Note: Graph Geodesics Intrinsic

Given a real valued function $f(x,y)$, its graph defines a surface which inherits an intrinsic geometry from the ambient $\mathbb{E}^3$. In this note, we compute the geodesic equations for this surface.

In this note we approach the calculation intrinsically, by expanding out the defining equation $\nabla_{\dot{\gamma}}\dot{\gamma}=0$ in coordinates. This involves some preliminary calculation of Christoffel symbols, which are recorded in generality in the note ‘2D Geometry Cheatsheet’. Here we focus on specializing this generality to the case of a surface $(x,y,f(x,y))$ in $\mathbb{R}^3$.

The Metric & Christoffel Symbols

From general calculations for a 2D Riemannian metric, we have the following form for the Christoffel symbols.

The Metric Components

Specializing to a surface parametrized by the function $F\colon (x,y)\mapsto (x,y,f(x,y))$, the metric is computed in coordinates $(x,y)$ by pullback. Specifically, let $T_x$ and $T_y$ be the $x$- and $y-$ partial derivatives of the parameterization

$$T_x = \langle 1,0, f_x\rangle\hspace{1cm}T_y=\langle 0,1, f_y\rangle$$

Then the components of the metric $g=\left(\begin{smallmatrix}E&F\ F&G\end{smallmatrix}\right)$ are the pairwise dot-products of these

$$E=T_x\cdot T_x\hspace{1cm} F=T_x\cdot T_y\hspace{1cm} G=T_y\cdot T_y$$

Below we compute these, and their derivatives (which show up in the Christoffel formulas referenced above).

$$ \begin{align*} E &= 1 + f_x^2 & F &= f_x f_y & G &= 1 + f_y^2 \
E_x &= 2 f_x f_{xx} & F_x &= f_{xx} f_y + f_x f_{xy} & G_x &= 2 f_y f_{xy} \
E_y &= 2 f_x f_{xy} & F_y &= f_{xy} f_y + f_x f_{yy} & G_y &= 2 f_y f_{yy} \end{align*} $$

With these in hand, we compute the three independent Christoffel symbols with upper index $x$. Under the symmetry exchanging variables $x$ and $y$ (and thus simultaneously exchanging $E$ and $G$) these determine all Christoffel symbols, which then determine the geodesic equations. All Christoffel symbols have the same denominator $2D$, so its helpful to compute this quickly here:

$$ \begin{align*} D &= EG-F^2\
&=(1+f_x^2)(1+f_y^2)-(f_xf_y)^2\
&=1+f_x^2+f_y^2+f_x^2f_y^2-f_x^2f_y^2\
&=1+f_x^2+f_y^2 \end{align*} $$

Computing $\Gamma_{xx}^x$

The numerator of this Christoffel symbol is $G E_x - F \left(2 F_x - E_y\right)$. Simplifying with what we know, $$ \begin{align*} G E_x - F \left(2 F_x - E_y\right) &= \left(1 + f_y^2\right) 2 f_x f_{xx} - f_x f_y \left(2 f_{xx} f_y + 2 f_x f_{xy} - 2 f_x f_{xy}\right) \
&= \left(1 + f_y^2\right) 2 f_x f_{xx} - f_x f_y \left(2 f_{xx} f_y\right) \
&= 2 f_x f_{xx} + 2 f_x f_y^2 f_{xx} - 2 f_x f_y^2 f_{xx} \
&= 2 f_x f_{xx} \end{align*} $$

This gives the full Christoffel symbol upon division by $2D$:

$$\Gamma_{xx}^x = \frac{2f_xf_{xx}}{2 D}=\frac{f_xf_{xx}}{1+f_x^2+f_y^2}$$

Computing $\Gamma_{xy}^x$

The numerator of this Christoffel symbol is $G E_y - F G_x $. Simplifying with what we know,

$$ \begin{align*} G E_y - F G_x &= \left(1 + f_y^2\right) 2 f_x f_{xy} - f_x f_y \cdot 2 f_y f_{xy} \
&= 2 f_x f_{xy} + 2 f_x f_y^2 f_{xy} - 2 f_x f_y^2 f_{xy} \
&= 2 f_x f_{xy} \end{align*} $$

This gives the full Christoffel symbol upon division by $2D$:

$$\Gamma_{xy}^x = \frac{ 2 f_x f_{xy}}{2 D}=\frac{f_xf_{xy}}{1+f_x^2+f_y^2}$$

Computing $\Gamma_{xx}^y$

The numerator of this Christoffel symbol is $G\left(2 F_y - G_x\right) - F G_y $. Simplifying with what we know,

$$ \begin{align*} G\left(2 F_y - G_x\right) - F G_y &= \left(1 + f_y^2\right) \left(2 f_{xy} f_y + 2 f_x f_{yy} - 2 f_y f_{xy}\right) - f_x f_y \cdot 2 f_y f_{yy} \
&= \left(1 + f_y^2\right)(2 f_x f_{yy}) - 2 f_x f_y^2 f_{yy} \
&= 2 f_x f_{yy} + 2 f_x f_y^2 f_{yy} - 2 f_x f_y^2 f_{yy} \
&= 2 f_x f_{yy} \end{align*} $$

This gives the full Christoffel symbol upon division by $2D$:

$$\Gamma_{yy}^x = \frac{ 2 f_x f_{yy}}{2 D}=\frac{f_xf_{yy}}{1+f_x^2+f_y^2}$$

The Geodesic Equations

We again reference the 2D geometry ‘cheatsheet’ where the geodesics of a general metric are calculated to satisfy the following ODEs:

With our calculation of the Christoffel symbols $\Gamma_{xx}^x,\Gamma_{xy}^x$ and $\Gamma_{yy}^x$, we have everything to write down the first equation. Noting that all share the same denominator $1+f_x^2+f_y^2$ we combine fractions to

$$\ddot{x}+\frac{\dot{x}^2\left(f_xf_{xx}\right)+ 2\dot{x} \dot{y} \left(f_xf_{xy}\right) + \dot{y}^2 \left(f_xf_{yy}\right)}{1+f_x^2+f_y^2}=0$$

Each term of the fraction contains a multiple of $f_x$, so factoring this out and moving to the other side gives

$$\ddot{x}=-f_x\frac{f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2}{1+f_x^2+f_y^2}$$

This entire situation is symmetric under the exchanges of $x$ and $y$, and performing this swap gives the second equation. Of interest is that the entire fraction is itself invariant under this interchange, so the only changes are to the second derivative and the single-partial-derivative prefactor.

$$\ddot{y}=-f_y\frac{f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2}{1+f_x^2+f_y^2}$$