Hamilton-Jacobi Theory

Consider the canonical transformation $ζ = ζ (η, t)$ . From the coordinates $(q, p)$ at $t$ to $(Q_0,P₀) at $t = 0$ . If we can claim, $ζ$ can transform the coordinates to constant variables $(Q_{0} \land P_{0})$ at $t = 0$ . Then, we can invert the transformation. So, $q_{i} = q_{i} (Q_{0}, P_{0}, t)$ and $p_{i} = p_{i} (Q_{0}, P_{0}, t)$ . Thus, we get the solutions of the problems.

Special Case

$ζ = ζ (η, t)$ . Suppose under the transformation, $H \to K = 0$ . Then, ${\dot{Q}}_{i} = \frac{\partial K}{\partial P_{i}} = 0 \Rightarrow Q_{i} = C_{i, q} = β_{i}$ . Then, ${\dot{P}}_{i} = - \frac{\partial K}{\partial Q_{i}} = 0 \Rightarrow P_{i} = C_{i, p} = α_{i}$ .

Consider a type 2 generating function, $F = F_{2} (q, P, t) - \sum Q_{i} P_{i}$ . The new Hamiltonian, $K = H (q, p, t) + \frac{\partial F_{2}}{\partial t}$ . Thus, $H (q, p, t) = - \frac{\partial F_{2}}{\partial t}$ .

Hence, $H (q, p, t) + \frac{\partial F_{2} (q, α, t)}{\partial t} = 0$ .

$p_{i} = \frac{\partial F_{2} (q, α, t)}{\partial q_{i}}$ and $Q_{i} = β_{i} = \frac{\partial F_{2} (q, α, t)}{\partial α_{i}} = \frac{\partial F_{2} (q, α, t)}{\partial P_{i}}$ .

Hamilton-Jacobi Equation

\begin{matrix} (1) & H (q, p, t) + \frac{\partial F_{2} (q, α, t)}{\partial t} = 0 \end{matrix}

This is also written as,

\begin{matrix} (2) & H (q, p, t) + \frac{\partial S (q, α, t)}{\partial t} = 0 \end{matrix}

With, $p_{i} = \frac{\partial S (q, α, t)}{\partial q_{i}}$ and $Q_{i} = β_{i} = \frac{\partial S (q, α, t)}{\partial α_{i}} = \frac{\partial S (q, α, t)}{\partial P_{i}}$ .

So we can invert to get, $q_{i} = q_{i} (α, β, t)$ . And we then get $p_{i} = p_{i} (α, β, t)$ .

So, $H (q, p, t) = H (q, \frac{\partial S}{\partial q}, t) + \frac{\partial S}{\partial t} = 0$ . In principle, $S = S (q, α, \dots, α_{n + 1}, t)$ . Where $α_{n + 1}$ is not related to the momenum, just a constant from the integration. But this can be written as $S + C$ we can simplify to $S = S (q, α_{1}, α_{n}, t)$ . We only care about this solution. Thus, we have only $n$ non-additive constants at most.

\begin{aligned} \frac{d S}{d t} & = \sum \frac{\partial S}{\partial q_{i}} {\dot{q}}_{i} + \frac{\partial S}{\partial α_{i}} {\dot{α}}_{i} + \frac{\partial S}{\partial t} \\ = \sum \frac{\partial S}{\partial q_{i}} {\dot{q}}_{i} + \frac{\partial S}{\partial t} \\ = \sum p_{i} {\dot{q}}_{i} + \frac{\partial S}{\partial t} \\ = \sum p_{i} {\dot{q}}_{i} - H (q, p, t) \\ (3) & = L (q, p, t) = L (q, \dot{q}, t) . \\ (4) & S = \int L d t + C . \end{aligned}

Thus, $S$ is the action integral of the Lagrangian. $S$ is also called Hamilton’s Principal Function.

Time Invariant Hamiltonian

$H (q, p) = C = E$ .

Then, $H (q, \frac{\partial S}{\partial q}) + \frac{\partial S (q, α, t)}{\partial t} = 0$ .

$S (q, α, t) = W (q, α) - E (α) t$ . Then, $W (q, α)$ is Hamilton’s Characteristic Function. So, the Time-Independent Hamilton-Jacobi Equation,

\begin{aligned} (5) & H (q, \frac{\partial W}{\partial q}) & = E (α) . \\ (6) & p_{i} & = \frac{\partial W}{\partial q_{i}}, \\ (7) & Q_{i} & = β_{i} = \frac{\partial S}{\partial α_{i}} . \end{aligned}

Example - 1D SHO

$H = \frac{1}{2 m} (p^{2} + m^{2} ω^{2} q^{2})$ where $ω^{2} = \frac{k}{m}$ . So, $S = S (q, α, t) = W (q, α) - E (α) t$ .

We can write $S (q, E, t) = W (q, E) - E t$ without loss of generality since $α$ is constant.

