\[% UTILITIES %
\newcommand{\half}{\frac{1}{2}}%
\renewcommand{\vec}[1]{\mathbf{#1}}%
\def\dd#1{\,\text{d}}%
\def\ddt#1{\frac{\text{d} #1}{\text{d}t}}%
\def\df#1{\frac{\text{d}}{\text{d} #1}}%
\def\ddf#1#2{\frac{\text{d} #1}{\text{d} #2}}%
\def\dddf#1#2#3{\frac{\text{d}^{#3} #1}{\text{d} #2^{#3}}}%
\def\cp{\times}%
% BASIC PHYSICS %
\newcommand{\PHacceleration}[1][x]{{\color{red} \vec{#1}}}%
\newcommand{\PHmass}[1][m]{{\color{green} #1}}%
\newcommand{\PHlength}[1][l]{{\color{maroon} #1}}%
\newcommand{\PHenergy}[1][E]{{\color{blue} #1}}%
% DISTANCE-LIKE %
\newcommand{\CMdistance}[1][x]{\PHlength[#1]}%
\newcommand{\CMposition}[1][x]{\vec{#1}}%
\newcommand{\CMvelocity}[1][x]{\ddt{\CMposition[#1]}}%
\newcommand{\CMacceleration}[1][x]{\dddf{\CMposition[#1]}{t}{2}}%
% MASS-LIKE %
\newcommand{\CMmass}[1][m]{\PHmass[m]}%
% PHYSICAL CONSTANTS %
\newcommand{\CMg}{\PHacceleration[g]}%\]
In order to motivate a more general solution to systems which involve oscillatory motion, first consider the double pendulum in which a mass is suspended on a chain of two conjoined pendula.
The Double Pendulum
Consider a double pendulum, consisting of two bobs, one hung below the
other. Each has length \(a\), and bobs of mass \(m\), so the
potential energy is
\[\PHenergy[V] = -\CMmass \CMg \CMdistance[a] \cos(\theta) - \CMmass \CMg \CMdistance[a] \left( \cos(\theta) + \cos(\phi) \right)\]
and the kinetic energy is
\[\PHenergy[T] = \half \CMmass a^2 \ddt{\theta}^2 + \half \CMmass a^2 \left( \ddt{\theta} + \ddt{\phi} \right)^2\]
Then,
\[\begin{split}\begin{aligned}
L & = \half \CMmass a^2 \dot{\theta}^2 + \half \CMmass a^2 (\dot{\theta}^2 + \dot{\phi}^2 + 2 \dot{\theta} \dot{\phi}) + \CMmass \CMg \CMdistance[a] (2 \cos(\theta) + \cos(\phi) ) \\
&\approx \half \CMmass \CMdistance[a]^2 \dot{\theta}^2 + \half \CMmass \CMdistance[a]^2 (\dot{\theta}^2 + \dot{\phi}^2 + 2 \dot{\theta} \dot{\phi}) - \CMmass \CMg \CMdistance[a] \left(\theta^2 + \half \phi^2 \right)
\end{aligned}\end{split}\]
The equations of motion from the Lagrange equations are
\[\begin{split}\begin{aligned}
2 \ddot{\theta} + \ddot{\phi} + \frac{2 \CMg}{a} \theta & =0 \\
\ddot{\theta} + \ddot{\phi} + \frac{\CMg}{a} \phi &= 0
\end{aligned}\end{split}\]
Each of these equations has a form comparable to that of a harmonic oscillator, \(\ddot{x} + \omega^2 x = 0\); attempting a trial solution
\[\begin{split}\begin{bmatrix}
\theta \\ \phi
\end{bmatrix}
=
\begin{bmatrix}
c_{\theta} e^{i \omega t} \\ c_{\phi} e^{i \omega t}
\end{bmatrix}\end{split}\]
For \(c_{\theta}\), \(c_{\phi}\) complex constants, then
\[\ddot{\theta} = - \omega^2 \theta, \qquad \ddot{\phi} = - \omega^2 \phi\]
So
\[\begin{split}\begin{aligned}
\label{eq:39}
\qty( - 2 \omega^2 + \frac{2g}{a} ) c_{\theta} - \omega^2 c_{\phi} &= 0 \\
\label{eq:38}
- \omega^2 c_{\theta} + \qty( -\omega^2 + \frac{g}{a} ) c_{\phi} &= 0
\end{aligned}\end{split}\]
Which can be expressed in matrix notation
\[\begin{split}\label{eq:37}
\begin{bmatrix}
2 \frac{g}{a} - 2 \omega^2 & - \omega^2 \\ - \omega^2 & \frac{g}{a} - \omega^2
\end{bmatrix}
\begin{bmatrix}
c_{\theta} \\ c_{\phi}
\end{bmatrix} = 0\end{split}\]
This implies that the determinant of the matrix must be zero, so
\[\begin{split}\label{eq:40}
\begin{vmatrix}
2 \frac{g}{a} - 2 \omega^2 & - \omega^2 \\ - \omega^2 & \frac{g}{a} - \omega^2
\end{vmatrix} = 2 \qty( \frac{g}{a} - \omega^2 )^2 - \omega^4 = 0\end{split}\]
This has two solutions,
\[\label{eq:41}
\omega^2 = \frac{g}{a} ( 2 \pm \sqrt{2})\]
which are the normal frequencies for the system, and the coordinates \(c_i\) are the normal modes. To find these we substitute the normal frequencies into equation,
