Geodesic on a Riemannian manifold with a random metric tensor

Given a metric tensor $g_{\mu\nu}$ on a Riemannian manifold, it's possible to write the geodesic equations using: $$\frac{d^2x^a}{ds^2} + \Gamma^{a}_{bc}\frac{dx^b}{ds}\frac{dx^c}{ds} = 0$$ where: $$\Gamma^a_{bc} = \frac{1}{2} g^{ad} \left( g_{cd,b} + g_{bd,c} - g_{bc,d} \right)$$ are the Christoffel symbols. In the euclidean metrics, the tensor $g_{\mu\nu}$ is given by the identity matrix $I.$ Suppose we have: $$g_{\mu\nu} = \begin{pmatrix} 1 & \eta_1(t) & \eta_2(t) \\ 0 & 1 & \eta_3(t) \\ \\ 0 & 0 & 1 \end{pmatrix}$$ with: $\eta_k(t)$ random variables normally distribuited with variance: $\sigma_1,\sigma_2,\sigma_3$, and zero means, how can I write down the geodesic equations? Thanks in advance.

• Shouldn't $g_{\mu\nu}$ be symmetric? – 23rd Sep 30 '13 at 14:42
• @Landscape: this is the problem. I have $g_{\mu\nu}$ not symmetric. I don't know if the usual geodesic equation using the Christoffel symbols is still correct also in this case. – Riccardo.Alestra Sep 30 '13 at 14:44

Geodesic is a shortest path in metric space. If $g$ is not symmetric, then triangle inequality does not hold :
$$g=I + E_{12}$$ where $E_{12}$ is $2\times 2$ matrix whose only nonzero element is $(1,2)$-entry. And its value is $1$. Then $$\overline{(0,0)(1,1)} > \overline{(0,0)(1,0)} + \overline{(1,0)(1,1)}$$