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

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Solution 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.

Solution 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.

Solution 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 $~\eta$

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Updated on August 01, 2022

Comments

  • Deepanshu Chaudhary
    Deepanshu Chaudhary over 1 year

    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.

    • Jonas
      Jonas almost 2 years
      Hello! It is preferable to type out screenshots or images of text; for formulae, one can use MathJax. Thanks!
  • Lost
    Lost almost 2 years
    @Michael Seifert Great Answers! Where does Michael Seifert's answer fit in the above argument? The above answer doesn't seem to explicitly make any restriction on the family of perturbations. I would guess that "However, the time averaged η˙ is still small. " has an implicit "smoothness assumption.
  • Andrew
    Andrew almost 2 years
    @Lost I think Michael and I are ultimately saying something very similar, but in different languages. I would summarize Michael's argument as, "we are assuming that we can take a smooth limit as the strength of the perturbation goes to zero, where the perturbative solution will vanish." As Michael says, this assumption is essentially restricting our attention to situations where this happens. That's connected to what I'm saying about how "you can check you made a self-consistent approximation." The smoothness in Michael's answer should guarantee that the perturbations remain small(...)
  • Andrew
    Andrew almost 2 years
    (...) which in turn implies that the effect of any large derivatives (coming from a large frequency component, for example) must average out of the solution. I'd have to think about it more to say something more precise. But I think these issues are related.
  • Michael Seifert
    Michael Seifert almost 2 years
    @Lost: In the formalism I'm describing in my answer, both $\eta$ and $\dot{\eta}$ remain small in magnitude instantaneously, not just in a time-averaged sense. Any solution for which $\dot{\eta}$ remains large in the limit $\lambda \to 0$ is ruled out by the smoothness assumption (otherwise that second derivative of $q_i(t;\lambda)$ wouldn't exist). I can't immediately see whether or not this time-averaging criterion is equivalent to the smoothness assumption; I'll have to think about it further.
  • Andrew
    Andrew almost 2 years
    I think maybe a subtlety is that $\dot\eta$ isn't dimensionless, so saying it is "big" or "small" isn't really meaningful. The reason I brought up that subtlety is that you can have "small" oscillations even for "large" frequencies. For simple harmonic motion, if the max amplitude is $A$, then the max amplitude of the time derivative is $A \omega$, so even if $A$ is "small" then $A \omega$ could be "large" if $\omega$ is large. A more careful definition should use a dimensionless quantity though. (...)
  • Andrew
    Andrew almost 2 years
    (...) Without thinking about it very long, a natural dimensionless ratio to use would be $\dot{\eta}/\omega$ where $\omega$ is the frequency of oscillation -- it wouldn't surprise me if Michael's answer is building in a dimensionless ratio like that somehow. I don't know if what I'm saying generalizes outside of simple harmonic motion, but at least in that case, since $1/\omega$ is a frequency-domain version of a time integral, this ratio also implies some kind of averaging in the time-domain.