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Action (and the principle of least of which)

  1. Nov 3, 2005 #1
    Hello,
    I recently began an Analytical Mechanics course and find it impossible to understand how to derive Euler-Lagrange formula from the principle of least action.
    There is actually one step I can't understand, how we get from:

    I WOULD HAVE USED LaTeX HERE BUT IT'S NOT POSSIBLE TO SEE IT IN "PREVIEW POST"!
    I DON'T TRUST I CAN SUBMIT WITHOUT ERRORS.


    Can anyone look at the image and explain how they got to the expression below? and I'm not sure I know what exactly is the meaning of δ (delta that is)...
    mechanics.jpg
    Thanks in advance.
     
  2. jcsd
  3. Nov 3, 2005 #2

    Stingray

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    When varying the action, you're varying the particle's path: [itex]q(t) \rightarrow q(t) + \delta q(t)[/itex]. The delta just denotes the difference beween the reference trajectory and the perturbed one. If the reference Lagrangian is [itex]L(q,\dot{q})[/itex], the perturbed one is obviously [itex]L(q + \delta q ,\dot{q}+ \delta \dot{q})[/itex]. The first order Taylor expansion of this is just
    [tex]
    L(q,\dot{q}) + \frac{ \partial L(q, \dot{q}) }{\partial q} \delta q + \frac{ \partial L(q, \dot{q}) }{\partial \dot{q} } \delta \dot{q}
    [/tex]

    [itex]\delta S[/itex] is the difference between the perturbed action (the time integral of the above equation) and the reference one, which is just the equation you're asking about.
     
  4. Nov 3, 2005 #3
    Thank you!! This was a huge leap in understanding.

    Now, one importand thing remains: I can't see why the variation of "Action" should be zero for the action itself to be minimal.

    I mean, the action is a definite integral (just a number). With respect to what do we want it to be minimal? time? path?
    And after we decided that, is this "delta" the thing we use to derive a definite integral?

    Thanks again.
     
  5. Nov 3, 2005 #4

    StatusX

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    This can be a little confusing. One way to understand it is to think of the definite integral as a function from the set of functions to the real numbers (also called a functional). With a normal function you plug in a number and get a number, but with a functional you plug in a function and get a number.

    (Another way to think of this is that a function is like an infinite dimensional vector. Instead of n components, like a vector in n dimensional space would have, a function f(x) has a component for each point, which is just it's value at that point. Crudely, you could say the function f(x)=1 is like the vector (1,1,1,1,1,...). (this isn't really true since there are more points then you can list, even with dots at the end, but you get the idea). So then a functional is just a function from this infinite dimensional function space to the real numbers.)

    Finding a path of extremal action is just finding a function (or a point in function space) that is an extreme value of the functional (the definite integral). This is fundamentally no different from finding the extreme values of a normal function. At these extrema, small changes in the function (corresponding to small displacements in function space), do not produce any first order change in the value of the integral. Loosely speaking, if you change the path a little, the action changes very little.
     
    Last edited: Nov 3, 2005
  6. Nov 4, 2005 #5

    Andrew Mason

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    The purpose is to find the difference of the path integral of the Lagrangian over two slightly different paths that begin and end on the same point.

    [tex]S = \int_{t_1}^{t_2} L(x(t), \dot x(t))dt[/tex]

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

    The difference is:

    [tex]\delta S = \int_{t_1}^{t_2} L(x(t)+\epsilon, \dot x(t)+\dot{\epsilon})dt - L(x(t), \dot x(t))dt[/tex]

    The integrand is just the partial derivative of L with respect to [itex]x [/itex] times [itex]\epsilon[/itex] plus the partial derivative of L with respect to [itex]\dot x[/itex] times [itex]\dot\epsilon[/itex]:

    [tex]\delta S = \int_{t_1}^{t_2} \left(\frac{\partial L}{\partial x}\epsilon + \frac{\partial L}{\partial\dot{x}}\dot{\epsilon}\right)dt[/tex]

    AM
     
  7. Nov 7, 2005 #6

    dextercioby

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    It would have been much simpler if that [itex] \delta [/itex] was the Gâteaux derivative... But I guess it isn't. It's something fishy and mathematically not accurate, i guess...

    Daniel.
     
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