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

Gauss-Bonnet term extrinsic curvature

  1. Jul 16, 2012 #1
    Gauss-Bonnet term extrinsic curvature calculations?

    In General Relativity if one wants to calculate the field equation with surface term, must use this equation:
    S=[itex]\frac{1}{16\pi G}[/itex][itex]\int\sqrt{-g} R d^{4} x[/itex]+[itex]\frac{1}{8\pi G}[/itex][itex]\int\sqrt{-h} K d^{3} x[/itex]
    The second term is so-called Gibbons-Hawking boundary term and K is extrinsic curvature.
    If one is about to use another Lagrangian, for instance Gauss-Bonnet term, must calculate the new extrinsic curvature, K, associated with this new Lagrangian.
    I want to know is there a standard method for calculating K?
    I would be grateful if anybody can help me in learning this procedure. Please introduce references if you know some.
     
    Last edited: Jul 16, 2012
  2. jcsd
  3. Jul 16, 2012 #2

    Bill_K

    User Avatar
    Science Advisor

    According to Wikipedia, what Gibbons-Hawking-York call "extrinsic curvature K" is the trace of the second fundmental form. Nonstandard terminology! But if this is correct, I can answer your question.

    To describe a surface embedded in a larger manifold you need two sets of coordinates, hence two types of indices. Let xi be the coordinates in the outer manifold and uα coordinates in the submanifold, the surface. A basic quantity that transforms between them is

    xiα = ∂xi/∂uα

    This is a hybrid quantity, sometimes called a bitensor, with both kinds of indices present. The covariant derivative of such an object requires both kinds of Christoffel symbols,

    xiα;β = ∂2xi/∂uα∂uβ + (outer)Γijkxjxk - (inner)Γδαβxiδ

    It's easy to convince oneself that xiα;β represents a set of vectors orthonormal to the surface. Hence there must exist a surface tensor bαβ such that

    xiα;β = bαβni

    where ni is the normal vector. bαβ is called the second fundamental form.

    The curvature you are after is the trace of bαβ.
     
  4. Jul 16, 2012 #3

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    The second fundamental form was defined to me as the vector-valued one-form (on the tangent bundle to the hypersurface) [itex]\nabla n[/itex] where n is the normal vector field to the hypersurface in question, and the nabla operator is the covariant derivative operator on your manifold (restricted to directions in the hypersurface obviously). In this way, the second fundamental form (which is also sometimes called the extrinsic curvature!) measures the amount by which the normal field varies from point to point on the hypersurface.

    Alternatively, the second fundamental form can be found, after some calculation, to be:
    [tex]K=\frac{1}{2} L_n g'[/tex]

    Where L is the Lie derivative, and g' is the induced 3-metric on the hypersurface.

    This equation is the one found in Wald.
     
  5. Jul 17, 2012 #4

    Bill_K

    User Avatar
    Science Advisor

    Sorry sourena, I think I misunderstood your question! You're not asking, "How do I calculate the extrinsic curvature", you're asking, "What if the Lagrangian is something else besides R, then what will K be?".

    I don't have an immediate answer, but here are some thoughts. For a general Lagrangian L which is a function of fields φ and their derivatives φμ,

    δL = (∂L/∂φ)δφ + (∂L/φμ)δφμ

    When we vary the action W = ∫L d4x,

    δW = ∫δL d4x = ∫((∂L/∂φ)δφ + (∂L/φμ)δφμ)d4x

    we need to integrate the second term by parts:

    δW = ∫((∂L/∂φ - ∂μ(∂L/φμ) )δφ d4x + ∫∂μ((∂L/φμ)δφ)d4x

    and write the total divergence as a surface term:

    ∫((∂L/φμ)δφ) d∑μ

    What you want then is to find a K that produces this, i.e. such that

    δK/δφ = (∂L/φμ)δφ nμ

    where nμ is the normal to the surface.

    Is this getting warm, do you think?

    (For relativity, L = R and φ = gμν)
     
    Last edited: Jul 17, 2012
  6. Jul 18, 2012 #5
    Dear Bill_K and Matterwave

    First of all, thank you so much for your time and attention. I value it a great deal.

    Yes Bill_k, this is what I want to do. To be more precise I'm going to explain what exactly I want:

    Consider the action

    SM=(1/2k2)[itex]\int d^{D}x\sqrt{-g}[/itex]{R-2[itex]\Lambda[/itex]+[itex]\alpha[/itex] LGB}

    where LGB=R2-4 RabRab+RabcdRabcd

    Varying the action with respect to metric gives

    [itex]\delta[/itex]SM=(1/2k2)[itex]\int d^{D}x\sqrt{-g}[/itex][itex]\delta[/itex]gab(Gab+[itex]\Lambda[/itex]gab+2[itex]\alpha[/itex]Hab)-(1/k2)[itex]\int d^{D-1}x\sqrt{-h}[/itex]na(ga[cgd]b+2[itex]\alpha[/itex]Pabcd)[itex]\nabla[/itex]d[itex]\delta[/itex]gbc,

    where hab=gab-nanb is induced metric,

    Pabcd=Rabcd+2Rb[cgd]a-2Ra[cgd]b+Rga[cgd]b,

    and Hab=RRab-2RacRcb-2RcdRabcd+RacdeRbcde-(1/4)(R2-4RcdRcd+RcdesRcdes) the Lovelock tensor.

    From the second term in [itex]\delta[/itex]SM one is able to calculate the following action

    S[itex]\Sigma[/itex]=[itex]\int d^{D-1}x\sqrt{-h}[/itex](K+2[itex]\alpha[/itex]{J-2[itex]\hat{G}[/itex]abKab})

    Please look at the following paper, page 2

    http://arxiv.org/abs/hep-th/0208205

    I want to learn the procedure by which I can calculate the extrinsic curvature, J and so forth from the boundary term.

    I hope I could explain what exactly I want.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gauss-Bonnet term extrinsic curvature
  1. Extrinsic curvature (Replies: 5)

Loading...