Sunday 1 December 2019

variational principle - Independent Variables of a Lagrangian




Let us consider a particle in one spatial dimension $x$ and one temporal dimension $t$. Its Lagrangian $L$ is given by


\begin{eqnarray*} L &=& T- V \\ &=& \frac{1}{2} m\dot{x}^2 - V(x) \\ &=& L(x, \dot{x}) \end{eqnarray*}


But if I write $ L = \frac{1}{2} m [\frac{d}{dt}({x})]^2 - V(x)$, then it seems to me that $L$ depends only on $x$.


So, my question is why it is said that $L$ is a function of $x$ and its derivative $\dot{x}$ ?




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