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Deriving Lagrange's Equations

  1. Sep 9, 2010 #1
    I am having a problem with deriving Lagrange's Equations.

    It's the derivation starting from Hamilton's Principle, the part where you consider the deviation from the path:

    Suppose [tex]q=q(t)[/tex] is the function for which the action is a minimum, and now consider a deviation from [tex]q[/tex] where we replace it with [tex]q(t)+\delta q(t)[/tex] and then consider [tex]\delta S[/tex]:

    [itex]\delta S = \int^{t_2}_{t_1} L(q+\delta q,\ \dot{q}+\delta \dot{q},\ t)\ dt-\int^{t_2}_{t_1}L(q,\ \dot{q},\ t)\ dt[/tex]

    And then the argument is made that [tex]\delta S = 0[/tex] as a necessary condition for [tex]S[/tex] to have an extremum.

    So I don't entirely follow why this last part is. Hand & Finch write as the reason for this:

    So now it seems sort of apparent to me mathematically why we must have this condition, but intuitively I'm still lost. I can see obviously in either case that if we have a maximum why it can't have [tex]\delta S>0[/tex] and likewise if it's a minimum it can't be that [tex]\delta S<0[/tex], but the other cases don't seem to follow immediately for me. Can anyone perhaps give an intuitive reason why this is? I'm sure I'm overlooking something simple.
     
  2. jcsd
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