# Partial time derivative of the on-shell action

I have a few questions about differentiating the on-shell action.

Here is what I currently understand (or think I do!):

1. Given that a system with Lagrangian $$\mathcal{L}(\mathbf{q}, \dot{\mathbf{q}}, t)$$ has the coordinate $$\mathbf{q}_1$$ at time $$t_1$$, and the coordinate $$\mathbf{q}_2$$ at time $$t_2$$, there exists a unique 'extremal path' $$\gamma(t_1, \mathbf{q}_1, t_2, \mathbf{q}_2; t)$$ which makes the action functional $$\mathcal{S}[\mathbf{q}(t)] = \int_{t_1}^{t_2} \mathcal{L}(\mathbf{q}, \dot{\mathbf{q}}, t)\text{d}t$$ stationary. In other words, $$\gamma$$ satisfies the Euler-Lagrange equations, $$\left.\left(\frac{\partial \mathcal{L}}{\partial q} -\frac{\text{d}}{\text{d}t}\frac{\partial \mathcal{L}}{\partial \dot{q}} \right)\right|_{q(t) = \gamma(t)} = \mathbf{0},$$ and has $$\gamma(t_1, \mathbf{q}_1, t_2, \mathbf{q}_2; t_1) = \mathbf{q}_1$$ and $$\gamma(t_1, \mathbf{q}_1, t_2, \mathbf{q}_2; t_2) = \mathbf{q}_2$$.

2. Moreover, the existence of this function allows the velocity, momentum etc. to be defined at the endpoints, e.g. the momentum at $$(t_2, \mathbf{q}_2)$$ is $$\mathbf{p}_2 = \left.\frac{\partial \mathcal{L}}{\partial \dot{\gamma}(t)}\right|_{t=t_2},$$ where $$\dot{\gamma} \equiv \partial \gamma(t_2, \mathbf{q}_2; t) /\partial t$$.

3. Ignoring $$t_1$$ and $$\mathbf{q}_2$$ for simplicity, this allows the on-shell action (see here) to be defined as $$s(t_2, \mathbf{q}_2) = \int_{t_1}^{t_2} \mathcal{L}(\gamma(t_2, \mathbf{q}_2; t), \dot{\gamma}(t_2, \mathbf{q}_2; t), t)\, \text{d} t. \tag{1}$$ Importantly, $$s$$ is a function of $$t_2$$, $$\mathbf{q}_2$$, and not a functional. It can therefore be differentiated as any other function.

4. It is shown in Landau that

$$\frac{\partial s}{\partial t_2} = -\mathcal{H}_2, \quad \frac{\partial s}{\partial \mathbf{q}_2} = \mathbf{p}_2, \tag{2}$$

but I don't follow the argument given.

I would like to derive the equations (2) by directly differentiating (1). I have read several answers which derive this in a different way (here, here and here), but I still have some questions. Firstly, here is my attempt at differentiating with respect to $$\mathbf{q}_2$$.

