# Numerically solving geodesic differential equations with a priori knowledge of the Riemann curvature tensor

The geodesic differential equations are given as

\begin{align} \frac{d^2 x^j}{ds^2} + \Gamma^{\phantom{h}j}_{h\phantom{j}k}\frac{dx^h}{ds}\frac{dx^k}{ds} = 0, \end{align}

where the $\Gamma^{\phantom{h}j}_{h\phantom{j}k}$ is the Christoffel symbol of second kind.

According to T. A. Moore - A general relativity workbook,

the Riemann curvature tensor quantifies the relative variation of initially parallel geodesics. Since such geodesics only variate relative to each other in a curved space, the Riemann tensor will be zero everywhere in flat space.

If one can conclusively distinguish flat space vs. curved space, can such information have any usefulness when attempting to solve numerically the geodesic differential equations

• Couldn't you find a change of coordinates $y(x)$, solving numerically the equation $(Dy)^T(Dy) = g$, which should have a solution because the metric is flat? Then the geodesics would be $d^2y/ds^2=0$. Is this the kind of thing you're looking for, or am I misunderstanding the question? Otherwise, all flatness means is that the rhs satisfies a set of linear PDEs. It might also help to write the geodesic equations in the Hamiltonian form and use a symplectic method, but that wouldn't use flatness. – Kirill Jan 4 '16 at 22:58
• Are you asking whether information about the Riemann tensor could help you find better or faster integrators for the given differential equations? – Wolfgang Bangerth Jan 5 '16 at 23:11
• @WolfgangBangerth Yes, as I am supposing that the term with the Christoffel symbol will become zero. Then in the end we end up with the Laplace equation. But all this depends if the Christoffel term becomes zero. – imranal Jan 6 '16 at 5:28
• I think I still don't understand. If the Christoffel symbols are zero, then indeed the equation is just the Laplace equation and that means that the geodesic is a straight line (as expected). Or is your question "if the Riemann curvature tensor is zero, does this imply that the Christoffel symbols are zero"? – Wolfgang Bangerth Jan 6 '16 at 14:40