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Nonconservative lagrangian systems

  1. Jul 9, 2007 #1
    Hi,

    I'm in the process of self-teaching myself Lagrangian and Hamiltonian dynamics. In my readings, I've found that some velocity dependent forces (e.g. the Lorentz force) can be derived from a classical Lagrangian whereas others such as friction cannot.

    Do there exist necessary and sufficient conditions to decide when velocity dependent forces are obtainable from a Lagrangian, in the same way that velocity independent forces can be obtained from a lagrangian iff they are the gradient of a potential function?

    James
     
  2. jcsd
  3. Jul 9, 2007 #2

    pervect

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    Staff Emeritus
    Science Advisor

    If all the forces on a system can be derived from a potential, you can use Lagrange's equation.

    If you have forces that can't be derived from a potential, one has the option of re-writing Lagrange's equation (Goldstein, Classical mechanics, pg 23-24)

    [tex]
    \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_j}} \right) - \frac{\partial L}{\partial q_j} = Q_j
    [/tex]

    where [itex]Q_j[/itex] is a generalized force that can include forces not derivable from a potential. In many cases, [itex]Q_j[/itex] can be specified via a dissipation function [itex]\mathcal{F}[/itex] so that [tex]Q_j = \frac{\partial \mathcal{F}}{\dot{q_j}}[/tex]

    One then needs to specify L and [itex]\mathcal{F}[/itex] to get the equations of motion.
     
  4. Jul 9, 2007 #3
    Hi pervect,

    Thanks for the reply.

    I think we need to distinguish between deriving forces from the Lagrangian and adding them directly to the RHS of the E-L equations.

    For the Lorentz force, we start with the Lagrangian [itex] L = (1/2)m \dot{\mathbf{r}} \cdot \dot{\mathbf{r}} + e V(\mathbf{r},t) + e \dot{\mathbf{r}}\cdot \mathbf{A}(\mathbf{r},t)[/itex] and then apply the standard, unmodified E-L equation [itex]\frac{d}{dt}\frac{\partial L}{\partial \dot{\mathbf{r}}} - \frac{\partial L}{\partial \mathbf{r}}[/itex] to obtain the velocity-dependent Lorentz force [itex]m\ddot{\mathbf{r}} = e[\mathbf{E} + \dot{\mathbf{r}}\times \mathbf{B}][/itex]. So we can say that these equations of motion are obtainable directly from a Lagrangian.

    If we consider linear frictional forces, on the other hand, we cannot obtain these directly from a Lagrangian, we must add them in later to the RHS of the E-L equations as you describe.

    My question: what determines whether a velocity-dependent force is directly obtainable from a Lagrangian?

    Thanks

    James
     
  5. Jul 11, 2007 #4

    StatusX

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    The second to last line in the derivation of Lagrange's equations is:

    [tex] \frac{d}{dt}\left(\frac{\partial T}{\partial \dot q_i} \right) - \frac{\partial T}{\partial q_i} = Q_i [/tex]

    where Q_i is the generalized force. Then if there is a function V(q_1,...,q_n,t) with:

    [tex] \frac{\partial V}{\partial q_i} = - Q_i [/tex]

    Then you get lagrange's equations (note this holds in one coordinate system iff it holds in all of them). But more generally, as long as you can find a function [itex]V(q_1,...,q_n,\dot q_1,...,\dot q_n,t)[/itex] with:

    [tex]Q_i = \frac{d}{dt}\left(\frac{\partial V}{\partial \dot q_i} \right) - \frac{\partial V}{\partial q_i} [/tex]

    Then you can just take it as your V when you define L=T-V and you get the correct equations. It can be shown that the lorentz force is of this form with:

    [tex] V = q \phi - q \vec v \cdot \vec A [/tex]
     
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