Action is defined as,

$$S ~=~ \int L(q, q', t) dt,$$

but my question is what variables does $S$ depend on?

Is $S = S(q, t)$ or $S = S(q, q', t)$ where $q' := \frac{dq}{dt}$?

In Wikipedia I've read that $S = S(q(t))$ and I think that suppose, $q$ and $t$ are considered as independent coordinates. Then $S$ should depend on $q'$ also because, for the typical Lagrangian

$$L ~=~ \frac{q'^2}{2} - V(q).$$

  • $\begingroup$ If you like this question you may also enjoy this Phys.SE post. $\endgroup$
    – Qmechanic
    Oct 2, 2011 at 21:16

1 Answer 1


1) Firstly, the Lagrangian $L(q(t),v(t),t)$ at some time $t$ is a function of:

  1. the instantaneous position $q(t)$ at the time $t$;
  2. the instantaneous velocity $v(t)$ at the time $t$; and
  3. the time $t$ (also known as explicit time-dependence).

2) Secondly, the (off-shell) action

$$\tag{1} S[q]~:=~ \left. \int_{t_i}^{t_f}\! dt \ L(q(t),v(t),t)\right|_{v(t)=\dot{q}(t)} $$

is a functional of the full position curve/path $q:[t_i,t_f] \to \mathbb{R}$ for all times $t$ in the interval $[t_i,t_f]$.

3) Thirdly, if one imposes boundary conditions (B.C.), e.g. Dirichlet B.C.,

$$\tag{2} q(t_i)~=~q_i \qquad \text{and}\qquad q(t_f)~=~q_f, $$

then there is also a notion of a (Dirichlet) on-shell action $^1$

$$\tag{3} S(q_f,t_f;q_i,t_i)~:=~S[q_{\rm cl}]$$

where $q_{\rm cl}:[t_i,t_f] \to \mathbb{R}$ is the classical path, which satisfies Euler-Lagrange equations with the Dirichlet B.C. (2). The on-shell action $S(q_f,t_f;q_i,t_i)$ is a function of

  1. the initial time $t_i$;
  2. the initial position $q_i$;
  3. the final time $t_f$; and
  4. the final position $q_f$.


$^1$ See also e.g. MTW Section 21.1. For the on-shell action $S(q_f,t_f;q_i,t_i)$ to be well-defined, there should exist a unique classical path with the B.C. (2). (Here the words on-shell and off-shell refer to whether the Euler-Lagrange equations are satisfied or not.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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