Triple Pendulum
Consider the triple pendulum made out of three masses $m_1 = 2m$ and $m_2 = m_3 = m$ and attached on rods of negligible mass and the same length $\ell$ as depicted in the figure below and parameterized with the angles $\theta$, $\phi$ and $\chi$.
For small angles $\mid \theta \mid, \mid \phi \mid, \mid \chi \mid\ll 1$ the Lagrangian is given (up to an irrelevant constant) by,
$$L=\frac12 m \ell^2\left(2{\dot{\theta}}^2+\left({\dot{\theta}}+{\dot{\phi}}\right)^2+\left({\dot{\theta}}+{\dot{\chi}}\right)^2\right)-\frac12gm\ell\left(4\theta^2+\phi^2+\chi^2\right)\tag{1}$$
Show that we can perform a change of coordinates $(\theta, \phi, \chi)\to(q_1,q_2,q_3)$ so that the kinetic term is diagonalized and normalized, $T=({\dot q_1}^2+{\dot q_2}^2+{\dot q_3}^2)$.
In terms of the vector ${\bf {q}}=(q_1, q_2, q_3)$, show that the Lagrangian can be expressed as follows,
$$L=\frac12{\bf{\dot q}\cdot\bf{\dot q}}-\frac12{\bf qkq}\tag{2}$$
and find an expression for the matrix $\bf k$.
So in $(1)$, I will let ${\dot{q_1}}^2=2m{\ell}^2{\dot{\theta}}^2$, ${\dot{q_2}}^2=m{\ell}^2(\dot{\theta}+{\dot{\phi}})^2$ & $\,{\dot{q_3}}^2=m{\ell}^2(\dot{\theta}+{\dot{\chi}})^2$.
Inverting these in favour of the angles gives
$$\dot \theta=\frac{\dot{q_1}}{\sqrt{2m}{\ell}}\tag{a}$$
$$\dot \theta +\dot \phi=\frac{\dot{q_2}}{\sqrt m\ell}\tag{b}$$
$$\dot \theta +\dot \chi=\frac{\dot{q_3}}{\sqrt m\ell}\tag{c}$$
Insertion of $(\mathrm{a})$, $(\mathrm{b})$ & $(\mathrm{c})$ into the first (kinetic) term of eqn $(1)$ gives
$$T=\frac12m\ell^2\left[2\frac{{\dot{q_1}}^2}{2m\ell^2}+\frac{{\dot{q_2}}^2}{m\ell^2}+\frac{{\dot{q_3}}^2}{m\ell^2}\right]=\frac12\left[{\dot{q_1}}^2+{\dot{q_2}}^2+{\dot{q_3}}^2\right]=\frac12{\bf{\dot q}\cdot\bf{\dot q}}\tag{3}$$
So I can correctly get the kinetic, ($T$) part of eqn $(2)$, now using the same method to get the potential part ($V$), from $(\mathrm{a})$, $(\mathrm{b})$ & $(\mathrm{c})$ it follows that
$$\theta=\frac{{q_1}}{\sqrt{2m}{\ell}}\tag{d}$$
$$\phi=\frac{{q_2}}{\sqrt{m}{\ell}}-\frac{{q_1}}{\sqrt{2m}{\ell}}=\frac{1}{\sqrt{m}\ell}\left(q_2-\frac{q_1}{\sqrt2}\right)\tag{e}$$
$$\chi=\frac{{q_3}}{\sqrt{m}{\ell}}-\frac{{q_1}}{\sqrt{2m}{\ell}}=\frac{1}{\sqrt{m}\ell}\left(q_3-\frac{q_1}{\sqrt2}\right)\tag{f}$$
From $(\mathrm{d})$, $(\mathrm{e})$ & $(\mathrm{f})$ it follows that
$$\theta^2=\frac{{{q_1}^2}}{2m{\ell^2}}\tag{g}$$
$$\phi^2=\frac{1}{m{\ell^2}}\left(q_2-\frac{q_1}{\sqrt2}\right)\left(q_2-\frac{q_1}{\sqrt2}\right)=\frac{1}{m\ell^2}\left({q_2}^2-\sqrt 2q_1q_2+\frac12{q_1}^2\right)\tag{h}$$
