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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 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 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 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 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 at 17:00

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