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 free particle

  1. Jun 11, 2008 #1
    In section 4 of Landau and Lifgarbagez they derive the expression for the kinetic energy by expanding the Lagrangian around v+e. The resulting expression has a term which must be a total time derivative so that the equations of motion are unaffected. The text claims that the term dL/d(v^2) v.e must be linear in v to be a total time derivative, but I don't understand why this is.
     
  2. jcsd
  3. Jun 17, 2008 #2
    I just read that section.
    I think it would have helped if they stated that the 2nd term is a total time derivative 'of a function of coordinates and time' ...
    df(x,t) / dt = df/dx * dx/dt + df/dt (partial d's now)
    Since f does not depend on the velocities, df/dx and df/dt don't, and the overall dependence of df/dt on v=dx/dt is linear.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Lagrangian for free particle
Loading...