### Near the surface of the earth

Figure 5.2 shows a double pendulum, with two equal masses, $m$ and rods of equal length $r$ with motion restricted to the plane. The only coordinates that change are the angles, $\theta, \phi$.

Nb. We could have different masses and lengths, but it is still far more complicated than a single pendulum. In fact, in general, the motion will be chaotic. Using Newton's Law to examine this system would be very hard indeed.

We first consider the system near the surace of the earth. The kinetic and potential energies for the mass at $P$ are the same as for the single pendulum, that is $$\begin{align*}T_P &= \tfrac{1}{2} mr^2\dot{\theta}^2\\U_P &= -mgr \cos\theta\end{align*}$$

For the mass at $Q$, we note that $$\begin{align*}d_2 &= r(\sin \theta + \sin \phi)\\h_2 &= r(\cos \theta + \cos \phi)\end{align*}$$ and taking the time-derivatives, $$\begin{align*}\dot{d_2} &= r(\dot{\theta} \cos \theta + \dot{\phi} \cos \phi)\\\dot{h_2} &= -r(\dot{\theta} \sin \theta + \dot{\phi} \sin \phi)\end{align*}$$

So the kinetic energy of the mass at $Q$ is $$T_Q = \tfrac{1}{2} m r^2 \left(\dot{\theta}^2 + \dot{\phi}^2 + 2\dot{\theta} \dot{\phi} \cos(\theta - \phi)\right)$$

The potential energy is also straightforward $$U_Q = -mgh_2 = -mgr(\cos\theta + \cos \phi)$$

So the *total* kinetic and potential energies are $$\begin{align*}T &= mr^2\dot{\theta}^2 + \tfrac{1}{2} m r^2 \dot{\phi}^2 + mr^2 \dot{\theta} \dot{\phi} \cos(\theta – \phi)\\U &= -2mgr \cos\theta -mgr \cos \phi\end{align*}$$ and the Lagrangian is $$L = mr^2\dot{\theta}^2 + \tfrac{1}{2} m r^2 \dot{\phi}^2 + mr^2 \dot{\theta} \dot{\phi} \cos(\theta – \phi) + 2mgr \cos\theta + mgr \cos \phi$$

Apart from the energy, or Hamiltonian, of the system, there aren't any conserved quantities, due to the gravitational potential, which does change under small variations in the coordinates $\theta,\phi$. So, no conservation of angular momentum or, equivalently, no rotational symmetry.

### Free space

Suppose we move the system out into space where there is no gravitational potential. Then, the Lagrangian reduces to just the kinetic energy $$L = mr^2\dot{\theta}^2 + \tfrac{1}{2} m r^2 \dot{\phi}^2 + mr^2 \dot{\theta} \dot{\phi} \cos(\theta – \phi)$$

Now, we do have rotational symmetry. That is, the action doesn't change if we vary the coordinates, $$\begin{align}\theta &\to \theta + \epsilon\\\phi &\to \phi + \epsilon\end{align}$$

To see this, note that none of $ \dot{\theta}, \dot{\phi}, \cos(\theta – \phi)$ change under the transformation, hence the Lagrangian doesn't change, and so the action doesn't change.

The Noether charge is the sum $$\pi_\theta f_\theta + \pi_\phi f_\phi = \pi_\theta + \pi_\phi$$ since, from the transformation, $$f_\theta = f_\phi = 1$$

The conjugate momenta are $$\begin{align*}\pi_\theta &= 2mr^2\dot{\theta} + mr^2 \dot{\phi} \cos(\theta – \phi)\\\pi_\phi &= mr^2\dot{\phi} + mr^2 \dot{\theta} \cos(\theta – \phi)\end{align*}$$ and so the conserved quantity - the total angular momentum of the system - is $$\pi_\theta + \pi_\phi = 2mr^2\dot{\theta} + mr^2\dot{\phi} + mr^2 (\dot{\theta} + \dot{\phi}) \cos(\theta – \phi)$$

The terms in the Euler-Lagrange equations are $$\begin{align*}\frac{\mathrm{d} \pi_\theta}{\mathrm{d} t} &= 2mr^2\ddot{\theta} + mr^2\ddot{\phi}\cos(\theta–\phi) - mr^2(\dot{\theta}-\dot{\phi})\dot{\phi}\sin(\theta–\phi)\\ \frac{\partial L}{\partial \theta} &= -mr^2\dot{\theta}\dot{\phi}\sin(\theta–\phi)\\ \frac{\mathrm{d} \pi_\phi}{\mathrm{d} t} &= mr^2\ddot{\phi} + mr^2\ddot{\theta}\cos(\theta–\phi) - mr^2\dot{\theta}(\dot{\theta}-\dot{\phi})\sin(\theta–\phi)\\ \frac{\partial L}{\partial \phi} &= +mr^2\dot{\theta}\dot{\phi}\sin(\theta–\phi)\end{align*}$$ so the equations of motion become $$\begin{align*}2\ddot{\theta} + \ddot{\phi}\cos(\theta–\phi) + \dot{\phi}^2\sin(\theta–\phi) &= 0\\\ddot{\phi} + \ddot{\theta}\cos(\theta–\phi) - \dot{\theta}^2\sin(\theta–\phi) &= 0\end{align*}$$

Even in free space, these equations would have been harder to derive from Newton's Law.