If we have a system and we know all the degrees of freedom, we can find the Lagrangian of the dynamical system. What happens if we apply some non-conservative forces in the system? I mean how to deal with the Lagrangian, if we get any external non-conservative forces perturbs the system?


We have a mass $m$ that is attached with a massless spring.

We could write the Lagrangian as $L= \frac{1}{2} m \dot x ^2 + mgx - \frac{1}{2} k x^2$.

What happens to this equation if we consider any non-conservative forces?


More generally, Lagrange equations$^1$ read

$$ \frac{d}{dt}\frac{\partial (T-U)}{\partial \dot{q}^j}-\frac{\partial (T-U)}{\partial q^j}~=~Q_j-\frac{\partial{\cal F}}{\partial\dot{q}^j}+\sum_{\ell=1}^m\lambda^{\ell} a_{\ell j}, \qquad j~\in \{1,\ldots, n\}, \tag{L}$$


  • $q^1,\ldots ,q^n,$ are $n$ generalized position coordinates;

  • $T$ is the kinetic energy;

  • $U$ is a generalized potential;

  • ${\cal F}$ is the Rayleigh dissipation function for friction forces;

  • $Q_1,\ldots ,Q_n,$ are the remaining parts of the generalized forces, which are not described by the generalized potential $U$ or the Rayleigh dissipation function ${\cal F}$;

  • $\lambda^1,\ldots ,\lambda^m$, are $m$ Lagrange multipliers for $m$ semi-holonomic constraints $$ \sum_{j=1}^n a_{\ell j}(q,t)\dot{q}^j+a_{\ell t}(q,t)~=~0, \qquad \ell~\in \{1,\ldots, m\}. \tag{SHC}$$ One may think of the last term on the right-hand side of eq. (L) as the generalized constraint forces for the semi-holonomic constraints (SHC). All other constraints are assumed to be holonomic.

For a discussion of conservative & non-conservative forces, see also e.g. this Phys.SE post.


  1. H. Goldstein, Classical Mechanics; Chapter 1 & 2.


$^1$ We distinguish between Lagrange equations (L) and Euler-Lagrange equations $$ \frac{d}{dt}\frac{\partial L}{\partial \dot{q}^j}-\frac{\partial L}{\partial q^j}~=~0, \qquad j~\in \{1,\ldots, n\}.\tag{EL} $$ In contrast to the Lagrange equations (L), the EL equations are by definition always assumed to be derived from a stationary action principle. We should stress that it is not possible to apply the stationary action principle to derive the Lagrange equations (L) unless all generalized forces have generalized potentials $U$. See also e.g. this and this Phys.SE posts.

| cite | improve this answer | |
  • $\begingroup$ Good answer. Also googling Lagrange multiplier should turn up related info. $\endgroup$ – Lewis Miller Sep 30 '16 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy