# Hamilton-Jacobi Theory

Consider the canonical transformation $$\zeta = \zeta(\eta,t)$$. From the coordinates $$(q,p)$$ at $$t$$ to \$(Q0,P0) at $$t=0$$. If we can claim, $$\zeta$$ can transform the coordinates to constant variables $$(Q_0\wedge P_0)$$ 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

$$\zeta = \zeta(\eta,t)$$. 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 $$Q_i = \beta_i = \frac{\partial F_2(q,\alpha,t)}{\partial \alpha_i} = \frac{\partial F_2(q,\alpha,t)}{\partial P_i}$$.

## Hamilton-Jacobi Equation

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

This is also written as,

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

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.

## Time Invariant Hamiltonian

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

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}

## Example - 1D SHO

$$H = \frac{1}{2m}\left(p^2 + m^2\omega^2q^2\right)$$ 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, $$S_i(q_i,\alpha,t) = W_i(q_i,\alpha) - E_i(\alpha)t$$.

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

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, $$S(q,\alpha,t) = W(q_1,\cdots,q_s,\alpha) - E(\alpha)t + \sum_{i=s+1}^n\gamma_iq_i$$.

## Example

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*}

Created: 2024-05-30 Thu 21:21

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