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I've been trying to calculate $d\phi/dt$ of photons orbiting a Kerr black hole (Kerr metric in Boyer-Lindquist coordinates) on the equatorial plane, both counter and along with its rotation. So I used that $u_\mu u^\mu=0$, which applies for the four-velocities of photons and, because I wanted a circular orbit, I said $r = \mathrm{const}$. The same goes for $\theta$ (we have one specific plane, $\theta=\pi/2$)

My problem is that I reach a result expressed in $$g_{ab}$$ which I can't seem to bring to a simpler form (for example with $\Delta$):

$$ \frac{d\phi }{dt}_\mathrm{with} = \frac{\sqrt{g_{t\phi }^{2}-g_{tt}g_{\phi \phi }} - g_{t\phi }}{g_{\phi \phi }} $$ $$ \frac{d\phi }{dt}_\mathrm{counter} = -\frac{\sqrt{g_{t\phi }^{2}-g_{tt}g_{\phi \phi }} - g_{t\phi }}{g_{\phi \phi }}$$

Is my approach completely wrong or am I missing something? Should I use l and e instead?

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  • $\begingroup$ Have you actually tried substituting the BL metric elements into those equations? You seem to have stopped short! This is probably something you want to do in a computer algebra package, like wxMaxima wxmaxima-developers.github.io/wxmaxima/index.html $\endgroup$
    – m4r35n357
    Jun 7, 2021 at 9:04
  • $\begingroup$ yes, i've been trying i use mathematica but it just can't bring it to a simpler form $\endgroup$ Jun 7, 2021 at 9:16
  • $\begingroup$ OK, just checking, been there done that - that is pretty much what it is! If you keep delta unexpanded you might miss simplifications, but if you expand delta, you end up with a mess. The CAS will help you avoid errors, but rarely gives you an answer in the form you want without a lot of meddling. Good luck! $\endgroup$
    – m4r35n357
    Jun 7, 2021 at 9:38
  • $\begingroup$ thank you so much! $\endgroup$ Jun 7, 2021 at 11:03

1 Answer 1

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You are right and you only need to develop the calculations with BL values and obtain the simplifications.

You have $$ g_{\phi\phi}(\frac{d\phi}{dt})^2+2g_{t\phi}\frac{d\phi}{dt}+g_{tt}=0\ \ \ \ \ \ [1] $$

With $ m=\frac{GM}{c^2} $, and:

$ g_{\phi\phi}=r^2+a^2+\frac{2ma^2}{r} $

$ g_{t\phi}=-\frac{2ma}{r}c $

$ g_{tt}=-\left (1-\frac{2m}{r}\right )c^2 $

By multiplying $ [1] $ by $ r^2 $, you have: $$ (r^4+a^2r^2+2ma^2r)(\frac{d\phi}{dt})^2-4marc\frac{d\phi}{dt}-(r^2-2mr)c^2 = 0\ \ \ \ \ \ [2] $$ After some calculations and simplifications, the discriminant of $ [2] $ is written as:

$ 4c^2r^2(r^2-2mr+a^2) $ or $ 4c^2r^2\Delta $ with $ \Delta=r^2-2mr+a^2 $.

Then the two solutions for $ \frac{d\phi}{dt} $ are:

$$ \frac{d\phi}{dt}_{with}=c\frac{2ma+r\sqrt{\Delta}}{r^3+a^2r+2ma^2} $$ and $$ \frac{d\phi}{dt}_{counter}=c\frac{2ma-r\sqrt{\Delta}}{r^3+a^2r+2ma^2} $$

Hoping to have answered the question,

Best regards.

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