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EL Equations for the modified electromagnetic field Lagrangian

  1. Apr 16, 2012 #1
    Hi,

    I'm trying to work through something and it should be quite simple but somehow i've gotten a bit confused.

    I've worked through the Euler Lagrange equations for the lagrangian:

    [tex]
    \begin{align*}
    \mathcal{L}_{0} &= -\frac{1}{4}(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu}) \\
    &= \frac{1}{4}F_{\mu\nu}F^{\mu\nu}
    \end{align*}
    [/tex]

    getting:

    [tex]\Box A_{\nu} - \partial^{\nu}\partial_{\mu}A^{\mu} = 0[/tex]

    I'm ok with this.

    Then considering the modified lagrangian:

    [tex]\mathcal{L}_{\xi} = \mathcal{L}_{0} + \frac{\lambda}{2}(\partial_{\sigma}A^{\sigma})^2[/tex]

    I'm trying to work out the EL equation components and as part of one of these calculations, I've to calculate:

    [tex]
    \begin{align*}
    \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} \left[ \frac{\lambda}{2} (\partial_{\sigma}A^{\sigma})^2 \right]

    &= \frac{\lambda}{2} \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} \left[ ( \partial_{\sigma}A^{\sigma} ) ( \partial_{\rho}A^{\rho} ) \right] \\

    &= \frac{\lambda}{2} \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} \left[ ( \partial_{\sigma}A_{\alpha} \eta^{\sigma \alpha} ) ( \partial_{\rho}A_{\beta} \eta^{\rho \beta} ) \right] \\

    & = \frac{\lambda}{2} \eta^{\sigma \alpha} \eta^{\rho \beta} \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} \left[ ( \partial_{\sigma}A_{\alpha} ) ( \partial_{\rho}A_{\beta} ) \right] \\

    & = \frac{\lambda}{2} \eta^{\sigma \alpha} \eta^{\rho \beta} \left[ ( \partial_{\sigma}A_{\alpha} ) \left( \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} ( \partial_{\rho}A_{\beta} ) \right) + \left( \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} ( \partial_{\sigma}A_{\alpha} ) \right) ( \partial_{\rho}A_{\beta} ) \right] \\

    & = \frac{\lambda}{2} \eta^{\sigma \alpha} \eta^{\rho \beta} \left[ ( \partial_{\sigma}A_{\alpha} ) \delta^{\mu}_{\rho} \delta^{\nu}_{\beta} + \delta^{\mu}_{\sigma} \delta^{\nu}_{\alpha} ( \partial_{\rho}A_{\beta} ) \right] \\

    & = \frac{\lambda}{2} \eta^{\sigma \alpha} \eta^{\rho \beta} ( \partial_{\sigma}A_{\alpha} ) \delta^{\mu}_{\rho} \delta^{\nu}_{\beta}
    +
    \frac{\lambda}{2} \eta^{\sigma \alpha} \eta^{\rho \beta} \delta^{\mu}_{\sigma} \delta^{\nu}_{\alpha} ( \partial_{\rho}A_{\beta} ) \\

    & = \frac{\lambda}{2} \eta^{\sigma \alpha} \eta^{\mu \nu} ( \partial_{\sigma}A_{\alpha} )
    +
    \frac{\lambda}{2} \eta^{\mu \nu} \eta^{\rho \beta} ( \partial_{\rho}A_{\beta} ) \\

    & = \frac{\lambda}{2} \eta^{\mu \nu} \left[ ( \partial_{\sigma}A^{\sigma} )
    +
    ( \partial_{\rho}A^{\rho} ) \right] \\

    & = \lambda \eta^{\mu \nu} ( \partial_{\sigma}A^{\sigma} ) \\

    \end{align*}
    [/tex]

    Now I was hoping to get:

    [tex]
    \lambda \partial^{\nu} A^{\mu}
    [/tex]

    as ultimately I need the EL equations to give me:

    [tex]
    \begin{align*}
    \frac{\partial \mathcal{L}_{\xi}}{\partial A_{\nu}} - \partial_{\mu} \left( \frac{\partial \mathcal{L}_{\xi}}{\partial (\partial_{\mu} A_{\nu})} \right)
    &=\Box A^{\nu} - \partial^{\nu} ( \partial_{\mu} A^{\mu} ) - \lambda \partial^{\nu}(\partial_{\mu} A^{\mu}) \\
    &= \Box A^{\nu} - ( 1 + \lambda ) \partial^{\nu} ( \partial_{\mu} A^{\mu} ) \\
    &= 0
    \end{align*}
    [/tex]

    Can anyone show me where i've gone wrong? I didn't stick this in the homework section as it's not homework. I'm just trying to work through the through missing steps from the text i'm reading.

    Thanks in advance
     
    Last edited: Apr 16, 2012
  2. jcsd
  3. Apr 16, 2012 #2
    edit... now i'm finished stating the question.
     
    Last edited: Apr 16, 2012
  4. Apr 16, 2012 #3

    Bill_K

    User Avatar
    Science Advisor

    Irrational, You've got exactly what you need. :smile: Well, up to a minus sign anyway. You've got λ ημν(∂σAσ). So plug this into the Euler-Lagrange equation:
    -∂μ(λ ημν(∂σAσ)) = -λ ∂ν(∂σAσ) = -λ ∂ν(∂μAμ)
     
  5. Apr 16, 2012 #4
    you have no idea how thick i feel right now. thanks for filling in the gap.
     
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