Why do we ignore the second-order terms in the following expansion?

Question

Consider the expansion done for the kinetic energy of a system executing small oscillations as done in Goldstein:

A similar series expansion can be obtained for the kinetic energy. Since the generalized coordinates do not involve the time explicitly, the kinetic energy is a homogeneous quadratic function of the velocities (cf. Eq. (1.71)): $$T=\frac{1}{2}m_{ij}\dot{q}_i\dot{q}_j=\frac{1} {2}m_{ij}\dot{\eta}_i\dot{\eta}_j\tag{6.5}$$ The coefficients $m_{ij}$ are in general functions of the coordinates $q_k$, but they may be expanded in a Taylor series about the equilibrium configuration: $$m_{ij}(q_1,...,q_n)=m_{ij}(q_{01},...,q_{0n})+\bigg(\frac{\partial m_{ij}}{\partial q_k}\bigg)_0 \eta_k+...$$ As 6.5 is already quadratic in the $\dot{\eta}_i$'s, the lowest nonvanishing approximation to $T$ is obtained by dropping all but the first term in the expansions of $m_{ij}$. Denoting the constant values of the $m_{ij}$ functions at equilibrium by $T_{ij}$, we can therefore write the kinetic energy as $$T=\frac{1}{2}T_{ij}\dot{\eta}_i\dot{\eta}_j\tag{6.6}$$

What does the italicized part of the above quote mean? How is the order of $\dot\eta$ the same as $\eta$?

As far as I know, if a function is small that doesn't guarantee that it's time derivative is also small. I have referred to many books and online lectures and none seem to explain this clearly.

Answer 1

This bothered me as an undergraduate too, and it wasn't until much later that I found out the rigorous mathematical way of understanding perturbation theory. The basic answer is that "we are restricting our attention to perturbations with this property." For the details, my answer will be loosely cribbed from Wald's General Relativity, which gives a better mathematical description of what we're really doing here than most classical mechanics texts do.

Let's suppose we want to solve a differential equation $\mathcal{E}[q_i(t)] = 0$, where $\mathcal{E}$ stands for some non-linear operator on the functions $q_i(t)$. Let us assume that there exists a family of exact solutions $q_i(t; \lambda)$ to the equations of motion, parameterized by a parameter $\lambda$, with the following properties:

• For all $\lambda$, $\mathcal{E}[q_i(t; \lambda)] = 0$;
• $q_i(t; 0) = q_{0i}(t)$, where $q_{0i}(t)$ is our "background solution"; and
• $q_i(t; \lambda)$ depends smoothly $\lambda$ and $t$.

In some sense, $\lambda$ measures the "size" of the perturbation away from the background solution. In particular, since all of our solutions are exact, we can say that $$ \left.\frac{d}{d\lambda} \mathcal{E}[q_i(t; \lambda)] \right|_{\lambda = 0} = 0, $$ and it is not too hard to see that this equation will be a linear equation in the functions $$ \gamma_{i}(t) \equiv \left. \frac{d q_i(t;\lambda)}{d\lambda}\right|_{\lambda = 0}. $$ For sufficiently small $\lambda$, the quantity $q_{0i}(t) + \lambda \gamma_i(t)$ will be a good approximation to $q_i(t;\lambda)$, allowing us to study solutions that are "close" to our background solution. The $\eta_i(t)$ used by Goldstein would be equal to $\lambda \gamma_i(t)$ in this language.

Once you have put all of this into place, then it is fairly straightforward to show that $\dot{\eta}_i(t)$ must go to zero as $\lambda \to 0$, since $$ \frac{d \eta_i}{dt} = \frac{d}{dt} \left[ \lambda \left. \frac{d q_i(t, \lambda)}{d \lambda} \right|_{\lambda = 0}\right] = \lambda \left.\frac{d^2 q_i(t,\lambda)}{dt d\lambda} \right|_{\lambda = 0} $$ and the smoothness assumption on the family $q_i(t;\lambda)$ ensures that this second derivative exists.

To see why the smoothness assumption is what saves us here from the pathology you're considering, consider the one-parameter family of functions $$ f(t; \lambda) = \lambda \sin (t/\lambda). $$ It is certainly the case that as $\lambda \to 0$, we have $\eta \to 0$ but $\dot{\eta} \not\to 0$. But this family of functions is not smooth in $\lambda$, since $$ \frac{df}{d\lambda} = \sin(t/\lambda) - \frac{t}{\lambda}\cos(t/\lambda) $$ and this is not well-defined at $\lambda = 0$. The assumption of a "smooth family of solutions" means that we have eliminated such pathology in the first place, though, and so we don't have to worry about such cases when we're trying to linearize our equations.

Answer 2

You are right that $\eta$ being small at one moment in time then $\dot\eta$ need not be small at that time. But, the assumption is actually that $\eta$ is always small. This does generally imply that $\dot\eta$ is small, since if $\dot \eta \Delta t \gg 1$ in some small time window $\Delta t$, then $\eta$ would become large after a time $\Delta t$ (this is assuming $\eta$ is dimensionless, otherwise you would need to divide the left and right hand side of that inequality by some parameter to make both sides dimensionless).

There is a subtlety in that $\dot\eta$ could be large for a very short period of time; in other words, if $\dot\eta$ changes very quickly so that we can't choose a $\Delta t$ which is both small enough that $\dot\eta$ is approximately constant and large enough so that $\dot \eta \Delta t \gg 1$. This can happen for high frequency oscillations in simple harmonic motion. However, the time averaged $\dot\eta$ is still small. Therefore the effect of higher order terms in $\dot\eta$ have a small effect on the overall motion, even if they can be large for a brief time interval.

Another way to justify this is to check at the end that you've made a self-consistent approximation. In other words, you can proceed assuming that the higher order terms are small, derive the solution, and then check what effect adding the higher order terms add. If you self-consistently assume you have small oscillations, the higher order terms will not have a big effect on the frequency of oscillation.

Answer 3

for one generalized coordinate $~q~$ the kinetic energy is

$$T=\frac 12 \,m(q)\,\dot q^2$$

$q\mapsto q_0+\eta~$ where $~\eta~$ is small and $~q_0 ~$ is the equilibrium state $~q_0~=$ constant .

$$\dot q=\dot\eta\\ T\mapsto \frac 12 \,m(q)\,\dot\eta^2$$

take the Taylor expansion for $~m(q)~$ you obtain

$$m\mapsto m(q_0)+ \frac{dm}{dq}\bigg|_{q_0}\,\eta\\ T=\frac 12\,\left[ m(q_0)+ \frac{dm}{dq}\bigg|_{q_0}\,\eta\right]\dot\eta^2$$

so $~\eta\,\dot\eta^2\mapsto 0$

I think that if $~\dot\eta(t)~$ is small hence $~\eta\,\dot\eta^2~$ is also small. for example

\begin{align} \dot\eta(t)&=a\,\omega\cos(\omega\,t) \\ \eta(t)&=a\,\sin(\omega\,t)~\\ \eta\,\dot\eta^2&=a^3\,\omega^2\,\sin(\omega\,t)\,\cos^2(\omega\,t) \ll 1 \end{align}

where $~ a\,\omega \ll 1~$ because $~\dot\eta~$ is small!! not $~\eta$