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I'm trying to get some understanding in treating action $S$ as a function of coordinates. Landau and Lifshitz consider $\delta S$, getting $\delta S=p\delta q$, thus concluding that

$$\frac{\partial S}{\partial q}=p.$$

I'm now trying to understand this result, and consider the definition of action. I suppose that action as a function of coordinates $x$ will have $q=q(x,t)$ and $\dot q=\dot q(x,t)$, so $S(x)$ will look like:

$$S(x)=\int_{t_1}^{t_2}L(q(x,t),\dot q(x,t),t)dt.$$

Now I take the partial derivative with respect to $x$ to get $p$:

$$\frac{\partial S}{\partial x}=\int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\frac{\partial q}{\partial x}+\frac{\partial L}{\partial \dot q}\frac{\partial \dot q}{\partial x}\right)dt.$$

And now I'm stuck. I see the momentum $\frac{\partial L}{\partial \dot q}=p$ and force $\frac{\partial L}{\partial q}=F$ inside the integral, but I can't seem to get, how to extract the momentum, so that all the other things cancelled.

Am I on the right track? What should be the next step?

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Comments to the question (v1):

  1. The main point is that Ref. 1 is considering the (Dirichlet) on-shell action function $$ \tag{1} S(q_f,t_f;q_i,t_i)~:=~I[q_{\rm cl}], $$ not the (off-shell) action functional $$\tag{2} I[q]~:=~ \left. \int_{t_i}^{t_f}\! dt \ L(q(t),v(t),t)\right|_{v(t)=\dot{q}(t)}. $$ See e.g. this Phys.SE post.

  2. The Lagrangian canonical/conjugate momentum is defined as $$\tag{3} p(t)~:=~\frac{\partial L(q(t),v(t),t)}{\partial v(t)},$$ as OP mentions.

  3. The statement that OP wants to prove is:

    $$\tag{4} \frac{\partial S}{\partial q_f}~=~p(t_f) , \qquad \frac{\partial S}{\partial q_i}~=~-p(t_i).$$

  4. Equation (4) is e.g. proven in my Phys.SE answer here.

References:

  1. L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, 1976; $\S$ 43.
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  • $\begingroup$ Further hint: What OP calls $x$ is $q_f$. $\endgroup$ – Qmechanic Nov 1 '14 at 19:40
  • $\begingroup$ Thanks, you made me understand some of my misconceptions, so now I've finished my proof and put it to an answer. $\endgroup$ – Ruslan Nov 1 '14 at 19:54
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Thanks to Qmechanic's answer, I've understood that $\partial S/\partial x=p(t_2)$, while I was under the illusion that it would somehow equal $p(t)$.

Now follows the finalization of my attempts.

First, consider the second part of the expression inside the integral for $\partial S/\partial x$ in the OP,

$$A=\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot q}\frac{\partial \dot q}{\partial x}dt.$$

Integrating by parts, namely using $u=\frac{\partial L}{\partial\dot q}$, $dv=\frac{\partial\dot q}{\partial x}dt$, we get, with $du=\frac d{dt}\frac{\partial L}{\partial\dot q}dt=\frac{\partial L}{\partial q}dt$ and $v=\frac\partial{\partial x}\int\dot q dt=\frac{\partial q}{\partial x}$,

$$A=\left.p\frac{\partial q}{\partial x}\right|_{t_1}^{t_2}-\int_{t_1}^{t_2}\frac{\partial q}{\partial x}\frac{\partial L}{\partial q}dt.$$

Substituting this into the integral in the OP, we get the integrals cancel, thus

$$\left.\frac{\partial S}{\partial x}=p\frac{\partial q}{\partial x}\right|_{t_1}^{t_2}=p(t_2)\frac{\partial q(x,t_2)}{\partial x}-p(t_1)\frac{\partial q(x,t_1)}{\partial x}.$$

But by definition of $x$, $q(x,t_2)\equiv x$, and because first point is fixed, we have $q(x,t_1)=q^{(1)}=\text{const}(x)$, thus the result is what was to be proved:

$$\frac{\partial S}{\partial x}=p(t_2).$$

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  • $\begingroup$ This is surprisingly confusing to me. Could you just explain for me why $q(x,t_2)=x$? My answer would have just boiled down to $\delta S = \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot q}\delta \dot{q} \right)dt = \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\delta q-\frac{d}{dt} \frac{\partial L}{\partial \dot q}\delta q \right)dt + \frac{\partial L}{\partial \dot q} \delta q |^{t_2}_{t_1} = p(t_2) \delta q(t_2) - p(t_1) \delta q(t_1) = p(t_2) \delta q(t_2)$ $\endgroup$ – Brian Moths Nov 1 '14 at 20:03
  • $\begingroup$ It's because we are considering action as a function of co-ordinates. In this case it's a function of the ending point of the trajectory, which is $q(t_2)=q^{(2)}$. But as the trajectory itself is a function of ending point, then it's $q(x,t)$. And as we consider $x$ to be the ending point, then $q(x,t_2)=q^{(2)}\equiv x$. As for your potential answer: I've said in the OP that I wanted to avoid variation of $S$ (because I understand ordinary calculus much better than that of variations). Now, though, the derivation appears mostly the same in form... $\endgroup$ – Ruslan Nov 2 '14 at 5:39

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