# 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

${\displaystyle L={\frac {1}{2}}\sum _{i,j}T_{ij}{\dot {q}}_{i}{\dot {q}}_{j}-V\left(q_{1},\ldots q_{n}\right).}$

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

We can expand the potential energy to give

${\displaystyle V\left(q_{1},\ldots q_{n}\right)=V\left(q_{0,1},\ldots q_{0,n}\right)+\sum _{i}\left({\frac {\partial V}{\partial q_{i}}}\right)_{q_{0,i}}\eta _{i}+{\frac {1}{2}}\sum _{i,j}\left({\frac {\partial ^{2}V}{\partial q_{i}\partial q_{j}}}\right)_{q_{0,i}}\eta _{i}\eta _{j}+\cdots .}$

The first term is a constant with respect to ${\displaystyle \eta _{i}}$ and constant terms do not affect the motion. The second term is zero, because ${\displaystyle q_{0,i}}$ is a point of equilibrium so we are left with

${\displaystyle L={\frac {1}{2}}\sum _{i,j}\left(T_{ij}{\dot {\eta }}_{i}{\dot {\eta }}_{j}-V_{ij}\eta _{i}\eta _{j}\right)}$

where

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

yielding the equations of motion

${\displaystyle \sum _{j}\left(T_{ij}{\ddot {\eta }}_{j}-V_{ij}\eta _{j}\right)=0}$

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

${\displaystyle \eta _{i}=Ca_{i}e^{-i\omega t}}$

so we have

${\displaystyle \sum _{j}\left(V_{ij}a_{j}-\omega ^{2}T_{ij}a_{j}\right)=0.}$

This is a matrix equation such that

${\displaystyle {\vec {\vec {A}}}\cdot {\vec {a}}=0}$ with

${\displaystyle {\vec {a}}=\left[{\begin{matrix}a_{1}\\a_{2}\\\vdots \\a_{j}\end{matrix}}\right]}$

and

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

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

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

${\displaystyle {\vec {\vec {V}}}{\vec {a}}=\omega ^{2}{\vec {\vec {T}}}{\vec {a}}.}$

The matrix ${\displaystyle {\vec {\vec {V}}}}$ is symmetric and real. The matrix ${\displaystyle {\vec {\vec {T}}}}$ should be positive definite (because a negative kinetic energy doesn't make sense). Technical issue: If ${\displaystyle {\vec {\vec {T}}}}$ 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 ${\displaystyle {\vec {\vec {V}}}}$. Either way, you can drop the null space from both sides of the equation.

Assuming that ${\displaystyle {\vec {\vec {T}}}}$ is invertable we have

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

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.

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 ${\displaystyle q_{0,i}(t)}$ satisfy

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {q}}_{i}}}-{\frac {\partial L}{\partial q_{i}}}=0,}$

and let's look at

${\displaystyle q_{i}(t)=q_{0,i}(t)+\eta _{i}(t)}$

where ${\displaystyle \eta _{i}}$ is small. Let's expand the entire Lagrangian to find the equations of motion for the deviations ${\displaystyle \eta _{i}}$. We have

${\displaystyle L\left(q_{1},\ldots q_{n}\right)=L\left(q_{0,1},\ldots q_{0,n};{\dot {q}}_{0,1},\ldots {\dot {q}}_{0,n}\right)+\sum _{i}\left[\left({\frac {\partial L}{\partial q_{i}}}\right)_{q_{0,i}}\eta _{i}+\left({\frac {\partial L}{\partial {\dot {q}}_{i}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}\right]+}$

${\displaystyle {\frac {1}{2}}\sum _{i,j}\left[\left({\frac {\partial ^{2}L}{\partial q_{i}\partial q_{j}}}\right)_{q_{0,i}}\eta _{i}\eta _{j}+\left({\frac {\partial ^{2}L}{\partial {\dot {q}}_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}{\dot {\eta }}_{j}+2\left({\frac {\partial ^{2}L}{\partial q_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}\eta _{i}{\dot {\eta }}_{j}\right]+\cdots .}$

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

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\eta }}_{i}}}-{\frac {\partial L}{\partial \eta _{i}}}=0}$

to give

${\displaystyle {\frac {d}{dt}}\left\{\left({\frac {\partial L}{\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}+\sum _{i}\left[\left({\frac {\partial ^{2}L}{\partial {\dot {q}}_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}+\left({\frac {\partial ^{2}L}{\partial q_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}\eta _{i}\right]\right\}}$

${\displaystyle -\left({\frac {\partial L}{\partial q_{j}}}\right)_{q_{0,i}}-\sum _{i}\left[\left({\frac {\partial ^{2}L}{\partial q_{i}\partial q_{j}}}\right)_{q_{0,i}}\eta _{i}+\left({\frac {\partial ^{2}L}{\partial q_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}\right]=0}$

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

${\displaystyle {\frac {d}{dt}}\left\{\sum _{i}\left[\left({\frac {\partial ^{2}L}{\partial {\dot {q}}_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}+\left({\frac {\partial ^{2}L}{\partial q_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}\eta _{i}\right]\right\}}$

