# Insights Orbital Precession in the Schwarzschild and Kerr Metrics - Comments

Tags:
1. Aug 11, 2016

2. Sep 3, 2016

### john baez

Nice post! There's a typo:

I guess you need something like two pound signs to do math here, not a dollar sign. Also, there are a number of bad line breaks - Wordpress blogs are unforgiving when you hit the carriage return.

What I'd really love to see is a study of orbits in a Kerr metric that describes a 'super-extremal' black hole, one with $J^2 > M^2$. These could get pretty weird!

3. Sep 4, 2016

### Dr.AbeNikIanEdL

Thanks for the Post!

Could someone explain where the conclusion about the photon orbits come from, i.e. that they exist for vanishing denominators of $L$ and $\Gamma$?

4. Sep 4, 2016

### m4r35n357

Good idea! I went away and commented out the horizon calculations in my own minimal simulator (I started a thread for it but struggling to find it ATM!), and tried a fairly complex bound 3+1D orbit with a = 1.1 and a = 1.5. Both produced "valid" orbits. It seems on the face of it that the equations of motion do not care about the existence of horizons ;)

For info, the equations I use are from this paper for the Kerr-deSitter spacetime, and previously the seminal Wilkins paper. Both are full 3+1D solutions based on Hamilton-Jacobi analysis.

5. Sep 4, 2016

### john baez

Thanks! You may know about this article; it's not about the geodesics just the geometry of the Kerr solution, with a reasonable amount of detail on the 'ring singularity' and the closed timelike curves it gives rise to - those were my main concerns this weekend:
I just noticed this:
You may know all this stuff, but it seems nice.

6. Sep 5, 2016

### m4r35n357

The first one is new to me (reading now, thanks), but I have seen the latter, very nice derivations for 2+1D and 3+1D. I wonder why the authors stopped short of providing equations for $\frac {d t} {d \tau}$ and $\frac {d \phi} {d \tau}$ though.

This is an issue ATM because I think I have found a disagreement between Wilkins and Kraniotis et. al. in the $\Theta$ potential . . . my eyes say so and so does Maxima.

[EDIT] pardon this but I've just showed in Maxima that they are equal after all. Since the Kerr-deSitter $\Theta$ potential is simpler, I shall use it for Kerr ($\Lambda = 0$) simulations in preference to the form in Wilkins.

Last edited: Sep 5, 2016
7. Sep 18, 2016

### Markus Hanke

Very nicely, clearly, and succinctly presented - thank you for this :)