# Question in Lagrangian formalism

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?

• 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 – Oswald Jul 3 '16 at 8:20
• 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. – DanielSank Jul 3 '16 at 15:32

• but why are these $q_i$is in the equation @lemon – Boris Jul 3 '16 at 8:25
• $\frac{d}{dt}\big( \frac{\partial L}{\partial \dot{q_i}}\big)-\frac{\partial L}{\partial {q_i}}=F_i$ for what are these 'i' in this equation @lemon – Boris Jul 3 '16 at 8:28
• $q_1=\theta,\,q_2=\phi$. 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... – lemon Jul 3 '16 at 8:31