2
$\begingroup$

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 $\frac{\partial{L}}{\partial \dot q_i}$?

$\endgroup$
0

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 $-\frac {\partial V}{\partial q_i}$ $\endgroup$ Dec 7, 2018 at 16:23
  • $\begingroup$ Yes.it is.Because $$L=f(\dot q)-V(q)$$ $\endgroup$ Dec 7, 2018 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.