Non-conservative forces in Lagrangian mechanics

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 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.
• 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 $T$? – usumdelphini Jan 7 at 8:42
• @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 $m\rightarrow 0$). – jacob1729 Jan 7 at 10:18