What's the partial derivative of the action wrt. velocity?

Question

Consider a point mass on $3$D euclidean space that has a trajectory $\boldsymbol q(t)$ satisfying the Euler-Lagrange equations (on-shell/classical mechanics). Its action may be computed as $$S[\boldsymbol q]=\int_{t_0}^{t}\mathcal L(\boldsymbol q(t'), \dot{\boldsymbol q}(t'),t')\,\mathrm dt'\tag{1}$$ From the above we have $$\frac{\partial S}{\partial t}+\frac{\partial S}{\partial \boldsymbol q}\dot{\boldsymbol q}+\frac{\partial S}{\partial \dot{\boldsymbol q}}\ddot{\boldsymbol q}=\frac{\mathrm dS}{\mathrm dt}=\mathcal L(\boldsymbol q(t), \dot{\boldsymbol q}(t),t)\tag{2}$$ From this relation I've been able to deduce that $$\frac{\partial \mathcal L}{\partial \dot{\boldsymbol q}}=\boldsymbol p(t)=\frac{\partial S}{\partial \boldsymbol q}\qquad \text{and}\qquad \frac{\partial S}{\partial t}=\mathcal L(\boldsymbol q, \dot{\boldsymbol q},t)-\frac{\partial S}{\partial \boldsymbol q}\dot{\boldsymbol q}=-H(t).\tag{3}$$ So since the lagrangian generally does not depend on the acceleration, this makes me think that $$\frac{\partial S}{\partial \dot{\boldsymbol q}}=\frac{\partial \mathcal L}{\partial \ddot{\boldsymbol q}}=0\tag{4}$$ Is this reasoning correct? I'm not sure if I saw this written: $$\frac{\partial S}{\partial \dot{q}_j}=Q_j\tag{5}$$ where $Q_j$ is the generalized force and so I wasn't sure what was right.

Answer 1

1. It does not make sense to (partially or functionally) differentiate the off-shell action $S[q]$ wrt. velocity.

2. The (Dirichlet) on-shell action $S(q_f,t_f;q_i,t_i)$ does not depend on velocity, cf. e.g. this Phys.SE post.