Hamilton-Jacobi Equation

A third formulation of classical mechanics that is equivalent to Newtonian mechanics and derivable from Hamiltonian and Lagrangian mechanics is the Hamilton-Jacobi formulation. It relates the action, S, to some of the other quantities described in Hamiltonian mechanics. It is also the classical theory with closest similarities to quantum mechanics. Derivation of the equations of this theory is largely taken from the following source:

http://galileoandeinstein.physics.virginia.edu/7010/CM_08_ActionEndPts.html

We know from our derivation of the least action principle that, after integration by parts, we are left with

(1)   \begin{equation*}\delta S=\frac{\partial L}{\partial \dot{q_i}}\delta q\Biggr|_{t_1}^{t_2}+\displaystyle\int\limits_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right)\delta q dt\end{equation*}

The second term on the right side, \displaystyle\int\limits_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right)\delta q dt, contains the expression that results from the principle of stationary action – \frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}. This expression equals zero so we can ignore it. That leaves us with

\delta S=\frac{\partial L}{\partial \dot{q_i}}\delta q\Biggr|_{t_1}^{t_2}

Now keep the spatial starting points and the starting and ending times of the configuration space trajectory the same but vary the spatial end point (which we’ll designate q_2) infinitesimally. We get:

\delta S(q_2,t_2)=\frac{\partial L}{\partial \dot{q_i}}\delta q\Biggr|_{t_1}^{t_2}=p_2\delta q_2=p\delta q

If we add up all of the little infinitesimal changes, we get the total change and we can drop the subscripts.

\delta S = p\delta q\quad\Rightarrow\quad \Delta S = p\Delta q where \Delta q is very small. Divide both sides of this equation by \Delta q \quad\Rightarrow\quad \frac{\Delta S}{\Delta q}=p.

Take the limit as \Delta q goes to zero of this last equation:

(2)   \begin{equation*}\displaystyle\lim_{\Delta q \to 0}\frac{\Delta S}{\Delta q}=p\quad\Rightarrow\quad \frac{\partial S}{\partial q_i}=p_i}\end{equation*}

Now let’s vary the time of the endpoint t_2 infinitesimally to t_2 + \delta t. That changes the path of the action slightly. Next, find the total derivative of the action (which is a function of position and time) with respect to time. This equals:

\frac{d S(q_i,t)}{dt}=\frac{\partial S}{\partial t} + \displaystyle\sum_i \frac{\partial S}{\partial q_i}\frac{\partial q_i}{\partial t}=\frac{\partial S}{\partial t} + \displaystyle\sum_i \frac{\partial S}{\partial q_i}\dot{q}

From the result we just obtained above, \frac{\partial S}{\partial q_i}=p_i. This leads to

\frac{d S(q_i,t)}{dt}=\frac{\partial S}{\partial t} + \displaystyle\sum_i p_i}\dot{q}

By the definition of the Hamiltonian

H=\displaystyle\sum_i p_i}\dot{q}-L \,\Rightarrow \, \displaystyle\sum_i p_i}\dot{q}=H+L

Thus,

\frac{d S(q_i,t)}{dt}=\frac{\partial S}{\partial t} + H+L

Recall that S=\displaystyle\int\limits_i L dt. By the fundamental theorem of calculus, \frac{dS}{dt}=L. Therefore,

\frac{d S(q_i,t)}{dt}=\frac{\partial S}{\partial t} + H+L becomes

L=\frac{\partial S}{\partial t} + H+L

Subtracting L from both sides, we get

(3)   \begin{equation*}\frac{\partial S}{\partial t}+H=0 \end{equation*}

Equation 3 is called the Hamilton-Jacobi equation.