Do I have the correct intuition behind the general dissipation function for Lagrangian Mechanics?

Question

I am reading these lecture notes, section 2.7.1.

The general dissipation function is given via a dissipative force:

$$ \vec{F}^{D} = -\mu(v)\frac{\vec{v}}{v} \tag{2.269} $$

The Professors references A. I. Lurie, Analytical Mechanics (Springer, 2002) and E. Minguzzi, European Journal of Physics 36, 035014 (2015) for this general dissipation function.

Lurie does not derive this relation and Minguizzi does not reference it.

Intuitively, this is indeed a velocity dependent function. If the environment is static then:

$$ \vec{v} = \dot{\vec{r}} \tag{2.270} $$

Then the equation is: $$ \vec{F}^{D} = -\mu(v)\frac{\dot{\vec{r}}}{v} $$

I can understand that the dissipation coefficient is assumed to be dependent on velocity, but I do not understand why velocity is divided by its magnitude other than to normalize it in the direction it is acting.

Is it correct to read Eq. 2.269 as "The dissipation force is equal to the negative speed dependent friction coefficient times the unit vector in the direction of movement."

Answer 1

The force $\mathbf{F}^D$, as modeled by the authors is in the opposite direction of motion, a kind of aerodynamic drag in a fluid at rest.

You can treat general forces, including non conservative forces, in a more general framework without the need of a dissipation function $D$, that can be useful only for some expression of these forces, see as an example https://physics.stackexchange.com/q/733986.

You can include any force $\mathbf{F}$ in Lagrange equation evaluating its virtual work, namely

$\delta W = Q_q \delta q = \mathbf{F} \cdot \dfrac{\partial \mathbf{r}}{\partial q } \, \delta q$,

so that Lagrange equations read

$\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot{q}}\right) - \dfrac{\partial L}{\partial q} = Q_q$.

An external forces dependent of the the velocity, like aerodynamic forces, can be written as the sum of two contributions, one in the (opposite direction) of the velocity of the point, the other orthogonal to it, namely

$\mathbf{F} = F_v \mathbf{\hat{v}} + F_{p} \mathbf{\hat{p}}$,

and the generalized force $Q_q$ reads

$Q_q = \dfrac{\partial \mathbf{r}}{\partial q} \cdot \left( F_v \mathbf{\hat{v}} + F_{p} \mathbf{\hat{p}} \right) = \dfrac{\partial \mathbf{v}}{\partial \dot{q}} \cdot F_v \mathbf{\hat{v}} + \dfrac{\partial \mathbf{r}}{\partial q} \cdot F_{p} \mathbf{\hat{p}} $.

Following this approach you can handle every external function with the Lagrangian formalism.

One particular case, I'd say the only one being a bit rude, where dissipation function is useful in practice, is the case where $\mathbf{F}$ is proportional to the velocity $\mathbf{v}$, $\mathbf{F} = - c \mathbf{v}$, so that we can write

$Q_q = \dfrac{\partial \mathbf{v}}{\partial \dot{q}} \cdot \mathbf{F}_v = - c \dfrac{\partial \mathbf{v}}{\partial \dot{q}} \cdot \mathbf{v} = - \dfrac{\partial}{\partial \dot{q}} \left( \dfrac{1}{2} c \mathbf{v} \cdot \mathbf{v} \right) = - \dfrac{\partial D}{\partial \dot{q}}$.