(relativity-general:notes:examples:schwarzschild-orbits)= # Planetary orbits in Schwarzschild spacetime Schwarzschild metrics {eq}`eq:schwarzschild-metrics` provides the metrics induced by a massive object in the vacuum spacetime around. The motion of a light object approximately doesn't affect the massive object generating the steady Schwarzschild metrics. The light object moves following the geodesics. Let the motion occurs in the equatorial plane, i.e. $\theta = \frac{\pi}{2}$, and $d \theta = 0$. Thus $$ - c^2 d \tau^2 = ds^2 = - \left(1 - \frac{r_s}{r} \right) c^2 dt^2 + \left(1 - \frac{r_s}{r} \right)^{-1} dr^2 + r^2 \, d\phi^2 \ . $$ **Geodesics.** With $A(r) = \frac{1}{2} \ln \left( 1 - \frac{r_s}{r} \right)$, $A'(r) = \frac{1}{2} \frac{1}{1 - \frac{r_s}{r}} \frac{r_s}{r^2}$ $$0 = \ddot{q}^k + \Gamma_{a b}^k \dot{q}^{a} \dot{q}^b$$ $$\begin{aligned} t : \qquad 0 & = \ddot{t} + \Gamma_{ab}^t \dot{q}^a \dot{q}^b = \\ & = \ddot{t} + 2 \Gamma_{tr}^t \dot{t} \dot{r} = \\ & = \ddot{t} + \frac{1}{1 - \frac{r_s}{r}} \frac{r_s}{r^2} \dot{t} \dot{r} \\ 0 & = \left(1 - \frac{r_s}{r} \right)^{-1} \left[ \left( 1-\frac{r_s}{r}\right) \ddot{t} + \frac{r_s}{r^2} \dot{t} \dot{r} \right] = \\ 0 & = \left(1 - \frac{r_s}{r} \right)^{-1} \dfrac{d}{d \tau} \left[ \left( 1-\frac{r_s}{r} \right) \dot{t} \right] \quad \rightarrow \quad \left( 1-\frac{r_s}{r} \right) \dot{t} =: \frac{E}{c} = \text{const.} \\ r : \qquad 0 & = \ddot{r} + \Gamma_{ab}^r \dot{q}^a \dot{q}^b = \\ & = \ddot{r} + \Gamma_{tt}^r \dot{t}^2 + \Gamma_{rr}^r \dot{r}^2 + \Gamma_{\theta \theta}^{r} \dot{\theta}^2 + \Gamma_{\phi \phi}^{r} \dot{\phi}^2 = && (\dot{\theta} = 0 )\\ & = \ddot{r} + \Gamma_{tt}^r \dot{t}^2 + \Gamma_{rr}^r \dot{r}^2 + \Gamma_{\phi \phi}^{r} \dot{\phi}^2 = \\ & = \ddot{r} + A' e^{4 A} c^2 \dot{t}^2 - A'(r) \dot{r}^2 - r \sin^2 \theta e^{2 A} \dot{\phi}^2 = \\ & = \ddot{r} + A' e^{4 A} c^2 \dot{t}^2 - A'(r) \dot{r}^2 - r \theta e^{2 A} \dot{\phi}^2 = \\ \phi : \qquad 0 & = \ddot{\phi} + \Gamma_{ab}^{\phi} \dot{q}^a \dot{q}^b \\ & = \ddot{\phi} + 2 \Gamma_{r \phi}^\phi \dot{r} \dot{\phi} + 2 \Gamma_{\theta \phi}^\phi \dot{\theta} \dot{\phi} = && ( \dot{\theta} = 0 ) \\ & = \ddot{\phi} + 2 \Gamma_{r \phi}^\phi \dot{r} \dot{\phi} = \\ & = \ddot{\phi} + 2 \frac{1}{r} \dot{r} \dot{\phi} \ , \\ 0 & = \frac{1}{r^2} \dfrac{d}{d \tau} \left( r^2 \dot{\phi} \right) \quad \rightarrow \quad r^2(\tau) \dot{\phi}(\tau) =: L = \text{const.