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# Archive for October 11th, 2009

## Hamiltonian treatment of rigid Spherical pendulum.

Posted by peeterjoot on October 11, 2009

For the spherical rigid pendulum of length $l$, we have for the distance above the lowest point

\begin{aligned}h = l (1 + \cos\theta)\end{aligned} \quad\quad\quad(157)

(measuring $\theta$ down from the North pole as conventional). The Lagrangian is therefore

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m l^2 (\dot{\theta}^2 + \sin^2\theta \dot{\phi}^2) - m g l (1 + \cos\theta)\end{aligned} \quad\quad\quad(158)

We can drop the constant term, using the simpler Lagrangian

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m l^2 (\dot{\theta}^2 + \sin^2\theta \dot{\phi}^2) - m g l \cos\theta\end{aligned} \quad\quad\quad(159)

To express the Hamiltonian we need first the conjugate momenta, which are

\begin{aligned}P_\theta &= \frac{\partial {\mathcal{L}}}{\partial {\dot{\theta}}} = m l^2 \dot{\theta} \\ P_\phi &= \frac{\partial {\mathcal{L}}}{\partial {\dot{\phi}}} = m l^2 \sin^2\theta \dot{\phi}\end{aligned} \quad\quad\quad(160)

We can now write the Hamiltonian

\begin{aligned}H = \frac{1}{{2 m l^2}} \left({P_\theta}^2 + \frac{1}{{\sin^2\theta}} {P_\phi}^2\right) + m g l \cos\theta\end{aligned} \quad\quad\quad(162)

Before going further one sees that there is going to be trouble where $\sin\theta = 0$. Curiously, this is at the poles, the most dangling position and the upright. The south pole is the usual point where we solve the planar pendulum problem using the harmonic oscillator approximation, so it is somewhat curious that the energy of the system appears to go undefined at this point where the position is becoming more defined. It seems almost like a quantum uncertainty phenomena until one realizes that the momentum conjugate to $\phi$ is itself proportional to $\sin^2 \theta$. By expressing the energy in terms of this $P_\phi$ momentum we have to avoid looking at the poles for a solution to the equations. If we go back to the Lagrangian and the Euler-Lagrange equations, this point becomes perfectly tractable since we are no longer dividing through by $\sin^2\theta$.

Examining the polar solutions is something to return to. For now, let’s avoid that region. For regions where $\sin\theta$ is nicely non-zero, we get for the Hamiltonian equations

\begin{aligned}\frac{\partial {H}}{\partial {P_\phi}} &= \dot{\phi} = \frac{1}{{ m l^2 \sin^2 \theta}} P_\phi \\ \frac{\partial {H}}{\partial {P_\theta}} &= \dot{\theta} = \frac{1}{{ m l^2 }} P_\theta \\ \frac{\partial {H}}{\partial {\phi}} &= -\dot{P}_\phi = 0 \\ \frac{\partial {H}}{\partial {\theta}} &= -\dot{P}_\theta = -\frac{\cos\theta}{ m l^2 \sin^3 \theta} {P_\phi}^2 - m g l \sin\theta\end{aligned} \quad\quad\quad(163)

These now expressing the dynamics of the system. The first two equations are just the definitions of the canonical momenta that we started with using the Lagrangian. Not surprisingly, but unfortunate, we have a non-linear system here like the planar rigid pendulum, so despite this being one of the most simple systems it does not look terribly tractable. What would it take to linearize this system of equations?

Lets write the state space vector for the system as

\begin{aligned}\mathbf{x} =\begin{bmatrix}P_\theta \\ \theta \\ P_\phi \\ \phi \\ \end{bmatrix}\end{aligned} \quad\quad\quad(167)

lets also suppose that we are interested in the change to the state vector in the neighborhood of an initial state

\begin{aligned}\mathbf{x} =\begin{bmatrix}P_\theta \\ \theta \\ P_\phi \\ \phi \\ \end{bmatrix}={\begin{bmatrix}P_\theta \\ \theta \\ P_\phi \\ \phi \\ \end{bmatrix}}_0+ \mathbf{z}\end{aligned} \quad\quad\quad(168)

The Hamiltonian equations can then be written

\begin{aligned}\frac{d\mathbf{z}}{dt} &=\begin{bmatrix}\frac{\cos\theta}{ m l^2 \sin^3 \theta} {P_\phi}^2 + m g l \sin\theta \\ \frac{1}{{ m l^2 }} P_\theta \\ 0 \\ \frac{1}{{ m l^2 \sin^2 \theta}} P_\phi \\ \end{bmatrix}\end{aligned} \quad\quad\quad(169)

Getting away from the specifics of this particular system is temporarily helpful. We have a set of equations that we wish to calculate a linear approximation for

\begin{aligned}\frac{dz_\mu}{dt} = A_\mu(x_\nu) \approx A_\mu(\mathbf{x}_0) + \sum_\alpha {\left. \frac{\partial {A_\mu}}{\partial {x_\alpha}} \right\vert}_{\mathbf{x}_0} z_\alpha\end{aligned} \quad\quad\quad(170)

