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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

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

I think we cannot find a function $Z(q, t)$ such that $$L = \frac{dZ(q, t)}{dt}?$$

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  • $\begingroup$ If you like this question you may also enjoy this Phys.SE post. $\endgroup$ – Qmechanic Oct 2 '11 at 21:16
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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 $q_f$.

4) To answer the last subquestion(v2): The Lagrangian is in general not a total time-derivative, cf. this Phys.SE post.

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$^1$ 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.)

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  • $\begingroup$ Why do you add at v(t) = qdot(t)? instead, why not just write qdot(t) in place of v(t) and write L(q, qdot, t)? $\endgroup$ – user5198 Oct 3 '11 at 14:53
  • $\begingroup$ @user5198: You can do that; it is equivalent. $\endgroup$ – Qmechanic Oct 3 '11 at 16:54
  • $\begingroup$ Well, when some students asked the same question our teacher would say that as path integral of the functional is being taken over time any dependence on time will be integrated out therefore the action S will not be function of qdot . though the above looked a very simple answer but it gave satisfaction to many of us. $\endgroup$ – drvrm Mar 19 '16 at 18:56

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