} \\ \theta : \qquad 0 & = \ddot{\theta} + \Gamma_{ab}^{\theta} \dot{q}^a \dot{q}^b \\ & = \ddot{\theta} + 2 \Gamma_{r \phi}^\theta \dot{\theta} \dot{\phi} + \Gamma_{\phi \phi}^\theta \dot{\phi}^2 = && ( \dot{\theta} = 0 ) \\ & = \ddot{\theta} - \sin \theta \cos \theta \Gamma_{\phi \phi}^\theta \dot{\phi}^2 = \\ & = \ddot{\theta} \ . \end{aligned}$$ Replacing the expression for $\dot{t}$ and $\dot{\phi}$ into the relation $u^\mu u_{\mu} = - c^2$ (from $ds^2/d \tau^2$) $$\begin{aligned} -c^2 & = - \left( 1 - \frac{r_s}{r} \right) c^2 \dot{t}^2 + \left( 1 - \frac{r_s}{r} \right)^{-1} \dot{r}^2 + r^2 \dot{\phi}^2 = \\ & = - \left( \right)^{-1} E^2 + \left( \right)^{-1} \dot{r}^2 + \frac{L}{r^2} \\ \end{aligned}$$ and thus $$ \dot{r}^2 = E^2 - c^2 \left( 1 - \frac{r_s}{r} \right) \left( 1 + \frac{L^2}{r^2 c^2} \right) \ . $$ Now, changing variable $u := \frac{1}{r}$ and using the relation $\phi(\tau)$ to write $$\dot{r} = \frac{d r}{d\tau} = \frac{d \phi}{d \tau} \frac{d}{d \phi} = \frac{L}{r^2} \frac{d r}{d \phi} = L u^2 \left( - \frac{1}{u^2} \right) u'(\phi) = - L u'(\phi) \ ,$$ the equation becomes $$ L^2 {u'}^2 = E^2 - c^2 ( 1 - r_s u ) \left( 1 + \frac{L^2}{c^2} u^2 \right) \ .$$ Differentiating both sides by $\phi$ $$0 = L^2 2 u' u'' - c^2 r_s u' \left( 1 + \frac{L^2}{c^2} u^2 \right) + c^2 ( 1 - r_s u) 2 \frac{L^2}{c^2} u u' $$ or dividing by $2 L^2 u'$, $$0 = u'' - \frac{c^2 r_s}{2 L^2} - \frac{3}{2}r_s u^2 + u$$ or, with $r_s = \frac{2 GM}{c^2}$ $$u'' + u = \frac{c^2 r_s}{2 L^2} + \frac{3}{2} r_s u^2 = \frac{GM}{L^2} + \frac{3 GM}{c^2} u^2 \ .$$ **Solution via perturbation method.** As $\frac{3 GM}{c^2} \to 0$, the equation becomes the equation of the Kepler orbits. Let the solution be $u(\phi) = u_0(\phi) + u_1(\phi)$, with $u_0(\phi)$ the solution of the equations of classical mechanics. $$u_0(\phi) = \frac{GM}{L^2}(1 + e \cos \phi) \ . $$ As $$0 = \underbrace{u_0'' + u_0 - \frac{GM}{L^2}}_{=0} + u_1'' + u_1 - \frac{3GM}{c^2} (u_0 + u_1)^2 \ ,$$ the perturbation $u_1(\phi)$ (of order $\frac{GM}{c^2}$ must satisfy (approximately, truncation) the equation $$\begin{aligned} u_1'' + u_1 & \approx \frac{3 GM}{c^2} u_0^2(\phi) = \\ & = \frac{3 (GM)^3}{c^2 L^4} \left( 1 + 2 e \cos \phi + e^2 \cos^2 \phi \right) = && ( \cos 2 x = 2 \cos^2 x - 1 ) = \\ & = 3 \left( \frac{(GM)^{3/2}}{ c L^2} \right)^2 \left( 1 + 2 e \cos \phi + \frac{e^2}{2} \left( 1 + \cos ( 2 \phi ) \right) \right) = \\ & = 3 \left( \frac{(GM)^{3/2}}{ c L^2} \right)^2 \left( 1 + \frac{e}{2} + 2 e \cos \phi + \frac{e^2}{2} \cos ( 2 \phi ) \right) \ , \end{aligned}$$ whose solution undergoes resonance, as the forcing $2 e \cos \phi$ has the same frequency as the natural frequency of the system.The general solution reads $$u_1(\phi) = A \cos \phi + B \sin \phi + 3 \left( \frac{(GM)^{3/2}}{c L^2} \right)^2 \left[ \left( 1 + \frac{e}{2} \right) + e \phi \sin \phi + \frac{e^2}{6} \cos 2 \phi \right]$$ Only the term $\propto \phi \sin \phi$ is unbounded. Combining this term with the unperturbed solution gives $$\begin{aligned} u(\phi) & \approx \frac{GM}{L^2} \left\{ ( 1 + e \cos \phi ) + 3 \left( \frac{GM}{cL} \right)^2 e \phi \sin \phi \right\} \ , \end{aligned}$$ and using small angle approximation $\cos(\alpha + x ) = \cos \alpha \cos x - \sin \alpha \sin x \approx \cos x - \alpha \sin x $, for $\alpha \sim 0$, $$\begin{aligned} u(\phi) & \approx \frac{GM}{L^2} \left\{ 1 + e \cos \phi + \underbrace{ 3 \left( \frac{ GM}{cL} \right)^2 \phi}_{"-\alpha"} e \sin \phi \right\} = \\ & \approx \frac{GM}{L^2} \left\{ 1 + e \cos \left( \phi - \left( \frac{GM}{cL} \right)^2 \phi \right) \right\} = \\ & = \frac{GM}{L^2} \left\{ 1 + e \cos \left[ \left( \phi - 3 \left(\frac{ GM}{c L} \right)^2 \right) \phi \right] \right\} \ . \end{aligned}$$ As $u:= \frac{1}{r}$, the perihelion occurs for $u(\phi_{perih}) = \max_{\phi} u(\phi)$, i.e. for $$\left( 1 - 3\left( \frac{GM}{c L} \right)^2 \right) \phi_{perih,n} = n 2 \pi \quad , \quad n \in \mathbb{Z} \ .$$ The **precession of the perihelion** can be evaluated as the difference between the angular coordinate of two successive revolutions (adding a $2pi$ to the first angle, mimicing an orbit with no precession) i.e. $$\begin{aligned} \Delta \phi & = \phi_{perih, n+1} - ( \phi_{perih, n} + 2 \pi ) = \\ & = \frac{2 \pi}{1 - 3 \left(\frac{GM}{cL}\right)^2} - 2 \pi = \\ & \approx 2 \pi \left( 1 + 3 \left( \frac{GM}{cL} \right)^2 - 1 \right) = 6 \pi \left( \frac{GM}{cL} \right)^2 \ . \end{aligned}$$ ```{dropdown} General solution The general solution has the expression $$u_1(\phi) = \underbrace{A \cos \phi + B \sin \phi}_{u_{1,homo}(\phi)} + \underbrace{C + D \phi \cos \phi + E \phi \sin \phi + F \cos (2 \phi)}_{u_{1,part}(\phi)} \ .$$ The coefficients of the particular solution are evaluated by coefficient matching $$\begin{aligned} ( \phi \cos \phi )' & = \cos \phi - \phi \sin \phi \\ ( \phi \cos \phi )'' & = - 2 \sin \phi - \phi \cos \phi \\ ( \phi \sin \phi )' & = \sin \phi + \phi \cos \phi \\ ( \phi \sin \phi )'' & = 2 \cos \phi - \phi \sin \phi \\ ( \cos ( 2 \phi ))'' & = - 4 \cos ( 2 \phi ) \\ \end{aligned}$$ then $$\begin{aligned} \cos \phi & : \quad 0 = A - A - 2 E + 2 \alpha e \\ \sin \phi & : \quad 0 = B - B + 2 D \\ \phi \cos \phi & : \quad 0 = D - D \\ \phi \sin \phi & : \quad 0 = E - E \\ 1 & : \quad 0 = - C + \alpha \left( 1 + \frac{e}{2} \right) \\ \cos 2 \phi & : \quad 0 = 4 F - F + \alpha \frac{e^2}{2} \ , \end{aligned}$$ so that $$\begin{aligned} E & = \alpha e \\ D & = 0 \\ C & = \alpha \left( 1 + \frac{e}{2} \right) \\ F & = - \frac{1}{6} \alpha e^2 \ . \end{aligned}$$ ```