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The most general Lagrange equation for classical mechanics systems is of the form:

$$\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \right) - \frac{\partial \mathcal{L}}{\partial q_i} + \frac{\partial \mathcal{R}}{\partial \dot{q}_i} = Q_i $$

where $\mathcal R$ is the Rayleigh dissipation function and $Q_i$ are Non-Constraint Forces.

As I understand the Rayleigh dissipation function $\mathcal R$ is used when we have some dissipation, either via friction or damping or something similar.

But I don't understand the usage of $Q_i$.

I understand that we can use the standard version of the EL equations whenever

$$\mathcal R=0 ,Q_i=-\frac{\partial V}{\partial q_i}$$

What are some examples of systems where we have to include the non-constrained forces $Q_i$ and that it is not already covered by the Rayleigh dissipation function $\mathcal R$?

Qmechanic
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User198
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1 Answers1

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  1. The Rayleigh dissipation function ${\cal R}(v)$ is mainly thought of being a function of velocities, while the generalized force $Q_i(q,\dot{q},t)$ is completely arbitrary.

  2. More generally, a generalized force $Q_i(q,\dot{q},t)$ does not necessarily have a generalized/velocity-dependent potential $U(q,\dot{q},t)$ or a Rayleigh dissipation function$^1$ ${\cal R}(q,\dot{q},t)$.

  3. Example: $Q_i=M_{ij}\dot{q}^j$, where $M_{ij}$ is a constant matrix.

    • Case $M_{ij}$ symmetric matrix: $\exists {\cal R}=-\frac{1}{2}\dot{q}^iM_{ij}\dot{q}^j.$

    • Case $M_{ij}$ antisymmetric matrix: $\exists U=-\frac{1}{2}q^iM_{ij}\dot{q}^j.$

    • Case $M_{ij}$ neither symmetric nor antisymmetric matrix: ${\cal R}$ and $U$ do not exist$^2$.

  4. See also e.g. this & this related Phys.SE posts.

$^1$ If we define an exterior derivative $\mathrm{d}=\mathrm{d}\dot{q}^i\frac{\partial}{\partial\dot{q}^i}$ using the generalized velocities, and define a 1-form $\omega:=Q_i\mathrm{d}\dot{q}^i$, it will in general not be an exact 1-form $\omega=-\mathrm{d}{\cal R}$.

$^2$ The 1-form $\omega:=Q_i\mathrm{d}\dot{q}^i=M_{ij}\dot{q}^j\mathrm{d}\dot{q}^i$ is closed $0=\mathrm{d}\omega=M_{ij}\mathrm{d}\dot{q}^j\wedge\mathrm{d}\dot{q}^i$ iff $M_{ij}$ is a symmetric matrix.

Qmechanic
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