SICM Exercise 1.10
Solution to exercise 1.10 of Structure and Interpretation of Classical Mechanics by Gerald Jay Sussman and Jack Wisdom.
Higher-derivative Lagrangians
🛈 Note
Derive Lagrange’s equations for Lagrangians that depend on accelerations. In particular, show that the Lagrange equations for Lagrangians of the form \(L(t, q, \dot{q}, \ddot{q})\) with \(\ddot{q}\) terms are
\[ D^2(\partial_3 L \circ \Gamma[q])D\eta - D(\partial_2 L \circ \Gamma[q])\eta + \partial_1 L \circ \Gamma[q] = 0 \]In general, these equations, first derived by Poisson, will involve the fourth derivative of \(q\). Note that the derivation is completely analogous to the derivation of the Lagrange equations without accelerations; it is just longer. What restrictions must we place on the variations so that the critical path satisfies a differential equation?
The variation of the \(\Gamma\) function is
\[ \delta_\eta \Gamma[q] = (0, \eta(t), D\eta(t), D^2\eta(t)) \]The variation of the action function is \begin{equation*} \delta_\eta S[q](t_1, t_2) = \int_{t_1}^{t_2} (DL \circ \Gamma[q]) \delta_\eta \Gamma[q] = \int_{t_1}^{t_2} \left((\partial_1 L \circ \Gamma[q])\eta + (\partial_2 L \circ \Gamma[q])D\eta + (\partial_3 L \circ \Gamma[q])D^2\eta \right) \end{equation*}
We can integrate the second and third term by parts. For the third term we have to integrate by parts twice.
\[\int_{t_1}^{t_2} (\partial_2 L \circ \Gamma[q])D\eta = {(\partial_2 L \circ \Gamma[q])\eta \bigg\rvert_{t_1}^{t_2}} - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q])\eta \]\[\int_{t_1}^{t_2} (\partial_3 L \circ \Gamma[q])D^2\eta = {(\partial_3 L \circ \Gamma[q])D\eta \bigg\rvert_{t_1}^{t_2}} - \int_{t_1}^{t_2} D(\partial_3 L \circ \Gamma[q])D\eta = {(\partial_3 L \circ \Gamma[q])D\eta \bigg\rvert_{t_1}^{t_2}} - {D(\partial_3 L \circ \Gamma[q])\eta \bigg\rvert_{t_1}^{t_2}} + \int_{t_1}^{t_2} D^2(\partial_3 L \circ \Gamma[q])\eta \]
So we can write the action as
\[ \delta_\eta S[q](t_1, t_2) = {(\partial_2 L \circ \Gamma[q])\eta \bigg\rvert_{t_1}^{t_2}} + {(\partial_3 L \circ \Gamma[q])D\eta \bigg\rvert_{t_1}^{t_2}} - {D(\partial_3 L \circ \Gamma[q])\eta \bigg\rvert_{t_1}^{t_2}} + \int_{t_1}^{t_2} \left((\partial_1 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q])\eta + D^2(\partial_3 L \circ \Gamma[q])D\eta \right) \]The action is zero if
\[ (\partial_1 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q])\eta + D^2(\partial_3 L \circ \Gamma[q])D\eta = 0 \]\[ (\partial_3 L \circ \Gamma[q])D\eta \bigg\rvert_{t_1}^{t_2} = 0 \]
So in addition to the requirements of the action that contains derivatives of only the first order, \(D\eta\) must be zero at \(t_1\) and \(t_2\).