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The lagrangian is defined as

$$L = T - V$$

where $T$ is kinetic energy and $V$ potential energy.

Then the euler-lagrange-equation is

$$ \frac{d}{dt} \frac{\partial{L}}{\partial \dot q_i} = \frac{\partial{L}}{\partial q_i}. $$

Now what is the physical meaning of $\frac{\partial{L}}{\partial \dot q_i}$?

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Actually it means the generalized momentum of the system. Because of $$V=V ( q )$$ so $$L ( q,\dot q )=\frac { 1} {2 } m \sum {{\dot q} ^2}- {V (q)}$$ then if we make a partial differentiation with respect to particular $\dot q$,that is , $\dot {q_i}$ : $$\frac {\partial L}{\partial \dot q_i}=m {\dot q_i}$$ So the Euler-Lagrange equation says us that $$\frac {d (m \dot q_i)}{dt}=-\frac {\partial V(q)}{\partial q_i}$$ so $$m\ddot{ {q_i}}=-\frac { \partial V } { \partial q_i}$$ as desired.

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  • $\begingroup$ L is called the Langrangian,not $\frac {d }{dt} \frac {\partial L}{\partial {\dot q_i}}$.it has a physical meaning that it is the generalized forced acted on the system. And if you make a partial differentiation of the R.H.S of the EL equ.,you will find only $-\frac {\partial V}{\partial q_i}$ $\endgroup$ – Raihan Amin Dec 7 '18 at 16:23
  • $\begingroup$ Yes.it is.Because $$L=f(\dot q)-V(q)$$ $\endgroup$ – Raihan Amin Dec 7 '18 at 16:48

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