Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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 pricipal 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 pricipal function we have $$ \frac{d}{dt}S(\gamma(t),t)=L(\gamma(t),\dot\gamma(t),t) $$ 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 leave of a foliation of phase space, which corresponds to the choice of initial condition $\gamma_q(t_0)=q_0$ from above.

share|improve this answer
add comment

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), $$ \tag{13} S(q_f,E,t_f)-S(q_i,E,t_i)~\stackrel{(8)+(12)}{=}~\frac{m}{2} \cdot \frac{(q_f-q_i)^2}{t_f-t_i}~\stackrel{(7)}{=}~ S(q_f,t_f;q_i,t_i).\qquad $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.