# Friction in Lagrangian formulation

We know the Lagrange equations are: $$\frac{\partial \mathcal{L}}{\partial q_i}-\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q_i}}\right)=0.$$ Then, when we add friction in there, we rewrite it in terms of the Rayleigh dissipation function as $$\frac{\partial \mathcal{L}}{\partial q_i}-\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q_i}}\right)=\frac{\partial \mathcal{F}}{\partial \dot{q_i}}.$$ However, this is assuming that you can write the friction force as $F_f=-k\dot{q_i}$. How would you do it for a case where the friction force is not proportional to velocity? For example, in the case of a block sliding on a flat surface the friction would be $F_f=-m\ddot{q}$, which is proportional to the acceleration not the velocity.

• Most people will tel you that you can't find Lagrangians for general friction etc. but there exist papers that claim to do so in various ways, see e.g. arxiv.org/abs/1210.2745 or journals.tubitak.gov.tr/physics/issues/fiz-04-28-4/… – Luboš Motl Nov 30 '15 at 17:49
• The problem with friction is that it changes the degrees of freedom of the system. When two bodies stick they share a DOF but have an unknown sticking force. When they slide their DOF are independent but the sliding force is known. – ja72 Nov 30 '15 at 19:27
• Could you please explain why in the case of a block sliding on a flat surface, the friction is given by $F_f = - m \ddot{x}$? @Josh_P – AlQuemist Nov 30 '15 at 19:35
• @Khwārazmi Im considering a one dimensional case. Sum the forces in the x-dir and set it equal to $m\ddot{x}$ assuming the mass has an initial velocity... – Josh Pilipovsky Nov 30 '15 at 22:27
• It seems the question (v5) is confusing the expression for the friction force $F_f$ & Newton's 2nd law $F=ma$. – Qmechanic Dec 1 '15 at 8:58