\[\begin{split}\label{eq:42}
\frac{g}{a}
\begin{bmatrix}
2-4-2 \sqrt{2} & -2-\sqrt{2} \\ -2 -\sqrt{2} & 1-2-\sqrt{2}
\end{bmatrix}
\begin{bmatrix}
c_{\theta} \\ c_{\phi}
\end{bmatrix} = 0\end{split}\]
These turn out to give two copies of the same equation relating the coefficients, so clearly only the relative relation of them is fixed,
\[\label{eq:43}
\frac{c_{\theta}}{c_{\phi}} = - \frac{2 + \sqrt{2}}{2(1+\sqrt{2})} = - \frac{(2+\sqrt{2})(1-\sqrt{2})}{2(1+\sqrt{2})(1-\sqrt{2})} = -\frac{1}{\sqrt{2}}\]
Thus
\[\begin{split}\begin{bmatrix}
c_{\theta} \\ c_{\phi}
\end{bmatrix} \propto
\begin{bmatrix}
-1 \\ \sqrt{2}
\end{bmatrix}\end{split}\]
and using the negative solution
\[\begin{split} \begin{bmatrix}
c_{\theta} \\ c_{\phi}
\end{bmatrix} \propto
\begin{bmatrix}
1 \\ \sqrt{2}
\end{bmatrix}\end{split}\]
Giving a general solution
\[\begin{split} \begin{bmatrix} \theta \\ \phi \end{bmatrix}
= \alpha_1 \begin{bmatrix} -1 \\ \sqrt{2} \end{bmatrix} e^{i \omega_1 t} + \alpha_2 \begin{bmatrix} 1 \\ \sqrt{2} \end{bmatrix} e^{i \omega_2 t}\end{split}\]
This can be rewritten in matrix form too,
\[\begin{split}\label{eq:46}
\begin{bmatrix} \theta \\ \phi \end{bmatrix} =
\begin{bmatrix} - 1 & 1 \\ \sqrt{2} & \sqrt{2} \end{bmatrix}
\begin{bmatrix} \alpha_1 e^{i \omega_1 t} \\ \alpha_2 e^{i \omega_2 t} \end{bmatrix}\end{split}\]
This can be inverted, giving
\[\begin{split}\label{eq:47}
\begin{bmatrix} \alpha_1 e^{i \omega_1 t} \\ \alpha_2 e^{i \omega_2 t} \end{bmatrix} =
\begin{bmatrix} - \half & \half \sqrt{2} \\ \half & \half \sqrt{2} \end{bmatrix}
\begin{bmatrix} \theta \\ \phi \end{bmatrix} =
\begin{bmatrix} \xi_1 \\ \xi_2 \end{bmatrix}\end{split}\]
For \(\xi_i\) the normal coordinates of the system, these cause
the Lagrange equations to completely decouple.
General Theory of Small Oscillations
Consider a system with time-independent constraints; this is in
equilibrium if
\[ \begin{align}\begin{aligned}\pdv{V}{q_i} = 0\\Furthermore, a stable equilibrium has\end{aligned}\end{align} \]
\[\pdv[2]{V}{q_i}{q_j} > 0 \quad \forall i, j\]
Denoting the equilibrium value of each coordinate \(q^{*}_i\), we
can introduce a small perturbation, \(\eta_i\), such that
\[ \begin{align}\begin{aligned} \label{eq:48}
q_i = q^{*}_i + \eta_i\\Assuming small displacements we can use Taylor’s theorem to expand the\end{aligned}\end{align} \]
potential about \(q_i = q_i^{*}\):
\[ \begin{align}\begin{aligned} \label{eq:49}
V = V^{*} + \sum \eta_i \pdv{V}{q_i} + \sum_{i,j} \half \qty( \pdv[2]{V}{q_i}{q_j} ) \eta_i \eta_j\\to the second-order. The potential thus has the form of the second\end{aligned}\end{align} \]
derivative term,
\[ \begin{align}\begin{aligned} \label{eq:50}
V = \sum_{i,j} \half V_{,ij} \eta_i \eta_j\\Using the comma notation for derivatives. The matrix :math:`\mat{V}`\end{aligned}\end{align} \]
has components \(V_{,ij}\), and the set of displacements
\(\eta_i\) forms a vector \(\vec{\eta}\), so
\[ \begin{align}\begin{aligned} \label{eq:51}
V = \half \trans{\eta} \mat{V} \eta\\Since :math:`\mat{V}` doesn’t depend upon the coordinates, just the\end{aligned}\end{align} \]
equilibrium values, it is constant. The kinetic energy has the form
\[ \begin{align}\begin{aligned} \label{eq:52}
T = \sum_{i,j} \half m_{ij} \dot{q}_i \dot{q}_j\\Expanding about the equilibrium we find\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\dot{q}_i = \dot{\eta}_i\\and\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}m_{ij}(q_1, \dots, q_n) = m_{ij}(q_1^{*}, \dots, q_n^{{*}}) + \cdots\\Then\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:53}
