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Lagrange's equations

  1. Aug 2, 2006 #1

    quasar987

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    I am trying to understand why the Euler-Lagrange's equations in generalized coordinates are equivalent to Newton's equations of motion. But there is a link missing in my reasoning...


    1° It is easy to show that Newton's equations in cartesian coordinates are equivalent to Lagrange's equations. I.e. one [itex]\Leftrightarrow[/itex] the other.

    2° Here is the missing link. The central theorem in the theory of calculus of variation says that... (see http://en.wikipedia.org/wiki/Euler-Lagrange_equations#Statement). Roughly, for a path [itex]\vec{r}(t)[/itex] which extremizes the action, the Lagrangian [itex]L(\vec{r}(t),\dot{\vec{r}}(t),t)[/itex] satisfies the Euler-Lagrange's equations. But the opposite need not be true. The goal of 2° is to show that in the case of mechanics, the converse is true.

    The fact that the converse need not be true implies that even tough Newton's cartesian equations are equivalent to Euler-Lagrange's cartesian equations, I cannot conclude that the action is extremizedfor the physical path (x(t), y(t), z(t)). If however, it is possible to show something like the fact that for a sufficiently small time interval [itex]\Delta t[/itex], the action

    [tex]S=\int_t^{t+\Delta t}L(x(t),y(t),z(t),\dot{x}(t),\dot{y}(t),\dot{z}(t),t)dt[/tex]

    always has one and only one extremum, then it is easy to conclude that the extremizing path in question must be the physical path. If it were not, there would be a contradiction because we'd assume that the physical path does not "minimizes" the action but since it is the physical path, it satisfies Newton's equations and hence also Euler-Lagrange's. But the path that minimizes the action also satisfies Euler-Lagrange's equ (by the central thm of calculus of variation) and thus Newton's (since they are related by a [itex]\Leftrightarrow[/itex] relation) too. By the unicity of the physical path, it must be that the physical path is the one which extremizes the action, since we know for a fact that such a path always exists for sufficiently small time intervals.


    But I know nearly nothing about calculus of variation. Is this existence and uniqueness of the extremizing function trivial?? Or is it plain and simply wrong? If so, how to achieve the goal of 2°? Or ultimately, forget 1° & 2° and give another rigorous outline leading to the undeniable equivalence of the original Newton's equations and the generalized Euler-Lagrange's equations.

    For completeness, given 2°, i.e. given the fact that the cartesian Euler-Lagrange equations imply the cartesian Hamilton principle, here is the rest of the argument....


    3° Given a complete set of generalized coordinates, write the "physical Lagrangian" as a function of these new coordinates. Since the action is extremized for the path in cartesian coordinates, it is also extremized for the path written in these new coordinates, hence, by virtue of the central theorem of calculus of variation, we can write the Euler-Lagrange's equations is generalized coordinates.

    4° By solving them for each generalized coordinates [itex]q_i(t)[/itex], we can recreate the physical path (x(t), y(t), z(t)) by substituting their explicit time-dependant form into the transformation equations

    [tex]x=x(q_1,...,q_N), \ \ y=y(q_1,...,q_N), \ \ z=z(q_1,...,q_N)[/tex]


    Thanks for reading.
     
    Last edited: Aug 2, 2006
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  3. Aug 2, 2006 #2

    vanesch

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    I'm probably missing entirely the subtlety of what your problem is, but "extremizing" is only saying: "is stationary up to second order in small variations of the path" ; it doesn't mean: "has the global extreme value".

    In that case, the condition "is stationary up to second order in small variations of the path" is - I thought - entirely equivalent to th E-L equations (in that the first-order variation is an integral of exactly these E-L equations times an arbitrary function, so: if the integral vanishes identically, then the E-L must be satisfied, and if the E-L are satisfied, then this first order variation is 0).

    Now, it could be that there are several paths obeying this first-order stationarity for small perturbations. But that would mean that there are several solutions to the E-L equation too. Normally, that shouldn't happen, given that they are second-order differential equations, and that we give two boundary conditions. But there might be mathematical subtleties that sometimes allow for multiple solutions. I'm thinking of cases where there are bifurcations or the like.
    Nevertheless, there is still equivalence between the set of solutions of the E-L equation, and this first-order stationarity of the action, I would think.
    I don't see how you can satisfy the E-L equations, and not have first order stationarity, and I also fail to see how you can have first order stationarity, and not satisfy the E-L equations, given that they are related by an integral with an arbitrary function.
     
