# Why is action a functional of $q$ only?

Action of a particle is written as $$S[q]=\int dt\hspace{0.2cm} L(q(t),\dot{q}(t),t).$$ How can I understand why $S$ is a functional of $q$, and not that of $\dot{q}$? Assuming $L=\frac{1}{2}m\dot{q}^2-V(q)$ and $V(q)=\alpha q^2$ (as an example), how can I understand that $S=S[q]$ and $S\neq S[q,\dot{q}]$?

1. The notation $q$ in the functional $S[q]$ stands for the whole parametrized curve/path $q:[t_i,t_f]\to \mathbb{R}$, not just a single position. In particular, the parametrized path already carries all the information about the derivative $\dot{q}\equiv\frac{dq}{dt}$, so there is no need to include it as an extra argument in the action.

2. In contrast, the Lagrangian $L(q(t),v(t),t)$ at some time $t$ is a function of

• the instantaneous position $q(t)$ at the time $t$;

• the instantaneous velocity $v(t)$ at the time $t$; and

• the time $t$ (also known as explicit time-dependence).

• But the Lagrangian is also parameterized by $q(t)$, the whole path. Isn't it? But we still write $L=L(q,\dot{q})$. @Qmechanic – SRS Jan 8 '18 at 17:47