T = \sum_{i,j} \half T_{ij} \dot{\eta}_i \dot{\eta}_j\\having defined :math:`T_{ij} = m_{ij}(q^*_1, \dots, q^{*}_n)`, which is\end{aligned}\end{align} \]
a constant matrix, so we have
\[ \begin{align}\begin{aligned} \label{eq:54}
T = \half \trans{\dot{\eta}} \mat{T} \dot{\eta}\\and\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:55}
L = T-V = \sum_{i,j} \half \qty[ T_{ij} \dot{\eta}_i \dot{\eta}_j - V_{ij} \eta_i \eta_j] = \half \trans{\dot{\eta}} \mat{T} \dot{\eta} - \half \trans{\eta} \mat{V} \eta\\The Lagrange equations have the form\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\dv{t} \pdv{L}{\dot{\eta}_k} = \pdv{L}{\eta_k}\\Taking this in bits,\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\pdv{L}{\dot{\eta}_k} = \half \sum_{i,j} T_{ij} \qty[ \pdv{\dot{\eta}_i}{\dot{\eta}_k} + \dot{\eta}_i \pdv{\dot{\eta}_j}{\dot{\eta}_k}]\\The generalised coordinates are independent, so\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\pdv{\dot{\eta}_i}{\dot{\eta}_k} = \delta_{ik}\\Thus\end{aligned}\end{align} \]
\[\label{eq:56}
\pdv{L}{\dot{\eta}_k} = \half \sum_{i,j} T_{ij} \qty[ \delta_{ik} \dot{\eta}_j + \dot{\eta}_i \delta_{jk}] = \sum_j T_{kj} \dot{\eta}_j\]
Similarly,
\[\label{eq:57}
\pdv{L}{\eta_k} = - \sum_j V_{,kj} \eta_j\]
This gives the Lagrange equations in the form
\[ \begin{align}\begin{aligned} \label{eq:58}
\sum_j (T_{ij} \ddot{\theta}_j + V_{,ij} \eta_j) = 0 \equiv \mat{T} \ddot{\vec{\eta}} + \mat{V} \vec{\eta}_j = \vec{0}\\As in the double pendulum case, we can find solutions of the form\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\eta = \vec{c} e^{i \omega t}\\Then :math:`\ddot{\eta} = - \omega^2 \eta`, and so\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:59}
\qty( \mat{V} - \omega^2 \mat{T} ) \eta = \vec{0}\\To satisfy the equation we need\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:60}
\abs{\mat{V}-\omega^2 \mat{T}} = 0\\which is a characteristic equation, and this can be approached as an\end{aligned}\end{align} \]
eigenvalue equation, and once the normal frequencies are found we
substitute them back in turn, and find the vectors \(\vec{c}\) by
solving
\[ \begin{align}\begin{aligned}\mat{V} \vec{c} = \omega^2 \mat{T} \vec{c}\\for each :math:`\omega^2`. THis is akin to finding the eigenvectors of\end{aligned}\end{align} \]
a matrix, and the vectors specify the normal modes of the oscillation.
In general, from the fact that \(\mat{V}\) and \(\mat{T}\) are
symmetric, and that \(\omega^2\) has real solutions, that
\(\vec{c}\) can be chosen to be orthonormal. The general solution
takes the form
\[ \begin{align}\begin{aligned}\begin{split} \label{eq:80}
\begin{pmatrix}
\eta_1 \\ \vdots \\ \eta_n
\end{pmatrix}
= \sum_j \alpha_j \vec{c}^{(j)} \exp( i \omega_j t) = \sum_j \alpha_j
\begin{pmatrix}
c_1^{(j)} \\ \vdots c_n^{(j)}
\end{pmatrix}
\exp(i \omega_j t)\end{split}\\which can be expressed more compactly, using Einstein notation,\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:81}
\eta_i = c_i^j \alpha_j \exp(i \omega_j t)\\for :math:`c_i^j = c_i^{(j)}`. We can then define normal coordinates,\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\xi_j = \alpha_j \exp(i \omega_j t)\\which correspond to the oscillation of the system at a single\end{aligned}\end{align} \]
frequency, and can be found by inverting equation ([eq:80]), so
\[ \begin{align}\begin{aligned} \label{eq:82}
\xi_j = (c_i^j)^{-1} \eta_i\\The Lagrangian then completely decouples into the sum of independent\end{aligned}\end{align} \]
harmonic oscillators,
\[ \begin{align}\begin{aligned} \label{eq:87}
L = \sum_j C_j \qty[ \dot{\xi}_j^2 - \omega^2_j \xi_j^2]\\for :math:`C_j` a normalisation constant.\end{aligned}\end{align} \]
