Christoffel Symbols and Geodesic Equations in 2D

Ready-to-use formulas for geodesics on any surface given a metric.

Lots of computations start with the dreaded task of computing christoffel symbols. This short note does the calculation once and for all for 2-dimensional Riemannian metrics, so I don’t have to keep repeating it.

Using these one can directly compute the geodesic equations, by expanding out the definition $\nabla_{\dot{\gamma}}\dot{\gamma}$ for a curve $\gamma(t)=(u(t),v(t))$.

In general plugging in the Christoffel symbols here is quite messy, and its best to rather wait until you’ve computed them and do it indivdually in each special case. We record two common and useful special cases here:

Notation

Setting some notation, we will use coordinates named $x,y$ in which our metric can be written as

$g=\begin{pmatrix}E&F\\ F&G\end{pmatrix}$

or as a line element, $ds^2=Edx^2+2Fdxdy+Gdy^2$. We will often have use to discuss the inverse of the metric tensor, and so it will be useful to have a nice notation for working with it mid-computation. Here, we will use script letters to represent the components of the inverse

$g^{-1}=\begin{pmatrix}\mathcal{E}&\mathcal{F}\\ \mathcal{F}&\mathcal{G}\end{pmatrix}$

To expand this out in terms of the original $E,F,G$ it is useful to give a name to the metric determinant:

$D=EG-F^2$

in which case we can write

$\mathcal{E}=\frac{G}{D}\hspace{1cm}\mathcal{F}=\frac{-F}{D}\hspace{1cm}\mathcal{G}=\frac{E}{D}$

General Christoffel Symbols

General christoffel symbols can be computed as functions of the metric tensor via the identity

$\Gamma_{k\ell}^i=\frac{1}{2}\sum_m g^{im}\left(\partial_\ell g_{mk}+\partial_kg_{m\ell}-\partial_mg_{k\ell}\right)$

Our goal here is to expand this out in two dimensions (where $i,k,\ell, m\in\{x,y\}$). A priori there are eight possible Christoffel symbols here (three choices of $x,y$), but the symbols enjoy a symmetry in their lower indices $\Gamma_{xy}^\ast = \Gamma_{yx}^\ast$ which reduces the number of independent possibilites to six. By the notational symmetry swapping all $x$‘s and $y$‘s, it then suffices to compute three Christoffel symbols directly from the definition: we choose $\Gamma_{xx}^x,\Gamma_{xx}^y$ and $\Gamma_{xy}^x$.

$\Gamma_{xx}^x$ Computation

$$\begin{align*} \Gamma_{xx}^x &= \frac{1}{2} \sum_m g^{xm} \left( \partial_x g_{mx} + \partial_x g_{mx} - \partial_m g_{xx} \right) \\ &= \frac{1}{2} \sum_m g^{xm} \left( 2 \partial_x g_{mx} - \partial_m g_{xx} \right) \\ &= \frac{1}{2} \left[ g^{xx} \left( 2 \partial_x g_{xx} - \partial_x g_{xx} \right) + g^{xy} \left( 2 \partial_x g_{yx} - \partial_y g_{xx} \right) \right] \\ &= \frac{1}{2} \left[ g^{xx} \partial_x g_{xx} + g^{xy} \left( 2 \partial_x g_{xy} - \partial_y g_{xx} \right) \right] \\ &= \frac{1}{2} \left[ \mathcal{E} \partial_x E + \mathcal{F} \left( 2 \partial_x F - \partial_y E \right) \right] \\ &= \frac{1}{2} \left[ \mathcal{E} E_x + \mathcal{F} \left( 2 F_x - E_y \right) \right] \end{align*}$$

Rewriting in terms of $E,F,G$ and $D=EG-F^2$:

$\Gamma_{xx}^x=\frac{G E_x - F \left( 2 F_x - E_y \right)}{2D}$

Swapping all instances $x$ and $y$ (which also swaps $E$ and $G$) gives the Christoffel symbol $\Gamma_{yy}^y$:

