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Signs in the Field-Theoretic Euler-Lagrange Equation

  1. Mar 8, 2014 #1
    So I have this book that considers the problem of a flexible vibrating string, taking [itex]\phi(x,t)[/itex] as the string's displacement from equilibrium. It then writes a Lagrangian density in terms of this [itex]\phi[/itex], takes [itex]\delta \mathcal{S} = 0[/itex], and eventually concludes that [itex]\frac{\partial}{\partial t}(\frac{\partial \mathcal{L}}{\partial \dot{\phi}}) + \frac{\partial}{\partial x}(\frac{\partial \mathcal{L}}{\partial \phi'}) = 0[/itex]. Notice that the time-varying and space-varying terms have the same sign.

    Two pages later, it considers a scalar field [itex]\phi(x^0,\mathbf{x})[/itex] with a Lagrangian density [itex]\mathcal{L}=\mathcal{L}(\phi,\partial_\mu\phi)[/itex], and concludes that [itex]\frac{\partial\mathcal{L}}{\partial\phi}-\partial_\mu(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)})=0[/itex]. Now, unless I am having some massive brain fart on how covariant and contravariant work, the time-varying and space-varying terms have opposite signs. Right?

    What gives? Why are the signs different between these two situations?
     
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  3. Mar 8, 2014 #2

    vanhees71

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    In
    [tex]\partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}[/tex]
    the temporal and spatial components have the same sign. How do you come to the conclusion that might be not so?

    The same is true for the four-dimensional divergence.
    [tex]\partial_{\mu} A^{\mu}=\frac{\partial A^{\mu}}{\partial x^{\mu}}=\frac{\partial A^0}{\partial t} + \vec{\nabla} \cdot \vec{A},[/tex]
    where I've used natural units, [itex]c=1[/itex], and [itex]x^0=t[/itex].
     
  4. Mar 8, 2014 #3
    Thanks. It's clearly been too long since I've done anything with covariant and contravariant vectors. I need to go back and refresh before jumping back into this stuff.
     
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