\begin{align} \frac{\partial s}{\partial \mathbf{q}_2} &= \frac{\partial}{\partial \mathbf{q}_2} \int_{t_1}^{t_2} \mathcal{L}(\gamma(t_2, \mathbf{q}_2; t), \dot{\gamma}(t_2, \mathbf{q}_2; t), t)\, \text{d} t \\ &= \int_{t_1}^{t_2} \frac{\partial}{\partial \mathbf{q}_2} \mathcal{L}(\gamma(t_2, \mathbf{q}_2; t), \dot{\gamma}(t_2, \mathbf{q}_2; t), t)\, \text{d} t\\ &= \int_{t_1}^{t_2} \frac{\partial \mathcal{L}}{\partial \gamma}\cdot\frac{\partial \gamma}{\partial \mathbf{q}_2} +\frac{\partial \mathcal{L}}{\partial \dot{\gamma}}\cdot\frac{\partial \dot{\gamma}}{\partial \mathbf{q}_2} \text{d}t. \end{align} Now, $$\frac{\partial \dot{\gamma}}{\partial \mathbf{q}_2} =\frac{\partial}{\partial \mathbf{q}_2} \frac{\text{d}\gamma}{\text{d} t} = \frac{\text{d}}{\text{d} t} \frac{\partial \gamma}{\partial \mathbf{q}_2},$$ so we can integrate by parts to yield \begin{align} \frac{\partial s}{\partial \mathbf{q}_2} &= \left[ \frac{\partial \mathcal{L}}{\partial \dot{\gamma}} \cdot \frac{\partial \gamma}{\partial \mathbf{q}_2}\right]_{t_1}^{t_2} + \int_{t_1}^{t_2} \underbrace{\left(\frac{\partial \mathcal{L}}{\partial \gamma} - \frac{\text{d}}{\text{d} t} \frac{\partial \mathcal{L}}{\partial \dot{\gamma}}\right)}_{\mathbf{0}}\cdot\frac{\partial \gamma}{\partial \mathbf{q}_2} \text{d} t\\ &=\mathbf{p}_2\cdot\frac{\partial \gamma}{\partial \mathbf{q}_2}(t_2). \end{align} For (2) to be true, we ought to have $$\frac{\partial \gamma}{\partial \mathbf{q}_2}(t_2) = \mathbf{I}$$. Is it valid interchange the order of evaluation and differentiation to write $$\left.\frac{\partial \gamma(t_2, \mathbf{q}_2; t)}{\partial \mathbf{q}_2}\right|_{t=t_2} = \frac{\partial \gamma(t_2, \mathbf{q}_2; t_2)}{\partial \mathbf{q}_2} = \frac{\partial \mathbf{q}_2}{\partial \mathbf{q}_2} =\mathbf{I}?\tag{3}$$ If so, why? If not, then how else is it possible to arrive at equation (2) from here?

Secondly, here is my attempt at differentiating with respect to $$t_2$$. \begin{align} \frac{\partial s}{\partial t_2} &= \frac{\partial}{\partial t_2} \int_{t_1}^{t_2} \mathcal{L}(\gamma(t_2, \mathbf{q}_2; t), \dot{\gamma}(t_2, \mathbf{q}_2; t), t)\, \text{d} t \\ &= \mathcal{L}_2 + \int_{t_1}^{t_2} \frac{\partial}{\partial t_2} \mathcal{L}(\gamma(t_2, \mathbf{q}_2; t), \dot{\gamma}(t_2, \mathbf{q}_2; t), t)\, \text{d} t\\ &=\mathcal{L}_2 + \int_{t_1}^{t_2} \frac{\partial \mathcal{L}}{\partial \gamma}\cdot\frac{\partial \gamma}{\partial t_2} +\frac{\partial \mathcal{L}}{\partial \dot{\gamma}}\cdot\frac{\partial \dot{\gamma}}{\partial t_2} \text{d}t\\ &= \mathcal{L}_2 +\left[ \frac{\partial \mathcal{L}}{\partial \dot{\gamma}} \cdot \frac{\partial \gamma}{\partial t_2}\right]_{t_1}^{t_2} + \int_{t_1}^{t_2} \underbrace{\left(\frac{\partial \mathcal{L}}{\partial \gamma} - \frac{\text{d}}{\text{d} t} \frac{\partial \mathcal{L}}{\partial \dot{\gamma}}\right)}_{\mathbf{0}}\cdot\frac{\partial \gamma}{\partial t_2} \text{d} t\\ &=\mathcal{L}_2 + \mathbf{p}_2\cdot\frac{\partial \gamma}{\partial t_2}(t_2) \end{align} To get from the first to the second line I used Leibniz' rule for differentiating integrals. For equation (2) to be true, we ought to have $$\frac{\partial \gamma}{\partial t_2}(t_2) = -\dot{\mathbf{q}}_2.\tag{4}$$ Is this correct? If so, how can it be shown?

I would be very grateful for any help anyone is able to give!

• Eq. (3) follows from the boundary condition $$\gamma(t_2, \mathbf{q}_2; t\!=\!t_2)~=~\mathbf{q}_2. \tag{A}$$
• Eq. (4) follows by differentiating eq. (A) wrt. $$t_2$$: $$\left.\frac{\partial\gamma(t_2, \mathbf{q}_2; t)}{\partial t_2}\right|_{t=t_2} + \left. \frac{\partial\gamma(t_2, \mathbf{q}_2; t)}{\partial t}\right|_{t=t_2}~=~0.\tag{B}$$