Sunday, 30 November 2014

lagrangian formalism - How to show that $partial S/partial q=p$ without variation of $S$?


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?



Answer



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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