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A About the least action principle

  1. Apr 20, 2016 #1
    hello, everyone. When a vector field ##A_{\mu}## has the Lagrangian of the form as
    ##L=Const.{\times}F^{\mu\nu}F_{\mu\nu}##, where
    [itex]F_{\mu\nu}=({\partial}_{\mu}-{\alpha}{\partial}^{\rho}{\partial}_{\rho}{\partial}_{\mu})A_{\nu}-({\partial}_{\nu}-{\alpha}{\partial}^{\rho}{\partial}_{\rho}{\partial}_{\nu})A_{\mu}[/itex]. Now I will apply the least action principle to it. Which one of the following two choices is the right one?
    [itex]{\delta}S=\frac{{\partial}S}{{\partial}A_{\mu}}{\delta}A_{\mu}+\frac{{\partial}S}{{\partial}({\partial}_{\mu}A_{\nu})}{\delta}({\partial}_{\mu}A_{\nu})=0,[/itex]
    or
    [itex]{\delta}S=\frac{{\partial}S}{{\partial}A_{\mu}}{\delta}A_{\mu}+\frac{{\partial}S}{{\partial}({\partial}_{\mu}A_{\nu})}{\delta}({\partial}_{\mu}A_{\nu})+\frac{{\partial}S}{{\partial}({\partial}_{\mu}{\partial}_{\nu}{\partial}_{\rho}A_{\lambda})}{\delta}({\partial}_{\mu}{\partial}_{\nu}{\partial}_{\rho}A_{\lambda})=0.[/itex]
     
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  3. Apr 21, 2016 #2

    Orodruin

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    Which do you think and why?
     
  4. Apr 21, 2016 #3

    haushofer

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    What is that alpha in your field strength? I'd say F=dA.

    You can include higher derivatives in your functional derivative; if you use partial integrations, you'll see that every higher order derivative swaps a sign.
     
  5. May 16, 2016 #4
    Thanks for your attention. I was reading a paper on a model of generalized uncertainty principle which originated from quantum gravity effect. ##\alpha## represents the parameter of quantum gravity effect.
     
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