In Ref. 1, in section 3, they wrote: \begin{equation} L'(q,\dot{q},\ddot{q},t)~=~L(q,\dot{q},\ddot{q},t)-\frac{dS(q,\dot{q},t)}{dt}.\tag{14} \end{equation} Then the Hamilton-Jacobi equation is \begin{equation} \frac{\partial S}{\partial t}~=~-H_0,\tag{27} \end{equation} where $$H_0~=~p\dot{q}+\pi\ddot{q}-L,\tag{28}$$ where $$p~=~\dfrac{\partial S}{\partial q}\quad\text{and}\quad \pi~=~\dfrac{\partial S}{\partial \dot{q}}.\tag{29}$$

I was confused because here the action is a function of $\dot{q}$ ! Then I read this answer by Qmechanic, and I had this question: is this action a Hamilton's principle function? If yes, in Hamilton's principal function $S(q,\alpha,t)$, the $\alpha$ is an integration constant. But here we have $S(q,\dot{q},t)$; $\dot{q}$ is not an integration constant. If no, what is that action?

EDIT: I want to know if the function $S$ is considered as an action. If yes, I need an explanation on how an action could be a function of $\dot{q}$? And if no, how is this function related to the known action $S[q]$?


  1. B.M. Pimentel & R.G. Teixeira, Hamilton-Jacobi formulation for singular systems with second order Lagrangians, arXiv:hep-th/9512099.
  • $\begingroup$ The Lagrangian is almost never written in terms of $ \ddot{q} $ or anything higher. $\endgroup$ – QuantumFool Oct 8 '16 at 0:31
  • $\begingroup$ @QuantumFool "almost" never , but not in my research case, where I'm interested in higher order derivatives Lagrangian. $\endgroup$ – Milou Oct 8 '16 at 0:42
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/283166/2451 $\endgroup$ – Qmechanic Oct 10 '16 at 14:41
  • $\begingroup$ This could be an answer for that post if S is the action.. but I don't know if S is the action or just a function. If it's not the action, then this is not the answer for the question in the other post. $\endgroup$ – Milou Oct 10 '16 at 15:00
  • $\begingroup$ If it's not the action, there should be a relation between the S and the action of course.. they're not totally different $\endgroup$ – Milou Oct 10 '16 at 15:01

OP is asking about Ref. 1. This touches upon several topics:

  1. Second-order Lagrangian formulations and their corresponding Ostrogradsky Hamiltonian formulation. This is discussed in e.g. this Phys.SE post, where it is explained how to reduce to first-order formulation in the non-singular case.

  2. Singular Legendre transformations, and constraints. See e.g. Ref. 2 & 3.

  3. Caratheodory’s method of equivalent Lagrangians, and its connection to Hamilton-Jacobi theory. This is explained in Ref. 4 in the first-order case.

Now let us return to OP's title question. Let us stress that the Hamilton's principal function $S$ and the off-shell action functional $S[q]$ are different objects. (For first-order Lagrangians, this is e.g. explained in my Phys.SE answers here & here.) In particular, eq. (14) is part of Caratheodory’s method of equivalent Lagrangians. The $S$ appearing in eq. (14) is Hamilton's principal function, not the off-shell action functional. It depends on velocities because the theory is of second order.


  1. B.M. Pimentel & R.G. Teixeira, Hamilton-Jacobi formulation for singular systems with second order Lagrangians, arXiv:hep-th/9512099.

  2. P.A.M. Dirac, Lectures on QM, (1964).

  3. M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994.

  4. H.A. Kastrup, Canonical theories of Lagrangian dynamical systems in physics, Phys. Rep. 101 (1983) 1; Section 2.4.

  • $\begingroup$ Alright, but now I'm confused. 1. why do we use the off-shell action to determine the equations of motion instead of the on-shell one? 2. Doesn't "on/off-shell" determine whether any term satisfies the e.o.m or not? 3. Also when do we use the on-shell action instead of the off-shell one? 4. Is that only related to higher order derivative Lagrangians? 5. or does it have a connection with virtual or unknown paths? $\endgroup$ – Milou Nov 1 '16 at 0:21
  • $\begingroup$ 1. One needs the eoms to construct the on-shell action in the first place. 2. Yes. 3. Too broad question. 4. No. 5. Off-shell action also applies to virtual paths. $\endgroup$ – Qmechanic Nov 1 '16 at 8:12

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