Hamiltonian Mechanics

Overview

Consider a monogenic system without constraints and with $n$ independent generalized coordinates, ${q_{i}}_{i \in I}$ . Then, $\forall i \in I, \frac{d}{d t} (\frac{\partial L}{\partial {\dot{q}}_{i}}) = \frac{\partial L}{\partial q_{i}}$ .

Define $H (q, p, t) =$ the Hamiltonian $= p_{i} {\dot{q}}_{i} - L (q, \dot{q}, t)$ . Which gives us from the Legendre transformation, $\forall i \in I$ ,

\begin{aligned} {\dot{q}}_{i} & = \frac{\partial H}{\partial p_{i}}, \\ {\dot{p}}_{i} & = - \frac{\partial H}{\partial q_{i}}, \\ \frac{\partial H}{\partial t} & = - \frac{\partial L}{\partial t} . \end{aligned}

Now, we have $2 N$ first-order differential equations rather than $N$ second-order equations. These are Hamilton’s Canonical equations of motion.

In this representation, $q$ and $p$ are both ’coordinates’ – canonical coordinates in phase space.

Note, in the state space, each coordinate has its own trajectory that never touch (otherwise they would have to collapse to the same trajectory but they are independent). Hence, we have a state manifold – with a volume.

Hence, we get a Continuity equation in the state space and acts like an incompressible fluid.

Then,

\begin{aligned} 0 = δ I & = δ \int [\sum p_{i} {\dot{q}}_{i} - H (q, p, t)] d t . \\ f (q, p, t) = f (q, \dot{q}, p, \dot{p}, t) = f (η, \dot{η}, t) & = \sum p_{i} {\dot{q}}_{i} - H (q, p, t) = 0. \end{aligned}

Thus, with zero variation at the ends,

\begin{aligned} \frac{d}{d t} (\frac{\partial f}{\partial \dot{η}}) & = \frac{\partial f}{\partial η} . \\ {\dot{p}}_{j} + \frac{\partial H}{\partial q_{j}} & = 0, \\ {\dot{q}}_{j} - \frac{\partial H}{\partial p_{j}} & = 0. \end{aligned}

In Matrix Form

\begin{aligned} \underset{―}{η} & = (\begin{array}{c} | \\ q_{i} \\ | \\ | \\ p_{i} \\ | \end{array}) \\ \frac{\partial H}{\partial \underset{―}{η}} & = (\begin{array}{c} | \\ \frac{\partial H}{\partial q_{i}} \\ | \\ | \\ \frac{\partial H}{\partial p_{i}} \\ | \end{array}) = (\begin{array}{c} | \\ - {\dot{p}}_{i} \\ | \\ | \\ {\dot{q}}_{i} \\ | \end{array}) \\ \dot{\underset{―}{η}} & = \underset{―}{J} \frac{\partial H}{\partial \underset{―}{n}} \\ \underset{―}{J} & = (\begin{array}{c} 0 & 0 & \dots & 1 & 0 & \dots \\ 0 & 0 & \dots & 0 & 1 & \dots \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ \dots & - 1 & 0 & \dots & 0 & 0 \\ \dots & 0 & - 1 & \dots & 0 & 0 \end{array}) \\ = (\begin{array}{c} 0 & I \\ - I & 0 \end{array}) . \\ \dot{\underset{―}{η}} & = (\begin{array}{c} 0 & I \\ - I & 0 \end{array}) (\begin{array}{c} \frac{\partial H}{\partial q} \\ \frac{\partial H}{\partial q} \end{array}) \\ = (\begin{array}{c} \frac{\partial H}{\partial p} \\ - \frac{\partial H}{\partial p} \end{array}) \\ = (\begin{array}{c} \dot{q} \\ \dot{p} \end{array}) . \end{aligned}

Liouville’s Theorem

Starting from a cloud of points. The trajectories between two times form a sort of cylinder – most likely will have irregular shape. Then, we can have a volume of this – How does the Volume change?

Theorem: If you look at the trajectory of the cloud the volume of the cloud remains invariant.

So, we need the divergence of the $\dot{q}$ and $\dot{p}$ coordinates to be zero. $\frac{\partial}{\partial p} \dot{p} + \frac{\partial}{\partial q} \dot{q} = 0$ , where $\frac{\partial}{\partial η} \dot{η}$ is shorthand for $\sum_{i} \frac{\partial}{\partial η_{i}} {\dot{η}}_{i}$ . Then, $\frac{\partial}{\partial p} \dot{p} + \frac{\partial}{\partial q} \dot{q} = \sum_{i} \frac{\partial^{2} H}{\partial q_{i} \partial p_{i}} + - \frac{\partial^{2} H}{\partial p_{i} \partial q_{i}} = 0$ .

