8.4 Hamilton-Jacobi Equation

Consider the case of a canonical transformation of the second type

\begin{align} F&=F_2(q_i,P_i,t)-\sum_iQ_iP_i\\ \frac{dF}{dt}&=\sum_i\left(\die{F_2}{q_i}\dot q_i+\die{F_2}{P_i}\dot P_i- Q_i\dot P_i-P_i\dot Q_i\right)+\die{F_2}{t} \tag{8.26}\end{align}

Substituting into Equation 8.4:

\begin{align} \sum_i\dot q_ip_i-H&=\sum_i \dot Q_iP_i-K+\sum_i\left(\die{F_2}{q_i}\dot q_i+\die{F_2}{P_i}\dot P_i- Q_i\dot P_i-P_i\dot Q_i\right)+\die{F_2}{t}\\ \sum_i\left(p_i-\die{F_2}{q_i}\right)\dot q_i -H&=\sum_i\left(\die{F_2}{P_i}- Q_i \right)\dot P_i+\die{F_2}{t} -K \tag{8.27}\end{align}

And we obtain the transformation equations:

\begin{align} p_i&=\die{F_2}{q_i}\\ Q_i&=\die{F_2}{P_i}\\ K&=H+\die{F_2}{t} \tag{8.28}\end{align}

As a notational convention, we will use \(S(q_i,P_i,t)\) instead of \(F_2\) (as will see that it is equal to the action). We can also take \(S(q_i,P_i,t)\) as the generator of a canonical transformation with the above equations, since this will guarantee that the canonical equations are satisfied. In fact, we can choose any form that we wish for \(S\), since the above equations guarantee that the canonical equations are satisfied. Let’s then choose the particular case that gives \(K=0\). This is a nice choice, since K will then be independent of all the variables, and all \(Q_i\) and \(P_i\) are cyclic:

\begin{align} \dot Q_i&=0\\ \dot P_i&=0 \tag{8.29}\end{align}

We have thus changed the problem of dynamics to one of finding a transformation that gives all cyclic coordinates. Note that this is not necessarily more straightforward mathematically (in fact, it is usually much harder).

Imposing that \(K=0\) gives us:

\begin{align} H(q_i,p_i,t)+\die{S}{t}&=0\\ H(q_i,\die{S}{q_i},t)+\die{S}{t}&=0 \tag{8.30}\end{align}

which is a partial differential equation for \(S\) and is called the “Hamilton-Jacobi equation”. The function \(S\) is called “Hamilton’s principal function”. In the transformed coordinates, the solutions are trivial constants:

\begin{align} P_i&=\alpha_i\\ Q_i&=\beta_i =\die{S}{\alpha_i} \tag{8.31}\end{align}

Is is clear that by combining this with the transformation equations, the problem is solved if all of the \(\alpha_i\) and \(\beta_i\) are known. Typically, one does not need to refer to \(P_i\) and \(Q_i\) in the Hamilton-Jacobi formalism, since these are constants of motion. To highlight this, they are usually called \(\alpha_i\) and \(\beta_i\):

\begin{align} S&=S(q_i,\alpha_i,t)\\ H(q_i,\die{}{q_i}S(q_i,\alpha_i,t),t)+\die{}{t}S(q_i,\alpha_i,t)&=0\\ \alpha_i &=\text{const.}\\ \beta_i &=\die{S}{\alpha_i} \tag{8.32}\end{align}

Connection to the Lagrangian and action

Noting that \(S\) depends on the \(q_i\) and the constants \(\alpha_i\), we can ask how \(S\) changes with time along the path of the system (the path where the \(P_i\) are constant and equal to \(\alpha_i\)):

\begin{align} S&=S(q_i,\alpha_i,t)\\ \therefore\frac{dS}{dt}&=\sum_i\die{S}{q_i}\dot q_i +\die{S}{t} \tag{8.33}\end{align}

Using the transformation equations:

\begin{align} p_i&=\die{S}{q_i}\\ K&=H+\die{S}{t}=0\\ \frac{dS}{dt}&=\sum_ip_i\dot q_i -H=L\\ S&=\int L dt+ \text{const} \tag{8.34}\end{align}

It is thus clear that \(S\) is equal to the action within an additive constant.

If the Hamiltonian does not depend on time, then \(H\) is a constant of the motion, call it \(E\). If this is the case, one can assume a form for \(S\) where the time variable is separated out:

\begin{align} H(q_i,\die{S}{q_i},t)+\die{S}{t}&=0\\ S(q_i,\alpha_i,t)&=W(q_i,\alpha_i)-Et\\ \therefore H(q_i,\die{W}{q_i})&=E \tag{8.35}\end{align}

Where \(W\) is called “Hamilton’s characteristic function”.

Example 8-5

Use the Hamilton-Jacobi formalism to solve the simple harmonic oscillator problem

The Hamiltonian is given by:

\begin{align*} H&=\frac{p^2}{2m}+\frac{k}{2}q^2 \end{align*}

We can write this in terms of Hamilton’s principal function to get the Hamilton-Jacobi equation:

\begin{align*} H(q,\die{S}{q},t)+\die{S}{t}&=0\\ \frac{1}{2m}\left(\die{S}{q}\right)^2+\frac{k}{2}q^2+\die{S}{t}&=0 \end{align*}

Since the Hamiltonian does not depend on time, we can try a solution using separation of variables:

\begin{align*} S(q,\alpha,t)=W(q,\alpha)-\alpha t \end{align*}

which we substitute back into the Hamilton Jacobi equation:

\begin{align*} \frac{1}{2m}\left(\die{W}{q}\right)^2+\frac{k}{2}q^2=\alpha \end{align*}

Since the left side only depends on \(q\) the partial derivatives can be made into total derivatives. It is also clear that the constant \(\alpha\) is equal to energy, since it is equal to the Hamiltonian.

\begin{align*} \frac{1}{2m}\left(\frac{dW}{dq}\right)^2+\frac{k}{2}q^2&=\alpha\\ W&=\int \sqrt{2m\alpha-m^2\omega^2q^2}dq\\ S&=-\alpha t+\int \sqrt{2m\alpha-m^2\omega^2q^2}dq\\ \end{align*}

We also know that we have a second constant of motion, \(\beta\):

\begin{align*} \beta &=\die{S}{\alpha}\\ &=-t+\int \frac{2m}{2\sqrt{2m\alpha-m^2\omega^2q^2}}dq\\ &=-t+\frac{1}{\omega}\sin^{-1}\left(q\sqrt{\frac{m\omega^2}{2\alpha}} \right) \end{align*}

Which we can invert to get \(q\):

\begin{align*} q(t)=\sqrt{\frac{2\alpha}{m\omega^2}}\sin\left(\omega(\beta+t)\right) \end{align*}

Similarly, the momentum is given by:

\begin{align*} p&=\die{S}{q}=\die{}{q}\left(-\alpha t+\int \sqrt{2m\alpha-m^2\omega^2q^2}dq\right)\\ &=\sqrt{2m\alpha-m^2\omega^2q^2}\\ &=\sqrt{2m\alpha}\sqrt{1-\sin^2\left(\omega(\beta+t)\right)}\\ &=\sqrt{2m\alpha}\cos\left(\omega(\beta+t)\right) \end{align*}