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Velocity dependent potentials in the Lagrangian(Goldstein)

  1. Mar 15, 2014 #1
    Hi, I'm studying classical mechanics via Goldstein's book, but I don't get the chapter about velocity dependent potentials. He writes:

    'Lagrange's equations can be put in the form [itex] \frac{d}{dt}(\frac{\partial{L}}{\partial{\dot{q}}})-\frac{\partial{L}}{\partial{q}}=0[/itex](eq. 1.57) even if there is no potential function [itex]V[/itex] in the usual sense, providing the generalized forces are obtained from a function [itex]U(q, \dot{q})[/itex] by the prescription: [itex]Q=- \frac{\partial{U}}{\partial{q}}+\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )[/itex](1.58). In such case eqs. 1.57(or plus another term [itex] \frac{dF}{dt}[/itex]) still follow from [itex] \frac{d}{dt}(\frac{\partial{T}}{\partial{\dot{q}}})-\frac{\partial{T}}{\partial{q}}=Q[/itex] with the Lagrangian given by [itex]L = T - U [/itex]. Here U may be called a "generalized potential, or "velocity-dependent potential"....."


    (not an exact quotation)

    Then he gives an example of an electromagnetic force potential(which is velocity depedent), which he then calls U and writes it like [itex]L=T-U[/itex] and later he writes: "Note that if not all the forces acting on the system are derivable from a potential, then Lagrange's equations can always be written in the form [itex] \frac{d}{dt}(\frac{\partial{L}}{\partial{\dot{q}}})-\frac{\partial{L}}{\partial{q}}=Q[/itex] where L contains the potential of the conservative force as before, and Q represents the forces not arising from a potential....."

    I don't really understand what the difference between U, V and Q is, and why U should be derivable from 1.58, what does 1.58 even mean? It looks like the normal Langrangian equation but only with the potential U and Q. Should V also be obtained from 1.58 ? But I think V should be obtained when I integrate the force. And when I can do the same with U, then why he writes it as U and not as V. And should a force not derivable from a potential still be obtained from 1.58?? Then in the next page he adds a dissipation function to the Lagrangian. And should this dissipation function also be derivable from 1.58?? I'm not sure but the whole derivation of the Lagrangian seems a bit like magic to me :((

    Greets
     
  2. jcsd
  3. Mar 15, 2014 #2
    The idea is that you can in general divide the forces in two categories: (1) Potential forces, i.e. such that the force can be written as [itex]-\vec{\nabla} V[/itex], where V(q) is a scalar function, and non-potential forces.

    If your system only contains potential forces then [itex]Q_i=-\partial V/\partial q_i[/itex] and [itex]\partial V/\partial \dot q_i=0[/itex] and you can define the Lagrangian, [itex]L=T-V[/itex] such that:
    $$
    \frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}=0.
    $$

    If your system also contains non-potential forces, then the expression for the Lagrange equation reads:
    $$
    \frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}=Q_i,
    $$
    where L contain all the potential forces and [itex]Q_i[/itex] contains the non-potential ones. However if some of the these non-potential forces satisfy Eq. (1.58), let's call them [itex]Q_i^{(1.58)}[/itex], then you can write:
    $$
    \frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}=Q_i^{(1.58)}+Q_i^{(non\;1.58)}=\frac{d}{dt}\frac{\partial U}{\partial \dot q_i}-\frac{\partial U}{\partial q_i}+Q^{(non\;1.58)}_i,
    $$
    which implies:
    $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}-\frac{d}{dt}\frac{\partial U}{\partial \dot q_i}+\frac{\partial U}{\partial q_i}=Q_i^{(non\;1.58)}.
    $$
    If you now define [itex]L'=L-U=T-V-U[/itex] you obtain a new Lagrangian which still satisfies the Lagrange equation (1.53). The difference is that now the Lagrangian contains information on both the potential forces and the non-potetial forces which satisfy Eq. 1.58. All the left forces are included on the right-hand-side.

    Of course you can't obtain the forces [itex]Q_i^{(1.58)}[/itex] by simply integrate [itex]U(q,\dot q)[/itex] since Eq. 1.58 is not just a derivative with respect to space.


    The last part is about dissipation function. This is a very simple topic. The idea is that some (not necessarily all of them) of the forces [itex]Q_i^{(non\;1.58)}[/itex] may be written as:
    $$
    Q_i^{(non\;1.58)}=-\frac{\partial \mathcal{F}}{\partial \dot q_i},
    $$ and in this case the Lagrange equations can be rewritten as Eq. 1.70. Be careful, that, in this case, you are not adding anything to the Lagrangian, you are just rewriting the forces in a different form.
     