Canonical Transformations

Motivation: Hamiltonian does not explicitly depends on time or coordinates. Then, $H = H (p)$ . So, $H$ must be a constant from $\frac{d H}{d t} = - \frac{\partial H}{\partial t} = 0$ . Then, ${\dot{p}}_{i} = - \frac{\partial H}{\partial q_{i}}$ . So $p_{i} = α_{i}$ is a constant. Thus, the hamiltonian can be expressed in terms of constants, $H (α_{1}, \dots, α_{n})$ . Thus, energy and momentum are conserved. Further, ${\dot{q}}_{i} = \frac{\partial H}{\partial p_{i}} = \frac{\partial H}{\partial α_{i}} = ω_{i}$ is a constant. Therefore, $q_{i} = ω_{i} t + q_{i}^{0}$ .

So, we want to find the Canonical Transformation to cast the Hamiltonian into this form. A point transformation in the state space: $Q_{i} = Q_{i} (q, p, t)$ and $P_{i} = P_{i} (q, p, t)$ . Then, $H \to K$ or $\hat{H}$ , transformation to the Camiltonian. We still require that ${\dot{P}}_{i} = \frac{\partial K}{\partial Q_{i}}$ and ${\dot{Q}}_{i} = \frac{\partial K}{\partial P_{i}}$ .

Generating Function Formalism

Introduce $\underset{―}{η} = (\begin{matrix} q \\ p \end{matrix}) \to \underset{―}{ζ} = (\begin{matrix} Q \\ P \end{matrix})$ . So, $0 = δ I = δ \int_{t_{1}}^{t_{2}} d t [\sum_{i} p_{i} {\dot{q}}_{i} - H (q, p, t)] \Rightarrow 0 = δ \int_{t_{1}}^{t_{2}} d t λ [\sum_{i} P_{i} {\dot{Q}}_{i} - K (Q, P, t) + \frac{d F (η, ζ, t)}{d t}]$ . So, we want the integrands to be the same.

Thus,

\begin{aligned} \sum_{i} p_{i} {\dot{q}}_{i} - H (q, p, t) & = λ (\sum_{i} P_{i} {\dot{Q}}_{i} - K (Q, P, t) + \frac{d F}{d t}) . \end{aligned}

For $λ \neq 1$ , we get an extendend Canonical Transformation. For $λ = 1$ , we get an Canonical Transformation. When $ζ = ζ (η)$ we get a restricted canonical transformation. When $ζ = ζ (η, t)$ we get a traditional Canonical Transformation. The $F$ function is the generating function.

Suppose $λ = 1$ .

\begin{aligned} \frac{d F}{d t} & = \sum_{i} p_{i} {\dot{q}}_{i} - H (q, p, t) - \sum_{i} P_{i} {\dot{Q}}_{i} + K (Q, P, t) \\ = \sum_{i} (p_{i} {\dot{q}}_{i} - P_{i} {\dot{Q}}_{i}) - (H (q, p, t) + K (Q, P, t)) \end{aligned}

Type I, $F = F_{1} (q, Q, t)$ . Then,

\begin{aligned} \sum_{i} \frac{\partial F_{1}}{\partial q_{i}} {\dot{q}}_{i} + \frac{\partial F_{1}}{\partial Q_{i}} {\dot{Q}}_{i} + \frac{\partial F_{1}}{\partial t} & = \sum_{i} p_{i} {\dot{q}}_{i} - H (q, p, t) - \sum_{i} P_{i} {\dot{Q}}_{i} + K (Q, P, t) \\ = \sum_{i} (p_{i} {\dot{q}}_{i} - P_{i} {\dot{Q}}_{i}) - (H (q, p, t) + K (Q, P, t)) . \\ \Rightarrow p_{i} & = \frac{\partial F_{1}}{\partial q_{i}}, \\ P_{i} & = - \frac{\partial F_{1}}{\partial Q_{i}}, \\ K & = H + \frac{\partial F_{1}}{\partial t} . \end{aligned}

