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Obtaining Euclidean action from Minkowski action

  1. Jan 15, 2016 #1
    The Euclidean classical action ##S_{\text{cl}}[\phi]## for a scalar field ##\phi## is given by

    \begin{equation}
    S_{\text{cl}}[\phi]=\int d^{4}x\ \bigg(\frac{1}{2}(\partial_{\mu}\phi)^{2}+U(\phi)\bigg).
    \end{equation}

    This can be obtained from the action ##S_{\text{Mn}}[\phi]## in Minkowski space for the same scalar field ##\phi## as given by

    \begin{equation}
    S_{\text{Mn}}[\phi]=\int d^{4}x\ \bigg(\frac{1}{2}(\partial_{\mu}\phi)^{2}-U(\phi)\bigg).
    \end{equation}

    To transform the action ##S_{\text{Mn}}[\phi]## in Minkowski space to the Euclidean classical action ##S_{\text{cl}}[\phi]##, one needs to perform a Wick rotation. I need someone to show the derivation.
     
  2. jcsd
  3. Jan 15, 2016 #2

    samalkhaiat

    User Avatar
    Science Advisor

    Continue from real [itex]x^{0}[/itex] to [itex]x^{0} = - i x_{4}[/itex], and from [itex]S_{Min}[/itex] so obtained get [itex]S_{Euc}[/itex] using [itex]S_{Euc} = - i S_{Min}[/itex]:
    [tex]S_{M} = \int dx^0 \ d^3x \ \left( (1/2) (\frac{\partial \phi}{\partial x^{0}})^{2} - (1/2) (\nabla \phi)^{2} - U(\phi) \right) [/tex]
    [tex]S_{M} = -i \int dx_{4} \ d^3x \ \left( - (1/2) (\frac{\partial \phi}{\partial x_{4}})^{2} - (1/2) (\nabla \phi)^{2} - U(\phi) \right) [/tex]
    [tex]S_{M} = i \int dx_{4} \ d^3x \ \left( (1/2) (\frac{\partial \phi}{\partial x_{4}})^{2} + (1/2) (\nabla \phi)^{2} + U(\phi) \right) = i S_{E}[/tex]
    [tex]S_{E} = \int d^{4}x_{E} \ \left( (1/2) (\frac{\partial \phi}{\partial x_{\mu}})^{2} + U(\phi) \right)[/tex]
     
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