8.2: Function of Endpoint Time

What about the action as a function of the final point arrival time?

Since \(\begin{equation}
S=\int_{t_{1}}^{t_{2}} L d t, \text { the total time derivative } d S / d t_{2}=L\left(q^{(2)}, t_{2}\right)
\end{equation}\),

the value of the Lagrangian at the endpoint. Remember we are defining the action at a point as that from integrating along the true path up to that point.

Landau denotes \(\begin{equation}
t_{2} \text { by just } t, \text { so he writes } d S / d t=L
\end{equation}\)

and we’ll be doing this, but it’s crucial to keep in mind that the endpoint position and time are the variables here!

If we now allow an incremental time increase, \(\begin{equation}
t_{2} \rightarrow t_{2}+d t
\end{equation}\), with the final coordinate position as a free parameter, the dynamical path will now continue on, to an incrementally different finishing point.

This will give (with t understood from now on to mean \(\begin{equation}
t_{2}, \text { and } q_{i} \text { means } \left.q_{i}^{(2)}\right)
\end{equation}\)

\begin{equation}
\frac{d S\left(q_{i}, t\right)}{d t}=\frac{\partial S}{\partial t}+\sum_{i} \frac{\partial S}{\partial q_{i}} \dot{q}_{i}=\frac{\partial S}{\partial t}+\sum_{i} p_{i} \dot{q}_{i}
\end{equation}

Putting this together with \(\begin{equation}
d S / d t=L
\end{equation}\) gives immediately the partial time derivative

\begin{equation}
\partial S / \partial t=L-\sum_{i} p_{i} \dot{q}_{i}=-H
\end{equation}

and therefore, combining this with the result \(\begin{equation}
\partial S / \partial q_{i}=p_{i}
\end{equation}\) from the previous section,

\begin{equation}
d S\left(q_{i}, t\right)=\sum_{i} p_{i} d q_{i}-H d t
\end{equation}

This, then, is the total differential of the action as a function of the spatial and time coordinates of the end of the path.