Geodesic over a plane with a "conformal" metric Let be $ds^2=G(x)(dx_1^2+\cdots+dx_n^2)$ a metric over $\mathbb{R}^n$ given by a scalar function $G:\mathbb{R}^n\rightarrow\mathbb{R}$.
Are the geodesic associated to this metric the curves $\gamma$ which satisfies: $\gamma'' =\nabla G$ ? 
If yes, how could you prove it?

The Christoffel symbols for the conformal metric $G\, dx^2= e^{\phi} dx^2$ are 
$$\Gamma^i_{jk} = \frac{1}{2}(\delta_{ij} \frac{\partial \phi}{\partial x_k}+\delta_{ik} \frac{\partial \phi}{\partial x_j}-\delta_{jk} \frac{\partial \phi}{\partial x_i })$$ and the system for the geodesics is
$$\frac{d^2 x_i}{dt^2}=\sum_{jk} \Gamma^{i}_{jk} \frac{d x_j}{dt} \cdot \frac{d x_k}{dt}$$
So there seems to be a discrepancy with your formula. I would just try the case $n=2$ and see what happens. There will be some denominators with $G$ appearing, since $\phi = \log G$. 
