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Classical mechanics derived Lagrange's Equations

  1. Dec 9, 2004 #1


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    At the beginning of this semester, when my prof in classical mechanics derived Lagrange's Equations, he introduced the concept of "generalised coordinates" to develop the ideas. These coordinates he called

    [tex] {q}_j \ \ \text{and} \ \ \dot{q}_j [/tex]

    j = 1, 2, ... , 3N

    N is the number of particles in the system. He wrote down the Lagrangian as

    [tex] L(q, \dot{q}, t) [/tex]

    But I was always confused by this. Those are not independent variables, I thought. Surely the genearlised position and velocity coordinates are both functions of time! He sort of glossed over this in class. The only thing I got from his explanation was that we consider them to be arbitrary, independent parameters until will solve the equations of motion and arrive at a solution for each, seeing that they are related. But to me this was more a description of what we were doing than a justifcation for it. Can anybody explain what's going on, especially in light of these observations (which just represent my understanding of what's going on...they may have errors)?

    1. [itex] q [/itex] is always related to [itex] \dot{q} [/itex] by the diff. eq. [itex] \dot{q} = dq/dt [/itex]. So they are never independent. In fact, for all t for which [itex] q [/itex] and [itex] \dot{q} [/itex] are both defined, there is one value of the latter for every value of the former. So would it be correct to say that [itex] \dot{q} [/itex] is a function of q? Is this true in general for a function and its time derivative?

    2. I understand how it might be "safest" or most "rigorous" and "general" NOT to make any assumptions about the nature of the generalised coordinates (just to pretend they are arbitrary, independent parameters that describe the system), but is all this care really necessary in light of 1? We know that q and q' are functions of time, and no solution of the equations of motion obtained by Lagrangian methods will result in q' independent of q...that would be totally unphysical, wouldn't it?

    Totally confused... :frown:
  2. jcsd
  3. Dec 9, 2004 #2


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    What he meant (I`m sure) was that [itex]q_i[/itex] and [itex]q_j[/itex] are independent when [itex]i \not=j[/itex]. That doesn't mean they are not functions of time, it means that changing one coordinate [itex]q_i[/itex] does not affect the others.

    A system where each coordinate can vary independently is called a holonomic system. In a nonholomonic system the coordinates cannot vary independently.

    For example, a sphere rolling down a perfectly rough surface needs only 5 coordinates to specify its configuration. 2 for the position of its center of mass and 3 for its orientation. But these 5 coordinates cannot all vary independently. When the sphere rolls, at least 2 coordinates must change. So this is a nonholomonic system.
    Last edited: Dec 9, 2004
  4. Dec 9, 2004 #3


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    Let's hope everything would be a bit more clear after reading my post.
    Lagrange's function for a simple classical nonrelativistic system is defined by the relation:
    [tex] L:=L(q^{i}(t),\dot{q}^{i}(t),t}) [/tex],where L is called "Lagrange function"and its variables are called generalized coodinates,generalized velocities and time.As u can see,it is essentially a function of time,but the time dependence is both implicit (through the generalized coordinates and velocities) and explicit.Now,the Lagrange function depends explicitely on the generalized coordinates and velocities chosen as INDEPENDENT VARIABLES BY DEFINITON OF THE LAGRANGE FUNCTION.To be actually fair,these variables are the coordinates of a 6N dimentional manifold called the "space of configurations".Viewed geometrically,a dynamical system is given in terms of this manifold.As coordinates of a geometrical entity/space,they are independent.So,the Lagrangian is a function defined on the tensor product of this space with the R as a mapping into R.Just the same way the Hamiltonian is defined on the phase space (or a tensor product of it with R) applying on R.
    Of course,for a system,the velocity is just the time derivative of the "r" (position) vector,but that has nothing to do with the Lagrange's function.For this special function,all these variables are independent,for the geometrical reason imposed earlier.
    When the system is put in time independent conditions (the forces acting on the particles are conservative),then the Lagrangian takes a familiar form as the difference between the KE and the PE.Note,that the PE term includes all explicit generalized coordinates dependence of the Lagrange's function.
    Both in Lagrangian and Hamiltonian dynamics the existance of nonconservative forces in the system poses a big problem,as the receipt of constructing the Lagrange (and hence its Legendre transformation wrt to canonical momenta) function is no longer valid.U won't be able to "paint" the lagrangian as the diff.between the KE and the PE,as there will not longer be a a conservative force which would yield a PE.In those cases,the method is to find the Lagrangian which would yield the Newton equations.So it basically uses the equivalence between the 2 formulations,but the other implication.Newton implies Lagrange.
    This was a necessary divagation.Now,u managed to find the function of Lagrange for a system.It looks badly or not,it's less relevant.What is relevant is that with it u can construct a functional,called Lagrangian action (defined on the space of configurations (i'm assuming the Lagrangian does not depend on time explicitelly).This lagrangion action is actually a number (it's the integral of the lagrangian wrt to time) and the variational principle (equivalent logically to Newton's second law) imposes that this actional reach an extremum value when computed in the configurations space.
    From this principle u find Lagrange's eq.of motion whose solutions,geometrically,would represent the trajectory of a point (specified by the initial conditions of the system) in the configurations space.Quite similar to the Hamiltonian case.
    To resume,there are geometrical justifications to every question about CM that u may have.For that:

    I would advise u to put your hand really quickly on VI Arnold' book,as it would definitely be useful in studying CM in both Lagrange and Hamilton formulation and would naturally provide u with info.to understanding QM from a geometrical point of view.

