Note: Graph Geodesics Extrinsic

$$\newcommand{\RR}{\mathbb{R}}$$

This note continues our story of deriving the geodesic equation for surfaces in $\mathbb{R}^3$ described as the graph of a function $z=f(x,y)$. This third derivation represents a substantial simplification by taking the extrinsic viewpoint, where we work entirely in Euclidean 3-space and describe geodesics as curves whose acceleration is at all times normal to the surface. This is by far the most efficient of the three approaches to date!

To set some notation: if $f\colon\RR^2\to\RR$ is a twice differentiable function we parameterize its graph $S$ by the embedding $F\colon\RR^2\to\RR^3$, $$F(x,y)=(x,y,f(x,y))$$

Acceleration & Surface Normal

Exactly as in a multivariable calculus course, we can find the normal $N$ to the surface $S$ by finding two tangents to the parameterization and taking their cross product:

$$T_u = \langle 1,0, f_x\rangle\hspace{1cm}T_v=\langle 0,1,f_y\rangle$$

$$N=T_u\times T_v = \left|\begin{matrix} \hat{i} &\hat{j}&\hat{k}\
1 & 0& f_x\ 0 & 1 & f_y \end{matrix}\right|=\langle -f_x,-f_y,1\rangle$$

The second quantity we need is the acceleration $\alpha$ of a particle traveling along the surface. Such a particle can be specified by giving $u,v$ as a function of time, as its trajectory is

$$t\mapsto \left(x(t),y(t),f(x(t),y(t))\right)$$

The velocity $\mathrm{vel}$ is the first time derivative

$$\begin{align} \mathrm{vel}(t)&=\langle \dot{x},\dot{y},\frac{d}{dt}f\rangle\
&=\langle \dot{x},,\dot{y},, f_x \dot{x}+f_y\dot{y}\rangle \end{align}$$

The acceleration is then the second time derivative

$$\begin{align}\mathrm{acc} &= \frac{d}{dt}\mathrm{vel}(t)\
&=\frac{d}{dt}\langle \dot{x},,\dot{y},, f_x \dot{x}+f_y\dot{y}\rangle\
&= \left\langle \ddot{x},,\ddot{y},,\frac{d}{dt}\left(f_x \dot{x}+f_y\dot{y}\right)\right\rangle \end{align} $$

Where

$$\begin{align} \frac{d}{dt}\left(f_x \dot{x}+f_y\dot{y}\right)&=\frac{d f_x}{dt}\dot{x}+f_x\ddot{x}+\frac{df_y}{dt}\dot{y}+f_y\ddot{y}\
&=\left(f_{xx}\dot{x}+f_{xy}\dot{y}\right)\dot{x}+f_x\ddot{x}+\left(f_{yx}\dot{x}+f_{yy}\dot{y}\right)\dot{y}+f_y\ddot{y}\
&=f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2+f_x\ddot{x}+f_y\ddot{y} \end{align}$$

This last quantity is a bit of a mouthful so we will often denote it $\alpha$, and write

$$\mathrm{acc} = \langle \ddot{x},\ddot{y},\alpha\rangle$$

The Geodesic Condition

A curve on the surface $S$ is a geodesic only if it does not turn relative to the surface. Of course, it must bend (and thus accelerate) in $\mathbb{R}^3$ as it is constrained to the surface! What’s constrained to be zero is its intrinsic acceleration: the projection of this Euclidean acceleration onto the tangent plane of the surface. Thus for a geodesic, the Euclidean acceleration must be perpendicular to the surface: it must be parallel to the surface normal at each point. To get an equation out of this we take advantage of the cross product, and note the geodesic condition implies

$$\mathrm{acc}\times N=0$$

We can expand this out into three equations (one for each component) since we know these vectors explicitly:

$$\begin{align} \mathrm{acc}\times{N}&= \left|\begin{matrix} \hat{i} &\hat{j}&\hat{k}\
\ddot{x} & \ddot{y} & \alpha\
-f_x &-f_y & 1 \end{matrix}\right|\
&= \hat{i}\left(\ddot{y}+\alpha f_y\right)-\hat{j}\left(\ddot{x}+\alpha f_x\right)+\hat{k}\left(-\ddot{x}f_y+\ddot{y}f_x\right) \end{align}$$

Setting the first two of these to zero yields the relations

$$\ddot{x}=-f_x\alpha\hspace{1cm}\ddot{y}=-f_y\alpha\hspace{1cm}f_y\ddot{x}=f_x\ddot{y}$$

And any functions $u,v$ satisfying these two automatically satisfy the third equation, as

$$f_y\ddot{x}=f_y(-f_x\alpha)=f_x(-f_y\alpha)=f_x \ddot{y}$$

Thus the first two equations specify a geodesic.

$$\ddot{x}=-f_x\left(f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2+f_x\ddot{x}+f_y\ddot{y}\right)$$ $$\ddot{y}=-f_y\left(f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2+f_x\ddot{x}+f_y\ddot{y}\right)$$

Algebraic Simplification

All that remains is some algebraic simplification: we would like equations directly giving $\ddot{u}$ and $\ddot{v}$, but at the moment both of these terms also appear inside of $\alpha$ so are mixed up in the two equations. To simplify, it is useful to remember the third equation (which we know to be satisfied):

$$f_y\ddot{x}=f_x\ddot{y}$$

This proves useful as it will let us replace terms containing $\ddot{y}$ with $\ddot{x}$ and vice versa. Indeed, focusing on the equation for $\ddot{u}$ we can write

$$\begin{align} \ddot{x}&=-f_x\left(f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2+f_x\ddot{x}+f_y\ddot{y}\right)\
&=-f_x\left(f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2\right)- f_xf_x\ddot{x}-f_xf_y\ddot{y} \end{align}$$

But now replacing $f_xf_y\ddot{y}$ with the equivalent $f_yf_y\ddot{x}$ we have

$$\begin{align} \ddot{x}&=-f_x\left(f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2\right)- f_xf_x\ddot{x}-f_yf_y\ddot{y}\
&=-f_x\left(f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2\right)- (f_x^2+f_y^2)\ddot{x}\
\end{align}$$

Moving this second term to the other side allows us to solve for $\ddot{y}$:

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

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

The situation is symmetric under the exchange of $x$ and $y$ so analogous reasoning gives

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