$\Gamma_{yy}^y=\frac{E G_y - F \left( 2 F_y - G_x \right)}{2D}$

$\Gamma_{xx}^y$ Computation

$$\begin{align*} \Gamma_{xx}^y &= \frac{1}{2} \sum_m g^{ym} \left( \partial_x g_{mx} + \partial_x g_{mx} - \partial_m g_{xx} \right) \\ &= \frac{1}{2} \sum_m g^{ym} \left( 2 \partial_x g_{mx} - \partial_m g_{xx} \right) \\ &= \frac{1}{2} \left[ g^{yx} \left( 2 \partial_x g_{xx} - \partial_x g_{yx} \right) + g^{yy} \left( 2 \partial_x g_{yx} - \partial_y g_{xx} \right) \right] \\ &= \frac{1}{2} \left[ g^{yx} \left( \partial_x g_{xx} \right) + g^{yy} \left( 2 \partial_x g_{yx} - \partial_y g_{xx} \right) \right] \\ &= \frac{1}{2} \left[ \mathcal{F} \partial_x E + \mathcal{G} \left( 2 \partial_x F - \partial_y E \right) \right] \\ &= \frac{1}{2} \left[ \mathcal{F} E_x + \mathcal{G} \left( 2 F_x - E_y \right) \right] \end{align*}$$

Rewriting in terms of $E,F,G$ and $D=EG-F^2$:

$\Gamma_{xx}^y=\frac{E\left( 2 F_x - E_y \right)-FE_x}{2D}$

Swapping all instances $x$ and $y$ gives the Christoffel symbol $\Gamma_{yy}^x$:

$\Gamma_{yy}^x=\frac{G\left( 2 F_y - G_x \right)-FG_y}{2D}$

$\Gamma_{xy}^x$ Computation

$$\begin{align*} \Gamma_{xy}^x &= \frac{1}{2} \sum_m g^{xm} \left( \partial_y g_{mx} + \partial_x g_{my} - \partial_m g_{xy} \right) \\ &= \frac{1}{2} \left[ g^{xx} \left( \partial_y g_{xx} + \partial_x g_{xy} - \partial_x g_{xy} \right) + g^{xy} \left( \partial_y g_{yx} + \partial_x g_{yy} - \partial_y g_{xy} \right) \right] \\ &= \frac{1}{2} \left[ g^{xx} \left( \partial_y g_{xx} \right) + g^{xy} \left( \partial_x g_{yy} \right) \right] \\ &= \frac{1}{2} \left[ \mathcal{E} \partial_y E + \mathcal{F} \partial_x G \right] \\ &= \frac{1}{2} \left[ \mathcal{E} E_y + \mathcal{F} G_x \right] \end{align*}$$

Rewriting in terms of $E,F,G$ and $D=EG-F^2$:

$\Gamma_{xy}^x=\frac{G E_y -F G_x}{2D}$

Swapping all instances $x$ and $y$ gives the Christoffel symbol $\Gamma_{yx}^y$:

$\Gamma_{yx}^y=\frac{E G_x -F E_y}{2D}$

Expanding the Geodesic Equation

The geodesic equation stipulates that the acceleration along the curve is zero. Given a curve $\gamma(t)=(x(t),y(t))$ its first derivative is calculable by usual differentiation

$\dot{\gamma}= \dot{x}\partial_x+\dot{y}\partial_y$

But the second derivative is now a derivative not of a function, but of a vector field along a curve. This requires covariant differentiation - specifically the generalization $D/dt$ of the usual covariant derivative $\nabla$ of vector fields to fields along curves. Computing

$$\begin{align*} \frac{D \dot{\gamma}}{dt} &= \frac{D}{dt} \left( \dot{x} \partial_x + \dot{y} \partial_y \right) \\ &= \frac{D}{dt} \dot{x} \partial_x + \frac{D}{dt} \dot{y} \partial_y \\ &= \left( \frac{D\dot{x}}{dt} \partial_x + \dot{x} \frac{D}{dt} \partial_x \right) + \left( \frac{D\dot{y}}{dt} \partial_y + \dot{y} \frac{D}{dt} \partial_y \right) \\ &= \ddot{x} \partial_x + \dot{x} \left( \nabla_{\dot{\gamma}} \partial_x \right) + \ddot{y} \partial_y + \dot{y} \left( \nabla_{\dot{\gamma}} \partial_y \right) \end{align*}$$

where we have used linearity and the Leibniz rule, together with the fact that $D/dt$ agrees with the time derivative along coordinate functions and the covariant derivative in direction $\gamma$ along vector fields. Continuing we expand occurances of $\nabla_{\dot{\gamma}}$ using the tensorality of the covariant derivative: for any vector field $V$,

$$\begin{align*} \nabla_{\dot{\gamma}} V &= \nabla_{\dot{x}\partial_x+\dot{y}\partial_y}V\\ &= \nabla_{\dot{x}} \partial_x + \nabla_{\dot{y}} \partial_y V \\ &= \nabla_{\dot{x}} V \partial_x + \nabla_{\dot{y}} V \partial_y \\ &= \dot{x} \nabla_{\partial_x} V + \dot{y} \nabla_{\partial_y} V \end{align*}$$

