I also understand that $\frac{\partial S}{\partial t_1} = -L(t_1)$.
This is only true when the integrand is fixed. When the integrand is not fixed, we can not use the usual rule for differentiation with respect to a lower limit of an integral.
This is why
$$
\frac{\partial S}{\partial t_1} = - L(t_1)\tag{$L(x(t),\dot x(t))$ fixed!}\;,
$$
when the integrand is fixed because the path is fixed. (Note that the functional form of $L$ is always fixed for a given problem.)
But
$$
\frac{\partial S_{cl}}{\partial t_1} = H(t_1) \tag{$L(x(t),\dot x(t))$ not fixed}\;,
$$
when the integrand changes due to the change in classical path required for the different initial and final actions.
I will explain this in detail below.
I have two following questions mostly related to each other.
Q1: we ask how much S changes when we change $t_1$. This basically means we're not changing the path in variation, but we're just saying what if we had started the path from $t1+\delta h$ which means the remaining path is the same. Correct ?
Q2: I wonder how we derived $\frac{\partial S}{\partial t_1} = H(t1)$. My idea is as follows - we know that $H = vp - L$ from which we have: $L = vp - H$.
As an introduction, perhaps it is best to begin with an appeal to authority. And what greater authority than Dick Feynman, who posed the following question in his incomparable textbook "Quantum Mechanics and Path Integrals":
In his book, and in the above-quoted passage, Dick used the notation "$E$" for the Hamiltonian. However, we have been using the letter $E$ to denote the total mechanical energy (which is not always the same as the Hamiltonian, but often is). We have also been using $t_1$ for the initial time and $t_2$ for the final time rather than $t_a$ and $t_b$. So, for consistency, I have rewriten the relevant equations below using our notation:
$$
H=\dot x p - L = \dot x \frac{\partial L}{\partial \dot x} - L
$$
$$
H(t_2) = -\frac{\partial S_{cl}}{\partial t_2}
$$
$$
H(t_1) = +\frac{\partial S_{cl}}{\partial t_2}\;.
$$
Dick also provides us with a Hint: "A change in the time of an end point requires a change in path, since all paths must be classical paths."
As another introductory matter, consider the usual rule for differentiation with respect to an endpoint. If we define
$$
F(t_1, t_2) = \int_{t_1}^{t_2}f(t')dt'\;,
$$
then for fixed $f(t)$ (that is, for a fixed integrand function) we have:
$$
\frac{\partial F}{\partial t_1} = -f(t_1)
$$
and
$$
\frac{\partial F}{\partial t_2} = +f(t_2)\;.
$$
Thus, consistent with the hint we were given, the integrand (the Lagrangian evaluated on a specific classical path) must not be fixed or else we would simply arrive at:
$$
\frac{\partial S_{cl}}{\partial t_1} \stackrel{??!}{=} -L(t_1)\tag{No, this is wrong!}
$$
and
$$
\frac{\partial S_{cl}}{\partial t_2} \stackrel{??!}{=} +L(t_2)\tag{No, this is wrong!}
$$
To explain how I will proceed, another picture may be helpful:
With reference to the above picture, we have:
- The classical path from $(t_1, x_1)$ to $(t_2, x_2)$ is $q(t)$
- The classical path from $(t_1+\delta t, x_1)$ to $(t_2, x_2)$ is r(t). (Note that $r(t)$ is also the path from $(t_1, x_1-\delta x)$ to $(t_2,x_2)$)
- The deviation from one path to the other is $\eta(t) = r(t)-q(t)$. This deviation will be "small" in the same sense that $\delta t$ and $\delta x$ are "small." That is, we will be content with only working to linear order in $\eta$ and $\delta t$ and $\delta x$ since we will take the limit $\delta t\to 0$ in the end.
Note that we have chosen $\eta(t)$ such that $\eta(t_1)=-\delta x$ and $\eta(t_2) = 0$. We made this choice because we are here considering only $\frac{\partial S_{cl}}{\partial t_1}$. (That is, here we are only going to show that $\frac{\partial S_{cl}}{\partial t_1}=H(t_1)$.)
The initial classical action of interest is:
$$
S_{cl}^{\text{init}}=\int_{t_1}^{t_2}L(q(t'),\dot q(t'))dt'\;.
$$
And the final classical action of interest is:
$$
S_{cl}^{\text{final}}=\int_{t_1+\delta t}^{t_2}L(r(t'),\dot r(t'))dt'\;.
$$
Thus, the relevant "small" change in the action $\delta S_{cl}$ that matters to us is:
$$
\delta S_{cl} = \int_{t_1+\delta t}^{t_2}L(r(t'),\dot r(t'))dt' - \int_{t_1}^{t_2}L(q(t'),\dot q(t'))dt'\;, \tag{1}\;.
$$
Continuing, we re-write our equation as:
$$
\delta S_{cl} = \int_{t_1+\delta t}^{t_2}\left[L(r,\dot r) - L(q,\dot q)\right]dt' - \int_{t_1}^{t_1+\delta t}L(q,\dot q)dt'
$$
$$
\approx \int_{t_1+\delta t}^{t_2}\left[L(r,\dot r) - L(r-\eta,\dot r-\dot \eta)\right]dt' -\delta t L(t_1)\;,
$$
where we pause here to note that the last term above $-\delta t L(t_1)$ would be the only contribution to the change in the case that the integrand was fixed.
By expanding $L(q-\eta,\dot q - \dot \eta)$ to first order in $\eta$, we find:
$$
\delta S_{cl}\approx \int_{t_1+\delta t}^{t_2}\left[\eta(t')\left.\frac{\partial L}{\partial x}\right|_{x=r}+\dot \eta \left.\frac{\partial L}{\partial \dot x}\right|_{x=r}\right]dt'-\delta t L(t_1)\;.
$$
$$
=\int_{t_1+\delta t}^{t_2}
\left[
\eta\frac{\partial L}{\partial x}
+\frac{d}{dt}\left(\frac{\partial L}{\partial \dot x}\eta\right)
-\eta\frac{d}{dt}\frac{\partial L}{\partial \dot x}
\right]dt'
-\delta t L(t_1)
$$
$$
=\int_{t_1+\delta t}^{t_2}
\left[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot x}\eta\right)
\right]dt'
-\delta t L(t_1)\;,
$$
where, in the last step, we used the Euler-Lagrange equation, since we are on the classical path from $t_1+\delta t$ to $t_2$.
Now, the integral in the first term above is trivial, since it is the integral of a total derivative. Performing the integration, and using $\eta(t_1+\delta t)\approx-\delta x$ and $\eta(t_2)=0$ we find
$$
\delta S_{cl} \approx \delta x\frac{\partial L}{\partial \dot x}-\delta t L\;.
$$
Or, taking the limit as $\delta t \to 0$ and ignoring terms of higher than linear order in $\delta t$, and using $\dot x=\lim_{\delta t\to 0}\frac{\delta x}{\delta t}$, we find
$$
\frac{\partial S_{cl}}{\partial t_1}=\left.
\dot{x} \frac{\partial L}{\partial \dot {x}}\right|_{x=q(t_1)} - L(t_1)=H(t_1)\;,
$$
which is what we set out to prove.
The similar proof that
$$
-\frac{\partial S_{cl}}{\partial t_2} = \dot x \frac{\partial L}{\partial \dot x}(t_2) - L(t_2)
$$
is left as an exercise to the interested reader.
