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Deriving wave equation from Lagrangian density

  1. May 9, 2014 #1
    Hi,
    This is a worked example in the text I'm independently studying. I hope this isn't too much to ask, but I am stupidly having trouble understanding how one step leads to the other, so was hoping someone could give me a little more of an in-depth idea of the derivation. Thanks.

    1. The problem statement, all variables and given/known data

    Consider waves on a string of mass m and length l. Let us define the mass density p=m/l, tension T and displacement from the equilibrium ψ(x,t). The kinetic energy T can then be written as T=(1/2)∫[itex]^{l}_{0}[/itex]dxp([itex]\partial[/itex]ψ/[itex]\partial[/itex]t)^2 and the potential energy V=(1/2)∫[itex]^{l}_{0}[/itex]dxT([itex]\partial[/itex]ψ/[itex]\partial[/itex]x)^2. The action is then

    S|ψ(x,t)|=∫dt(T-V)=∫dtdxL(ψ,[itex]\partial[/itex]ψ/[itex]\partial[/itex]t,[itex]\partial[/itex]ψ/[itex]\partial[/itex]x)

    where

    L(ψ,[itex]\partial[/itex]ψ/[itex]\partial[/itex]t,[itex]\partial[/itex]ψ/[itex]\partial[/itex]x)=p/2([itex]\partial[/itex]ψ/[itex]\partial[/itex]t)^2 - T/2([itex]\partial[/itex]ψ/[itex]\partial[/itex]x)^2

    is the Lagrangian density. We then have immediately

    0=δS/δψ=[itex]\partial[/itex]L/[itex]\partial[/itex]ψ - (d/dx)[itex]\partial[/itex]L/[itex]\partial[/itex]([itex]\partial[/itex]ψ/[itex]\partial[/itex]x) - (d/dt)[itex]\partial[/itex]L/[itex]\partial[/itex]([itex]\partial[/itex]ψ/[itex]\partial[/itex]t)

    =0 + T([itex]\partial[/itex]^2ψ/[itex]\partial[/itex]x^2) - p([itex]\partial[/itex]^2ψ/[itex]\partial[/itex]t^2)

    Of which the wave equation falls out effortlessly.
     
  2. jcsd
  3. May 18, 2014 #2
    Hi sciencegem

    I know exactly where this has come from as I am also independently studying the same text. I think I can give you some pointers to help follow this example through. I take the expressions for the kinetic and potential energies as given. The expressions for the action and Lagrangian density follow directly from the definitions with no calculation involved beyond plugging the expressions for the kinetic and potential energy in.

    The key step is the expression for the functional derivative of the action following the words "we then have immediately". Here the authors have used a different version of the Euler-Lagrange equation than the one they derived in example 1.3 of the book. The one they previously derived is for a function of a single variable and its first derivative. Here we have two independent variables (x and t) and their first derivatives. Deriving this version of the Euler-Lagrange equation involves a double integral hence the expression for the functional derivative of the action involves the Lagrangian density rather then the Lagrangian.

    We then simply plug in the expression we previously derived for the Lagrangian density and the rest follows from simple algebra.

    I hope that is somewhat helpful. The details of deriving the Euler-Lagrange equation for several independent variables are available fairly easily from a google search.
     
  4. May 31, 2014 #3
    Thank you Darktobz , that was very helpful!
     
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