Our linear approximation is thus

\begin{aligned}\frac{d\mathbf{z}}{dt} &\approx{\begin{bmatrix}\frac{\cos\theta}{ m l^2 \sin^3 \theta} {P_\phi}^2 + m g l \sin\theta \\ \frac{1}{{ m l^2 }} P_\theta \\ 0 \\ \frac{1}{{ m l^2 \sin^2 \theta}} P_\phi \\ \end{bmatrix}}_0+{\begin{bmatrix}0 & -\frac{{P_\phi}^2 (1 + 2 \cos^2 \theta)}{m l^2 \sin^4 \theta} +m g l \cos\theta & \frac{2 \cos\theta}{ m l^2 \sin^3 \theta} P_\phi & 0 \\ \frac{1}{{m l^2}} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & \frac{-2 P_\phi \cos\theta}{m l^2 \sin^3\theta} & \frac{1}{{ m l^2 \sin^2 \theta}} & 0 \\ \end{bmatrix}}_0\mathbf{z}\end{aligned} \quad\quad\quad(171)

Now, this is what we get blindly trying to set up the linear approximation of the state space differential equation. We see that the cyclic coordinate $\phi$ leads to a bit of trouble since no explicit $\phi$ dependence in the Hamiltonian makes the resulting matrix factor non-invertible. It appears that we would be better explicitly utilizing this cyclic coordinate to note that $P_\phi = \text{constant}$, and to omit this completely from the state vector. Our equations in raw form are now

\begin{aligned}\dot{\theta} &= \frac{1}{{ m l^2 }} P_\theta \\ \dot{P}_\theta &= \frac{\cos\theta}{ m l^2 \sin^3 \theta} {P_\phi}^2 + m g l \sin\theta \\ \dot{\phi} &= \frac{1}{{ m l^2 \sin^2 \theta}} P_\phi \end{aligned} \quad\quad\quad(172)

We can treat the $\phi$ dependence later once we have solved for $\theta$. That equation to later solve is just this last

\begin{aligned}\dot{\phi} &= \frac{1}{{ m l^2 \sin^2 \theta}} P_\phi \end{aligned} \quad\quad\quad(175)

This integrates directly, presuming $\theta = \theta(t)$ is known, and we have

\begin{aligned}\phi - \phi(0) &= \frac{P_\phi}{ m l^2} \int_0^t \frac{d\tau}{\sin^2 \theta(\tau)} \end{aligned} \quad\quad\quad(176)

Now the state vector and its perturbation can be redefined omitting all but the $\theta$ dependence. Namely

\begin{aligned}\mathbf{x} =\begin{bmatrix}P_\theta \\ \theta \\ \end{bmatrix}\end{aligned} \quad\quad\quad(177)

\begin{aligned}\mathbf{x} =\begin{bmatrix}P_\theta \\ \theta \\ \end{bmatrix}={\begin{bmatrix}P_\theta \\ \theta \\ \end{bmatrix}}_0+ \mathbf{z}\end{aligned} \quad\quad\quad(178)

We can now write the remainder of this non-linear system as

\begin{aligned}\frac{d\mathbf{z}}{dt} &=\begin{bmatrix}\frac{\cos\theta}{ m l^2 \sin^3 \theta} {P_\phi}^2 + m g l \sin\theta \\ \frac{1}{{ m l^2 }} P_\theta \\ \end{bmatrix}\end{aligned} \quad\quad\quad(179)

and make the linear approximation around $\mathbf{x}_0$ as

\begin{aligned}\frac{d\mathbf{z}}{dt} &\approx{\begin{bmatrix}\frac{\cos\theta}{ m l^2 \sin^3 \theta} {P_\phi}^2 + m g l \sin\theta \\ \frac{1}{{ m l^2 }} P_\theta \\ \end{bmatrix}}_0+{\begin{bmatrix}0 & -\frac{{P_\phi}^2 (1 + 2 \cos^2 \theta)}{m l^2 \sin^4 \theta} +m g l \cos\theta \\ \frac{1}{{m l^2}} & 0 \\ \end{bmatrix}}_0\mathbf{z}\end{aligned} \quad\quad\quad(180)

This now looks a lot more tractable, and is in fact exactly the same form now as the equation for the linearized planar pendulum. The only difference is the normalization required to switch to less messy dimensionless variables. The main effect of allowing the trajectory to have a non-planar component is a change in the angular frequency in the $\theta$ dependent motion. That frequency will no longer be $\sqrt{{\left\lvert{\cos\theta_0}\right\rvert} g/l}$, but also has a $P_\phi$ and other more complex trigonometric $\theta$ dependencies. It also appears that we can probably have hyperbolic or trigonometric solutions in the neighborhood of any point, regardless of whether it is a northern hemispherical point or a southern one. In the planar pendulum the unambiguous sign of the matrix terms led to hyperbolic only above the horizon, and trigonometric only below.