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Mysterious Lagrangian mechanics

  1. Aug 3, 2006 #1


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    In the theory of Lagrangian mechanics, it is said that given a system of particles, write the Lagrangian T-U in cartesian coordinates, convert to generalized coordinates, solve the Lagrange equations, and re-convert to cartesian if you feel like it.

    But there's something weird going on, and this weirdness is well embodied by the double pendulum problem (http://scienceworld.wolfram.com/physics/DoublePendulum.html). The potential energy chosen does not include the tensions? Why is that?

    On the other hand, I suspect that if one choses as generalized coordinates [itex]\theta_1, \ \theta_2, \ l_1, \ l_2[/itex] "together" with the constraints [tex]\dot{l}_1=\dot{l}_2=0[/tex], then the values of the tensions can be found from the E-L equations of [itex]l_1, \ l_2[/itex], and substituted in the E-L of [itex]\theta_1, \ \theta_2[/itex] to find the same equations [itex]\theta_1(t), \ \theta_2(t)[/itex] as if one had solve only 2 E-L equations with a potential energy taking only gravity into consideration.

    What's going on here? The REAL potential energy for this system is the one where the tensions appear. How do we know what "restricted potential energy" must be used given a certain choice of generalized coordinates?

    I'm not looking for a quick rule of thumb as much as a real explanation of what's going on. Thanks all.
    Last edited: Aug 3, 2006
  2. jcsd
  3. Aug 4, 2006 #2
    Tension is a force, not a form of potential energy. The only way there could be energy associated with tension is if it did work. In this case, the only work that could be done by tension which is not already covered by including how both angles are relevant to find the kinetic energy of the second mass would be if the rods holding the masses could contract. However, the problem is generally formulated with rigid rods, so this is not an issue.

    One way to think of this is that in any problem where you have written the Lagrangian in terms of the minimum possible number of coordinates, you will find that the effects of constraint forces will already be included in the Lagrangian. Here, tension acts as a constraint.
  4. Aug 4, 2006 #3


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    1) I tought the potential energy function was just a function such that when you take its gradient, you get the force field. In this case, it would be a function [itex]U(x_1,y_1,x_2,y_2)[/itex] such that [itex]-\nabla_1 U = \vec{F}_{\rightarrow 1}[/itex] and [itex]-\nabla_2 U = \vec{F}_{\rightarrow 2}[/itex]. (where [itex]\nabla_i = \frac{\partial}{dx_i}\hat{x} + \frac{\partial}{dy_i}\hat{y}[/itex]) Is this not all a potential energy function is?

    2) Why do you say tension does no work? For simplicity, consider the pendula starting from the position in which they are illustrated in the physicsworld article. As the angle [itex]\theta_2[/itex] diminishes a little, there is a component of the tension T in the direction of motion. Hence, it does work.
  5. Aug 4, 2006 #4
    elastic double pendulum


    You could solve your problem by considering an elastic double pendulum.

    You would simply assume that the length of the pendulum part is not fixed. Then you would need to include the elastic potential in the total potential energy. You would also have two additional variables: the length of the two parts of the pendulum. The calculations would be a little bit longer too.

    In principle, you would be able in this way to calculate the tension, which you could not in the "non-elastic" version. Taking the limit of rigid rods (elastic constant very large) would provide you the solution of the initial problem with the additional information on the internal tension.



    Your question illustrates one of the interrest of the method: it determines the motion without considering any forces that "do not work", typically the so-called "reaction forces". Doing so, it also eliminates unneeded variables (like the fixed lenght of a pendulum). Of course, an engineer may sometimes be interrested to know the internal forces too. But the Lagragian method, then, has already prepared part of the solution.
    Last edited: Aug 4, 2006
  6. Aug 4, 2006 #5
    Well, that's because what you're talking about only works for so called conservative forces. This means that
    (*) the work done (by the force) by going around on a closed path (i.e. you end where you started) will =0.
    The force field is a vector field and so one can use a result from vector analysis: For a simply connected (no holes) domain of your vector field, condition (*) is equivalent to the requirement you stated: that there exists a potential ("energy") function for which the gradient will be the force (Assuming you're working in [tex]\mathbb{R}^n[/tex]).
    Think about it: Have you ever tried using Lagrangians for friction forces? Now you know why. Of course there's a way to do it, but that's a different story.

    No. The tension is always in the direction of the rods, the motion always tangential to the rod (otherwise the length couldn't be a constant). Just like a planet going around the sun.
  7. Aug 5, 2006 #6


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    i) I'm certain that I have found a form for U that takes the tensions into account and that satisfies [itex]-\nabla_1 U = \vec{F}_{\rightarrow 1}[/itex] & [itex]-\nabla_2 U = \vec{F}_{\rightarrow 2}[/itex]. Are you saying that is not possible?

    ii) How do I know then what forces I must include in the potential function, and most importantly, why those and not the others?

    Thanks y'all for participating! :smile:
    Last edited: Aug 5, 2006
  8. Aug 5, 2006 #7
    No, that's not what I meant. I only meant to say that the method you presented was not omnipotent. Why don't you post your potential function so we can discuss it?

    lalbatros already mentioned this: You need to think of the benefits of the Lagrangian method. Why do you do it that way?
    In many physical situations there are constraint forces acting on the objects of interest. Therefore it can be horribly complicated to analyze a given situation only by considering forces (I'm sure you can easily make up an example of your own). One would rather like to include those constraints as limitations to the motion, for example by stating a certain relation between some coordinates (In your example (double pendulum) the distance between the first and the second ball must be a constant and so must be the distance between the first ball and the suspension point). Such an information can then be used to make a more fortunate choice of coordinates than the cartesian ones: one can adapt the coordinates to the degree of freedom (number of independent coordinates needed to describe the motion) your system has (For the double pendulum this was done by choosing two angles, [itex]\theta_1,\theta_2[/itex] instead of four cartesian coordinates). Additionally, energies are much more easily "accesible" than forces, that's the interesting thing about Lagrange.
    This was meant to lighten the idea behind the thing up a little bit. It should be clear to you that this won't work for every type of force (as mentioned earlier, consider frictional forces). Within the limitations of this model: What forces do you need to include in your potential function?
    - no constraint forces, they're not doing any work (you might want to read up on this: look for d'Alemberts principle).
    Note that constraint forces do not have a continuous potential function: Take for example a string pendulum in the gravitational field of the earth. If you simply let it swing, there surely is a tension force in the string. But if you hold up your object a little tiny bit all tension's gone, as the string is no longer stretched. The tiniest shift of position (in some vertical direction) will result in the vanishing of your force. That's certainly not the idea of a potential. Also, with a potential there is the idea of doing work by moving in the force field. How would you do that in this situation?
    Another example: A mass sitting on a table: What's the potential for the normal forces?
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