I'm trying to get some understanding in treating action S as a function of coordinates. Landau and Lifshitz consider δS, getting δS=pδq, thus concluding that
∂S∂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 ˙q=˙q(x,t), so S(x) will look like:
S(x)=∫t2t1L(q(x,t),˙q(x,t),t)dt.
Now I take the partial derivative with respect to x to get p:
∂S∂x=∫t2t1(∂L∂q∂q∂x+∂L∂˙q∂˙q∂x)dt.
And now I'm stuck. I see the momentum ∂L∂˙q=p and force ∂L∂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 ∂S/∂x=p(t2), 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 ∂S/∂x in the OP,
A=∫t2t1∂L∂˙q∂˙q∂xdt.
Integrating by parts, namely using u=∂L∂˙q, dv=∂˙q∂xdt, we get, with du=ddt∂L∂˙qdt=∂L∂qdt and v=∂∂x∫˙qdt=∂q∂x,
A=p∂q∂x|t2t1−∫t2t1∂q∂x∂L∂qdt.
Substituting this into the integral in the OP, we get the integrals cancel, thus
∂S∂x=p∂q∂x|t2t1=p(t2)∂q(x,t2)∂x−p(t1)∂q(x,t1)∂x.
But by definition of x, q(x,t2)≡x, and because first point is fixed, we have q(x,t1)=q(1)=const(x), thus the result is what was to be proved:
∂S∂x=p(t2).
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