Advanced Classical Mechanics/Small Oscillations and Perturbed Motion

In Linear Motion, we argued that all sufficiently small oscillations are harmonic. In this section we will exploit this result in several ways to understand


 * 1) The motion of systems with many degrees of freedom near equilibrium,
 * 2) The motion of systems perturbed from known solutions, and
 * 3) The motion of systems with Lagrangians perturbed from systems with known solutions.

All three of these points are applications of perturbation theory, and they all start with the harmonic oscillator.

Normal Modes
The modes of oscillation of systems near equilibrium are called the normal modes of the system. Understanding the frequencies of the normal modes of the system is crucial to design a system that can move (even if it isn't meant to). Let's look at a system with many degrees of freedom; we have

$$ L = \frac{1}{2} \sum_{i,j} T_{ij} \dot q_i \dot q_j - V \left (q_1, \ldots q_n \right ). $$

Let $$q_{0,i}$$ be an equilibrium position and expand about this point $$ q_i = q_{0,i} + \eta_i $$ so $$\dot q_i = \dot \eta _i$$.

We can expand the potential energy to give

$$ V \left (q_1, \ldots q_n \right ) = V \left (q_{0,1}, \ldots q_{0,n} \right ) + \sum_{i} \left ( \frac{\partial V}{\partial q_i} \right )_{q_{0,i}} \eta_i + \frac{1}{2} \sum_{i,j} \left ( \frac{\partial^2 V}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i\eta_j + \cdots. $$

The first term is a constant with respect to $$\eta_i$$ and constant terms do not affect the motion. The second term is zero, because $$q_{0,i}$$ is a point of equilibrium. If we drop all terms higher than the third term shown, we are left with

$$ L = \frac{1}{2} \sum_{i,j} \left ( T_{ij} \dot \eta_i \dot \eta_j - V_{ij} \eta_i \eta_j \right ) $$

where

$$ T_{ij} = T_{ij} \left ( q_{0,1}, \ldots q_{0,n} \right ) $$ and $$ V_{ij} = \left ( \frac{\partial^2 V}{\partial q_i \partial q_j} \right )_{q_{0,i}}, $$

yielding the equations of motion

$$ \sum_{j} \left ( T_{ij} \ddot \eta_j - V_{ij} \eta_j \right ) = 0 $$

This is a linear differential equation with constant coefficients. We can try the solution

$$ \eta_i = C a_i e^{-i \omega t} $$

so we have

$$ \sum_{j} \left ( V_{ij} a_j - \omega^2 T_{ij} a_j \right ) = 0. $$

This is a matrix equation such that

$$ \vec \vec A \cdot \vec a = 0 $$ with

$$ \vec a = \left [ \begin{matrix} a_1 \\ a_2 \\ \vdots \\ a_j \end{matrix} \right ] $$

and

$$ \vec \vec A = \left [ \begin{matrix} V_{11} - \omega^2 T_{11} & V_{12} - \omega^2 T_{12} & \cdots \\ V_{21} - \omega^2 T_{21} & V_{22} - \omega^2 T_{22} & \cdots \\ \vdots & & \end{matrix} \right ] $$

This equation only has a solution is $$\det \vec \vec A=0$$. This gives a $$n$$th-degree polynomial to solve for $$\omega^2$$. We will get $$n$$ solutions for $$\omega^2$$ that we can substitute into the matrix equation and solve for $$a_j$$.

Is this guaranteed to work? Yes, it turns out. Look at the equation in terms of matrices we have

$$ \vec \vec V \vec a = \omega^2 \vec \vec T \vec a. $$

The matrix $$\vec \vec V$$ is symmetric and real. The matrix $$\vec \vec T$$ should be positive definite (because a negative kinetic energy doesn't make sense). Technical issue: If $$\vec \vec T$$ has a null space, the degrees of freedom corresponding to the null space are massless and cannot be excited unless they are in the null space of $$\vec \vec V$$. Either way, you can drop the null space from both sides of the equation.

Assuming that $$\vec \vec T$$ is invertable we have

$$ \left ( \vec\vec T \right ) ^{-1} \vec \vec V \vec a = \omega^2 \vec a $$

and we have a standard eigenvalue equation. In most examples, the kinetic energy matrix will be diagonal, so it is straightforward to construct the quotient matrix and diagonize it.

Perturbations about Steady Motion
Let's say I have some solution to the equations of motion and I would like to look at small deviations from the solution. Let's $$ q_{0,i}(t) $$ satisfy

$$ \frac{d}{dt} \frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i} = 0, $$

and let's look at

$$ q_i(t) = q_{0,i}(t) + \eta_i(t) $$

where $$\eta_i$$ is small. Let's expand the entire Lagrangian to find the equations of motion for the deviations $$\eta_i$$. We have

$$ L \left (q_1, \ldots q_n \right ) = L \left (q_{0,1}, \ldots q_{0,n} ; \dot q_{0,1}, \ldots \dot q_{0,n} \right ) + \sum_{i} \left [ \left ( \frac{\partial L}{\partial q_i} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial L}{\partial \dot q_i} \right )_{q_{0,i}} \dot \eta_i \right ] + $$

$$ \frac{1}{2} \sum_{i,j} \left [ \left ( \frac{\partial^2 L}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i\eta_j + \left ( \frac{\partial^2 L}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \dot\eta_i \dot \eta_j + 2 \left ( \frac{\partial^2 L}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \eta_i \dot \eta_j \right ] + \cdots. $$

