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


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 $t=t_2$:


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 $\frac{\partial S}{\partial t_2}$?


  1. L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, 1976; $\S$ 43

1 Answer 1


When you write this:

$$D_{t_2} S=-x_2^2\csc^2t_2.\tag6$$

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:

$D_{t_2} S=-x_2^2\csc^2t_2+\frac{\partial S}{\partial x_2}\frac{\partial x_2}{\partial t_2}=-x_2^2\csc^2t_2+2x_2\cot t_2\frac{dx_2}{dt_2}$

  • $\begingroup$ Comment to the answer (v1): Consider explaining the prime notation. Note that $S(q_2,t_2)$ does not depend on $\dot{q}_2$ by definition. $\endgroup$
    – Qmechanic
    Commented Nov 3, 2014 at 13:45
  • $\begingroup$ Oh, my bad... I'll just omit that part, then (it changes nothing). $\endgroup$ Commented Nov 3, 2014 at 16:03
  • $\begingroup$ But $dx_2/dt_2=0$ as these are independent constants — aren't they? $\endgroup$
    – Ruslan
    Commented Nov 4, 2014 at 4:48
  • $\begingroup$ If at the beginning you say $x_2 = a sin{t_2}$, how come, they're independent? $\endgroup$ Commented Nov 4, 2014 at 16:02
  • $\begingroup$ They are independent because we put them independently into $S(x_2,t_2)$. I.e. we can evaluate $S$ at arbitrary point $(x_2,t_2)$. When I said $x_2=a\sin t_2$, I used it to find $a$, not $x_2$. $\endgroup$
    – Ruslan
    Commented Nov 4, 2014 at 16:03

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