Then,

\begin{aligned} 0 = H + \frac{\partial S}{\partial t} & = \frac{1}{2 m} ({(\frac{\partial W}{\partial q})}^{2} + m^{2} ω^{2} q^{2}) - E \\ E & = \frac{1}{2 m} ({(\frac{\partial W}{\partial q})}^{2} + m^{2} ω^{2} q^{2}) \\ \frac{\partial W}{\partial q} & = \sqrt{2 m E - m^{2} ω^{2} q^{2}} = \sqrt{2 m E} \sqrt{1 + \frac{m ω^{2} q^{2}}{2 E}} . \\ W (q, E) & = \sqrt{2 m E} \int d q \sqrt{1 - \frac{m ω^{2} q^{2}}{2 E}} \\ S (q, E) & = \sqrt{2 m E} \int d q \sqrt{1 - \frac{m ω^{2} q^{2}}{2 E}} - E t \\ Q = β & = \frac{\partial S}{\partial E} \\ = \sqrt{\frac{m}{E}} \int \frac{d q}{\sqrt{1 - \frac{m ω^{2} q^{2}}{2 E}}} - t \\ = \frac{1}{ω} arcsin (q \sqrt{\frac{m ω^{2}}{2 E}}) - t, \\ q & = \sqrt{\frac{2 E}{m ω^{2}}} \sin ω (β + t) . \\ p = \frac{\partial S}{\partial q} & = \frac{\partial W}{\partial q} \\ = \sqrt{2 m E} \sqrt{1 - \frac{m ω^{2} q^{2}}{2 E}} \\ = \sqrt{2 m E} \sqrt{1 - \frac{m ω^{2} {(\sqrt{\frac{2 E}{m ω^{2}}} \sin ω (β + t))}^{2}}{2 E}} \\ = \sqrt{2 m E} \cos (ω t + θ_{0}) . \end{aligned}

Where $θ_{0} = ω β$ .

For a Seperable H-J equation: $S (q, α, t) = S_{i} (q_{i}, α, t) + S^{'} (q \neq q_{i}, α, t)$ .

If we have $S (q, α, t) = \sum_{i} S_{i} (q_{i}, α, t)$ then we get a completely seperable HJ equation. Then, $S_{i} (q_{i}, α, t) = W_{i} (q_{i}, α) - E_{i} (α) t$ .

Special Case

Suppose $H = H (q, p)$ with a cyclic coordinate, suppose $q_{1}$ is cyclic (i.e. ignorable). Cyclic implies the conjugate momentum is constant. Then, $S (q, α, t) = W (q \neq q_{1}, α) - E (α) t + γ_{1} q_{1}$ .

Consider: $H (q, p) = E$ , first $s$ of $n$ generalized coordinates are cyclic. Then, $S (q, α, t) = W (q_{s + 1}, \dots, q_{n}, α) - E (α) t + \sum_{i = 1}^{s} γ_{i} q_{i}$ .

If the first are not cyclic. Then, $S (q, α, t) = W (q_{1}, \dots, q_{s}, α) - E (α) t + \sum_{i = s + 1}^{n} γ_{i} q_{i}$ .

Example

Consider the central force problem where the particle is constrained to move in a plane.

\begin{aligned} (8) & H (q, p, t) & = \frac{p_{r}^{2}}{2 m} + \frac{p_{θ}^{2}}{2 m r^{2}} + V (r), \\ (9) & = \frac{p_{r}^{2}}{2 m} + \frac{p_{θ}^{2}}{2 m r^{2}} - \frac{k}{r}, \\ (10) & = \frac{p_{r}^{2}}{2 m} + \frac{p_{θ}^{2}}{2 I} - \frac{k}{r} . \end{aligned}

Note $H (q, p, t)$ is not a function of $θ$ , and so $θ$ is ignorable/ cyclic.

Then,

\begin{aligned} 0 & = H (q, p, t) + \frac{\partial S}{\partial t} \\ S (r, θ, α_{r}, α_{θ}, t) & = W (r, θ, α_{r}, α_{θ}) - E (α_{r}, α_{θ}) t . \end{aligned}

Let $α_{θ} = E$ be one of the constants.

\begin{aligned} 0 & = H (q, p, t) + \frac{\partial S}{\partial t} \\ S (r, θ, α_{r}, α_{θ}, t) & = W (r, α, E) + α θ - E t . \\ P_{θ} & = α . \\ \frac{p_{r}^{2}}{2 m} + \frac{p_{θ}^{2}}{2 I} + V (r) & = E \\ \frac{1}{2 m} ({(\frac{\partial W}{\partial r})}^{2} + \frac{1}{r^{2}} {(α)}^{2}) + V (r) & = E \\ (\frac{\partial W}{\partial r}) & = \sqrt{2 m E - \frac{α^{2}}{r^{2}} + V (r)} . \\ S & = \int d r \sqrt{2 m E - \frac{α^{2}}{r^{2}} + V (r)} + α θ - E t . \\ β_{E} & = \frac{\partial S}{\partial E} \\ β_{θ} & = \frac{\partial S}{\partial α} . \end{aligned}