  4. Mar 15, 2014 #3

    stevendaryl

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    I don't have Goldstein, but I do remember someone mentioning that you can get damped motion in some cases from a Lagrangian with an explicit time-dependence:

    [itex]L = (\frac{1}{2} m (\frac{dx}{dt})^2 - V(x)) e^{+\lambda t}[/itex]

    Then the lagrangian equations of motion give:

    [itex]m \frac{d^2 x}{dt^2} = F - \lambda \frac{dx}{dt}[/itex]

    where [itex]F = -\dfrac{\partial V}{\partial x}[/itex]. The term [itex]- \lambda \frac{dx}{dt}[/itex] is a viscous drag force that works to slow the particle down.

    That seems like a trick with no physical significance to me. The more physical explanation of dissipative forces would involve transfer of energy from the macroscopic object to the many randomly moving microscopic particles that make up the viscous medium. I know that the careful analysis of this is pretty complicated (Einstein wrote a paper about it the same year as his relativity paper), and the use of time-dependent Lagrangians is just a simplification to make the problem tractable, but I am bothered by the fact that I don't see how the simplification relates to the more fundamental derivation.
     
  5. Mar 15, 2014 #4
    I honestly don't know if, doing this simplification, you are relating to some kind of fundamental property. I think that, if you mathematically manage to include dissipative forces into the Lagrangian, you are greatly simplifying the problem basically because you can use all the mathematics related to Lagrangian systems. I don't see deeper reasons to that.
     
  6. Mar 15, 2014 #5
    Don't you get a mass factor on the viscous term?
     
  7. Mar 15, 2014 #6

    stevendaryl

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    You're right, I was being sloppy. So maybe the trick only works for particles of the same mass.
     
  8. Mar 15, 2014 #7
    I don't think that trick works at all, since the equation of motion we get describes different physics.

    Edit: Wait, if we have λ/m instead in the lagrangian it works.
     
    Last edited: Mar 15, 2014
  9. Mar 15, 2014 #8
    Hi, thanks for the long answer !

    I just summerize it to see if I understood it now(and I drop the indices).

    There are potential forces which only depend on position, and there are potential forces which depend on velocity(which makes sense).
    (In the 3D case [itex]-\frac{\partial{V(q)}}{\partial{q}}[/itex] would be [itex]-[/itex]∇[itex]V(q)[/itex] right??)

    Position dependent potential force:

    [itex]Q = - \frac{\partial V(q)}{\partial q} + \frac{d}{dt}(\frac{\partial V(q)}{\partial \dot{q}})[/itex]
    (the far right term vanishes of course)

    So the lagrange equation will look like
    [itex] \frac{d}{dt}(\frac{\partial T}{\partial \dot{q}}) - \frac{\partial T}{\partial q} = - \frac{\partial V(q)}{\partial q} [/itex]
    And then you can define the lagrangian like [itex] L = T-V[/itex]
    (and you can add an V to the far left term because it vanishes anyway)

    Velocity dependent potential force(which is called [itex]U(q, \dot{q})[/itex]):

    [itex]Q = - \frac{\partial U(q,\dot{q})}{\partial q} + \frac{d}{dt}(\frac{\partial U(q,\dot{q})}{\partial \dot{q}})[/itex]
    (no term vanishes here)

    And the langrange equation will look like:

    [itex] \frac{d}{dt}(\frac{\partial T}{\partial \dot{q}}) - \frac{\partial T}{\partial q} = - \frac{\partial U(q,\dot{q})}{\partial q} + \frac{d}{dt}(\frac{\partial U(q,\dot{q})}{\partial \dot{q}}) [/itex]

    And then you can define the lagrangian like [itex] L = T - U[/itex]

    So when I have a mix of velocity dependent forces and position dependent forces I can write the lagrange equation like :

    [itex] \frac{d}{dt}(\frac{\partial T}{\partial \dot{q}}) - \frac{\partial T}{\partial q} = - \frac{\partial V(q)}{\partial q} - \frac{\partial U(q,\dot{q})}{\partial q} + \frac{d}{dt}(\frac{\partial U(q,\dot{q})}{\partial \dot{q}}).......[/itex]
    In which case I can write the lagrangian like [itex]L = T-V-U [/itex] And when I have a dissipation function [itex]- \frac{\partial F}{\partial \dot{q}}[/itex] it just get's into the Q like all the other potentials, but I can't make a simple looking lagrangian out of it of course.

    I hope that's right :)

    /edit: Extra question:

    Then where does this definition come into play:

    [itex]Q = F * \frac{\partial r}{\partial q}[/itex]

    He uses this in one of the following problems to derive a general force in polar coordinates. I know this definition is derived from some other stuff, but it's just plugging equations into one another and rewriting everything. Or is it just so general that it could be anything? Position potential, velocity potential, dissipation function ect.?



    Greets !
     
    Last edited: Mar 15, 2014
  10. Mar 16, 2014 #9
    You got it! The equation for Q that you wrote is just its definition, given in Eq. 1.48.
     
  11. Mar 16, 2014 #10
    Yeah, ok thank you :)

    Greets
     
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