# What variables does the action $S$ depend on?

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

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

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

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 $$q_f$$.

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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.)