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How does lagrangian mechanics explain loss of momentum conservation in presence of friction?
My try is this:
The lagrange equation would then include a generalized force term $Q_i$: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{q_i}} - \frac{\partial L}{\partial q_i} = Q_{i}$$ $$\implies \frac{d}{dt}p_i = Q_i + \frac{\partial L}{\partial q_i}$$ Now how do I show that $Q_i + \frac{\partial L}{\partial q_i}$ is never $0$?
Since $Q_i$ is generalized force, $$Q_i = \sum_{j}\vec{f_j}\cdot \frac{\partial \vec{r}_j}{\partial q_i}$$

Qmechanic
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  • Related:

    https://physics.stackexchange.com/q/147341/

    https://physics.stackexchange.com/q/96466/

    https://physics.stackexchange.com/q/391870/

    – Arturo don Juan May 13 '21 at 22:14
  • $Q_i = \sum_{j}\vec{f_j}\cdot \frac{\partial \vec{r}_j}{\partial q_i}\ne 0~$ because the friction force is not constraint force. thus even if $\frac{\partial L}{\partial q_i}=0$ you don’t obtain momentum conservation – Eli May 15 '21 at 16:11

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