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Lagrangian for a free particle

  1. Feb 2, 2014 #1
    Hello, this is probably one of those shoot yourself in the foot type questions.

    I am going through Landau & Lifshits CM for fun. On page 7 I do not understand this step:

    [itex] L' = L(v'^2) = L(v^2 + 2 \vec{v} \cdot \vec{\epsilon} + \epsilon^2) [/itex]

    where [itex] v' = v + \epsilon [/itex]. He then expands the expression in powers of [itex] \epsilon [/itex] (neglecting higher order terms) to get:

    [itex] L(v'^2) = L(v^2) + \frac{\partial L}{\partial v^2} 2\vec{v} \cdot \vec{\epsilon} [/itex]

    How did he arrive here? What expansion did he use? Taylor expansion?

    Thanks for any help or comments!
     
  2. jcsd
  3. Feb 5, 2014 #2

    Avodyne

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    Yes, Taylor expansion. Higher-order terms are neglected because they want to determine the infinitesimal change in L that follows from an infinitesimal change in v. (I'm guessing here because I don't have the book, but there's no other reason to do this calculation. I presume that the final result will be the Euler-Lagrange equations.)
     
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