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In the Lagrangian formalism with a dissipative frictional force $F$, we can write

$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{q}_{k}}-\frac{\partial\mathcal{L}}{\partial q_{k}}=Q^{(nc)}_{k}$$

where I have indicated the generalised force

$$Q^{(nc)}_{k}(\mathbf{q} )=\frac{\partial r_{j}(\mathbf{q})}{\partial q_{k}}\ F_j(\mathbf{\dot{r}})$$

and '$nc$' stands for non-conservative.

Let us assume that the system is driven by some conservative forces $\bf Q^{(c)}$ such that $$Q_k^{(c)}=-\frac{\partial U}{\partial q_k}$$ where $U$ is the internal energy.

In the paper below, the system has no inertial forces ($Re=0$) and once they have determined $\bf q$, $\partial r_{j}/\partial q_j(\mathbf{q})$ and $F_j(\mathbf{r})$ they go straight onto solving the following force balance $$Q_k^{(c)}=Q_k^{(nc)}.$$ Why?

Reference

Polotzek, Katja, and Benjamin M Friedrich. “A Three-Sphere Swimmer for Flagellar Synchronization.” New Journal of Physics 15, no. 4 (April 10, 2013): 045005. https://doi.org/10.1088/1367-2630/15/4/045005.

Fragment of interest

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The paper's argument is that low Reynolds number means that |inertial forces| $\ll$ |viscous forces|, i.e. that the forces $\sum_i {\bf F}_i=m{\bf a}\simeq{\bf 0}$ approximately balance. Equivalently, in Lagrange equations, this amounts to neglecting the kinetic term $T$, so that the generalized forces $\sum_i {\bf Q}_i\simeq{\bf 0} $ approximately balance.

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  • $\begingroup$ That's clear, but how does that affect the Euler Lagrange equation? Why is that equivalent to have the right hand side of EL equal to zero? $\endgroup$ – usumdelphini Jan 6 at 6:20
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jan 6 at 15:44
  • $\begingroup$ Thanks, this is what I suspected. Could you please expand a bit on why that corresponds to neglecting $T$? At what stage in the "construction" of the Lagrangian would you impose that inertia does not play a role and therefore neglect $T$? $\endgroup$ – usumdelphini Jan 7 at 8:42
  • $\begingroup$ @usumdelphini not an expert by any means, but kinetic terms are usually proportional to an inertia of some kind, so setting that to zero would get rid of it (eg $(1/2) mv^2$ becomes zero if $m\rightarrow 0 $). $\endgroup$ – jacob1729 Jan 7 at 10:18
  • $\begingroup$ Yes, if one wanted to do it formally it may be enough to work in the limit of vanishing inertial mass. That makes sense. Thanks! $\endgroup$ – usumdelphini Jan 7 at 10:34

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