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Homework Help: QFT Classical string intro

  1. Jul 11, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm starting QFT and many books I've started to read start with the introduction of a field in a classical string model

    with a Lagrange equation

    [tex]L(q,\dot{q}) = \sum[\frac{m}{2}\dot{q}_{j}^{2}-\frac{k}{2}(q_{j}-q_{j+1})^{2}][/tex]

    the equation of motion becomes

    [tex]m \ddot{q_j}-k(q_{j+1}-2q_j + q_{j-1}) = 0[/tex]

    my question is how do we get this equation of motion its a discrete derivative but I recognize this as the 3 point second derivative formula, but Lagrange formulation is only first order for the potential part

    [tex]f'' = \frac{f(x+h)-2f(x)+f(x-h)}{h^2}[/tex]

    any help?
     
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  3. Jul 11, 2008 #2

    Kurdt

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    If you remember from classical mechanics, the equations of motion formed from the Lagrangian will always give you [itex]n[/itex] second order differential equations for a system with [itex]n[/itex] degrees of freedom.

    [tex]\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}_i}\right) = \frac{\partial L}{\partial q_i}[/tex]
     
  4. Jul 11, 2008 #3
    I agree with you on that however the second order only appears on the

    [tex] \frac{d}{dt}(\frac{\partial L}{\partial \dot{q_i}}) [/tex]

    the potential term is only first order so how do I evaluate

    [tex] \frac{\partial L}{\partial q_i}[/tex] to give

    [tex]
    \frac{\partial}{\partial q_i}(-\frac{k}{2}(q_{j}-q_{j+1})^{2}) = k(q_{j+1}-2q_j + q_{j-1})
    [/tex]
     
  5. Jul 11, 2008 #4

    George Jones

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    What do you get for

    [tex] \frac{\partial L}{\partial q_i}[/tex]

    when

    [tex]
    L = \sum_j \left[\frac{m}{2}\dot{q}_{j}^{2}-\frac{k}{2} \left(q_{j}-q_{j+1} \right)^{2} \right] ?
    [/tex]
     
  6. Jul 11, 2008 #5
    i dont know thats exactly what I'm asking...

    [tex]

    \frac{\partial L}{\partial q_j} = -k(q_j-q_{j+1}) (1-\frac{\partial}{\partial q_j}q_{j+1})

    [/tex]
     
    Last edited by a moderator: Jul 11, 2008
  7. Jul 11, 2008 #6
    There are two terms that have [tex]q_j[/tex] in them because of the sum. You are taking the derivative of:
    [tex]\frac{\partial}{\partial q_i}(-\frac{k}{2}(q_{j}-q_{j+1})^{2}-\frac{k}{2}(q_{j-1}-q_{j})^{2})\equiv -k(q_{j}-q_{j+1})+k(q_{j-1}-q_{j}) \equiv k(q_{j-1}-2q_{j}+q_{j+1})[/tex].
    I hope that helps
     
  8. Jul 11, 2008 #7

    George Jones

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    Careful; in my post the derivative was with respect to [itex]i[/itex] and the sum is over [itex]j[/itex]. Let me expand on what badphysicist wrote. Use

    [tex]\frac{\partial q_j}{\partial q_i} = \delta_{ij}.[/tex]

    For example, [itex] \partial x / \partial x = 1[/itex] and [itex] \partial y / \partial x = 0.[/itex]

    After differentiating, there is still a sum over [itex]j[/itex], i.e.,

    [tex]\frac{\partial}{\partial q_i} \sum_j = \sum_j \frac{\partial}{\partial q_i},[/tex]

    but, because of the [itex]\delta[/itex]'s, all but a few terms will be zero.
     
  9. Jul 11, 2008 #8
    Oh yeah, I see the typo I made. Here's the corrected version..
    [tex]\frac{\partial}{\partial q_i}(-\frac{k}{2}(q_{i}-q_{i+1})^{2}-\frac{k}{2}(q_{i-1}-q_{i})^{2})\equiv -k(q_{i}-q_{i+1})+k(q_{i-1}-q_{j}) \equiv k(q_{i-1}-2q_{i}+q_{i+1})[/tex]
     
  10. Jul 12, 2008 #9
    thanks that definitely helps I didn't think of the sum thanks again
     
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