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Equivalence of DAlembert's principle and Action Principle

  1. Jun 8, 2007 #1
    Hello everyone.

    I'm reading Goldstein's Classical Mechanics (2nd Ed) and I have worked out the derivation of Euler-Lagrange equation of motion from DAlembert's Principle as described in Chapter 1 and also the Action Integral approach in Chapter 2.

    I want to understand the equivalence of the two approaches: the derivation from DAlembert's Principle works for forces of constraint that do no (virtual) work:

    [tex]\sum_{i = 1}^{N}{\bf f_{i}\bullet \delta{\bf r_{i}} = 0[/tex]

    But the derivation from the action principle does not make any explicit mention of the constraint forces. Since both eventually lead to the Euler-Lagrange condition [also refer to chapter 1 of http://www.ks.uiuc.edu/Services/Class/PHYS480/qm_PDF/QM_Book.pdf] [Broken] I think there should be some equivalence, but I can't see what it is.

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jun 8, 2007 #2
    By action principle, do you mean the lagrangian?

    If you want to add constriants for the lagrangian you simple include the lagrange multipliers method, or you can use the much easier matrix form of the augment-force matrix method, which side steps lagrange multipliers altogether and spits them out for you in the end.
  4. Jun 9, 2007 #3
    If you go through the steps leading to the derivation of the Euler-Lagrange condition for the extremal of the action integral ([itex]S = \int_{t_{1}}^{t_{2}}L dt[/itex]) to exist in the pdf file linked in my post, then (since this is a completely mathematical proof with no physics involved) you get to the Euler-Lagrange equations for the N particles

    [tex]\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_{i}}}\right) - \frac{\partial L}{\partial q_{i}} = 0[/tex]

    for i = 1, 2,...,N.

    Further by Hamilton's principle, the equations of motion (Newton's Second Law)

    [tex]{\bf F_{i}} = \dot{\bf {p_{i}}}[/tex]

    are equivalent to the Euler Lagrange Condition with

    [tex]L = \sum_{i = 1}^{N}\frac{1}{2}m\dot{q}_{i}^2 - V(q_{1},q_{2},\ldots,q_{N})[/tex]

    as the Lagrangian of the mechanical system of N particles. This is the derivation I was referring to, from the condition that the variation of the action integral is zero. The steps in the pdf file do not make any mention of the kind of constrains there are on the particles.

    The second derivation is from dAlembert's principle, as given in Goldstein, and starts from Newton's Second law. Here we assume that the forces of constraint are those for which work done on the system is zero.

    My question is: what is the physical equivalence of the two methods, which both lead to the same system of differential equations?
    Last edited: Jun 9, 2007
  5. Jun 9, 2007 #4
    At one point in the derivation from the action principle, you assume that your generalized coordinates are independent i. e. can be varied freely except at the begining and end points. If your system has constraints, that is not possible for any imaginable set of generalized coordinates because the constraints will connect some of them. Using independent generalized coordinates takes the constraints into account implicitly.

    Let me give a simple example - a pendulum in two dimensions - a material point of mass m, hanging on a massless string of length R. The constraint is that the distance of the mass to the point of hanging is always R=const:

    [tex]\sqrt{x^2+y^2} - R =0[/tex]

    In the usual Cartesian coordinates the constraint can be taken into account by adding a Lagrange multiplier to the original Lagrangian:

    [tex]L (x,y, \lambda; \dot{x}, \dot{y})= \frac{m}{2}(\dot{x}^2+\dot{y}^2) \, + mgy \,+ \, \lambda (\sqrt{x^2+y^2} - R )[/tex]

    The Lagrange multiplier lambda takes care of the non-independence of x and y. The Euler-Lagrange equations of that Lagrangian will produce the correct equations of motion for the mass m. There will be one extra equation for lambda which will give the force maintaining the constraint - the tension in the string.

    If you just blindly write L = K - U in those coordinates:

    [tex]L (x,y; \dot{x}, \dot{y})= \frac{m}{2}(\dot{x}^2+\dot{y}^2) + mgy [/tex]

    the Euler-Lagrange equations of that Lagrangian will NOT produce the correct equations of motion because the x,y coordinates are NOT independent under the constraint and the whole derivation of the Euler-Lagrange equations from the action principle fails.

    The second way to take the constraint into account is to use coordinates that are independent i.e. can be varied freely uder that constraint. In this case, use the angle of the pendulum with the vertical and the Lagrangian = kinetic - potential energy in those coordinates is:

    [tex]L (\theta; \dot{\theta})= \frac{m}{2}R^2\dot{\theta}^2 + mgR\, cos(\theta) [/tex]

    which will produce the correct equations of motion because the angle theta is independent variable under the constraint i.e. can be varied freely. Notice that one constraint reduced the number of independent coordinates by one, from (x,y) to theta.

    In the derivation from the action principle they assumed implicitly the q's are like theta not like (x,y). To formulate the action principle for dependent coordinates like (x,y) you will have to add lagrange multipliers to L, one multiplier for each constraint.
    Last edited: Jun 9, 2007
  6. Jun 10, 2007 #5
    smallphi, thanks for your post. I now have a better understanding of this.

    In fact Goldstein does say that if the constrains are holonomic, then it is possible to find sets of independent coordinates [itex]q_{j}[/itex] that contain constraint conditions implicitly in the transformation equations (that relate dependent coordinates to the generalized coordinates), so that [itex]q_{j}[/itex] is independent of [itex]q_{k}[/itex] for [itex]j \neq k[/itex]. So through your example, I now understand what this means.

    Thanks again :smile:
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