# Orbital Photon Speed at the equatorial plane of a rotating black hole

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?

• 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 Jun 7, 2021 at 9:04
• yes, i've been trying i use mathematica but it just can't bring it to a simpler form Jun 7, 2021 at 9:16
• 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! Jun 7, 2021 at 9:38
• thank you so much! Jun 7, 2021 at 11:03

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.