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I'm reading volume n. 1 of "Course of Theoretical Physics" by Landau,

Scope: Lagrange function of a material point in free motion, in an inertial system.

What: two inertial systems, K ad K', the second moving with velocity $ v' = v + \epsilon $. We have:

$ L' = L(v'^2) = L (v^2 + 2v\epsilon+\epsilon^2) = L(v^2)+\frac{\partial L}{\partial v^2} (2v\epsilon + \epsilon^2) + O(v\epsilon) = L(v^2)+\frac{\partial L}{\partial v^2}2v\epsilon + O(v\epsilon) $.

First: In the book the error estimation is not specified. Is my addition correct?

Second: The book proceedes by saying that $ \frac{\partial L}{\partial v^2}2v\epsilon $ is a total derivative with respect to time only if there's a linear dependence to the velocity $v$, therefore $\frac{\partial L}{\partial v^2}$ is independent to the velocity $v$. And this is equivalent to saying that the Lagrange function is proprortional to $v$, as following $L = av^2$, where $a$ is a constant. Why?

Can't get it. Any help?

dattiluca
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  • What is $L$, why was it decided that it is a function of $v^2$? Otherwise, this is just the application of the linear Taylor expansion $L(u+\delta u)=L(u)+\partial_uL(u)\delta u+o(\delta u)$ where $u=v^2$, $\delta u=2\epsilon v+o(\epsilon)$, with the usual abuse of notation replacing $u$ with $v^2$ everywhere. Are you sure that a big $O$ is used? As inventor of the Landau notation the author was surely more careful to use small $o$ where appropriate. – Lutz Lehmann Jul 19 '23 at 11:16
  • L is the lagrangian $L(q,\dot{q},t)$. At the beginning of the chapter, it says that for a material point in free motion in an inertial system, the lagrangian is only function of the modulo of the velocity vector, to the power of 2. $\epsilon$ is set to be near 0. – dattiluca Jul 19 '23 at 11:38
  • By the way,he's not the Laundau you are referring to. The author is Lev, not Edmund Landau. – dattiluca Jul 19 '23 at 11:41
  • Then a better notation would be $L(q,\dot q,t)=K(|\dot q|^2)$, which is justified in that the laws of the free particle are time, translation and rotation invariant. // Mixing coordinate and function symbols can be done consistently, more often it leads to confusion in chain rules, higher order derivatives etc. – Lutz Lehmann Jul 19 '23 at 12:26

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