${\displaystyle -\sum _{i}\left[\left({\frac {\partial ^{2}L}{\partial q_{i}\partial q_{j}}}\right)_{q_{0,i}}\eta _{i}+\left({\frac {\partial ^{2}L}{\partial q_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}\right]=0.}$

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

${\displaystyle \sum _{i}\left[\left({\frac {\partial ^{2}L}{\partial {\dot {q}}_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\ddot {\eta }}_{i}-\left({\frac {\partial ^{2}L}{\partial q_{i}\partial q_{j}}}\right)_{q_{0,i}}\eta _{i}\right]=0.}$

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

${\displaystyle L=L_{0}+L_{1}}$

where ${\displaystyle L_{1}}$ is considered to be small compared to ${\displaystyle L_{0}}$ Let's say I have some solution to the equations of motion for ${\displaystyle L_{0}}$ and I would like to look at small deviations from the solution induced by the change in the Lagrangian. Let's say ${\displaystyle q_{0,i}(t)}$ satisfy

${\displaystyle {\frac {d}{dt}}{\frac {\partial L_{0}}{\partial {\dot {q}}_{i}}}-{\frac {\partial L_{0}}{\partial q_{i}}}=0,}$

and let's look at

${\displaystyle q_{i}(t)=q_{0,i}(t)+\eta _{i}(t)}$

where ${\displaystyle \eta _{i}}$ is small. Let's expand the entire Lagrangian to find the equations of motion for the deviations ${\displaystyle \eta _{i}}$. We have

${\displaystyle L\left(q_{1},\ldots q_{n}\right)=L\left(q_{0,1},\ldots q_{0,n};{\dot {q}}_{0,1},\ldots {\dot {q}}_{0,n}\right)+\sum _{i}\left[\left({\frac {\partial L}{\partial q_{i}}}\right)_{q_{0,i}}\eta _{i}+\left({\frac {\partial L}{\partial {\dot {q}}_{i}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}\right]+}$

${\displaystyle {\frac {1}{2}}\sum _{i,j}\left[\left({\frac {\partial ^{2}L}{\partial q_{i}\partial q_{j}}}\right)_{q_{0,i}}\eta _{i}\eta _{j}+\left({\frac {\partial ^{2}L}{\partial {\dot {q}}_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}{\dot {\eta }}_{j}+2\left({\frac {\partial ^{2}L}{\partial q_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}\eta _{i}{\dot {\eta }}_{j}\right]+}$

${\displaystyle L_{1}\left(q_{0,1},\ldots q_{0,n};{\dot {q}}_{0,1},\ldots {\dot {q}}_{0,n}\right)+\sum _{i}\left[\left({\frac {\partial L_{1}}{\partial q_{j}}}\right)_{q_{0,i}}\eta _{i}+\left({\frac {\partial L_{1}}{\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}\right]+\cdots .}$

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

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\eta }}_{i}}}-{\frac {\partial L}{\partial \eta _{i}}}=0}$

to give

${\displaystyle {\frac {d}{dt}}\left\{\left({\frac {\partial L_{0}}{\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}+\left({\frac {\partial L_{1}}{\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}+\sum _{i}\left[\left({\frac {\partial ^{2}L_{0}}{\partial {\dot {q}}_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}+\left({\frac {\partial ^{2}L_{0}}{\partial q_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}\eta _{i}\right]\right\}}$

${\displaystyle -\left({\frac {\partial L_{0}}{\partial q_{j}}}\right)_{q_{0,i}}+\left({\frac {\partial L_{1}}{\partial q_{j}}}\right)_{q_{0,i}}+\sum _{i}\left[\left({\frac {\partial ^{2}L_{0}}{\partial q_{i}\partial q_{j}}}\right)_{q_{0,i}}\eta _{i}+\left({\frac {\partial ^{2}L_{0}}{\partial q_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}\right]=0}$

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

${\displaystyle {\frac {d}{dt}}\left\{\sum _{i}\left[\left({\frac {\partial ^{2}L_{0}}{\partial {\dot {q}}_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}+\left({\frac {\partial ^{2}L_{0}}{\partial q_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}\eta _{i}\right]\right\}}$

${\displaystyle -\sum _{i}\left[\left({\frac {\partial ^{2}L_{0}}{\partial q_{i}\partial q_{j}}}\right)_{q_{0,i}}\eta _{i}+\left({\frac {\partial ^{2}L_{0}}{\partial q_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\dot {\eta }}_{i}\right]=\left({\frac {\partial L_{1}}{\partial q_{j}}}\right)_{q_{0,i}}-{\frac {d}{dt}}\left({\frac {\partial L_{1}}{\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}.}$

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

${\displaystyle \sum _{i}\left[\left({\frac {\partial ^{2}L_{0}}{\partial {\dot {q}}_{i}\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}{\ddot {\eta }}_{i}-\left({\frac {\partial ^{2}L_{0}}{\partial q_{i}\partial q_{j}}}\right)_{q_{0,i}}\eta _{i}\right]=\left({\frac {\partial L_{1}}{\partial q_{j}}}\right)_{q_{0,i}}-{\frac {d}{dt}}\left({\frac {\partial L_{1}}{\partial {\dot {q}}_{j}}}\right)_{q_{0,i}}.}$

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