Plugging this in for both $V=\partial_x$ and $V=\partial_y$, yields the following, where in the second line we have used the symmetry that $\nabla_{\partial_x}\partial_y=\nabla_{\partial_y}\partial_x$ for coordinate vector fields.

$$\begin{align*} \frac{D\dot{\gamma}}{dt} &= \ddot{x} \partial_x + \dot{x} \left( \dot{x} \nabla_{\partial_x} \partial_x + \dot{y} \nabla_{\partial_y} \partial_x \right) + \ddot{y} \partial_y + \dot{y} \left( \dot{x} \nabla_{\partial_x} \partial_y + \dot{y} \nabla_{\partial_y} \partial_y \right) \\ &= \ddot{x} \partial_x + \ddot{y} \partial_y + \dot{x}^2 \nabla_{\partial_x} \partial_x + 2 \dot{x} \dot{y} \nabla_{\partial_y} \partial_x + \dot{y}^2 \nabla_{\partial_y} \partial_y \end{align*}$$

Finally, it has come time for the Christoffel symbols to make their appearance, as the the coefficients of covariantly differentiating coordinate fields:

$$\begin{align*} \nabla_{\partial_x} \partial_{x} &= \Gamma_{xx}^x \partial_{x} + \Gamma_{xx}^y \partial_{y} \\ \nabla_{\partial_x} \partial_{y} &= \Gamma_{xy}^x \partial_{x} + \Gamma_{xy}^y \partial_{y} \\ \nabla_{\partial_y} \partial_{y} &= \Gamma_{yy}^x \partial_{x} + \Gamma_{yy}^y \partial_{y} \end{align*}$$

Plugging these in,

$$\begin{align*} \frac{D\dot{\gamma}}{dt} &= \ddot{x} \partial_x + \ddot{y} \partial_y \\ &\quad + \dot{x}^2 \left( \Gamma_{xx}^x \partial_x + \Gamma_{xx}^y \partial_y \right) \\ &\quad + 2 \dot{x} \dot{y} \left( \Gamma_{xy}^x \partial_x + \Gamma_{xy}^y \partial_y \right) \\ &\quad + \dot{y}^2 \left( \Gamma_{yy}^x \partial_x + \Gamma_{yy}^y \partial_y \right) \end{align*}$$

and re-grouping terms by basis vector,

$$\begin{align*} \frac{D\dot{\gamma}}{dt} &= \left( \ddot{x} + \dot{x}^2 \Gamma_{xx}^x + 2\dot{x} \dot{y} \Gamma_{xy}^x + \dot{y}^2 \Gamma_{yy}^x \right) \partial_x \\ &\quad + \left( \ddot{y} + \dot{x}^2 \Gamma_{xx}^y + 2\dot{x} \dot{y} \Gamma_{xy}^y + \dot{y}^2 \Gamma_{yy}^y \right) \partial_y \end{align*}$$

This gives a full expression for the acceleration along a curve $\gamma$. Because geodesics have vanishing acceleration, setting both components to zero gives a pair of coupled nonlinear differential equations specifying geodesics:

$$\left\{\begin{matrix} \ddot{x} + \dot{x}^2 \Gamma_{xx}^x + 2\dot{x} \dot{y} \Gamma_{xy}^x + \dot{y}^2 \Gamma_{yy}^x=0\\ \ddot{y} + \dot{x}^2 \Gamma_{xx}^y + 2\dot{x} \dot{y} \Gamma_{xy}^y + \dot{y}^2 \Gamma_{yy}^y =0 \end{matrix}\right\}$$

