How to solve the ODE for unstable circular orbits?

Question

I'm reading Wald's book and on page 139 there is an ODE which I can't solve. $$\frac{1}{2}\dot{r}^2 + \frac{1}{2}\left(1-\frac{2M}{r}\right)\left(\frac{L^2}{r^2}+K\right) = \frac{1}{2}E^2$$ $$L = r^2 \dot{\phi}$$ I know that K = 1. For stable orbits it's really easy, we just make $\dot{r} = 0$. But what happens if it's unstable? Is there any code available to plot this?

Answer 1

To get a circular orbit of radius $r^*$ in the dynamic to $$ \frac12\dot r^2+V_L(r)=\frac12E^2, $$ or as kinetic equation $$ \ddot r+{V_L}'(r)=0, $$ one needs that $r^*$ be a stationary value of $V_L$, ${V_L}'(r^*)=0$. This binds $r^*$ and $L$ (even if not always uniquely). The stability property is then determined by the sign of ${V_L}''(r^*)$, for stability $r^*$ has to be a strict minimizer of $V_L$.

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One possible next step is to parametrize $r(t)=1/u(\phi(t))$ which results in $\dot r=Lu'$. The potential for the dynamic in $u$ is then polynomial $$ L^2\frac12u'^2+\frac12(1-2Mu)(L^2u^2+K)=\frac12E^2. $$ The derivative of the potential is $$ P_L'(u)=-M(L^2u^2+K)+(1-2Mu)L^2u=-3ML^2u^2+L^2u-MK\\ =-\frac{L^2}{12M}(6Mu-1)^2+\frac{L^2}{12M}-MK $$ with real roots for $L^2>12M^2K$ at $$ U_\pm=\frac{L\pm\sqrt{L^2-12KM^2}}{6ML} $$ As the parabola of the derivative is open below, the second derivative at the smaller root is positive, at the larger negative. The smaller $U_-$ corresponds to the larger $R_+=1/U_-$, so that indeed the circular orbit at $R_+$ is stable, as long as $R_+$ exists as real solution.

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This parametrization also leads directly to a solution of the plotting problem, the second order equation is $$ 0=u''(ϕ)+P_L'(u(ϕ))=u''-3ML^2u^2+L^2u-MK $$ This can be solved for $ϕ\in[0,n·2\pi]$, then plotted as real and imaginary part of $e^{iϕ}/u(ϕ)$. The corresponding times can be obtained from the solution of the coupled equation $t'(ϕ)=1/(Lu(ϕ)^2)$.