Gravity on a Surface

A ball sliding on the graph of a function, and the one term that separates it from a geodesic.

This note is a companion to the three on geodesics of a graph, and to the earlier one on gravity along a curve. Those three computed the same equations three different ways, for a particle with nothing pushing on it. Here we hang gravity on the particle and see what changes. It turns out to be one term.

The setup is the same as before. A surface is the graph of a function $z=f(x,y)$, a curve on it is given by $x$ and $y$ as functions of time,

$\gamma(t)=\left(x(t),\,y(t),\,f(x(t),y(t))\right)$

and a dot is a time derivative along that curve, so that $\dot{f}=f_x\dot{x}+f_y\dot{y}$ is the vertical component of the velocity and

$\|\dot\gamma\|^2=\dot{x}^2+\dot{y}^2+\dot{f}^2$

Gravity points down with strength $g$. The particle is confined to the surface but otherwise free, and there is no friction.

Set $g=0$ and this is exactly the geodesic equation of the previous three notes. So the whole story is one formula with a knob on it:

The Derivation

The variational note minimized the energy functional $\tfrac{1}{2}\int\|\dot\gamma\|^2dt$ and got the geodesic equations out. Adding gravity means adding a potential, so we minimize the action of

$\Lambda=\frac{1}{2}\left(\dot{x}^2+\dot{y}^2+\dot{f}^2\right)-gf$

A potential enters the Lagrangian additively, so none of the previous computation has to be redone. Every term we worked out before is still there, and the only new contribution comes from $-gf$.

Varying $x$, the derivatives are the old ones plus a single term:

$$\begin{align*} \frac{\partial\Lambda}{\partial\dot{x}}&=\dot{x}+\dot{f}f_x\\ \frac{\partial\Lambda}{\partial x}&=\dot{f}\left(f_{xx}\dot{x}+f_{xy}\dot{y}\right)-gf_x \end{align*}$$

The total time derivative of the first is exactly as before,

$\frac{d}{dt}\frac{\partial\Lambda}{\partial\dot{x}}=\ddot{x}+\ddot{f}f_x+\dot{f}\left(f_{xx}\dot{x}+f_{xy}\dot{y}\right)$

and setting the two equal, the $\dot{f}\left(f_{xx}\dot{x}+f_{xy}\dot{y}\right)$ cancels off both sides just as it did without gravity. What is left is

$\ddot{x}=-f_x\left(\ddot{f}+g\right)\hspace{1cm}\ddot{y}=-f_y\left(\ddot{f}+g\right)$

Compare this with the result of the variational note, which was $\ddot{x}=-f_x\ddot{f}$. Gravity has done exactly one thing: it replaced $\ddot{f}$ with $\ddot{f}+g$.

Solving for the Vertical Acceleration

As before these are not yet explicit, since $\ddot{f}$ contains $\ddot{x}$ and $\ddot{y}$. Differentiating $\dot{f}=f_x\dot{x}+f_y\dot{y}$ a second time gives the same expression as it did without gravity,

$\ddot{f}=A+f_x\ddot{x}+f_y\ddot{y}\hspace{1cm}A=f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2$

and substituting the two equations above,

$$\begin{align*} \ddot{f}&=A-f_x^2\left(\ddot{f}+g\right)-f_y^2\left(\ddot{f}+g\right)\\ &=A-\left(f_x^2+f_y^2\right)\left(\ddot{f}+g\right) \end{align*}$$

Adding $g$ to both sides collects everything onto the combination we actually want:

$\ddot{f}+g=A+g-\left(f_x^2+f_y^2\right)\left(\ddot{f}+g\right)$

$\implies \left(1+f_x^2+f_y^2\right)\left(\ddot{f}+g\right)=A+g$

$\implies \ddot{f}+g=\frac{A+g}{1+f_x^2+f_y^2}$

Putting that back into $\ddot{x}=-f_x(\ddot{f}+g)$ proves the theorem. The bookkeeping is identical to the geodesic case with $A$ replaced by $A+g$ throughout.

The Ball is Not the Gradient Flow

A ball released on a hillside does not run straight downhill. That would be the gradient flow $\dot{x}=-f_x$, $\dot{y}=-f_y$, which is first order: no momentum, always along the fall line, and it stops at the bottom of the first basin it finds. Ours is second order, so the ball overshoots the bottom, climbs the far side, and with no friction never settles at all.

Where the Ball Can Go

With no friction the total energy is conserved. Along a trajectory,

$E=\frac{1}{2}\|\dot\gamma\|^2+gf$

is constant, and since the kinetic term cannot be negative,

$f(x,y)\leq \frac{E}{g}$

everywhere along the motion. So the ball is trapped for all time in the part of the plane where $f$ stays under $E/g$, and it reaches the edge of that region exactly where it comes momentarily to rest.

In the bead note this was the dashed line the particle kept returning to. Here the reachable height is still fixed by where the ball started, but the set of points at that height is a curve instead of a pair of points, so there is room to wander around inside it.

Jacobi’s Trick

We have not really left the subject of the previous notes at all.

Fix the energy $E$. A trajectory of our ball is then a curve on which $\tfrac12\|\dot\gamma\|^2=E-gf$, and Jacobi’s principle says such trajectories are geodesics of a different metric on the same surface: the induced metric, scaled at each point by how much kinetic energy is available there. Writing $ds^2$ for the induced metric of the surface, the new one is

$d\hat{s}^2=2\left(E-gf\right)ds^2$

Here $g$ already means gravity, so I am writing both metrics as line elements rather than giving them names.

They are geodesics of $d\hat{s}^2$ up to reparameterization. That caveat is the same one the variational note spent a section on: length cannot see the schedule, so a statement about routes only ever pins a curve down to how it is traversed.

Notice two things about $d\hat{s}^2$. It is a conformal rescaling, so it changes lengths but not angles, and it shrinks distances wherever the ball is slow. And it degenerates exactly on the contour $f=E/g$: there the conformal factor is zero, so every curve lying on that contour has zero length and the metric stops being a metric. That contour is the boundary of the reachable region from the previous section, showing up again from the other side.

So gravity doesn’t take us out of geodesic geometry. It just changes which metric we are doing geodesic geometry in.