mmeent Jun 3 '20 at 12:13
• Past exam question – user3613025 Jun 3 '20 at 12:20

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
• 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). – baker_man Jun 3 '20 at 13:59
• 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. – user3613025 Jun 3 '20 at 14:02