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Problem with Lagrange's Equations

  1. Oct 10, 2011 #1

    Erland

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    I have problem with Lagrange's equations and their derivation, the way it is presented in Goldstein's "Classical Mechanics". I have never seen this problem mentioned anywhere, so I wonder if I am the only one who see this problem.

    To see the problem, consider a simple case where the motion of a system can be described by just one generalized coordinate [itex]q[/itex], plus time explicitly. Let [itex]x[/itex] be a cartesian coordinate of one particle of the system. Then [itex]x=x(q,t)[/itex].
    Then, [tex]\dot x=\frac{\partial x}{\partial q}\dot q+ \frac{\partial x}{\partial t}.[/tex]
    (In Goldstein we have several generalized coordinates, but the reasoning is the same.)

    Now, reasoning as Goldstein, we could without problem differentiate [itex]\dot x[/itex] wrt [itex]\dot q[/itex], obtaining [itex]\partial \dot x / \partial \dot q =\partial x/\partial q[/itex], and also plug in [itex]\dot q [/itex] in the expression for the total kinetic energy [itex]T[/itex] and the Lagrangian [itex]L=T-V[/itex], and then without problem differentiate [itex]L[/itex] wrt [itex]\dot q[/itex], obtaning Lagrange's Equations (one for each generalized coordinate):
    [tex]\frac d {dt}\frac{\partial L}{\partial\dot q}-\frac{\partial L}{\partial q}=0. [/tex]
    (Differentiating [itex]L[/itex] wrt [itex]q[/itex] seems to be no problem, either.)

    But how can we differentiate [itex]\dot x[/itex] (and then [itex]L[/itex]) wrt [itex]\dot q[/itex]? In the formula we obatined for [itex]\dot x[/itex], we have only one indepentent variable: [itex]t[/itex], and therefore it shouldn't make sense to differentiate wrt any other variable.

    To be able to differentiate wrt [itex]\dot q[/itex], we must actually view [itex]\dot x[/itex] (and similarly, [itex]L[/itex]) as a composite function, [itex]\dot x= f(q,\dot q,t) [/itex], where [itex]f(u_1,u_2,u_3)=g_1(u_1,u_3)u_2+g_2(u_1,u_3)[/itex] and [itex]g[/itex] is the function which gives the coordinate transformation: [itex]x=x(q,t)=g(q,t)[/itex], and [itex]f[/itex] and [itex]g[/itex] can be differentiated wrt any of their arguments. ([itex]g_1[/itex] and [itex]g_2[/itex] are the partial derivatives of [itex]g[/itex].)
    The expression [itex]\partial \dot x /\partial \dot q[/itex] will then make sense if we recognize it as the same as [itex]f_2(q,\dot q,t)[/itex], and then it will be equal to [itex]\partial x/\partial q[/itex], just as it should be.

    However, this will be consistent only if the choice of the function [itex]f[/itex] is unique. If there was another function [itex]h(u_1,u_2,u_3)\ne f(u_1,u_2,u_3)[/itex], which also satisfies [itex]\dot x = h(q,\dot q,t)[/itex], but for which [itex]h_2(q,\dot q,t)\ne f_2(q,\dot q,t)[/itex], then [itex]\partial \dot x/\partial \dot q[/itex] cannot be uniquely defined (and neither can [itex]\partial L/\partial \dot q[/itex]).

    This problem is not discussed in Goldstein, and I have not seen it mentioned anywhere (although I did not search very much). Am I really the only person who see this problem?

    Fortunately, the problem can be solved. It turns out that the function [itex]f[/itex] above must be unique. The reason is that the equation [itex]\dot x= f(q,\dot q,t) [/itex] must hold for all possible paths [itex]q(t)[/itex], and, for every triplet of values [itex](u_1,u_2.u_3)[/itex], it is always possible to find a path [itex]q(t)[/itex] such that [itex]q(t)=u_1[/itex] and [itex]q'(t)=u_2[/itex], for [itex]t=x_3[/itex]. This implies that [itex]f[/itex] is uniquely determined by the requirement [itex]\dot x= f(q,\dot q,t)[/itex], it must be the function given above. This argument can be generalized to any number of generalized coordinates, and it also follows that [itex]\partial L/\partial \dot q[/itex] will be meaningful and unique.

    But it is not trivial to find and prove this. I was very confused when I read Goldstein the first time, and I wondered how on Earth one could differentiate wrt [itex]\dot q[/itex]. It took a while until I found the "proof" above. I think it is the author's job to do this, not the reader's.
     
    Last edited: Oct 11, 2011
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  3. Oct 11, 2011 #2

    Bill_K

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    Erland, The reason you're confused is that you've made a crucial change in the notation. The aim is to describe the transformation between Cartesian coordinates and generalized coordinates. To do this you need to express x and v in terms of q and q·. The transformation of the position is given by x(q, t). The transformation of the velocity is v(q·, q, t):

    v = (∂x/∂q) q· + ∂x/∂t

    The mistake you've made is calling the left hand side x·, which it is not.
     
  4. Oct 11, 2011 #3

    Erland

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    But [itex]v=\dot x[/itex], by definition (at least according to Goldstein, 3rd ed, p. 18, eq. 1.46), so I don't understand what difference it makes to use v instead.
     
  5. Oct 11, 2011 #4

    Bill_K

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    The difference is the functional dependence. In the Lagrangian formulation, position and velocity are regarded as independent variables. x(q, t) and x· are functions of two variables q and t, and therefore do not depend on q·. But v(q·, q, t) is a function of three variables and does depend on q·.

    Similarly, when you get to the Lagrangian L(q, q·, t) one could argue it's all just a function of one variable q(t). But the functional form is crucial. One must treat q· as an independent variable and L as a function of all three variables.
     
  6. Oct 11, 2011 #5

    Erland

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    OK, but then the question arises: what does the functional dependence [itex]v=v(q, \dot q, t)[/itex] look like? The only way to determine this is to use the condition [itex]v(q(t),\dot q(t),t)=\dot x(t)[/itex]. But then, what if there is a different function [itex]w=(q, \dot q,t)[/itex] which also satisfies [itex]w(q(t),\dot q(t),t)=\dot x(t)[/itex], but for which
    [itex]\partial v/ \partial \dot q\ne \partial w/\partial\dot q[/itex]. Whch of the two latter quantities is then the correct one to use? There seems to be no way to decide this.
    It is therefore important to prove that [itex]v=v(q, \dot q, t)[/itex] is uniquely defined by the condition [itex]v(q(t),\dot q(t),t)=\dot x(t)[/itex]. I do that in my first post, but I think Goldstein (and other textbook writers) should do this job.
     
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