I'm following Ref. 1 in my reasoning, struggling with action as a function of time.
Consider a Lagrangian
$$L=\dot x^2-x^2.\tag1$$
Solving the corresponding equations of motion with initial position $x(0)=0$, we'll find that true motion is described by $x=a\sin t$. Now to find action as a function of time, we have to fix final position $x(t_2)=x_2$ and rewrite $x$ in terms of it. Since $x_2=a\sin t_2$, we have
$$x=x_2\csc t_2\sin t.\tag2$$
We can find the velocity
$$v=x_2\csc t_2\cos t\tag3$$
and with rewritten Lagrangian
$$L=x_2^2\csc^2t_2\cos2t\tag4$$
now compute the action as a function of time $t_2$:
$$S(t_2)=\int_0^{t_2}(x_2^2\csc^2t_2\cos2t)dt=x_2^2\cot t_2.\tag5$$
Now I'm trying to see the difference between $(43.4)$ and $(43.5)$ in Ref. 1, i.e. citing it,
From the definition of the action, its total time derivative along the path is
$$\text{d}S/\text{d}t=L.\tag{43.4}$$
Next, regarding $S$ as a function of co-ordinates and time, in the sense described above, and using formula $(43.3)$, we have
$$\frac{\text dS}{\text dt}=\frac{\partial S}{\partial t}+\sum_i \frac{\partial S}{\partial {q_i}}\dot q_i=\frac{\partial S}{\partial t}+\sum_ip_i\dot q_i.$$
A comparison gives $\partial S/\partial t=L-\sum p_i\dot q_i$ or
$$\partial S/\partial t=-H.\tag{43.5}$$
Formula $(43.3)$ mentioned in the citation states $\partial S/\partial q_i=p_i$.
And I see the only way to find a derivative of $(5)$:
$$D_{t_2} S=-x_2^2\csc^2t_2.\tag6$$
Now computing $L-pv=-H$ should give us $\partial S/\partial t_2$. We have
$$L-pv=x_2^2\csc t_2\cos2t-2x_2\csc t_2\cos t\cdot x_2\csc t_2\cos t=-x_2^2\csc^2t_2.\tag7$$
OK, so I understand that what I've found in $(6)$, according to $(43.5)$ in the citation, is partial derivative of $S$:
$$D_{t_2} S=\frac{\partial S}{\partial t_2}.\tag8$$
On the other hand, taking total derivative of $S(t_2)$ should give us back the Lagrangian at $t=t_2$:
$$L(t_2)=x_2^2(\cot^2t_2-1).\tag9$$
But how can we recover the Lagrangian $(9)$ from $(5)$? It seems it doesn't even matter whether we take a partial or total time derivative of $(5)$, the results are identical.
I've tried looking at the general expression for $S$ and how my "partial" derivative of it looks in general, denoting $x(t)=q(x_2,t_2,t)$:
$$\frac{\partial S}{\partial t_2}=L(x_2,\dot q(x_2,t_2,t_2),t_2)+p(t_2)\left.\frac{\partial q(x_2,t_2,t)}{\partial t_2}\right|_{t=t_2}-p(0)\frac{\partial q(x_2,t_2,0)}{\partial t_2}.\tag{10}$$
Here $\partial q/\partial t_2=0$ in the example Lagrangian $(1)$, the term with $p(0)$ vanishes. Comparing with the expression in citation right before $(43.5)$, I see that $p\dot q$ in Ref. 1 relates to $(10)$ as
$$p\frac {\partial q}{\partial t}=-p(t_2)\left.\frac{\partial q(x_2,t_2,t)}{\partial t_2}\right|_{t=t_2}.$$
And somehow subtracting this from my "paritial" derivative yields the Lagrangian. But from this I can only conclude that $\frac{\text{d}S}{\text{d}t_2}$ is an artificially defined entity, not something one could find by direct differentiation… What am I missing? How should one define $\frac{\text{d}S}{\text{d}t_2}$ and $\frac{\partial S}{\partial t_2}$?
References:
- L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, 1976; $\S$ 43
Best Answer
When you write this:
You obviously mean total time derivative by $D_{t2}$. Which means:
$D_{t2}S=dS/dt_2=\frac{\partial S}{\partial q}\frac{\partial q}{\partial t_2}+\frac{\partial S}{\partial t_2}\frac{\partial t_2}{\partial t_2}$
In our case: $q=x_2$
and you're missing this part: $\frac{\partial S}{\partial q}\frac{\partial q}{\partial t_2}$, which is not zero. So, we have:
$D_{t_2} S=-x_2^2\csc^2t_2+\frac{\partial S}{\partial x_2}\frac{\partial x_2}{\partial t_2}=-x_2^2\csc^2t_2+2x_2\cot t_2\frac{dx_2}{dt_2}$