Type II, $F = F_{2} (q, p, t) - \sum P_{i} Q_{i}$ . Then,

\begin{aligned} \sum_{i} \frac{\partial F_{2}}{\partial q_{i}} {\dot{q}}_{i} + \frac{\partial F_{2}}{\partial p_{i}} {\dot{p}}_{i} + \frac{\partial F_{2}}{\partial t} - \sum_{i} (P_{i} {\dot{Q}}_{i} + {\dot{P}}_{i} Q_{i}) & = \sum_{i} p_{i} {\dot{q}}_{i} - H (q, p, t) - \sum_{i} P_{i} {\dot{Q}}_{i} + K (Q, P, t) \\ \sum_{i} \frac{\partial F_{2}}{\partial q_{i}} {\dot{q}}_{i} + \frac{\partial F_{2}}{\partial p_{i}} {\dot{p}}_{i} + \frac{\partial F_{2}}{\partial t} - \sum_{i} ({\dot{P}}_{i} Q_{i}) & = \sum_{i} p_{i} {\dot{q}}_{i} - H (q, p, t) + K (Q, P, t) \\ P_{i} & = \frac{\partial F_{2}}{\partial q_{i}}, \\ Q_{i} & = \frac{\partial F_{2}}{\partial p_{i}}, \\ K & = H + \frac{\partial F}{\partial t} . \end{aligned}

So, $F = F_{2} (q, p, t) = \sum q_{i} p_{i}$ . So, $P_{i} = \frac{\partial F_{2}}{\partial q_{i}} = P_{i}$ . Then, $Q_{i} = \frac{\partial F_{2}}{\partial p_{i}} = q_{i}$ . This is the identity transformation.

Type 3: $F = F_{3} (P, Q, t) + \sum p_{i} q_{i}$ .

Type 4: $F = F_{4} (p, P, t) - \sum P_{i} Q_{i} + \sum p_{i} q_{i}$ .

Example

For a 1D HO, $H = \frac{1}{2 m} (p^{2} + k q^{2})$ . Let $ω^{2} = \frac{k}{m}$ . Then, $H = \frac{1}{2 m} (p^{2} + m^{2} ω^{2} q^{2})$ . Thus, the Hamiltonian is a constant.

$p = f (P) \cos Q$ and $q = \frac{f (P)}{m ω} \sin Q$ . So, $\frac{p}{q} = m ω \cot Q$ . Show that this gives a canonical transformation.

Note, $p = m ω q \cot Q$ .

Consider a Type 1. $F = F (q, Q)$ . So, $p = \frac{\partial F_{1}}{\partial q} = m ω q \cot Q$ . So, $F_{1} = \frac{1}{2} ω q^{2} \cot Q + h (Q)$ . Further, $P = - \frac{\partial F_{1}}{\partial Q} = \frac{m ω q^{2}}{2 \sin^{2} Q} - \frac{d h}{d Q}$ . Setting the last term to zero, $P = \frac{m ω q^{2}}{2 \sin^{2} Q}$ . So, $q = \sqrt{\frac{2 P}{m ω}} \sin Q$ and $p = \sqrt{2 m ω P} \cos Q$ . Then, $f (P) = \sqrt{2 m ω P}$ . $K = H + \frac{d F_{1}}{d t} = H = ω P$ . So, $P = \frac{E}{ω}$ . Then, $\dot{Q} = \frac{\partial K}{\partial P} = ω$ . $Q = ω t + β$ . $q = \sqrt{\frac{2 P}{m ω}} \sin Q$ and $p = \sqrt{2 P m ω} \sin Q$ .

Sympletic Approach

Restricted Canonical Transformations

Suppose we have $η \to ζ$ . So, $ζ = ζ (η)$ . So, for the $i-$th coordinate, $ζ_{i} = ζ (η)$ .

\begin{aligned} {\dot{ζ}}_{i} & = \sum_{j} \frac{\partial ζ_{i}}{\partial η_{j}} {\dot{η}}_{j} \end{aligned}

Then,

\begin{aligned} \dot{ζ} & = (\begin{array}{c} \frac{\partial ζ_{1}}{\partial η_{1}} & \frac{\partial ζ_{1}}{\partial η_{2}} & \dots \\ \frac{\partial ζ_{2}}{\partial η_{1}} & \frac{\partial ζ_{2}}{\partial η_{2}} & \dots \\ ⋮ & ⋮ & \dots \\ \frac{\partial ζ_{2 n}}{\partial η_{1}} & \frac{\partial ζ_{2 n}}{\partial η_{2}} & \dots \end{array}) (\begin{array}{c} {\dot{η}}_{1} \\ {\dot{η}}_{2} \\ ⋮ \end{array}), \\ = \frac{\partial ζ}{\partial η} \dot{η} = M \dot{η} = M J \frac{\partial H}{\partial η} . \end{aligned}

$M$ is the sympletic matrix. Recall, $J = (\begin{matrix} 0 & I \\ - I & 0 \end{matrix})$ . Because, this is a restricted canonical transformation, $K = H$ . Then, $H = H (ζ)$ . ( $\frac{\partial H}{\partial η_{i}} = \sum \frac{\partial H}{\partial ζ_{j}} \frac{\partial ζ_{i}}{\partial η_{i}}$ .) So, $\dot{ζ} = J \frac{\partial K}{\partial ζ} = M J {\frac{\partial ζ}{\partial η}}^{T} \frac{\partial H}{\partial ζ} = M J M^{T} \frac{\partial H}{\partial ζ}$ . Hence, $\dot{ζ} = J \frac{\partial H}{\partial ζ}$ . Thus, $J = M J M^{T} = M^{T} J M$ . Hence, this is a necessary and sufficient condition for the Canonical Transformation. Thus, the transformation is canonical iff $J = M J M^{T}$ .

