Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hilbert-Einstein action, total derivative vanish

  1. Mar 30, 2015 #1
    I'm looking at the deriviation of Einstein's equation via applying the principle of least action to the Hilbert-Einstein action.

    I'm trying to understand the vanishing of a term because it is a total derivative: http://www.tapir.caltech.edu/~chirata/ph236/2011-12/lec33.pdf, equation 19.

    My question is that, I've seen some sources that explain it via the generalized Stoke's theorem in differential forms which is a generalization of Gauss theorem, fundamental theorem of calculus etc.

    I'm wondering if, once the identity in equation 19 has been made, a covariant derivative expression being expressed as a solely partial derivative expression, can this term vanishing be justified via Gauss's theorem which states that the divergence of a vector field integrated over a volume is equal to the vector integrated over the corresponding surface of the volume? If the vector fields goes to zero fast enough at infinity.

    I'm not looking for rigor, just an argument that merely suffices.

    Thanks your help is greatly appreciated.
  2. jcsd
  3. Mar 30, 2015 #2


    User Avatar
    Science Advisor

    Yes, I don't see a question here
    [tex]\int_{D} d^{4} x \ \sqrt{-g} \ \nabla_{\mu} V^{\mu} = \int_{D} d^{4} x \ \partial_{\mu} ( \sqrt{-g} V^{\mu} ) = \int_{\partial D} d \sigma_{\mu} \ \sqrt{-g} \ V^{\mu} \left( = \int d S \ n_{\mu} V^{\mu} \ \sqrt{-g} = 0 \right),[/tex] if the vector field tends to zero sufficiently fast at infinity.
  4. Mar 31, 2015 #3
    Thanks. And many sources say that this term goes to zero if the variation of the metric ##\delta g^{uv}## vanishes at infinity.
    I don't understand how this relates?
  5. Mar 31, 2015 #4


    User Avatar
    Science Advisor

    I don’t know which term that is. If you are talking about the variation of the Einstein-Hilbert action, then I believe you encounter such thing [tex]g^{\mu \nu} \delta R_{\mu \nu} = \nabla_{\sigma} ( g^{\mu \sigma} \delta \Gamma^{\rho}_{\mu \rho} - g^{\mu \nu} \delta \Gamma^{\sigma}_{\mu \nu} ) .[/tex] So, the vector field in this context is [tex]V^{\sigma} = g^{\mu \sigma} \delta \Gamma^{\rho}_{\mu \rho} - g^{\mu \nu} \delta \Gamma^{\sigma}_{\mu \nu} .[/tex] Therefore, the identity [tex]\sqrt{-g} \nabla_{\sigma} V^{\sigma} = \partial_{\sigma} ( \sqrt{-g} \ V^{\sigma} ) ,[/tex] allows you to write [tex]\int d^{4} x \ \sqrt{-g} \ g^{\mu \nu} \ \delta R_{\mu \nu} = \int d^{4} x \ \partial_{\sigma} ( \sqrt{-g} \ g^{\mu \sigma} \delta \Gamma^{\rho}_{\mu \rho} - \sqrt{-g} \ g^{\mu \nu} \delta \Gamma^{\sigma}_{\mu \nu} ) = 0 .[/tex]
  6. Apr 1, 2015 #5
    And this requires ##V^{\sigma} = g^{\mu \sigma} \delta \Gamma^{\rho}_{\mu \rho} - g^{\mu \nu} \delta \Gamma^{\sigma}_{\mu \nu} \to 0 ## fast enough, So I think thats that all tensors in this expression must ##\to 0 ## fast enough, but the variation of the metric tensor vanishing here plays no role in ##V^{\sigma} \to 0## does it?

    Wikipedia looks like it has the same working as above, but makes the final comment on the variation of the metric vanishing which I don't understand : http://en.wikipedia.org/wiki/Einstein–Hilbert_action , under Variation of the Riemann tensor, the Ricci tensor, and the Ricci scalar, bottom two lines.

  7. Apr 1, 2015 #6


    User Avatar
    Science Advisor

    Yes it does. The variation of the connection, [itex]\delta \Gamma[/itex], is brought about by the variation of the metric tensor, i.e. [itex]\delta \Gamma \propto \delta g[/itex].
  8. Apr 2, 2015 #7
    Ah okay thanks. I thought they may not because of palatini's formulation, where the variations with respect to the metric and connection are taken seperately.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Hilbert-Einstein action, total derivative vanish
  1. Einstein Hilbert Action (Replies: 13)