I am aware that the Lagrangian $L=T-V$ where $T$ is the kinetic energy and $V$ is the potential energy when $L$ depends on, for example, $r, \dot{r}, t$. My question is, does this still hold when the Lagrangian does not explicitly depend on time, i.e. I have $L(r,\dot{r})$ instead?
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Kyle Kanos
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Janitt
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Possible duplicates: https://physics.stackexchange.com/q/50075/2451 and links therein. – Qmechanic Feb 27 '19 at 12:40
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Of course. Indeed most Lagrangians do not depend explicitly on time, v.g. for the harmonic oscillator $$ L(x,\dot{x})=\frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2 $$ doesn't depend explicitly on $t$.
ZeroTheHero
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It's also interesting to note that is the lagrangian does not explicitly depent on time, energy is conserved. – Ballanzor Feb 27 '19 at 13:07
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1@Ballanzor actually the Hamiltonian is conserved, which need not always coincide with the energy. – ZeroTheHero Feb 27 '19 at 13:09
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