Hamilton-Jacobi Theory

\(\zeta = \zeta(\eta,t)\). Suppose under the transformation, \(H\to K=0\). Then, \(\dot{Q}_i = \frac{\partial K}{\partial P_i} = 0\Rightarrow Q_i = \mathbb{C}_{i,q} = \beta_i\). Then, \(\dot{P}_i = -\frac{\partial K}{\partial Q_i} = 0\Rightarrow P_i = \mathbb{C}_{i,p} = \alpha_i\).

Consider a type 2 generating function, \(F = F_2(q,P,t) - \sum Q_iP_i\). 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, \(H(q,p,t) + \frac{\partial F_2(q,\alpha,t)}{\partial t} = 0\).

\(p_i = \frac{\partial F_2(q,\alpha,t)}{\partial q_i}\) and \(Q_i = \beta_i = \frac{\partial F_2(q,\alpha,t)}{\partial \alpha_i} = \frac{\partial F_2(q,\alpha,t)}{\partial P_i}\).

\begin{equation} H(q,p,t) + \frac{\partial F_2(q,\alpha,t)}{\partial t} = 0 \end{equation}

This is also written as,

\begin{equation} H(q,p,t) + \frac{\partial S(q,\alpha,t)}{\partial t} = 0 \end{equation}

With, \(p_i = \frac{\partial S(q,\alpha,t)}{\partial q_i}\) and \(Q_i = \beta_i = \frac{\partial S(q,\alpha,t)}{\partial \alpha_i} = \frac{\partial S(q,\alpha,t)}{\partial P_i}\).

So we can invert to get, \(q_i = q_i(\alpha,\beta,t)\). And we then get \(p_i = p_i(\alpha,\beta,t)\).

So, \(H(q,p,t) = H\left(q,\frac{\partial S}{\partial q},t\right) + \frac{\partial S}{\partial t} = 0\). In principle, \(S = S(q,\alpha,\cdots,\alpha_{n+1},t)\). Where \(\alpha_{n+1}\) is not related to the momenum, just a constant from the integration. But this can be written as \(S + \mathbb{C}\) we can simplify to \(S = S(q,\alpha_1,\alpha_n,t)\). We only care about this solution. Thus, we have only \(n\) non-additive constants at most.

\begin{align} \frac{dS}{dt} &= \sum \frac{\partial S}{\partial q_i}\dot{q}_i + \frac{\partial S}{\partial\alpha_i}\dot{\alpha}_i + \frac{\partial S}{\partial t} \nonumber \\ &= \sum \frac{\partial S}{\partial q_i}\dot{q}_i + \frac{\partial S}{\partial t} \nonumber \\ &= \sum p_i\dot{q}_i + \frac{\partial S}{\partial t} \nonumber \\ &= \sum p_i\dot{q}_i - H(q,p,t) \nonumber \\ &= \mathcal{L}(q,p,t) = \mathcal{L}(q,\dot{q},t). \\ S = \int\mathcal{L}dt + \mathbb{C}. \end{align}

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

\(H(q,p) = \mathbb{C} = E\).

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

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

\begin{align} H\left(q,\frac{\partial W}{\partial q}\right) &= E(\alpha). \\ p_i &= \frac{\partial W}{\partial q_i}, \\ Q_i &= \beta_i = \frac{\partial S}{\partial\alpha_i}. \end{align}

\(H = \frac{1}{2m}\left(p^2 + m^2\omega^2q^2\right)\) where \(\omega^2 = \frac{k}{m}\). So, \(S = S(q,\alpha,t) = W(q,\alpha) - E(\alpha)t\).

