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Why does Lagrangian Density only depend on field variables and their first derivative

  1. May 28, 2012 #1
    Hi, guys,

    Why do we assume Lagrangian Density only depend on field variables and their first derivative?

    Currently, I am reading Ashok Das's Lectures on Quantum Field Theory.

    He says (when he is talking about Klein-Gordon Field Theory):

    "In general, of course, a Lagrangian density can depend on higher
    order derivatives. However, for equations which are at most second
    order in the derivatives, the Lagrangian density can depend at the
    most on the first order derivatives of the field variables. These are
    the kinds of equations we will be interested in and correspondingly
    we will assume this dependence of the Lagrangian density on the
    field variables through out."


    I am a bit confused about what he says.
     
  2. jcsd
  3. May 28, 2012 #2

    Pengwuino

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    Gold Member

    Re: Why does Lagrangian Density only depend on field variables and their first deriva

    If you look at the Euler-Lagrange equation for Lagrangians with at most, 1st order derivatives, you'll see that you can only get 2nd order derivatives out as your equations of motions. For example, if you have the classical Newtonian point-particle lagrangian

    [itex]L = \frac{1}{2}{\dot {\vec{x}}}^2 + V(\vec{x})[/itex]

    you have a Lagrangian that depends only on the position and first-time derivative of the position. The Euler-Lagrange equations of a Lagrangian with at most, first-time derivatives will only introduce 1 more order of time derivatives. What comes out, of course, is the equations of motion as Newton's 2nd law, [itex] \nabla {V(\vec{r})} = m{\ddot{\vec{x}}}[/itex] which is 2nd order in time derivatives.

    You can construct Lagrangians with 2nd or higher-order time derivatives and the Euler Lagrange equations will produce equations of even higher-order time derivatives. As they said, you can move beyond Lagrangians with first-order derivatives, but you want to start at first-order. When you setup the Klein-Gordon field, you're essentially doing the same thing, except with space-time coordinates of a field instead of simply time and you're now dealing with a density (since they're fields) instead of a point-particle model.
     
    Last edited: May 28, 2012
  4. May 28, 2012 #3
    Re: Why does Lagrangian Density only depend on field variables and their first deriva

    Don't worry about not understanding what he says. He's basically saying "I feel like it's a waste of time to explain why."

    One thing you might consider is that the energy relation is: p^2/2m, or sqrt[p^2+m^2], and noting that momentum is the derivative, so your equation will have 1st or 2nd derivatives in the fields, since:

    d/dt[wave function]=H[wave function]

    where H is the Hamiltonian.
     
  5. May 29, 2012 #4

    dextercioby

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    Re: Why does Lagrangian Density only depend on field variables and their first deriva

    For field theories, you can assume whatever order of the derivatives you want, even infinite (so-called non local theories), it's that the na:ive quantum theory of those fields makes no sense (hamiltonian not bounded from below), that's why the field equations must be second order the most.
     
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