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The Schwarzschild orbit is $U'' + U = 1 + \varepsilon \cdot U^2$. I've only seen turning-point analysis that demonstrates shift of perihelion. Is there an exact solution?

Mauricio
  • 6,886

2 Answers2

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Closed-form analytic solutions for bound timelike geodesics around a Schwarzschild black hole were first (to the best of my knowledge) given by Charles G. Darwin (grandson of) in 1959 in this paper (unfortunately paywalled).

In terms of $u =1/r$, and making use of the fact that energy $E$ and angular momentum $L$ are constants of motion, the Schwarzchild geodesic equation can be written in first order form:

\begin{align} \left( \frac{du}{d\phi}\right)^2 &= 2M u^3 -u^2 + \frac{2M}{L}u + \frac{E^2-1}{L^2}\\ &=2M (u-u_1)(u-u_2)(u-u_3), \end{align}

where $M$ is the mass of the Schwarzschild solution (also units such that $G=c=1$).

For bound orbits, $0<u_1\leq u_2\leq u_3$, and the solution is given by

$$ u= u_1 + (u_2-u_3) sn^2(\xi(\phi),k),$$

where $sn$ is the Jacobi elliptic sine function, elliptic parameter $k= (u_2-u_1)/(u_3-u_1)$, and

$$\xi(\phi) = \sqrt{\frac{M(u_3-u_1)}{2}}\phi + \xi_0.$$

TimRias
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Exact solutions for this equation are generally not available, and the orbits near a Schwarzschild black hole usually rely on numerical methods and perturbation theory.