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Let there exist a Kerr metric $ds^2$ in Boyer–Lindquist coordinates such that $$ds^2=-\left(1-\frac{r_sr}{\Sigma}\right)dt^2+\left(\frac{\Sigma}{\Delta}\right)dr^2+\Sigma d \theta^2+\left(r^2+a^2+\frac{r_sra^2}{\Sigma}\sin^2(\theta)\right)\sin^2(\theta)d \phi^2-\left(\frac{2r_sra\sin^2(\theta)}{\Sigma}\right)dt d \phi,$$ such that $$a=\frac{J}{Mc}$$ $$\Sigma=r^2+a^2\cos^2(\theta)$$ $$\Delta=r^2-r_sr+a^2$$ $$r_s=2M$$ $$L=\frac{1}{2}\left[-\left(1-\frac{r_sr}{\Sigma}\right)t'^2+\left(\frac{\Sigma}{\Delta}\right)r'^2+\Sigma \theta'^2+\left(r^2+a^2+\frac{r_sra^2}{\Sigma}\sin^2(\theta)\right)\sin^2(\theta)\phi'^2-\left(\frac{2r_sra\sin^2(\theta)}{\Sigma}\right)t' \phi'\right]$$ where we are letting $$c=G=1$$hence the equation of $ds^2$. To find the geodesics of the Kerr metric $ds^2$ we need to apply the Euler-Lagrange equations $$ \frac{\partial L}{\partial x^\mu}-\frac{d}{d \tau}\left(\frac{\partial L}{\partial x'^\mu}\right)=0$$ where $$x^\mu=(x^0,x^1,x^2,x^3)=(t,r,\theta,\phi)$$ and $$\forall x'^\mu=\frac{d x^\mu}{d \tau}.$$ In order to obtain the system of ODE geodesic equations for the Kerr metric will I just solve $$\frac{\partial L}{\partial t},\frac{\partial L}{\partial t'},\frac{\partial L}{\partial r},\frac{\partial L}{\partial r'},\frac{\partial L}{\partial \theta},\frac{\partial L}{\partial \theta'},\frac{\partial L}{\partial \phi}, \frac{\partial L}{\partial \phi'}$$ then just plug them into the equations $$\frac{\partial L}{\partial t}-\frac{d}{d \tau}\left(\frac{\partial L}{\partial t'}\right)=0$$ $$\frac{\partial L}{\partial r}-\frac{d}{d \tau}\left(\frac{\partial L}{\partial r'}\right)=0$$ $$\frac{\partial L}{\partial \theta}-\frac{d}{d \tau}\left(\frac{\partial L}{\partial \theta'}\right)=0$$ $$\frac{\partial L}{\partial \phi}-\frac{d}{d \tau}\left(\frac{\partial L}{\partial \phi'}\right)=0$$ and numerically solve or is there something I am missing?

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    $\begingroup$ This might not directly address your question but these sources may help you: adsabs.harvard.edu/full/1987SvAL...13...99S en.wikipedia.org/wiki/Kerr_metric#Trajectory_equations $\endgroup$
    – Alwin
    Nov 17, 2021 at 4:47
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    $\begingroup$ You can do it without using symmetries of the spacetime, but unless you take steps to avoid it, in practice you will suffer what is known as "energy drift"; typically an orbiting solution will tend to spiral inwards over time. A symplectic integrator will not suffer this, but is extremely non-trivial to write in this case as the equations are not "separable". If you use the trajectory equations that @Alwin mentioned, it is possible to generate a "fake" hamiltonian to solve using a more basic symplectic integrator. I have used this technique very successfully in my personal projects. $\endgroup$
    – m4r35n357
    Nov 17, 2021 at 10:24
  • $\begingroup$ Incidentally, if you try to use the trajectory equations "as-is" you will get stuck dealing with the square roots! $\endgroup$
    – m4r35n357
    Nov 17, 2021 at 10:31
  • $\begingroup$ @m4r35n357 Wouldnt we arive at the trajectory equations by just applying the Euler-Lagrange equations on the Lagrangian $L$ and then using constants to manipulate the results? $\endgroup$
    – aygx
    Nov 17, 2021 at 11:40
  • $\begingroup$ If I understand you, then yes. But those constants need to be derived somehow! This is done using Hamilton-Jacobi analysis, which is beyond my skills, so I only speak here of using the equations. $\endgroup$
    – m4r35n357
    Nov 17, 2021 at 11:44

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