1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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.)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Lagrangian for a free particle
  1. Free Particle (Replies: 6)

  2. The free particle (Replies: 1)

  3. A free particle (Replies: 5)