Diagonal Metrics

A diagonal metric has $F=0$, which simplfies many of the Christoffel symbols. We compute three such simplifications here:

$$\begin{align} \Gamma_{xx}^x&=\frac{G E_x - F \left( 2 F_x - E_y \right)}{2(EG-F^2)}=\frac{G E_x}{2EG}=\frac{E_x}{2E}\\ \Gamma_{xx}^y&=\frac{E\left( 2 F_x - E_y \right)-FE_x}{2(EG-F^2)}=\frac{-EE_y}{2EG}=-\frac{E_y}{2G}\\ \Gamma_{xy}^x&=\frac{G E_y -F G_x}{2(EG-F^2)}=\frac{GE_y}{2EG}=\frac{E_y}{2E} \end{align}$$

Switching $x$ and $y$ as previously gives the other three. We summarize all in the table below

$\Gamma_{xx}^x=\frac{E_x}{2E}\hspace{1cm}\Gamma_{xx}^y= -\frac{E_y}{2G}$ $\Gamma_{yy}^x=-\frac{G_x}{2E}\hspace{1cm}\Gamma_{yy}^y=\frac{G_y}{2G}$ $\Gamma_{xy}^x=\frac{E_y}{2E}\hspace{1cm}\Gamma_{xy}^y=\frac{G_x}{2G}$

To get the geodesic equations, we simply plug these in and simplify. Note that all with a superscript $x$ (which occur in the $\ddot{x}$ equation) have denominator $2E$ and all terms for the $\ddot{y}$ equation share denominator $2G$.

The geodesics of a diagonal metric $ds^2=Edx^2+Gdy^2$ satisfy $\ddot{x}+\frac{E_x\dot{x}^2+2E_y\dot{x}\dot{y}-G_x\dot{y}^2}{2E}=0$ $\ddot{y}+\frac{G_y\dot{y}^2+2G_x\dot{x}\dot{y}-E_y\dot{x}^2}{2G}=0$

Conformally Flat Metrics

A conformally flat metric is a positive scalar multiple of the Euclidean metric $ds^2=\Psi (dx^2+dy^2)$. In the terminology above this is $F=0$ and $E=G$. This makes the Christoffel symbols easy to compute from the diagonal metric case

$\Gamma_{xx}^x=\frac{\Psi_x}{2\Psi}\hspace{1cm}\Gamma_{xx}^y= -\frac{\Psi_y}{2\Psi}$ $\Gamma_{yy}^x=-\frac{\Psi_x}{2\Psi}\hspace{1cm}\Gamma_{yy}^y=\frac{\Psi_y}{2\Psi}$ $\Gamma_{xy}^x=\frac{\Psi_y}{2\Psi}\hspace{1cm}\Gamma_{xy}^y=\frac{\Psi_x}{2\Psi}$

Plugging these in and simplifying gives the geodesic equations

$\ddot{x}+\frac{\Psi_x(\dot{x}^2-\dot{y}^2)+2\Psi_y\dot{x}\dot{y}}{2\Psi}=0$ $\ddot{y}+\frac{\Psi_y(\dot{y}^2-\dot{x}^2)+2\Psi_x\dot{x}\dot{y}}{2\Psi}=0$

Since the function $\Psi$ is always positive, it is often convenient to express it as an exponential, indeed as $\Psi=e^{2\psi}$ for some lowercase $\psi$. $ds^2=e^{2\psi}(dx^2+dy^2)$

This notation results in particularly simple forms of the Christoffel symbols: for instance

$\Gamma_{xx}^x=\frac{\Psi_x}{2\Psi}=\frac{\partial_x e^{2\psi}}{2e^{2\psi}}=\frac{e^{2\psi}2\psi_x}{2e^{2\psi}}=\psi_x$

The result of all such Christoffel symbols are summarized below:

$\Gamma_{xx}^x=\psi_x \hspace{1cm}\Gamma_{xx}^y= -\psi_y$ $\Gamma_{yy}^x=-\psi_x\hspace{1cm}\Gamma_{yy}^y=\psi_y$ $\Gamma_{xy}^x=\psi_y\hspace{1cm}\Gamma_{xy}^y=\psi_x$

The geodesic equations in this case become

$\ddot{x}+\psi_x(\dot{x}^2-\dot{y}^2)+2\psi_y\dot{x}\dot{y}=0$ $\ddot{y}+\psi_y(\dot{y}^2-\dot{x}^2)+2\psi_x\dot{x}\dot{y}=0$