  5. Dec 9, 2004 #4


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    It is indeed true that if you know q(t), then you also know q-dot. But a priori, you know neither, and a given particle can have any position or momentum, depending upon the initial values of these coordinates. So think of q and q-dot as independent from the point of view of having arbitrary launch conditions.

    Imagine you have a particle with certain position and velocity coordinates. The Lagrangian (or Hamiltonian) is a function that tells you what's different about the situation where these coordinates are slightly different. From this info (i.e., the derivatives), you can derive all the needed equations of motion.

    To you, Lagrangian is a new concept, so maybe it's helpful to think about something similar. A particle an a potential field has kinetic and potential energy. The total energy is a function of positions and their time derivatives. Though these are not independent in the sense you pointed out, it is still helpful to ask how the energy changes if for example one of the velocity components was changed without changing the position coordinates. (In fact these derivatives tell you the momenta.)

    I sympathize with your confusion. It points out a common problem in physics education and the way a human mind works. There are many conceptual leaps a student must learn. Once they are learned, they become "obvious", so a prof will, from the student's point of view, gloss over them, because she has forgotten that there was any difficulty.
  6. Dec 10, 2004 #5
    I had the same problem when I studied the Lagrangian formalism. In particular understanding why:

    [tex] \frac{d \dot{q}}{dq}=0[/tex]

    For if you pick any function for q whose first derivative to time is nonzero you contradict this formula... I thinks that's where the problem lies. The Euler-Langrange equations describe the path for wich the action is stationary. In this formalism you deal with all kinds of paths: q(t) is not determined but is associated with every possible path, so there is no definite connection between q and q'! For every value of q you can have any value for q' as they represent all possible routes. So in this formalism q and q' are independent.

    It is when you pick one path, that is what the Euler-Lagrange equations do, that the confusion arises. Now you have a solution q(t) with a time derivative q'(t) for wich dq'(t)/dq can be nonzero. But it is only when you realise that this solution is one of many paths you can appreciate the independence of these coördinates.
  7. Dec 10, 2004 #6


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    Exactly, but if I understand what people are trying to tell me here, you do not arrive at this solution until after you have picked that path, i.e. after you have already solved the equations of motion derived from Lagrange's equations. That means that you have already taken the derivative in question:

    under the assumption of total generality or "arbitraryness" of the two coordinates (for their values at [itex] t_0 [/itex] are completely arbitrary initial conditions that are totally independent of each other)...and subsequently used that result to arrive at a specific path.

    How am I doing so far?
  8. Dec 10, 2004 #7


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    The requirement that the variation of the action be zero picks the path. The technique of doing this is part of the mathematical field called the Calculus of Variations. Once you have the path you can derive the Euler equations.
  9. Dec 10, 2004 #8
    I don't see what you mean by this. You can derive the Euler Lagrange equations by 'extremizing' the action without knowing the path, right? All you need to know is how the Lagrangian depends on you generalised coördinates. :confused:
  10. Dec 11, 2004 #9


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    Let's be very simple minded. Suppose you wish to drive from Boston to Seattle. At any time on the trip, you can go in most any direction -- allowed by the road grid -- at most any speed, subject to the constraints of practicality. Thus direction and speed are quite independent, and we then call q(i) and dq(i)/dt independent -- we are talking empirical. (Or, in a grocery store, you are free to buy carrots, chicken stock and chicken, and .......) The Lagrangian approach, simply provides a recipe to do what you want to do -- after the fact, the coordinates and velocities are no longer independent -- go East for 40 mile at 65mph, then turn south on Route 87, for 10 mile at 35mph. Or, broil chicken for 15min/side after browning it in a pan. Deglaze wih chicken stock .....

    The idea of independent variables is as old and honorable as any human intellectual acheivement. It's easy to get confused -- I've seen many, many students worry about the independence of generalized coordinates. But, when confronted with a step-by-step explanation of the logic and use of mathematical variables in the sciences, they usually begin to see the light. By the time constraints and Lagrange multipliers come around, students do see the light, and move on to more fundamental concerns -- will the Red Sox do it again next year?

    Reilly Atkinson
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