I'm trying to derive the HJ the easiest way I can but some issues come up.

$$\mathrm{dS}=\dfrac{\partial S}{\partial q}\mathrm{d}q+\dfrac{\partial S}{\partial t}\mathrm{d}t\Rightarrow\displaystyle{S=\int \mathrm{d}S=\int\dfrac{\partial S}{\partial q}\mathrm{d}q+\int\dfrac{\partial S}{\partial t}\mathrm{d}t}.\tag{1}$$

We also know that: $$S=\displaystyle{\int p\dfrac{\mathrm{d}q}{\mathrm{d}t}\mathrm{d}t-\int H \mathrm{d}t=\int p{d}q-\int H \mathrm{d}t}.\tag{2}$$

By identification: $$p=\dfrac{\partial S}{\partial q}\quad\text{and }\quad\dfrac{\partial S}{\partial t}=-H.\tag{3}$$ These two equations give he HJ equation.

But I'm not comfortable with this derivation.

  • I feel like I'm playing with the physical quantities but not in a "rigorous way" especially after reading this answer: Hamilton-Jacobi Equation. Is there a problem with that?

  • I don't see anywhere in this derivation where we assumed that the equations of motion were holding. I.e: we didn't impose $\delta S=0$ or $\frac{\partial L}{\partial q}=\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}}$ for example (while in the others derivation I saw (at least the ones which were not involving canonical transformations) the Euler-Lagrange equation was used). Isn't it a problem?


1 Answer 1


Superficially, it seems OP is conflating

  1. Hamilton's principal function$^1$ $S(q,\alpha\!=\!P,t)$ [which is a generating function of a type 2 canonical transformation (CT)] in OP's eqs. (1) & (3)

  2. with the (off-shell) Hamiltonian action functional$^1$ $$S_H[q,p]~=~\int\! dt~ L_H, \qquad L_H~=~p_k\dot{q}^k-H,\tag{2}$$ in OP's eq. (2).

For a derivation of the Hamilton–Jacobi (HJ) equation, see e.g. H. Goldstein, Classical Mechanics, chapter 10. (A prerequisite is the conditions for a type 2 CT, see e.g. H. Goldstein, Classical Mechanics, chapter 9.)


$^1$ It is imperative to understand the difference between a function and a functional, see e.g. this related Phys.SE post.

  • $\begingroup$ I didn't understand this idea of "on shell" and "off shell" action (might still be the case...). But I'm still confused about the idea of $S$ being two different things in (1) and (2). From what I understand, $S_H$ is the off shell (Hamiltonian) action and $\displaystyle{\int\mathrm{d}S}$ can be interpreted (or maybe we can't?) as the integral of the total differential of the Hamiltonian action. So we can do the identification. But by doing so, the $S$ in the HJ equation can't be interpreted as the On-Shell action since we derived it with the Off-Shell action. I'm missing something here... $\endgroup$
    – Syrocco
    Apr 30, 2020 at 17:57
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Apr 30, 2020 at 18:18

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