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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?

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  • $\begingroup$ 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
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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}

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  • $\begingroup$ 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

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