Note: Gravity Along Curve

This note records a simple calculation in lagrangian mechanics that is useful when making 2D animations. We consider a curve given by the graph of a function $u\mapsto (u,f(u))$ and a particle moving along the curve under the influence of a uniform downwards gravitational field.

PICTURE

The Lagrangian

The kinetic energy of a particle is $\frac{1}{2}m{v}^2$, where in ${\mathbb{R}}^2$ the square velocity is given in coordinates by $v^2={\dot{x}}^2+{\dot{y}}^2$. Confined to the curve $(u,f(u))$ this becomes

$$\begin{array}{rlrlr}K& =\frac{m}{2}({\dot{x}}^2+{\dot{y}}^2)& =\frac{m}{2}({\dot{u}}^2+{[f(u{)}^{\prime }]}^2)& =\frac{m}{2}({\dot{u}}^2+[f_u\dot{u}{]}^2)& =\frac{m}{2}(1+f_u^2){\dot{u}}^2\end{array}$$

Where we’ve written $f_u$ for the derivative. For a uniform gravitational field the potential is simply proportional to the height (by the constants $g$ giving strength of gravity and $m$ the particle’s mass). Thus we may take

$$V=my=mgf(u)$$

Putting these together gives the lagrangian for our system,

$$\mathcal{L}=K-V=\frac{m}{2}(1+f_u^2){\dot{u}}^2-mgf$$

The Calculus of Variations

Let $u(t)$ be the coordinate representation of a particle moving along this curve for $t\in I$. The action of such a trajectory is given by the functional

$$S[u]:={\int }_I\mathcal{L},dt$$

The physical trajectory of the particle is that which minimizes the action, which we find using the Euler-Lagrange equation:

$$\frac{\partial \mathcal{L}}{\partial u}=\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{u}}$$

Thus we need the derivatives of $\mathcal{L}$ with respect to $u,\dot{u}$ to get started:

$$\begin{array}{rlrl}\frac{\partial \mathcal{L}}{\partial u}& =\frac{\partial }{\partial u}[\frac{m}{2}(1+f_u^2){\dot{u}}^2-mgf]& =\frac{m}{2}\left[2f_u{f}_{uu}\right]{\dot{u}}^2-mg{f}_u& =m{f}_u[f_{uu}{\dot{u}}^2-g]\end{array}$$

$$\begin{array}{rlr}\frac{\partial \mathcal{L}}{\partial \dot{u}}& =\frac{\partial }{\partial \dot{u}}[\frac{m}{2}(1+f_u^2){\dot{u}}^2-mgf]& =m(1+f_u^2)\dot{u}\end{array}$$

Next we need the total time derivative of this latter quantity:

$$\begin{array}{rlrlr}\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{u}}& =\frac{d}{dt}m(1+f_u^2)\dot{u}& =m[\dot{u}\frac{d}{dt}(1+f_u^2)+(1+f_u^2)\frac{d}{dt}\dot{u}]& =m[\dot{u}(2f_u{f}_{uu}\dot{u})+(1+f_u^2)\ddot{u}]& =m[2f_u{f}_{uu}{\dot{u}}^2+(1+f_u^2)\ddot{u}]\end{array}$$

Both sides of the Euler-Lagrange equation are proportional to $m$, which then drops out of the equation yielding

$$f_u[f_{uu}{\dot{u}}^2-g]=2f_u{f}_{uu}{\dot{u}}^2+(1+f_u^2)\ddot{u}$$

To simplify, we solve for $\ddot{u}$:

$$\begin{array}{rlr}(1+f_u^2)\ddot{u}& =f_u[f_{uu}{\dot{u}}^2-g]-2f_u{f}_{uu}{\dot{u}}^2& =-(f_u{f}_{uu}{\dot{u}}^2+g{f}_u)\end{array}$$

$$⟹\ddot{u}=-\frac{f_u{f}_{uu}{\dot{u}}^2+g{f}_u}{(1+f_u^2)}$$

This proves the claimed theorem.