SHO Example

\begin{aligned} (\begin{array}{c} p \\ q \end{array}) & = (\begin{array}{c} \sqrt{2 m ω P} \cos Q \\ \sqrt{\frac{2 P}{m ω}} \sin Q \end{array}) . \end{aligned}

Then, $p = m ω q \cot Q$ . Also, $P = \frac{E}{ω}$ . Then, $P_{p} = 0, P_{q} = 0$ .

What about Time

Suppose we have time dependence, after many steps, you can show that this is still true. You can also use infinitesimal canonical transformations to show it.

Infinitesimal Canonical Transformation

Consider $η \to ζ (t)$ . So, $η \to ζ (t_{0})$ is also canonical and no longer involves time, a restricted canonical transformation. Hence, it satisfies $M J M^{T} = J$ . So, we must show $ζ (t_{0}) \to ζ (t)$ satsifies $M J M^{T} = J$ . Consider $ζ (t_{0} + n d t) \to ζ (t_{0} + (n + 1) d t)$ . So, $Q_{i} = q_{i} + δ q_{i}$ and $P_{i} = p_{i} + δ p_{i}$ . Then, $Q = q + δ q$ and $P = p + δ p$ . Hence, $ζ = η + δ η$ . This is nearly an identity transformation. We know $F_{2} = \sum q_{i} p_{i}$ is an identity transformation. So, $F_{2} = \sum q_{i} p_{i} + ε G (q, p, t)$ includes the small variation. Then, $Q_{i} = \frac{\partial F_{2}}{\partial p_{i}} = q_{i} + ε \frac{\partial G}{\partial P_{i}}$ . So, $P_{i} = \frac{\partial F_{2}}{\partial q_{i}} = p_{i} + ε \frac{\partial G}{\partial q_{i}}$ . Then, $Q_{i} = q_{i} + ε \frac{\partial G}{\partial p_{i}} + O (ε^{2})$ . Hence, $P_{i} - p_{i} = - ε \frac{\partial G}{\partial q_{i}}$ . So, $δ q_{i} = ε \frac{\partial G}{\partial p_{i}}$ and $δ p_{i} = - ε \frac{\partial G}{\partial q_{i}}$ . Therefore, $δ η = ε J \frac{\partial G}{\partial η}$ . So, $ζ = η + δ η = η + ε J \frac{\partial G}{\partial η}$ . We need to show $M J M^{T} = J$ .

Recall, $M = \frac{\partial ζ}{\partial η} = (\begin{matrix} \frac{\partial ζ_{1}}{\partial η_{1}} & \frac{\partial ζ_{1}}{\partial η_{2}} & \dots \\ ⋮ & ⋮ & ⋮ \end{matrix})$ . So, $M_{i}^{j} = \frac{\partial ζ_{i}}{\partial η_{j}}$ . So, $\frac{\partial}{\partial η} (η + δ η) = \frac{\partial}{\partial η} (η + ε J \frac{\partial G}{\partial η}) = I + ε \frac{\partial}{\partial η} (J \frac{\partial G}{\partial η}) = I + ε J \frac{\partial^{2} G}{\partial η \partial η}$ . The $\frac{\partial^{2} G}{\partial η^{2}}$ is a symmetric matrix. Further, $J$ is an antisymmetric matrix. Thus, we get an antisymmetric matrix $M$ . Hence, $M^{T} = I - ε \frac{\partial^{2} G}{\partial η \partial η} J$ . Therefore, $M J M^{T} = (I + ε J \frac{\partial^{2} G}{\partial η^{2}}) J (I - ε \frac{\partial^{2} G}{\partial η^{2}} J) = J + ε J G_{2} J - ε J G_{2} J - ε^{2} J G_{2} J G_{2} J = J + O (ε^{2})$ . Hence, for infinitesimal transformations we have the sympletic condition satisfied. Thus, by repeated infinitesimal transformations, we finally are able to transform, for fixed times $t^{'}, t_{0}$ , $η (t^{'}) \to ζ (t_{0}) \to ζ (t)$ . Therefore, the sympletic condition is necessary and sufficient for a canonical transformation.