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Varying The Gibbons-Hawking Term

  1. Jan 3, 2016 #1
    The Gibbons Hawking boundary term is given as ##S_{GHY} = -\frac{1}{8 \pi G} \int_{\partial M} d^dx \sqrt{-\gamma} \Theta##.
    I want to calculate its variation with respect to the induced boundary metric, ##h_{\mu \nu}##.

    The answer (given in eqns 6&7 of http://arxiv.org/pdf/hep-th/9902121v5.pdf) is ##\delta S_{GHY} = \frac{1}{16 \pi G} \int_{\partial M} d^dx \sqrt{-\gamma} (\Theta^{\mu \nu} - \Theta \gamma^{\mu \nu}) \delta \gamma_{\mu \nu}##

    My attempt to obtain this goes as follows:

    [tex]\delta (\sqrt{-\gamma} \Theta) = (\delta \sqrt{-\gamma}) \Theta + \sqrt{-\gamma} \delta \Theta[/tex]
    [tex] =\frac{1}{2} \sqrt{-\gamma} \gamma^{\mu \nu} \delta \gamma_{\mu \nu} \Theta+ \sqrt{-\gamma} \delta \Theta[/tex]
    [tex] = \frac{1}{2} \sqrt{-\gamma} \left( \Theta \gamma^{\mu \nu} + 2 \frac{\delta \Theta}{\delta \gamma_{\mu \nu}} \right) \delta \gamma_{\mu \nu}[/tex]

    I do not understand how to vary the extrinsic curvature, ##\Theta = \gamma_{\mu \nu} \Theta^{\mu \nu} = \gamma_{\mu \nu} \nabla^\mu N^\nu ##. Can anyone help me with this?

    There is a post that explains how to vary the normal vector (http://physics.stackexchange.com/qu...ibbons-hawking-york-boundary-term/10628#10628) that is probably useful although it is varying with respect to the full spacetime metric and not the induced boundary metric but the end result looks the same so I imagine the technique is correct. I just do not understand how it works!

    Thanks.
     
  2. jcsd
  3. Jan 8, 2016 #2
    I would suggest expanding the summation over mu and nu and then applying the variation operator.
     
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