# How to show that the geodesics of a metric are the solutions to a second-order differential equation?

On $\mathbb R^n$, let $\rho: \mathbb R^n\to\mathbb R$ be a smooth function, and $g$ be the metric given by scaling the usual flat metric by $e^{2\rho}$.

I want to know how to show that the geodesics of this metric are the solutions $\vec x(t)$ to the second-order differential equation $$\frac{d^2{\vec x}}{dt^2}+2(\vec{\text{grad}(\rho)}\cdot \frac{d\vec{x}}{dt})(\frac{d\vec{x}}{dt})-(\frac{d\vec{x}}{dt}\cdot\frac{d\vec{x}}{dt})\vec{\text{grad}(\rho)}=0.$$ (the gradient operation and dot products here are the ordinary Euclidean versions.)

I was also wondering when $n=2,3$, what is the (ordinary Euclidean) curvature of these curves?

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Equation (1) in my answer to math.stackexchange.com/questions/98113 may be helpful. –  Yuri Vyatkin Mar 26 '12 at 2:34
I retagged the question just to ensure that it is found under the right categories: there is no curvature involved but rather it is related to conformal geometry. –  Yuri Vyatkin Mar 27 '12 at 12:37

In $\mathbb{R}^n$ we have some coordinates, so let $x^i (t)$ be the coordinate presentation of the curve. A shorthand notation $\dot{x}^i (t_0) = \frac{d}{dt}x^i |_{t=t_0}$ simplifies the calculations, as usual. The geodesic equation then says that $$(\nabla_{\dot{x}}\dot{x})^k = \frac{d^2 x^k}{dt^2} + \dot{x}^i \dot{x}^j \Gamma^k_{ij} = 0 \tag{1}$$ where $x(t)$ has a unit speed parametrization.

Recall that the Christoffel symbols are given by $$\Gamma^k_{ij} = \frac{1}{2} g^{kl} (g_{il,j} + g_{lj,i} - g_{ij,l}) \tag{2}$$

Remark. In the standard Euclidean metric $g_{ij} = \delta_{ij}$ in $\mathbb{R}^n$ the Christoffel symbols vanish and equation (1) gives straight lines as its solutions.

Let's now look what happens with the Christoffel symbols if we replace $g$ with $\hat{g} = e^{2 \rho} g$. Observe that $$\hat{g}_{ij,l} = 2 \rho_l e^{2 \rho} g_{ij} + e^{2 \rho} g_{ij,l}$$ and substitute this into (2) to get \begin{align} \hat{\Gamma}^k_{ij} &= \frac{1}{2} e^{-2 \rho} g^{kl} ( 2 \rho_j e^{2 \rho} g_{il} + e^{2 \rho} g_{il,j} + 2 \rho_i e^{2 \rho} g_{lj} + e^{2 \rho} g_{lj,i} - 2 \rho_l e^{2 \rho} g_{ij} - e^{2 \rho} g_{ij,l}) \\ &= \Gamma^k_{ij} + \rho_j \delta^k{}_i + \rho_i \delta^k{}_j - \rho^k g_{ij} \tag{3} \end{align} where we have used metric $g$ to raise index $k$ in $\rho_k := \partial_k \rho$

This transformation affects equation (1) in the following way: \begin{align} (\hat{\nabla}_{\dot{x}}\dot{x})^k &= \frac{d^2 x^k}{dt^2} + \dot{x}^i \dot{x}^j \hat{\Gamma}^k_{ij} \\ &= \frac{d^2 x^k}{dt^2} + + \dot{x}^i \dot{x}^j (\Gamma^k_{ij} + \rho_j \delta^k{}_i + \rho_i \delta^k{}_j - \rho^k g_{ij}) = 0 \end{align}

Your equation is now obtained immediately from the above calculations since you start with the Euclidean metric $g_{ij} = \delta_{ij}$ for which the Christoffels vanish, as has been already noted, so you only need to take into account that $\operatorname{grad}(\rho)\cdot \dot{x} = \rho_i \dot{x}^i$ and $\dot{x} \cdot \dot{x} = \dot{x}^i \dot{x}_i$

Ah, well, $\operatorname{grad}(\rho)^i = \rho^i = g^{ij} \rho_j$, of course.

Indeed, the above considerations are very formal and straightforward. In order to get a deeper insight one needs to contemplate the conformal transformations of $\mathbb{R}^n$. In dimension $n=2$ the situation is controlled by the complex analysis, while in dimensions 3 and higher the Liouville's theorem states that all such transformations are compositions of translations, dilations, and inversions (so called Moebius transformations).

(The Einstein summation convention is used throughout this post)

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Thanks for your wonderful answer! I just saw this post, could you say something about that? Seems interesting. –  Vladimir Mar 27 '12 at 2:58
You are welcome. I am glad that it helps. –  Yuri Vyatkin Mar 27 '12 at 3:11
@Jack Regarding that post I would wait until someone gets Maple or other tool to compute all the required quantities. –  Yuri Vyatkin Mar 27 '12 at 3:13
:) But could you list a "outline" that how to consider "the behavior of all unit speed geodesics" in that problem? Just generally speaking. Thanks. –  Vladimir Mar 27 '12 at 3:24
@JackWitt It depends on what one understands by "the behavior of all unit speed geodesics". The local behavior is determined by the curvature at the point of consideration, the global one needs some analysis of ODE. I don't want to interfere with that question at the moment, sorry :-) –  Yuri Vyatkin Mar 27 '12 at 4:48