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Einstein gauss bonnet gravity

  1. Aug 28, 2011 #1
    Hello, in a paper http://www-library.desy.de/preparch/conf/theo-ws/workshop2004/data/Chatillon.pdf .

    says that.

    1) In 4d, it is a total derivative, then does not contribute to the equations of motion

    A total derivative respect to time does not contribute to equation of motion???, ¿or other parameter??

    2) Only [tex]R^2[/tex] order combination giving equations of motion
    with no derivatives of higher order than two and divergence free, like the Einstein tensor.

    ¿why einstein tensor gives equation of motion of second derivative of time?...I think that varying the action [tex] \int \sqrt{g_{uv}dx^u dx^v} [/tex] i get the equation of motion, but i don't understand but the einstein tensor too...
     
  2. jcsd
  3. Aug 28, 2011 #2

    Bill_K

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    alejandrito29, One way to derive the Einstein equations is to take as the action S = ∫√-g R, and vary this with respect to gμν. It turns out that δS/δgμν = √-g Gμν. It's generally true that for any action of the form S = ∫√-g L, where L is a scalar, that δS/δgμν will be a divergence-free tensor.

    The paper considers adding to the Einstein action a term α(R2 − 4RμνRμν +RμνρσRμνρσ)

    In four dimensions the variation of this additional term vanishes identically, so it makes no contribution to the equations of motion.
     
  4. Aug 28, 2011 #3
    very thank , but,
    1. ¿a divergence-free tensor does not contribute to the equations of motion?

    2. ¿the einstein tensor then does not contribute to the equations of motion???. I understand by Einstein equation [tex]G_{uv}= R_{uv}-\frac{1}{2}g_{uv}R=k T_{uv}[/tex] but by equation of motion [tex] \ddot{x} + \Gamma^v_{su} \dot{x}^s \dot{x}^u=0[/tex]
     
    Last edited: Aug 28, 2011
  5. Aug 29, 2011 #4

    Bill_K

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    ¡I didn't say that! A term T in the Lagrangian which can be written as a total divergence, T = Vμ, does not contribute to the equations of motion. Lagrangians are not unique. Two Lagrangians that differ by a total divergence will yield exactly the same equations of motion.
    You're confusing two things that have nothing to do with each other. S = ∫√-g R d4x is the action for the field, and its variation δS/δgμν yields the Einstein equations. ∫√gμνdxμdxν is a Lagrangian that can be used to describe geodesics, and its variation yields the equations of motion for a test particle.
     
  6. Aug 29, 2011 #5
    ok, very very thanks, but
    what is the equation of motion in general relativity? [tex] G_{uv}= k T_{uv}[/tex] or [tex]\ddot{x}+ \Gamma^u_{s v} \dot{x}^s \dot{x}^v[/tex] ???,

    or other??
     
  7. Aug 29, 2011 #6
    The first proposition you give is the Einstein's field equation.
    The second proposition, if vanishing, is the equation describing a parallel transported speed vector.
    Solutions of the field equation is of the following type:
    gab. va. vb = constant where v is the speed of the particle and where a, b = 0, 1, 2 and 3.
    If you prefer a formulation like in classical mechanics, with forces, then you have to write the equation Dv = forces to recover the solutions.
     
  8. Aug 29, 2011 #7

    WannabeNewton

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    If your lagrangian is of the form [tex]L = \frac{1}{2}g_{\mu \nu }\dot{x^{\mu }}\dot{x^{\nu }}[/tex] then you can use the euler lagrange equations [tex]\frac{\partial L}{\partial x^{\mu }} - \frac{\mathrm{d} }{\mathrm{d} \lambda }(\frac{\partial L}{\partial \dot{x^{\mu }}}) = 0[/tex] (where lambda is an affine parameter) to arrive at the geodesic equation [tex]\ddot{x}^{\mu } + \Gamma ^{\mu }_{\alpha \beta }\dot{x^{\alpha }}\dot{x^{\beta }} = 0[/tex] If you use the Einstein lagrangian [itex]L = (-g)^{1/2}R[/itex] then variation of the lagrangian will yield the vacuum field equations [itex]G^{\mu \nu } = 0[/itex] and the related bianchi identity [itex]\triangledown _{\mu }G^{\mu \nu } = 0[/itex].
     
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