Euler-Lagrange equations for a damped mass-spring system.

Question

Is there a way to use Euler-Lagrange equations to derive the following formula?

I can derive the term not due to damping, I guess the damping force should be somehow modeled using a generalized force. Can you please show me how?

My attempt was starting by

$$ F_{d,ij} = k_d \lVert v_{ij} \rVert u_{ij} $$

Where $v_{ij}$ is the relative velocity vector and $u_{ij}$ is the unit vector going from mass $i$ to $j$. But I'm missing how should I go from here to the generalized forces.

Thank you.

Answer 1

1. OP is essentially asking to find a velocity-dependent potential for coupled damped oscillators. This can be reduced to the question of finding a velocity-dependent potential $U({\bf x},\dot{\bf x},t)$ for the 1-particle force $$ {\bf f} ~=~-k_d{\bf x}\frac{{\bf x} \cdot\dot{\bf x}}{|{\bf x}|^2} ~=~-k_d{\bf x}\frac{d\ln |{\bf x}|}{dt}.$$ One may show, using methods e.g. outlined in my Phys.SE answer here that such velocity-dependent potential $U({\bf x},\dot{\bf x},t)$ does not exist. So OP's model cannot be obtained from a stationary action principle (under the assumption that we shouldn't modify the kinetic sector of the action).

2. However, it is still possible to describe OP's model via Lagrange equations $$ \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{\bf x}_i}\right)-\frac{\partial L}{\partial {\bf x}_i}~=~-\frac{\partial {\cal F}}{\partial \dot{\bf x}_i} $$ using the Rayleigh dissipation function$^1$ $$ {\cal F} ~=~ \frac{1}{2}\sum_{i<j} k_d^{ij} \frac{({\bf x}_{ij} \cdot\dot{\bf x}_{ij})^2}{|{\bf x}_{ij}|^2}, \qquad {\bf x}_{ij}~:=~{\bf x}_{i}-{\bf x}_{j}. $$ The Lagrangian itself is $$L~=~T-V, \qquad T~=~\frac{1}{2}\sum_i m_i |\dot{\bf x}_i|^2, \qquad V~=~\frac{1}{2}\sum_{i<j}k^{ij}_s (|{\bf x}_{ij}| - \ell^{ij}_0)^2. $$ See also this related Phys.SE post.

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$^1$ In this answer, we have slightly generalized OP's model to allow the spring and dissipation constants to depend on the particle pair $(i,j)$.

Answer 2

Let us derive the undamped case. The kinetic energy $T$ and the potential energy $V$ are $$ T = \frac{1}{2}m\left({\dot x_i}^2 + {\dot x_j}^2 \right) \qquad\text{and}\qquad V = \frac{1}{2}k_s (|x_j - x_i| - l_0)^2 \, . $$ Therefore, the Euler-Lagrange equations $\frac{\text d}{\text d t}\frac{\partial \mathcal{L}}{\partial \dot{\boldsymbol x}} = \frac{\partial \mathcal{L}}{\partial {\boldsymbol x}}$ deduced from the Lagrangian $\mathcal{L} = T-V$ are \begin{aligned} m \ddot x_i &= \phantom{-}\frac{x_j - x_i}{|x_j - x_i|} k_s (|x_j - x_i| - l_0) = \phantom{-}f|_{k_d=0} \, ,\\ m \ddot x_j &= -\frac{x_j - x_i}{|x_j - x_i|} k_s (|x_j - x_i| - l_0) = -f|_{k_d=0} \, . \end{aligned} The dissipative term with damping constant $k_d\neq 0$ is not included in the present Euler-Lagrange equations, but it can be done upon introducing a dissipation potential (see other answers)