  4. Aug 3, 2006 #3

    quasar987

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    Like I wrote, I don't know much about calculus of variation but from what I gathered reading all the related articles on wiki, given [itex]L(q_1(t),...,q_N(t), \dot{q}_1(t),...\dot{q}_N(t),t)[/itex], the stationarity of S means

    [tex]\frac{\delta S}{\delta q_i(t)}=0[/tex]

    I got confused by what this actually means, but now reading the article on the action, you might be right in thinking [itex]\delta S =0[/itex] simply means that the integral of the variation up to first order vanish.
     
  5. Aug 3, 2006 #4

    quasar987

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    Based on the proof given in the action article of wiki (http://en.wikipedia.org/wiki/Action_(physics)#Euler-Lagrange_equations_for_the_action_integral), I agree with that. But it would mean that the statement of the "central thm of calculus of variation" is in fact an [itex]\Leftrightarrow[/itex] one. I.e.not just "IF acton is extremized, THEN L satisfies E-L", but rather "Acton is extremized, iff L satisfies E-L"

    someone can confirm?
     
  6. Aug 3, 2006 #5

    vanesch

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    Yes, as in: d f / dx = 0 means that f(x0) is "stationary" up to first order in variations x0 + dx. Even "extremal" is strictly speaking not correct, as it could be just a horizontal point of inflexion (like x^3 has for x = 0).

    That said, in the case that there is only one solution to the E-L equations (the usual Newtonian solution to the mechanical problem at hand), and if we can show somehow that the action is bounded from below, this is proof that the usual solution does indeed *minimize* in an absolute sense, the action.

    Upon reflexion, it is not evident to see that action is bounded from below, considering the Lagrangian of the form L = T - V.
    Indeed, by considering trajectories in "high potential" regions, it would seem that we can make L as negative as one likes. However, to get there within the allowed time between the boundary conditions, one would have to go very fast, so this gives, during the voyage, a very positive contribution due to high T. Frankly, I don't know how to demonstrate easily that the action is bounded from below...
     
  7. Aug 3, 2006 #6

    quasar987

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    The correct word instead of "extremize" would be "criticalize" (berch! :p). Since by analogy with regular calculus, points such that [itex]f'(x)=0[/itex] are called critical points.

    I'm gonna take my math question of post #4 to the math forum.
     
  8. Aug 4, 2006 #7
    quasar987,

    I had difficulties to understand exactly what your question is.
    Can I assume that it is to prove the following:



    [tex]\mbox{f is an extremum of\ \ \ } J = \int_a^b F(x,f(x),f'(x))\, dx [/tex]

    [tex]\Longleftrightarrow[/tex]

    [tex]\frac{\partial F}{\partial f} - \frac{d}{dx} \frac{\partial F}{\partial f'} = 0[/tex]


    Assuming now this is indeed your question.
    If I remember well how the proof goes its way to the Lagrange equation, I think each step is an equivalence. Indeed:

    from extremum to derivative=0 is an equivalence
    all the subsequent steps are algebraic equivalences​

    Therefore, I would guess that this equivalence above is right.

    Of course, this does not mean anything else. In particular, it does not mean that the Lagrange function of a dynamic system is unique.
    Note also a related question that I posted recently: Can all differential equations be derived from a variational principle? Maybe you have an idea?

    Michel
     
    Last edited: Aug 4, 2006
  9. Aug 5, 2006 #8

    quasar987

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    You can look at the proof here if you which to refresh your memory:
    http://en.wikipedia.org/wiki/Action_(physics)#Euler-Lagrange_equations_for_the_action_integral

    or here for the general statement and proof of the said "central thm of calculus of variation":
    http://en.wikipedia.org/wiki/Euler-Lagrange#Multidimensional_variations

    You'll notice that the statement of the theorem uses the words "only if" and not "if and only if". But if you follow the proof from bottom to top, can you see a step that would prevent the statement from being an "if and only if" one? I can't.
     
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