Graph Geodesics via Extrinsic Geometry

Geodesics on graphs from the condition that acceleration is normal.

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 second 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. It is by far the shortest of the three derivations in this series.

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:

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

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

The second quantity we need is the acceleration $\alpha$ of a particle traveling along the surface. Such a particle can be specified by giving $x,y$ 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

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

It is worth pausing on what $\alpha$ actually is. The third coordinate of the curve is $f$ itself, so the third component of its acceleration can only be $\ddot{f}$: the vertical acceleration of the particle. The mouthful above is just the chain rule applied twice to $f(x(t),y(t))$, and nothing was really abbreviated at all.

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

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

Before turning the crank, it is worth seeing the claim. Below, a geodesic runs across the surface, and at the moving point sit the velocity $T$, the normal $\hat{N}$, and the acceleration — which lies along $\hat{N}$ and nowhere else.

The translucent panel is the plane spanned by $T$ and $\hat{N}$, and the dark curve is where it cuts the surface. This is the right plane to flatten out, and the only one: $T\cdot\hat{N}=0$ identically for a graph, since the velocity’s vertical component is $f_x\dot{x}+f_y\dot{y}$ and the normal is $\langle -f_x,-f_y,1\rangle$, so the dot product cancels term by term. The two are perpendicular unit vectors, and the panel can be drawn flat with no distortion at all.

(The plane one first reaches for — the vertical plane through the velocity — is the wrong one. It contains $\hat{N}$ only when the velocity runs straight up or down the fall line, and otherwise the acceleration points out of it. That special case is exactly the one-dimensional picture of ‘Gravity Along a Curve’, which is why a vertical slice served there and will not serve here.)

Aim the acceleration anywhere but along $\hat{N}$ and it acquires a component along $T$, which speeds the particle up or slows it down. That is one of the two ways to fail, and it is the one this flat picture settles. The other — turning sideways within the surface — points out of the page, and is easiest to see from directly above:

Looking down, the acceleration casts a shadow $\langle\ddot{x},\ddot{y}\rangle$ on the coordinate plane, and it stays pinned to the line through $\nabla f$ no matter how the path turns. That is the content of the two equations we are about to extract.

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 $x,y$ 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 $\ddot{x}$ and $\ddot{y}$, 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 $\ddot{x}$ 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

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

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

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

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

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

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