If neither the potential energy nor kinetic energy depends on time, then Lagrangian is explicitly independent of time I find this statement a little bit odd because velocity is distance over time or may be i missed something,please clarify.
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$\begingroup$ Position always depends on time. Therefore lagrangians always depend on time. Sometimes not explicitly, but always implicitly. $\endgroup$– QuantumBrickJun 24, 2016 at 13:01
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$\begingroup$ independent of time means :"remains constant as time changes". Even if velocity is the time derivative of position it can be fixed and thus independent of time. $\endgroup$– TofiJun 24, 2016 at 13:05
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2 Answers
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It seems relevant to mention the importance of distinguishing between explicit, implicit, and total time-dependence. The Lagrangian $L=L(q,v,t)$ depends implicitly on time via the position $q$ and the velocity $v$. The total time derivative of the Lagrangian $L=L(q,v,t)$ is $$\underbrace{\frac{dL}{dt}}_{\text{total $t$-derivative}}~=~\underbrace{\frac{\partial L}{\partial t}}_{\text{expl. $t$-derivative}} + \dot{q}\frac{\partial L}{\partial q} + \dot{v}\frac{\partial L}{\partial v}.$$ See also e.g. this Phys.SE post and links therein. In particular the Lagrangian can depend on time even if it does not depend explicitly on time.
answered Jun 24, 2016 at 13:44
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The Lagrangian only depends on the potential energy and the kinetic energy. What the statement you quoted means is that if both the potential and kinetic energies are constant w.r.t. time, then so is the Lagrangian. This makes a lot of sense. Usually, we have: $$\mathscr{L}=K(x,t)-P(x,t)$$ Where $K$ and $P$ are the kinetic and potential energies. But if those are independent of time (ie constant over time), then it becomes: $$\mathscr{L}=K(x)-P(x)$$ Which, naturally, you can see would result in a Lagrangian that is not a function of time; it is independent of time.
