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Action in the continuum limit for an elastic medium.

  1. Apr 7, 2013 #1
    1. The problem statement, all variables and given/known data
    https://www.physicsforums.com/attachment.php?attachmentid=57592&stc=1&d=1365348538
    I'm stuck at the second part, not really sure what to here to be honest.
    https://www.physicsforums.com/attachment.php?attachmentid=57590&stc=1&d=1365346819


    2. Relevant equations
    Usually the Lagrangian is the function inside the integral for the action.
    \begin{equation}
    L = T - U.
    \end{equation}
    Usually this becomes
    \begin{equation}
    L = \frac{p^{2}}{2m} - U.
    \end{equation}

    3. The attempt at a solution
    I figured for a single particle the kinetic energy is:
    \begin{equation}
    T = \sum_{i=1}^{3} \frac{1}{2} m (\delta_{t} u_{i})^{2}.
    \end{equation}
    The potential energy should be some force constant times the displacement squared:
    \begin{equation}
    U = \sum_{i=1}^{3} k (a \delta_{i}u_{i})^{2}.
    \end{equation}
    So for N\end particles the Lagrangian would become:
    \begin{equation}
    L_{discrete} = \sum_{n=1}^{N} \sum_{i=1}^{3} \left[ \frac{1}{2} m (\delta_{t} u_{i})^{2} - \frac{k}{3} (a \delta_{i}u_{i})^{2}\right].
    \end{equation}
    Where the factor 1/3 comes from the fact that I am counting each bond three times.
    Now this is where it gets a bit fuzzy for me; what exactly does it mean to take the continuum limit? I believe it has something to do with letting m and a approach zero in such a way that if N approaches infinity the mass stays constant, but I'm not exactly sure how to do that/what that means.

    I believe that I require the fact that this is a cubic lattice to be able to write the potential energy down the way I did. But how would change the potential energy if, say for instance, the particles were on a face centered cubic lattice? Or any other lattice for that matter.

    Any help would be appreciated.
     
    Last edited: Apr 7, 2013
  2. jcsd
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