Why do these unwanted partial derivatives vanish?

Question

Problem

Let $(q = 0,\; \dot q=0)$ be an equilibrium point for a dynamical system, that is, a solution of Lagrange's equations $d/dt(\partial L / \partial \dot q^k) = \partial L / \partial q^k$ for which $q$ and $\dot q$ are identically 0. Here $L = T- V$ where $V = V(q)$ and where $2T = g_{ij}(q)\dot q^i \dot q^j$ is assumed positive definite. Assume that $q = 0$ is a nondegenerate minimum for $V$; thus $\partial V / \partial q^k = 0$ and the Hessian matrix $Q_{jk} = (\partial^2 V / \partial q^j \partial q^k)(0)$ is positive definite. For an approximation of small motions near the equilibrium point one assumes $q$ and $\dot q$ are small and one discards all cubic and higher terms in these quantities.

Using Taylor expansions, show that Lagrange's equations in our
quadratic approximation become $$g_{kl}(0)\ddot q^l = -Q_{kl}q^l$$

(From The Geometry of Physics, Theodore Frankel, Problem 2.4(4), part (i).)

Solution so far

\begin{equation*} L(q, \dot q) = \tfrac{1}{2} g_{ij}(q) \dot q^i \dot q^j - V(q) \end{equation*} whence \begin{align*} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q^k}(q)\right) & = \frac{d}{dt}\left(\tfrac{1}{2} g_{ij}(q) \frac{\partial}{\partial \dot q^k} (\dot q^i \dot q^j)\right) \\ & = \frac{d}{dt}\big(g_{kl}(q) \dot q^l\big) \\ & = \frac{\partial g_{kl}}{\partial q^i}(q) \dot q^i \dot q^l + g_{kl}(q) \ddot q^l \\ \end{align*} and \begin{equation*} \frac{\partial L}{\partial q^k}(q) = \frac{1}{2} \frac{\partial g_{ij}}{\partial q^k}(q) \dot q^i \dot q^j - \frac{\partial V}{\partial q^k}(q) \end{equation*} so we have \begin{equation}\label{L} \frac{\partial g_{kl}}{\partial q^i}(q) \dot q^i \dot q^l + g_{kl}(q) \ddot q^l = \frac{1}{2} \frac{\partial g_{ij}}{\partial q^k}(q) \dot q^i \dot q^j - \frac{\partial V}{\partial q^k}(q) \\ \end{equation} Taylor expand about $(q = 0,\; \dot q = 0)$, ignoring cubic terms in $q$ and $\dot q$. \begin{align*} \frac{\partial g_{kl}}{\partial q^i}(q) \dot q^i \dot q^j &= \frac{\partial g_{kl}}{\partial q^i}(0) \dot q^i \dot q^l + \ldots \\ g_{kl}(q) \ddot q^l &= g_{kl}(0) \ddot q^l + \frac{\partial g_{kl}}{\partial q^i}(0) q^i \ddot q^l + \frac{\partial^2 g_{kl}}{\partial q^i \partial q^j}(0) q^i q^j \ddot q^l + \ldots \\ \frac{\partial g_{ij}}{\partial q^k}(q) \dot q^i \dot q^j &= \frac{\partial g_{ij}}{\partial q^k}(0) \dot q^i \dot q^j + \ldots \\ \frac{\partial V}{\partial q^k}(q) &= \frac{\partial V}{\partial q^k}(0) + \frac{\partial^2 V}{\partial q^l \partial q^k}(0)q^l + \frac{\partial^3 V}{\partial q^m \partial q^l \partial q^k}(0) q^l q^m + \ldots \\ &= 0 + Q_{kl}q^l + \frac{\partial^3 V}{\partial q^m \partial q^l \partial q^k}(0) q^l q^m + \ldots \\ \end{align*} Deleting the ignored terms and substituting, \begin{equation*} % g_{kl}(q) \ddot q^l \frac{\partial g_{kl}}{\partial q^i}(q) \dot q^i \dot q^l + g_{kl}(0) \ddot q^l + \frac{\partial g_{kl}}{\partial q^i}(0) q^i \ddot q^l + \frac{\partial^2 g_{kl}}{\partial q^i \partial q^j}(0) q^i q^j \ddot q^l % = \frac{1}{2} \frac{\partial g_{ij}}{\partial q^k}(q) \dot q^i \dot q^j = \frac{1}{2} \frac{\partial g_{ij}}{\partial q^k}(0) \dot q^i \dot q^j % - \frac{\partial V}{\partial q^k}(q) \\ - Q_{kl}q^l - \frac{\partial^3 V}{\partial q^m \partial q^l \partial q^k}(0) q^l q^m \end{equation*}

Questions

All the terms except $q_{kl}(0)\ddot q^l$ and $-Q_{kl}q^l$ are unwanted. Should I have avoided introducing them in the first place? How? Or are the unwanted terms zero, individually or taken together? How come?

It was suggested on IRC that the term in $q^i q^j \ddot q^l$ counts as a cubic term that can be discarded, but that doesn't seem above board (is it?) since it prejudges that $\ddot q$ isn't much bigger than $q$ and $\dot q$, and we don't know that yet.

Another suggestion was that a change of coordinates such that all the first derivatives $(\partial g_{ij}/\partial q^k)(0)$ are zero should help. But I don't think we can do that and still get the desired answer in terms of the original $g_{ij}(0)$.

Answer 1

From the problem description I get the impression that we're supposed to Taylor expand the Lagrangian, instead of the equation of motion to quadratic order. Somewhat crudely, if you want equations of motion to a certain order, you should expand the Lagrangian up to one order higher. If you do the quadratic expansion, the Lagrangian has the simple form $\frac{1}{2} g_{ij}(0) \dot{q}^i\dot{q}^j - Q_{ij} q^i q^j$, which produces the desired equation of motion.

Your equation of motion is actually correct, except at a higher order. You can check that if we keep the cubic terms in the Taylor expansion of the Lagrangian, you will also get equation of motion, except for the term containing $q^i q^j \ddot{q}^l$. This nonlinear equation of motion can still be used to justify dropping this term, because you can rearrange the equation into the form $(g(0) + O(q, \dot{q}))\ddot{q} = Q q + O(q, \dot{q})$, which shows that $\ddot{q}$ is on the same order as $q$. In fact you can show that this will always be the case, as long as you are expanding near an equilibrium, since the lowest order term containing $\ddot{q}$ will always be $g_{k l} \ddot{q}^l$, and the lowest order term not containing $\ddot{q}$ will be $Q_{k l} q^l$.