Now let's apply Lagrange's equations for the deviations

$$ \frac{d}{dt} \frac{\partial L}{\partial \dot \eta_i} - \frac{\partial L}{\partial \eta_i} = 0 $$

to give

$$ \frac{d}{dt} \left \{ \left ( \frac{\partial L}{\partial \dot q_j} \right )_{q_{0,i}} + \sum_{i} \left [ \left ( \frac{\partial^2 L}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \dot\eta_i + \left ( \frac{\partial^2 L}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \eta_i \right ] \right \} $$

$$ - \left ( \frac{\partial L}{\partial q_j} \right )_{q_{0,i}} - \sum_{i} \left [ \left ( \frac{\partial^2 L}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial^2 L}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \dot \eta_i \right ] = 0 $$

The two terms without $$\eta$$ actually cancel each other out, leaving the following equations of motion.

$$ \frac{d}{dt} \left \{ \sum_{i} \left [ \left ( \frac{\partial^2 L}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \dot\eta_i + \left ( \frac{\partial^2 L}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \eta_i \right ] \right \} $$

$$ - \sum_{i} \left [ \left ( \frac{\partial^2 L}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial^2 L}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \dot \eta_i \right ] = 0. $$

In steady motion, the partial derivatives are taken to be constant in time yielding the even simpler result

$$ \sum_{i} \left [ \left ( \frac{\partial^2 L}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \ddot\eta_i - \left ( \frac{\partial^2 L}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i \right ] = 0. $$

Again we have a linear differential equation with constant coefficients, and all of the results from the previous section carry over.

Perturbed Lagrangians
What about finding solutions to Lagrangians that are almost like ones that we have already solved? Let's say we have

$$ L = L_0 + L_1 $$

where $$L_1$$ is considered to be small compared to $$L_0$$ Let's say I have some solution to the equations of motion for $$L_0$$ and I would like to look at small deviations from the solution induced by the change in the Lagrangian. Let's say $$ q_{0,i}(t) $$ satisfy

$$ \frac{d}{dt} \frac{\partial L_0}{\partial \dot q_i} - \frac{\partial L_0}{\partial q_i} = 0, $$

and let's look at

$$ q_i(t) = q_{0,i}(t) + \eta_i(t) $$

where $$\eta_i$$ is small. Let's expand the entire Lagrangian to find the equations of motion for the deviations $$\eta_i$$. We have

$$ L \left (q_1, \ldots q_n \right ) = L \left (q_{0,1}, \ldots q_{0,n} ; \dot q_{0,1}, \ldots \dot q_{0,n} \right ) + \sum_{i} \left [ \left ( \frac{\partial L}{\partial q_i} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial L}{\partial \dot q_i} \right )_{q_{0,i}} \dot \eta_i \right ] + $$

$$ \frac{1}{2} \sum_{i,j} \left [ \left ( \frac{\partial^2 L}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i\eta_j + \left ( \frac{\partial^2 L}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \dot\eta_i \dot \eta_j + 2 \left ( \frac{\partial^2 L}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \eta_i \dot \eta_j \right ] + $$

$$ L_1 \left ( q_{0,1}, \ldots q_{0,n}; \dot q_{0,1}, \ldots \dot q_{0,n} \right ) + \sum_{i} \left [ \left ( \frac{\partial L_1}{\partial q_j} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial L_1}{\partial \dot q_j} \right )_{q_{0,i}} \dot \eta_i \right ] + \cdots. $$

Now let's apply Lagrange's equations for the deviations

$$ \frac{d}{dt} \frac{\partial L}{\partial \dot \eta_i} - \frac{\partial L}{\partial \eta_i} = 0 $$

to give

$$ \frac{d}{dt} \left \{ \left ( \frac{\partial L_0}{\partial \dot q_j} \right )_{q_{0,i}} + \left ( \frac{\partial L_1}{\partial \dot q_j} \right )_{q_{0,i}} + \sum_{i} \left [ \left ( \frac{\partial^2 L_0}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \dot\eta_i + \left ( \frac{\partial^2 L_0}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \eta_i \right ] \right \} $$

$$ - \left ( \frac{\partial L_0}{\partial q_j} \right )_{q_{0,i}} + \left ( \frac{\partial L_1}{\partial q_j} \right )_{q_{0,i}} + \sum_{i} \left [ \left ( \frac{\partial^2 L_0}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial^2 L_0}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \dot \eta_i \right ] = 0 $$

The two lowest orders terms without $$\eta$$ actually cancel each other out, leaving the following equations of motion.

$$ \frac{d}{dt} \left \{ \sum_{i} \left [ \left ( \frac{\partial^2 L_0}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \dot\eta_i + \left ( \frac{\partial^2 L_0}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \eta_i \right ] \right \} $$

$$ - \sum_{i} \left [ \left ( \frac{\partial^2 L_0}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial^2 L_0}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \dot \eta_i \right ] = \left ( \frac{\partial L_1}{\partial q_j} \right )_{q_{0,i}} - \frac{d}{dt} \left ( \frac{\partial L_1}{\partial \dot q_j} \right )_{q_{0,i}}. $$

Let's specialize and assume that the unperturbed motion is steady so the partial derivatives of the unperturbed Lagrangian are constant in time, to obtain

$$ \sum_{i} \left [ \left ( \frac{\partial^2 L_0}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \ddot\eta_i - \left ( \frac{\partial^2 L_0}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i \right ] = \left ( \frac{\partial L_1}{\partial q_j} \right )_{q_{0,i}} - \frac{d}{dt} \left ( \frac{\partial L_1}{\partial \dot q_j} \right )_{q_{0,i}}. $$

which is the equation of a coupled set of driven harmonic oscillators.