# Euler-Lagrange equations and friction forces

We can derive Lagrange equations supposing that the virtual work of a system is zero.

$$\delta W=\sum_i (\mathbf{F}_i-\dot {\mathbf{p}_i})\delta \mathbf{r}_i=\sum_i (\mathbf{F}^{(a)}_i+\mathbf{f}_i-\dot {\mathbf{p}_i})\delta \mathbf{r}_i=0$$

Where $\mathbf{f}_i$ are the constrainded forces and are supposed to do no work, which it's true in most cases. Quoting Goldstein:

[The principle of virtual work] is no longer true if sliding friction forces are present [in the tally of constraint forces], ...

So I understand that we should exclude friction forces of our treatmeant. After some manipulations we arrive to:

$$\frac{d}{dt}\frac {\partial T}{\partial \dot q_i}-\frac{\partial T}{\partial q_i}=Q_i$$

Further in the book, the Rayleigh dissipation function is introduced to include friction forces. So given that $Q_i=-\frac {\partial \mathcal{F}}{\partial \dot q_i}$ and $L=T-U$, we get:

$$\frac{d}{dt}\frac {\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}+\frac {\partial \mathcal{F}}{\partial \dot q_i}=0$$

Question: Isn't this an inconsistency of our proof, how do we know the equation holds? Or is it just an educated guess which turns out to be true?

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– jinawee Feb 2 '14 at 20:09
The condition $\sum _i F _i \delta r _i=0$ expresses equilibrium and does not apply to the dynamic case. It should be $\sum _i (F_i - m a_i) \delta r _i$, right? – pppqqq Feb 2 '14 at 20:14

The main point is that Goldstein is not saying we must exclude friction forces in our treatment, but we must place them in the tally of applied forces (that we keep track of in D'Alembert's principle) and not in the other bin of the remaining forces, see this and this Phys.SE posts.

Of course, there does not exist a generalized potential $U$ for the friction forces ${\bf F}=-k {\bf v}$, only the Rayleigh dissipation function, see this Phys.SE post and this mathoverflow post.

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It seems to me there is a conflict between your answer to your linked question (#20929) and Professor Moretti's "frictional" Lagrangian here, which can be written as kinetic energy minus a time- and velocity-dependent potential. I suspect the explicit time dependence is at the root of the matter. Thoughts? Thx. – Art Brown Feb 6 '14 at 6:52
@Art Brown: 1. Firstly, as V. Moretti already mentions himself in his answer, his Lagrangian does not originate from D'Alembert's principle. It is therefore strictly speaking not an answer to OP's question. D'Alembert's principle contains a kinematic and a dynamical term, which are additively written, while his Lagrangian mixes kinematic and dynamical quantities. 2. Secondly, his Lagrangian does not work if there are other velocity-dependent forces present, such as, e.g. the Lorentz force. – Qmechanic Feb 6 '14 at 10:30
3. Thirdly, his Lagrangian does not contradict the fact that the friction force ${\bf F}=-k {\bf v}$ does not have a generalized potential $U$ such that ${\bf F}=\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}$, cf. the post #20929. – Qmechanic Feb 6 '14 at 11:05
But, but, but, but... Ah, I thought item 3) was possible, but I see now why it doesn't work. Sorry for the bother... – Art Brown Feb 6 '14 at 17:51

Actually, at least for a single point subjected to a friction force $F= -\gamma v$ and other forces associated with a potential $U(t,x)$ there exists a Lagrangian: $${\cal L}=e^{t\gamma/m}\left(\frac{m}{2}\dot{x}^2 -U(t,x)\right)\:.$$ The point is that this Lagrangian is not of the form $T-U$, nevertheless it gives rise to the correct equation of motion, the same obtained by using the Rayleigh dissipation function you mentioned.

This Lagrangian however cannot be produced by direct application of the principle of virtual works you mention.

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But wouldn't my last equation give the same equations of motion? – jinawee Feb 2 '14 at 21:00
@jinawee Sure, just try! – Valter Moretti Feb 2 '14 at 21:16
Interesting. Do you know of some useful application of this or similar Lagrangian for description of dissipative motions? – Ján Lalinský Feb 2 '14 at 23:01
Unfortunately not! I use similar things just as examples in my lectures on analytic mechanics, but my research field is quantum relativistic theory, so I never entered these issues in depth. – Valter Moretti Feb 3 '14 at 15:17
v clever! Could you put this Lagrangian into the standard $T-U$ form by adding and subtracting $m \dot{x}^2 /2$ (so $U=L+T$ is a velocity- and time-dependent mess)? – Art Brown Feb 5 '14 at 5:34