Physlib.ClassicalMechanics.Pendulum.SimplePendulum.Hamiltonian

For a simple pendulum at time $t$ and angular position $x$ in a one-dimensional Euclidean space $E \cong \mathbb{R}^1$, the canonical momentum is defined as the linear equivalence between the space of angular velocities and the space of angular momenta (both represented by $E$). The mapping takes a velocity $v$ to the momentum $p$ via the gradient of the Lagrangian with respect to velocity: $$p = \nabla_v L(t, x, v) = I v$$ where $I$ is the moment of inertia. The inverse of this map is given by $v = \frac{1}{I} p$.

For a simple pendulum at time $t$ with angular position $x \in \mathbb{R}^1$ and angular velocity $v \in \mathbb{R}^1$, the canonical momentum is equal to $I v$, where $I$ is the moment of inertia of the pendulum.

The Hamiltonian $H(t, p, \theta)$ of a simple pendulum is a function of time $t$, canonical momentum $p$, and angular position $\theta$ (modeled in a one-dimensional Euclidean space). It is defined as the Legendre transform of the Lagrangian $L$: $$H(t, p, \theta) = \langle p, \dot{\theta} \rangle - L(t, \theta, \dot{\theta})$$ where the angular velocity $\dot{\theta}$ is obtained from the momentum $p$ via the inverse of the canonical momentum linear equivalence at time $t$ and position $\theta$.

The Hamiltonian $H(t, p, \theta)$ of a simple pendulum, as a function of time $t$, canonical momentum $p$, and angular position $\theta$ (represented in a one-dimensional Euclidean space), is equal to the kinetic energy expressed in terms of the momentum plus the potential energy $V(\theta)$. Specifically, it is given by: $$H(t, p, \theta) = \frac{1}{2I} \langle p, p \rangle + V(\theta)$$ where $I$ is the moment of inertia of the pendulum, $V$ is the potential energy function, and $\langle \cdot, \cdot \rangle$ denotes the standard inner product on the one-dimensional Euclidean space.

For any $n \in \mathbb{N} \cup \{\infty\}$, the Hamiltonian $H(t, p, \theta)$ of the simple pendulum is $n$-times continuously differentiable ($C^n$) with respect to time $t$, canonical momentum $p$, and angular position $\theta$ jointly.

For a simple pendulum with Hamiltonian $H(t, p, \theta)$ and potential energy $V(\theta)$, the gradient of the Hamiltonian with respect to the angular position $\theta$ at any time $t$ and momentum $p$ is equal to the gradient of the potential energy with respect to $\theta$: $$\nabla_{\theta} H(t, p, \theta) = \nabla V(\theta)$$ Physically, this gradient represents the negative of the torque acting on the pendulum.

For a simple pendulum with moment of inertia $I$, the gradient of the Hamiltonian $H(t, p, \theta)$ with respect to the canonical momentum $p$ at any time $t$ and angular position $\theta$ is given by $$\nabla_p H(t, p, \theta) = \frac{1}{I} p,$$ where $p$ and $\theta$ are represented as elements of a one-dimensional Euclidean space $\mathbb{R}^1$. This gradient corresponds to the angular velocity $\dot{\theta}$ recovered from the momentum.

For any trajectory of the angular position $\theta : \text{Time} \to \mathbb{R}^1$ of a simple pendulum, let $p(t)$ be the canonical momentum at time $t$ corresponding to the angular velocity $\dot{\theta}(t)$, defined by the linear equivalence $p(t) = I \dot{\theta}(t)$ (where $I$ is the moment of inertia). The Hamiltonian $H(t, p, \theta)$ evaluated along this trajectory and its canonical momentum is equal to the total energy $E(t)$ of the trajectory: $$H(t, p(t), \theta(t)) = E(t).$$ This equality holds for any trajectory $\theta$, regardless of whether it satisfies the pendulum's equation of motion.

For a momentum trajectory $p: \text{Time} \to \mathbb{R}^1$ and an angular position trajectory $\theta: \text{Time} \to \mathbb{R}^1$ of a simple pendulum, the Hamilton equation operator is the function mapping each time $t$ to the pair: $$\left( \frac{d\theta}{dt}(t) - \nabla_p H(t, p(t), \theta(t)), \ -\frac{dp}{dt}(t) - \nabla_\theta H(t, p(t), \theta(t)) \right)$$ where $H$ is the Hamiltonian of the simple pendulum, $\frac{d\theta}{dt}$ and $\frac{dp}{dt}$ are the time derivatives of the trajectories, and $\nabla_p H$ and $\nabla_\theta H$ are the gradients of the Hamiltonian with respect to momentum and position, respectively. The vanishing of this operator characterizes the trajectories that satisfy the classical Hamilton's equations.

Let $\theta : \text{Time} \to \mathbb{R}^1$ be a smooth ($C^\infty$) trajectory representing the angular position of a simple pendulum. Let $p(t)$ be the canonical momentum trajectory defined by $p(t) = I \dot{\theta}(t)$, where $I$ is the moment of inertia. Then, $\theta$ satisfies the pendulum's equation of motion $I \ddot{\theta}(t) = \tau(\theta(t))$ if and only if the Hamilton equation operator vanishes on the pair $(p, \theta)$, which corresponds to the satisfaction of Hamilton's equations: $$\frac{d\theta}{dt} = \frac{\partial H}{\partial p} \quad \text{and} \quad \frac{dp}{dt} = -\frac{\partial H}{\partial \theta}$$ where $H$ is the Hamiltonian of the system.

Let $\theta: \text{Time} \to \mathbb{R}^1$ be a smooth trajectory of a simple pendulum. The following formulations of the dynamics are equivalent: 1. **Newtonian Equation of Motion:** The pointwise rotational Newton law balancing the rate of change of angular momentum against the gravitational torque: $I \ddot{\theta}(t) = \tau(\theta(t))$. 2. **Scalar Equation of Motion:** The mass-independent second-order differential equation: $\ddot{\theta}(t) + \omega^2 \sin(\theta(t)) = 0$, where $\omega = \sqrt{g/\ell}$ is the angular frequency. 3. **Hamilton's Equations:** The vanishing of the Hamilton equation operator for the trajectory $\theta$ and its conjugate canonical momentum $p(t) = I \dot{\theta}(t)$. 4. **Lagrangian Variational Principle:** The vanishing of the variational gradient of the action integral $S[\theta] = \int L(t, \theta(t), \dot{\theta}(t)) \, dt$, where $L$ is the Lagrangian of the system. 5. **Hamiltonian Variational Principle:** The vanishing of the variational gradient of the phase-space action $S[p, \theta] = \int (\langle p(t), \dot{\theta}(t) \rangle - H(t, p(t), \theta(t))) \, dt$, evaluated at the pair of the canonical momentum $p$ and the trajectory $\theta$.