- #1
Hak
- 709
- 56
- Homework Statement
- While studying the calculation of the Lagrangian for a relativistic free particle, I came across this equation below
$$L((v')^2) = L(v^2) + \frac{\partial L(v^2)}{\partial (v^2)}\big ((v'^2) - v^2) \big)$$, with [tex]L'(v^2) = \frac{\partial L(v^2)}{\partial (v^2)}[/tex],
obtained by using the Taylor expansion for ##L## at the point ##(v')^2##.
- Relevant Equations
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My doubt arises over the definition of [tex]L'(v^2)[/tex]. If we are using ##x= v'^2##, shouldn't the derivative be made with respect to that very term? In essence, shouldn't it be: [tex]L'(v^2) = \frac{\partial L(v^2)}{\partial (v'^2)}[/tex]? In the article I read, [tex]L'(v^2) = \frac{\partial L(v^2)}{\partial (v^2)}[/tex] is assumed. Could you explain to me why the latter definition is right and mine is wrong? Thank you very much.