In lagrangian mechanics, where $L=T-U$ and the lagrangian formulation is $ \frac{d}{dt}\big( \frac{\partial L}{\partial \dot{q_i}}\big)-\frac{\partial L}{\partial {q_i}}=F_i$, where $F$ is the non-conservative force.

My question is if I want to find out the above equation for a given problem then the $q_i$ should be written for every term in which the system is expressed. Like if I want to write the equation for a pendulum then the $q_i$ will be the angle displacement.

So for example in a double pendulum there will be two angles $\phi ,\theta $ for the respective rods than the equation in lagrangian formalism will be $$\frac{dL}{dt}(\frac{\partial L}{\partial \dot{\phi}}+\frac{\partial L}{ \partial \dot{\theta}})+\frac{\partial L}{\partial \theta}+\frac{\partial L}{\partial \phi}=0$$ is this correct?

  • $\begingroup$ Even though you are right, The E-L equations for both the coordinates are individually $0$ too, you are just doing $0+0$, also $F_i$ in your equations is a non conservative force $\endgroup$ – Oswald Jul 3 '16 at 8:20
  • $\begingroup$ Every post on this site is a question so there's no need to remind the readers that the question is a question in the title. Please see our FAQ on writing good titles. $\endgroup$ – DanielSank Jul 3 '16 at 15:32

No, you get a separate Euler-Lagrange equation for each individual degree of freedom, i.e. a system of simultaneous equations. So in your example,

\begin{align} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right)-\frac{\partial L}{\partial \theta} &= 0, \,\mathrm{and} \\ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right)-\frac{\partial L}{\partial \phi} &= 0 \end{align}

  • $\begingroup$ but why are these $q_i $is in the equation @lemon $\endgroup$ – Boris Jul 3 '16 at 8:25
  • $\begingroup$ @Boris I don't understand what you're asking... $\endgroup$ – lemon Jul 3 '16 at 8:26
  • $\begingroup$ $\frac{d}{dt}\big( \frac{\partial L}{\partial \dot{q_i}}\big)-\frac{\partial L}{\partial {q_i}}=F_i$ for what are these 'i' in this equation @lemon $\endgroup$ – Boris Jul 3 '16 at 8:28
  • $\begingroup$ $q_1=\theta,\,q_2=\phi$. So set $i=1$ and you get the first equation I give above, and set $i=2$ and you get the second. They are not to be interpreted as Einstein summation... $\endgroup$ – lemon Jul 3 '16 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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