I was trying to derive equation of motion for test particle around a Kerr black hole. My work is as follows:

The Kerr metric is as follows

$$ \mathrm ds^2 = -\left(1-\dfrac{2Mr}{\rho^2}\right)\mathrm dt^2-\dfrac{4Ma}{\rho^2}\sin^2( \theta)\,\mathrm dt\,\mathrm d\phi+\dfrac{\rho^2}{\Delta}\,\mathrm dr^2+\rho^2\mathrm d\theta^2+\left((r^2-a^2)\sin^2(\theta)+\dfrac{2Mra^2}{\rho^2}\sin^4(\theta)\right)\mathrm d\phi^2$$

where \begin{align} \Delta&\triangleq r^2-2Mr+a^2\\ \rho^2&\triangleq r^2+a^2\cos^2(\theta) \end{align}

4-velocity vector is given as


and the basis killing vectors for the Kerr metric are given as follows $$\xi=(1,0,0,0)$$ $$\eta=(0,0,0,1)$$

hence, the conserved quantities are (say) $$\xi\cdot u=e$$ $$\eta\cdot u=l$$

which after little simplification is $$g_{tt}u^t+g_{t\phi} u^\phi = e$$ $$g_{\phi t}u^t+g_{\phi \phi} u^\phi = l$$

which can be solved for $u^t$ and $u^\phi$, both of them will be functions of $\theta$ and $\phi$.

Hence we have $$\boxed{u^t=F(r,\theta;e,l) \equiv F }$$ $$\boxed{u^\phi=G(r,\theta;e,l) \equiv G }$$

Third invariant quantity that can be created is by using the fact that for particles with $m\neq 0$ $u^\alpha u_\alpha =-1$

$$\Rightarrow u^\alpha u_\alpha = g_{tt}(u^t)^2+g_{rr}(u^{r})^2+g_{\phi\phi}(u^\phi)^2+g_{\theta \theta}(u^{\theta})^2+g_{t\phi}u^tu^\phi=-1$$

$$\Rightarrow g_{rr}(u^{r})^2+g_{\theta\theta}(u^\theta)^2+[g_{tt}F^2+g_{\phi\phi}G^2+g_{\phi t}FG] = -1$$

Renaming the terms in bracket as, say, $H(r,\theta; e,l) \equiv H$, hence

$$\boxed{ g_{rr}(u^{r})^2+g_{\theta\theta}(u^\theta)^2 + H = -1}$$

Now, we have got $3$ equations of motion but I don't understand how to integrate it (numerically or analytically). If we would have had $1$ more constraint involving either $r$ or $\theta$ we could have easily separated the equations and then integrated it but that it not the case as no other constraints are available (?).

So, just having these 3 equations is it possible to find the trajectory or do we require something more? If yes, then please provide what I am missing here to continue the calculation.

  • $\begingroup$ There certainly is no closed-form solution for position as a function of time in terms of elementary functions, since even the Newtonian problem has no such solution. If you'd be satisfied with a numerical solution, then why not just solve the geodesic equation directly using Runge-Kutta? Code here for the Schwarzschild spacetime: github.com/bcrowell/karl . Finding the conserved quantities can reduce the size of the system of ODEs, which would speed things up somewhat, but probably not enough to be worth all the effort. $\endgroup$
    – user4552
    Jun 23, 2018 at 12:40
  • $\begingroup$ Are you only interested in trajectories that stay outside the horizon? If you want to describe trajectories that cross the horizon, then you're going to have problems with this coordinate system, because there is a coordinate singularity. $\endgroup$
    – user4552
    Jun 23, 2018 at 12:42
  • $\begingroup$ Yes I am only interested for solution outside the horizon. I agree that geodetic equation can directly be used to calculate the trajectory of the free fall but I was trying this method as in the Schwarzschild metric since the metric is independent of $\theta$ and we have the same number and form of equations we got in this case and in that case that was enough to integrate and find closed turn solutions. So is there a way to do this this way? $\endgroup$ Jun 23, 2018 at 17:32
  • $\begingroup$ and in that case that was enough to integrate and find closed turn solutions. I don't understand. When you say "closed turn," did you mean to say "closed form?" If so, then a closed-form solution for what variable(s) in terms of what variable(s)? $\endgroup$
    – user4552
    Jun 24, 2018 at 12:29
  • $\begingroup$ Does this question still stand or have you resolved this issue? The short answer is no, these three conservation laws are not enough to determine the geodesic time-like trajectory of a test-particle in Kerr geometry.You need four conservation laws to determine the evolution of four coordinates. However, you can write down the geodesic equations in a reduced form where you would have only two differential equations with two variables. $\endgroup$ Dec 7, 2018 at 23:28


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