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In $ x^{\mu}= \{t,\rho,\theta,\phi\}$Boyer-Lindquist coordinates the Kerr solution takes a manageable form. Given some initial conditions $x^{\mu}(\lambda=0)$ and $\dfrac{dx^{\mu}}{d\lambda}(\lambda=0)$ one can numerically solve the geodesic equations for the parametrised curve $x^{\mu}(\lambda)$. My question is, how would one try to plot and animate the curves in some simple plot3d function? I have seen people “convert” the B-L coordinates to euclidean $x,y,z$ coordinates and animate the curve via the affine parameter $\lambda$.

This seems wrong to me, as the transformation $$x = \sqrt{\rho^2+a^2} \sin\theta \cos\phi$$ $$y = \sqrt{\rho^2+a^2} \sin\theta \sin\phi$$ $$ z =\rho \cos\phi$$ is only valid in the $m=0$ limit of Kerr, while the geodesics were calculated with some nonzero mass.

Obviously one can NOT globally establish euclidean flat coordinates on a curved Kerr manifold, so this scheme makes no sense from a theoretical view. What is the proper way of doing such a plot?

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  • $\begingroup$ Asking for "the" proper way of doing it seems misguided, for the reasons you state. It's like asking "what is the proper way to project the globe onto a piece of paper": there are some better ways and some worse ways, but each one has its advantages and its drawbacks. $\endgroup$ Jun 1, 2022 at 16:11
  • $\begingroup$ What are some reasonable ways of doing it? I have seen both the one above, as well as a similar one with the omission of the rotational parameter $a$. Which would be more adequate for, say, an observer at infinity? $\endgroup$
    – Johnny
    Jun 1, 2022 at 16:18
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    $\begingroup$ If you want to know what the observer sees with his eyes you have to raytrace, but if you want to project trajectories on a flat background the transformation you have is fine, there is no better one for that. It is not only valid for M=0, a is in units of M. That's just the transformation from pseudospherical to cartesian, for the true distances you have to integrate the metric tensor. The representation you quoted gives proper circumference along θ, but not around φ (you can't have both at the same time in noneuclidean spacetime) $\endgroup$
    – Yukterez
    Jun 1, 2022 at 16:47
  • $\begingroup$ Thanks @Yukterez ! It was actually your animations (notizblock.yukterez.net/viewtopic.php?t=81) that inspired my project! $\endgroup$
    – Johnny
    Jun 1, 2022 at 16:52
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    $\begingroup$ If you just transform the r,θ,φ to x,y,z as if were truly spherical coordinates (like setting a=0 in the transformation while having it nonzero on the black hole) the ring singularity would look like a point, while if you use the BH's a in the transformation its radius r=0→R=√(r²+a²)=a is the proper one, see physics.stackexchange.com/questions/471419/… $\endgroup$
    – Yukterez
    Jun 1, 2022 at 17:00

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Simply solving the geodesic equations with a "standard" integrator (e.g. RK4 and friends) suffers badly from energy drift in my experience. To get an accurate and time-stable solution you might be interested in my numerical solution here, or you could use the explicit equations here, if you understand them (I don't!).

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  • $\begingroup$ We used ‘scipy.integrate’ ’s adaptive RK45 and all 4 of the conserveved quantities (imgur.com/a/yfgzkJa) were conserved (up to numerical precision ) during a generic geodesic solution. $\endgroup$
    – Johnny
    Nov 21, 2022 at 13:35
  • $\begingroup$ Well I suppose that depends on how long the simulation is ;) I certainly encountered the problem myself with non-adaptive RK4, and the symplectic integrator solved it. $\endgroup$
    – m4r35n357
    Nov 21, 2022 at 16:16
  • $\begingroup$ @Johnny on closer inspection, that logarithmic scale makes things look better than they are - the error growth is linear, and will eventually become significant (BTW I use 80-bit or 128-bit floats so I would see it sooner!). Also, with the Wilkins equations, E, L and Q are constant by definition, so there can be no errors at all in these quantities. Not trying to convert you, just explaining my position and claims above. $\endgroup$
    – m4r35n357
    Nov 21, 2022 at 16:54
  • $\begingroup$ That’s true, but we specifically wanted to implement the full 4+4 system instead of imposing the conserved quantities. It served as a decent check of our numerics. $\endgroup$
    – Johnny
    Nov 21, 2022 at 19:52
  • $\begingroup$ Fair enough. I wanted the non-symplectic integration to work because at one time I was writing a "generic" geodesic equation solver, where you implement the metric with dual numbers, and generate the symbols using automatic differentiation, en.wikipedia.org/wiki/Dual_number#Differentiation. Energy drift spoiled that! $\endgroup$
    – m4r35n357
    Nov 22, 2022 at 9:45

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