1
$\begingroup$

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:

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

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:

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

$\endgroup$
  • $\begingroup$ 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 Nov 3 '14 at 13:45
  • $\begingroup$ Oh, my bad... I'll just omit that part, then (it changes nothing). $\endgroup$ – Chanto Nov 3 '14 at 16:03
  • $\begingroup$ But $dx_2/dt_2=0$ as these are independent constants — aren't they? $\endgroup$ – Ruslan Nov 4 '14 at 4:48
  • $\begingroup$ If at the beginning you say $x_2 = a sin{t_2}$, how come, they're independent? $\endgroup$ – Chanto Nov 4 '14 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 Nov 4 '14 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.