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Physical meaning of lagrangian

  1. Mar 11, 2010 #1
    i have some question about lagrangian ...here are those

    1) what is the physical meaning of L=0 .. from wiki ..i have found .."The Lagrangian, L, of a dynamical system is a function that summarizes the dynamics of the system." that means when L=0 ..the system has no dynamics ..as far as i know a free falling body has T=V in the midpoint ..that means it has no dynamics in the midpoint ..is not it ridiculous??

    2)in the actual path time integral of lagrangian is minimum ..we know L=T-V ..if V becomes larger then the potential become larger that means if potential increases ..it needs less time to go one point to another..that is so peculiar to me .. it should be V-T ,or anything else.
     
  2. jcsd
  3. Mar 11, 2010 #2

    Fredrik

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    The dynamics depends on what function the Lagrangian is, not on what value it has at a specific point in its domain. Let's look at an example. For a single non-relativistic particle in one dimension, the lagrangian L is defined by

    [tex]L(a,b)=\frac{1}{2}mb^2-V(a)[/tex]

    for all a and b in some subset of the real numbers. V is the potential function, and m is the mass of the particle. You use the lagrangian to construct the action S, which is a function that takes a curve x in spacetime to a number S[x]:

    [tex]S[x]=\int_{t_1}^{t_2}L(x(t),x'(t))dt[/tex]

    ("Spacetime" in this case is the set [itex]\mathbb R^2[/itex] and a "curve in spacetime" is a continuous function [itex]x:[t_1,t_2]\rightarrow\mathbb R^2[/itex]). In this approach to mechanics, the axiom that replaces Newton's second law says that the particle will move as described by the curve x that minimizes the action. This condition leads to the Euler-Lagrange equations for the physical system represented by the Lagrangian, and in this case there's only one:

    [tex]mx''(t)=-V'(x(t))[/tex]

    This is of course just Newton's 2nd law.
     
    Last edited: Mar 11, 2010
  4. Mar 11, 2010 #3

    Fredrik

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    I'll just add a derivation of the Euler-Lagrange equation for this simple Lagrangian, because I think this kind of stuff is fun, and because similar questions have come up before, and I expect them to come up again.

    Let [itex]x:[t_1,t_2]\rightarrow\mathbb R^2[/itex] be the curve that minimizes the action. Let [itex]\big\{x_\epsilon:[t_1,t_2]\rightarrow\mathbb R^2\big\}[/tex] be an arbitrary one-parameter family of curves such that [itex]x_0=x[/itex] and [tex]x_\epsilon(t_1)=x_\epsilon(t_2)[/itex] for all [itex]\epsilon[/itex].

    If it it's not 100% clear already, I mean that each [itex]x_\epsilon[/itex] is a curve, that the one with [itex]\epsilon=0[/itex] is the one that minimizes the action, and that all of these curves have the same endpoints as x.

    We have

    [tex]0=\frac{d}{d\epsilon}\bigg|_0 S[x_\epsilon]=\int_{t_1}^{t_2}\frac{d}{d\epsilon}\bigg|_0 L(x_\epsilon,x'_\epsilon(t))dt=\int_{t_1}^{t_2}\bigg(L_{,1}(x(t),x'(t))\frac{d}{d\epsilon}\bigg|_0 x_\epsilon(t)+L_{,2}(x(t),x'(t))\frac{d}{d\epsilon}\bigg|_0 x'_\epsilon(t) \bigg)dt[/tex]

    The notation [itex]L_{,i}[/itex] means the ith partial derivative of L. There's a trick we can use to rewrite the last term above.

    [tex]L_{,2}(x(t),x'(t))\frac{d}{d\epsilon}\bigg|_0 x'_\epsilon(t)=\frac{d}{dt}\bigg(L_{,2}(x(t),x'(t))\frac{d}{d\epsilon}\bigg|_0 x_\epsilon(t)\bigg)-\frac{d}{dt}\bigg(L_{,2}(x(t),x'(t))\bigg)\frac{d}{d\epsilon}\bigg|_0[/tex]

    We have

    [tex]L_{,1}(a,b)=\frac{d}{da}\Big(\frac 1 2 mb^2-V(a)\Big)=-V'(a)[/tex]

    [tex]L_{,2}(a,b)=\frac{d}{db}\Big(\frac 1 2 mb^2-V(a)\Big)=mb[/tex]

    so

    [tex]L_{,1}(x(t),x'(t))=V'(x(t))[/tex]

    [tex]L_{,2}(x(t),x'(t))=mx'(t)[/tex]

    This turns the equation into

    [tex]0=\int_{t_1}^{t_2}\bigg(-V'(x(t))\frac{d}{d\epsilon}\bigg|_0 x_\epsilon(t)+mx'(t)\frac{d}{d\epsilon}\bigg|_0 x'_\epsilon(t) \bigg)dt[/tex]

    [tex]=\int_{t_1}^{t_2}\bigg(-V'(x(t))\frac{d}{d\epsilon}\bigg|_0 x_\epsilon(t)+\frac{d}{dt}\bigg(mx'(t)\frac{d}{d\epsilon}\bigg|_0 x_\epsilon(t)\bigg)-mx''(t)\frac{d}{d\epsilon}\bigg|_0 x_\epsilon(t)\bigg)dt[/tex]

    The last step might look weird, but the only thing I'm doing there is to use the product rule for derivatives. The integral of the middle term is actually =0, because the assumption that all the curves have the same endpoints implies that

    [tex]\frac{d}{d\epsilon}\bigg|_0 x_\epsilon(t_1)=\frac{d}{d\epsilon}\bigg|_0 x_\epsilon(t_2)=0[/tex]

    All we have left after throwing away the zero term is

    [tex]0=\int_{t_1}^{t_2}\bigg(-V'(x(t))-mx''(t)\bigg)\frac{d}{d\epsilon}\bigg|_0 x_\epsilon(t) dt[/tex]

    But the fact that this is supposed to hold for arbitrary one-parameter families of curves that satisfy the necessary requirements means that we can choose that derivative to the right of the parentheses to be any continuous function of t that we want, and the integral is still supposed to be =0. This is only possible if the expression in parentheses is =0, so we end up with

    [tex]mx''(t)=-V'(x(t))=F(x(t))[/tex]

    as promised.
     
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