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I Covariance of Euler-Lagrange equations for fields

  1. Aug 4, 2017 #1
    [tex]
    \newcommand{\dep}[1]{\partial_{#1}}
    \newcommand{\parcial}[2]{\frac{\partial{#1}}{\partial{#2}}}
    \renewcommand{\d}{\text{d}}
    \newcommand{\ddt}{\frac{\text{d}}{\text{d}t}}
    \newcommand{\ppartial}[3]{\frac{\partial^2{#1}}{\partial{#2}\partial{#2}}}
    [/tex]

    I haven't found this problem solved around maybe because it's a simple thing that I'm not yet able to see.
    I have studied a bit of calculus of variations and I have succesfully derived Euler equations for the case of particles (i.e. for curves [itex]q: [ a , b ] \rightarrow \mathbb{R}^n[/itex]) when I have a lagrangian and action of the form:

    [tex]
    S[q] = \int_a^b L(t,q,\dot{q}) \d t
    [/tex]


    This yields the classical Euler-Lagrange equations


    [tex]
    0 = \parcial{L}{q^k}-\ddt\parcial{L}{\dot{q}^k}
    [/tex]



    I have been able to show that these equations share the same form for a change of coordinates in configuration space [itex]q^\mu \rightarrow x^\mu[/itex] which means that if E-L equations for [itex]q[/itex] holds then:

    [tex]
    0 = \parcial{L}{x^k} - \ddt \parcial{L}{\dot{x}^k}
    [/tex]
    This also happens if [itex]t[/itex] is involved in the coordinate transformation. Now I want to check that this covariance holds also for Euler Lagrange equations of fields, in which case we have:

    • [itex]q^\mu, x^\mu[/itex] are now space-time coordinates
    • [itex]\phi[/itex] is the scalar field involved

    Now the action over the field is

    [tex]
    S[\phi] = \int_\Omega \mathcal{L}(x,\phi,\dep{\mu}\phi) \d^4 x
    [/tex]

    The derived Euler Lagrange equations are:


    [tex]
    0 = \parcial{\mathcal{L}}{\phi} - \dep{\mu}\left(\parcial{\mathcal{L}}{(\dep{\mu}\phi)}\right)
    [/tex]
    For convenience I shall write out explicitly partial derivatives

    [tex]
    0 = \parcial{\mathcal{L}}{\phi} - \parcial{}{x^\mu} \left( \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)} \right)
    [/tex]

    Now consider a space-time transformation [itex]x^\mu \rightarrow q^\mu[/itex] which in principle is not Lorentz transformation. The transformation of the derivatives of the field follow simply from the chain rule:
    [tex]
    \parcial{\phi}{x^\mu} = \parcial{\phi}{q^\nu} \parcial{q^\nu}{x^\mu}
    [/tex]
    Hence:
    [tex]
    \parcial{}{\left(\parcial{\phi}{q^\sigma}\right)}\left(\mathcal{L}(\phi,\parcial{\phi}{x^\mu}) \right)
    =
    \parcial{}{\left( \parcial{\phi}{q^\sigma}\right)}
    \left(\mathcal{L}(\phi,\parcial{\phi}{q^\nu}\parcial{q^\nu}{x^\mu}) \right)
    =
    \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)} \parcial{}{\left(\parcial{\phi}{q^\sigma}\right)}
    \left(
    \parcial{\phi}{q^\nu}\parcial{q^\nu}{x^\mu}
    \right)
    =
    \parcial{\mathcal{L}}{\left(\parcial{\phi}{x^\mu}\right)} \parcial{q^\sigma}{x^\mu}
    [/tex]
    This yields:
    [tex]
    \begin{aligned}
    \parcial{}{q^\sigma}
    \left(
    \parcial{\mathcal{L}}{\left( \parcial{\phi}{q^\sigma}\right)}
    \right)
    && = &&
    \parcial{}{q^\sigma}
    \left(
    \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)}
    \right)
    \parcial{q^\sigma}{x^\mu}
    +
    \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)}
    \parcial{}{x^\alpha}
    \left(
    \parcial{q^\sigma}{x^\mu}
    \right)
    \parcial{x^\alpha}{q^\sigma}\\
    && = &&
    \parcial{}{x^\alpha}
    \left(
    \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)}
    \right)
    \parcial{q^\sigma}{x^\mu}
    \parcial{x^\alpha}{q^\sigma}
    +
    \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)}
    \ppartial{q^\sigma}{x^\alpha}{x^\mu}
    \parcial{x^\alpha}{q^\sigma}\\
    && = &&
    \parcial{}{x^\alpha}
    \left(
    \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)}
    \right)
    \delta_\mu^\alpha
    +
    \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)}
    \ppartial{q^\sigma}{x^\alpha}{x^\mu}
    \parcial{x^\alpha}{q^\sigma}\\
    \end{aligned}
    [/tex]
    This and the fact that [itex]\parcial{\mathcal{L}}{\phi}[/itex] is independent of coordinates yields that the latter extra term in Euler Lagrange equations for [itex]q[/itex] coordinates. Moreover, I tend not to trust second derivatives expressions of the coordinates to be covariant from differential geometry (I may be wrong for a general case though). Any ideas on how to continue from this point or possible mistakes?

    One possible thing I could be stepping on is the fact that lagrangian is a density and maybe I have to consider the jacobian of the transformation as a weight. However this seems a very big complication to treat a 4x4 determinant expression and could not be generalized (or at least have an equal treatment) for higher dimensions.
     
  2. jcsd
  3. Aug 9, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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