$$\chi^2=\frac{1}{m{\ell^2}}\left(q_3-\frac{q_1}{\sqrt2}\right)\left(q_3-\frac{q_1}{\sqrt2}\right)=\frac{1}{m\ell^2}\left({q_3}^2-\sqrt 2q_1q_3+\frac12{q_1}^2\right)\tag{i}$$
Now, insertion of $(\mathrm{g})$, $(\mathrm{h})$ & $(\mathrm{i})$ into the second term of $(1)$, $V=\frac12gm\ell\left(4\theta^2+\phi^2+\chi^2\right)$, transforms to
$$\begin{align}V&=\frac{gm\ell}{2m{\ell}^2}\left(4\frac{{q_1}^2}{2}+{q_2}^2-\sqrt 2q_1q_2+\frac12{q_1}^2+{q_3}^2-\sqrt 2q_1q_3+\frac12{q_1}^2\right)\\&=\frac12\frac{g}{\ell}\left(2{q_1}^2+{q_1}^2+{q_2}^2+{q_3}^2-\sqrt 2q_1q_2-\sqrt 2q_1q_3\right)\\&=\frac{g}{2\ell}\left(3{q_1}^2+{q_2}^2+{q_3}^2-\sqrt 2q_1\left(q_2+q_3\right)\right)\tag{4}\end{align}$$
I have really laboured the point with the algebra here, but I did so on purpose so it would be easy to identify errors.
This is as far as I can get, and it doesn't look anything like the second term of eqn $(2)$ – $\frac12{\bf qkq}$ . But I can tell you that the last expression (for $V$) is correct as I have the author's solution below:
I have some questions regarding this solution.
- Why is the authors eqn $(29)$ different from my eqn $(3)$ and what is meant by "the kinetic term is diagonal"?
- My equation $(4)$ matches with the author's equation $(33)$ which is good. But how on earth do you deduce the kinetic matrix (sometimes known as 'mass matrix'), $\bf k$ from this?
- Taking the matrix $\bf k$ as given for a moment, if ${\bf {q}}=(q_1, q_2, q_3)$ is a row vector, then the only way to compute $\bf qkq$ as matrix multiplication is to write it as ${\bf q}{\bf k}{\bf q}^T$ because $${\bf q}{\bf k}{\bf q}^T=\frac{g}{\ell}\begin{pmatrix}q_1 & q_2 & q_3\end{pmatrix}\begin{pmatrix}
3 & -1/\sqrt2 & -1/\sqrt2 \\
-1/\sqrt2 & 1 & 0 \\
-1/\sqrt2 & 0 & 1 \\
\end{pmatrix}\begin{pmatrix}q_1 \\ q_2 \\ q_3\end{pmatrix}
$$Which is in the format required for matrix multiplication (row times column) as from left to right we have matrices with (1×3) times (3×3) times (3×1) with the result being a (1×1) matrix (scalar) as required by equation $(33)$ or my $(4)$. So why is $(2)$ not being written as ${\bf q}{\bf k}{\bf q}^T$?
Best Answer
1.) Equation (29) is an intermediate form for the variables $\bf \tilde q$, in the next step they get transformed so that your and the reference expression in $\bf q$ are the same.
2.) This is called "polarization", this can be done for any quadratic form. $\sum_{i\le j} a_{ij}x_ix_j$ can be written as ${\bf x}^TA{\bf x}$ with $A_{ii}=a_{ii}$ and $A_{ij}=A_{ji}=\frac12a_{ij}$ for $i<j$.
3.) This is indeed some sloppiness with the operations, one could write $\bf q·kq$ similar to the kinetic energy term, where $\bf q$ is a column vector and the dot signifies the scalar product.