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

Then,

\begin{align*} 0 = H + \frac{\partial S}{\partial t} &= \frac{1}{2m}\left(\left(\frac{\partial W}{\partial q}\right)^2 + m^2\omega^2q^2\right) - E \\ E &= \frac{1}{2m}\left(\left(\frac{\partial W}{\partial q}\right)^2 + m^2\omega^2q^2\right) \\ \frac{\partial W}{\partial q} &= \sqrt{2mE - m^2\omega^2q^2} = \sqrt{2mE}\sqrt{1 + \frac{m\omega^2q^2}{2E}}. \\ W(q,E) &= \sqrt{2mE}\int dq\sqrt{1 - \frac{m\omega^2q^2}{2E}} \\ S(q,E) &= \sqrt{2mE}\int dq\sqrt{1 - \frac{m\omega^2q^2}{2E}} - Et \\ Q = \beta &= \frac{\partial S}{\partial E} \\ &= \sqrt{\frac{m}{E}}\int\frac{dq}{\sqrt{1-\frac{m\omega^2q^2}{2E}}} - t \\ &= \frac{1}{\omega}\text{arcsin}\left(q\sqrt{\frac{m\omega^2}{2E}}\right) - t, \\ q &= \sqrt{\frac{2E}{m\omega^2}}\sin\omega(\beta+t). \\ p = \frac{\partial S}{\partial q} &= \frac{\partial W}{\partial q} \\ &= \sqrt{2mE}\sqrt{1 - \frac{m\omega^2q^2}{2E}} \\ &= \sqrt{2mE}\sqrt{1 - \frac{m\omega^2\left(\sqrt{\frac{2E}{m\omega^2}}\sin\omega(\beta+t)\right)^2}{2E}} \\ &= \sqrt{2mE}\cos(\omega t+\theta_0). \end{align*}

Where \(\theta_0 = \omega\beta\).

For a Seperable H-J equation: \(S(q,\alpha,t) = S_i(q_i,\alpha,t) + S'(q\neq q_i,\alpha,t)\).

If we have \(S(q,\alpha,t) = \sum_i S_i(q_i,\alpha,t)\) then we get a completely seperable HJ equation. Then, \(S_i(q_i,\alpha,t) = W_i(q_i,\alpha) - E_i(\alpha)t\).

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, \(S(q,\alpha,t) = W(q\neq q_1,\alpha) - E(\alpha)t + \gamma_1q_1\).

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

If the first are not cyclic. Then, \(S(q,\alpha,t) = W(q_1,\cdots,q_s,\alpha) - E(\alpha)t + \sum_{i=s+1}^n\gamma_iq_i\).

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

\begin{align} H(q,p,t) &= \frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2} + V(r), \\ &= \frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2} - \frac{k}{r}, \\ &= \frac{p_r^2}{2m} + \frac{p_\theta^2}{2I} - \frac{k}{r}. \end{align}

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

Then,

\begin{align*} 0 &= H(q,p,t) + \frac{\partial S}{\partial t} \\ S(r,\theta,\alpha_r,\alpha_\theta,t) &= W(r,\theta,\alpha_r,\alpha_\theta) - E(\alpha_r,\alpha_\theta)t. \end{align*}

Let \(\alpha_\theta = E\) be one of the constants.

\begin{align*} 0 &= H(q,p,t) + \frac{\partial S}{\partial t} \\ S(r,\theta,\alpha_r,\alpha_\theta,t) &= W(r,\alpha,E) + \alpha\theta - Et. \\ P_\theta &= \alpha. \\ \frac{p_r^2}{2m} + \frac{p_\theta^2}{2I} + V(r) &= E \\ \frac{1}{2m}\left(\left(\frac{\partial W}{\partial r}\right)^2 + \frac{1}{r^2}\left(\alpha\right)^2\right) + V(r) &= E \\ \left(\frac{\partial W}{\partial r}\right) &= \sqrt{2mE - \frac{\alpha^2}{r^2} + V(r)}. \\ S &= \int dr\sqrt{2mE - \frac{\alpha^2}{r^2} + V(r)} + \alpha\theta - Et. \\ \beta_E &= \frac{\partial S}{\partial E} \\ \beta_\theta &= \frac{\partial S}{\partial\alpha}. \end{align*}