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Definition of Lagrange function

  1. Mar 9, 2004 #1
    I'm learning classical mechanics right now, and I have a question about the "initial definition" of the Lagrange function. In my book, it is only introduced as L=T-V, but in many cases this doesn't help a lot, since it's not obvious what T or V is. For example, how would I come to the Lagrangian of a particle in a el.-magn. field without ever having heard of the vector potential, B=rot(A) ?

    This question came to me when I wanted to find the (time-dependant) Lagrangian of a simple one-dim. movement with friction proportional to velocity:
    m*d2(x) = f(x) - c*x

    thanks for help

  2. jcsd
  3. Mar 9, 2004 #2
    sorry, the equation should be
    m*d2(x) = f(x) - c*d(x)

    d( ) and d2( ) are the derivatives, f a force function
  4. Mar 9, 2004 #3


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    "Lagrangian of a simple one-dim. movement with friction proportional..."

    Maybe I am talking stupidly (I am going by memory from some reading done about ten years ago), but isn't it the case that extremal methods (such as those involving a Lagrangian) are not useful for solving the dynamics of non-conservative systems (such as those involving friction)?

    If I am wrong about this, I am sure somebody here will correct me.
  5. Mar 9, 2004 #4
    The Lagragian is defined as that which appears under the action integral. It's usually given by L = Kinetic Energy - Generalized Potential. I don't know of a way to derive the Lagrangian for a particle in an EM field - sorry. I only know the Lagrangian itself. For a relativistic particle it's at


    For non-relativistic particle its

    L = T - U


    U = Phi - qv*A

    where Phi = Coulomb Potential, q = charge, v = velocity and A = vector potential. Plug this into Lagrange's equations and you get the Lorentz force
  6. Mar 9, 2004 #5


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    You can tabulate the work done by non-conservative forces in a Lagrangian formulation by including generalized forces.

    [tex] L = T^* - U [/tex]

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

    (That's just the notation I remember: qi = ith generalized coordinate, Qi = ith generalized force, T* = kinetic coenergy, U = potential energy, etc.)

    Back to the original question, here's a site where the Lagrangian of a particle is found starting with the Lorentz force law. I don't know if you can get around dealing with the magnetic vector potential. Hope that helps.
  7. Mar 10, 2004 #6


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    The electromagnetic potential is necessarily a 4-vector. The only literal answer to your question that I can think of is: "Someone can tell you what the Lagrangian is for a charge in an E&M field."
  8. Mar 11, 2004 #7
    Re: Re: Definition of Lagrange function

    Well, certainly you end up with the same potential, whatever you do. I meant, how would you come to it by only looking at the equation of movement. In my course, the potential U = Phi - qv*A had simply been "claimed", and we checked whether it is of the correct form for the Lagrange function.

    Well, I'm sure you're right. When attacking that little but frustrating problem I mentioned, I thaught it would be an exception to that rule, but now I think it's one of those excercises where you'd have to show that they cannot be solved:smile: .
  9. Mar 11, 2004 #8


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    Re: Re: Re: Definition of Lagrange function

    I've never been completely satisfied with the Lagrangian formulation as a fundamental formulation of mechanics. I think that you have to already have a good understanding of E&M in Newtonian mechanics. You would then know of course that the equation of motion can be written in component form as:

    d2xn/dt2 = (e/m) { En + εnml (dxm/dt) Bl }

    and that, in general:

    En = -∂nφ - ∂tAn


    Bl = εlkj∂kAj

    where the Latin indices represent Cartesian coordinate indices, ∂n is the partial derivative with respect to the coordinate with the given index, and εnml/εlkj is the Levi-Civita symbol.

    Putting the potential expressions into the equation of motion gives:

    d2xn/dt2 = (e/m) { ( -∂nφ - ∂tAn ) + εnml (dxm/dt) ( εlkj∂kAj ) }
    = (e/m) { -∂nφ - ∂tAn + εnmlεlkj (dxm/dt) ∂kAj }
    = (e/m) { -∂nφ - ∂tAn + ( δnkδmj - δnjδmk ) (dxm/dt) ∂kAj }
    = (e/m) { -∂nφ - ∂tAn + ( (dxm/dt) ∂nAm - (dxm/dt) ∂mAn ) }
    = (e/m) { -∂nφ + (dxm/dt) ∂nAm - ( ∂tAn + (dxm/dt) ∂mAn ) }
    = (e/m) { -∂nφ + (dxm/dt) ∂nAm - (dAn/dt) }


    d2xn/dt2 + (e/m) (dAn/dt) = (e/m) { -∂nφ + (dxm/dt) ∂nAm }


    (d/dt){ dxn/dt + (e/m) An } + (e/m) { ∂nφ - ∂n[ (dxm/dt) Am ] + ∂n[ (dxm/dt) ] Am } = 0


    (d/dt){ m dxn/dt + e An } + ∂n{ e φ - e (dxm/dt) Am } = - e ∂n[ (dxm/dt) ] Am

    I can't remember off the top of my head how to argue that the R.H.S. vanishes, so this is the one flaw that I leave to you. Anyway, assuming that this has been argued, we have:

    (d/dt){ m dxn/dt + e An } - ∂n{ -e φ + e (dxm/dt) Am } = 0

    which can be identified with the Euler-Lagrange equation:

    (d/dt){ ∂L/∂vn } - ∂n{ L } = 0

    to give

    ∂L/∂vn = m dxn/dt + e An


    L = -e φ + e (dxm/dt) Am + f(vi,t)

    where f(vi,t) is an arbitrary function of velocity and time. Therefore:

    L = f(vi,t) - e ( φ - (dxm/dt) Am )
    = (1/2)mv2 - e ( φ - v.A )

    Clearly then, e ( φ - v.A ) is the potential energy, and φ - v.A is the electromagnetic potential.
    Last edited: Mar 11, 2004
  10. Mar 12, 2004 #9
    It should be noted that the L in that equation may not obey Hamilton's principle.

    turin wrote
    What is is that you're calling "electromagnetic potential"?

    The 4-potential is defined as a 4-vector whose time component is proportional to the Coulomb potential and whose spatial portion is the vector potential .
  11. Mar 12, 2004 #10
    The Lag. is not always derived. It's like finding a closed formula for the integral of a new function. You guess at a definition and try it to see if it works. Maybe in some cases, if you are unlucky, you do too much trial-and-error work.
  12. Mar 12, 2004 #11


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    Yes, that's exactly what I meant.
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