# Hamilton-Jacobi Equation

In the Hamilton-Jacobi equation, we take the partial time derivative of the action. But the action comes from integrating the Lagrangian over time, so time seems to just be a dummy variable here and hence I do not understand how we can partial differentiate $S$ with respect to time? A simple example would also be helpful.

The action functional and Hamilton's principal function are two different mathematical objects related to the same physical quantity.

The action along a trajectory $$\gamma:[t_1,t_2]\rightarrow Q$$ is given by $$S[\gamma] = \int_{t_1}^{t_2}L(\gamma(t'),\dot\gamma(t'),t')dt'$$ whereas the principal function is the solution of the Hamilton-Jacobi equation $$H(q,\nabla S(q,t),t) + \frac{\partial S}{\partial t}(q,t) = 0$$

If you denote by $$\gamma_{q,t}$$ the solution of the Euler-Lagrange equations with $$\gamma_{q,t}(t_0)=q_0\\ \gamma_{q,t}(t)=q$$ then $$S(q,t):=S[\gamma_{q,t}]=\int_{t_0}^{t}L(\gamma_{q,t}(t'),\dot\gamma_{q,t}(t'),t')dt'$$ will solve the Hamilton-Jacobi equation.

On the flip side, for the principal function we have the following $$\frac{d}{dt}S(\gamma(t),t)=L(\gamma(t),\dot\gamma(t),t)$$ and thus $$S[\gamma]=S(\gamma(t_2),t_2)-S(\gamma(t_1),t_1)$$

Note that the last two equations only hold for trajectories with $$\frac{\partial L}{\partial\dot q}(\gamma(t),\dot\gamma(t),t) = \nabla S(\gamma(t),t)$$

Geometrically, the choice of integration constants of the principal function selects a leaf of a foliation of phase space, which corresponds to the choice of initial condition $$\gamma_q(t_0)=q_0$$ from above.

I) At least three different quantities in physics are customary called an action and denoted with the letter $S$.

1. The (off-shell) action $$\tag{1}S[q]~:=~ \int_{t_i}^{t_f}\! dt \ L(q(t),\dot{q}(t),t)$$ is a functional of the full position curve/path $q^i:[t_i,t_f] \to \mathbb{R}$ for all times $t$ in the interval $[t_i,t_f]$. See also this question. (Here the words on-shell and off-shell refer to whether the equations of motion (eom) are satisfied or not.)

2. If the variational problem $(1)$ with well-posed boundary conditions, e.g. Dirichlet boundary conditions $$\tag{2} q(t_i)~=~q_i\quad\text{and}\quad q(t_f)~=~q_i,$$ has a unique extremal/classical path $q_{\rm cl}^i:[t_i,t_f] \to \mathbb{R}$, it makes sense to define an on-shell action $$\tag{3} S(q_f;t_f;q_i,t_i) ~:=~ S[q_{\rm cl}],$$ which is a function of the boundary values. See e.g. MTW Section 21.1.

3. The Hamilton's principal function $S(q,\alpha, t)$ in Hamilton-Jacobi equation is a function of the position coordinates $q^i$, integration constants $\alpha_i$, and time $t$, see e.g. H. Goldstein, Classical Mechanics, chapter 10. The total time derivative $$\tag{4} \frac{dS}{dt}~=~ \dot{q}^i \frac{\partial S}{\partial q^i}+ \frac{\partial S}{\partial t}$$ is equal to the Lagrangian $L$ on-shell, as explained here. As a consequence, the Hamilton's principal function $S(q,\alpha, t)$ can be interpreted as an action on-shell.

II) Example: A non-relativistic free particle in 1 dimension.

1. The off-shell action is $$\tag{5} S[q]~=~ \frac{m}{2}\int_{t_i}^{t_f}\! dt \ \dot{q}(t)^2.$$

2. If we assume Dirichlet boundary conditions (2), the unique classical trajectory $q_{\rm cl}$ has constant velocity $$\tag{6}\dot{q}_{\rm cl}~=~\frac{q_f-q_i}{t_f-t_i}.$$ The Dirichlet on-shell action (3) is $$\tag{7} S(q_f,t_f;q_i,t_i) ~=~ \frac{m}{2} \cdot \frac{(q_f-q_i)^2}{t_f-t_i}.$$

3. The Hamilton's principal function, i.e. a solution to Hamilton-Jacobi equation, is $$\tag{8} S(q,E,t)~=~\pm\sqrt{2m E} q - Et,$$ where $E$ is an integration constant (=the total energy). Due to the interpretation of Hamilton's principal function as a type 2 generator of canonical transformations, the partial derivative $$\tag{9}Q~:=~ \frac{\partial S}{ \partial E}~\stackrel{(8)}{=}~\pm\sqrt{\frac{m}{2E}}q -t$$ must be a constant of motion. In other words, the position $q(t)$ is, as expected, an affine function of time $t$. This implies that the velocity is constant $$\tag{10} \dot{q} ~\approx~\pm\sqrt{\frac{2E}{m}},$$ where the "$\approx$" symbol means equality modulo eom. The total time derivative of the Hamilton's principal function (8) is equal to the Lagrangian (=the kinetic energy) on-shell $$\tag{11} \frac{dS}{dt}~\stackrel{(8)}{=}~ \pm\sqrt{2m E} \dot{q} -E ~\stackrel{(10)}{\approx}~E.$$

4. Let us now compare point 2 and 3. With the Dirichlet boundary conditions (2), the energy becomes $$\tag{12} E~=~ \frac{m}{2} \cdot \left(\frac{q_f-q_i}{t_f-t_i}\right)^2.$$ A comparison of eqs. (6) and (10) shows that we should use the plus (minus) branch of the solution (8) if $q_f\geq q_i$ ($q_f\leq q_i$), respectively. It is straightforward to check that the difference in the Hamilton's principal function becomes the on-shell action (7), $$\tag{13} S(q_f,E,t_f)-S(q_i,E,t_i)~\stackrel{(8)+(12)}{=}~\frac{m}{2} \cdot \frac{(q_f-q_i)^2}{t_f-t_i}~\stackrel{(7)}{=}~ S(q_f,t_f;q_i,t_i).\qquad$$

I think the other two answers are overkill. The simpler answer is that time $$t$$ is not a dummy variable. The integration of $$L$$ over time here is an indefinite integration, so if we have $$L(q,\dot{q},t)$$, and we want to integrate it over time $$t$$, the result is $$\int_{t_i}^tL(q,\dot{q},\tau)d\tau$$ here $$\tau$$ is the dummy variable but $$t$$ is not, and the result is a function of $$t$$.