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Palatini f(R) gravity and the variation

  1. Jan 16, 2013 #1
    Hi friends,

    going through Palatini gravity, I cannot do the variation for palatini f(R) gravity and get to the famous equation (Tsujikawa dark energy book equation 9.6):
    [tex]R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R =\frac{\kappa^2 T_{\mu\nu}}{F} - \frac{FR-f}{2F}g_{\mu\nu} + \frac{1}{F}(\nabla_\mu \nabla_\nu F - g_{\mu\nu} \Box F)- \frac{3}{2F^2}(\partial_\mu F\partial_\nu F - \frac{1}{2}g_{\mu\nu} (\nabla F)^2)[/tex]

    I tried but it does not work!
     
    Last edited by a moderator: Jan 17, 2013
  2. jcsd
  3. Jan 16, 2013 #2

    bcrowell

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    If you want your math to show up correctly, you need to surround it with itex tags, like this: [itex]a^2+b^2=c^2[/itex]. Click on the QUOTE button in my post to see how I did that.
     
  4. Jan 16, 2013 #3
    Thanks!
     
  5. Jan 16, 2013 #4

    bcrowell

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    If you edit your #1, we'll be able to read it.
     
  6. Jan 17, 2013 #5
    Hi friends,

    going through Palatini gravity, I cannot do the variation for palatini f(R) gravity and get to the famous equation (Tsujikawa dark energy book equation 9.6):

    R[itex]_{\mu\nu}[/itex]-[itex]\frac{1}{2}[/itex] g[itex]_{\mu\nu}[/itex] =
    [itex]\frac{\kappa^{2} T_{\mu\nu}}{F}[/itex] - [itex]\frac{F R -f}{2F}[/itex] g[itex]_{\mu\nu}[/itex] + [itex]\frac{1}{F}[/itex] ([itex]\nabla[/itex] [itex]_{\mu}[/itex] [itex]\nabla[/itex] [itex]_{\nu}[/itex] F - g[itex]_{\mu\nu}[/itex] d'lambert F) - [itex]\frac{3}{2 F ^{2}}[/itex] [ [itex]\partial[/itex] [itex]_{\mu}[/itex] F [itex]\partial[/itex][itex]_{\nu}[/itex] F - [itex]\frac{1}{2}[/itex] g[itex]_{\mu\nu}[/itex] ( [itex]\nabla[/itex] F)[itex]^{2}[/itex]]


    I tried but it does not work!
     
  7. Jan 17, 2013 #6

    Mentz114

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    [tex]R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R =\frac{\kappa^2 T_{\mu\nu}}{F} - \frac{FR-f}{2F}g_{\mu\nu} + \frac{1}{F}(\nabla_\mu \nabla_\nu F - g_{\mu\nu} \Box F)- \frac{3}{2F^2}(\partial_\mu F\partial_\nu F - \frac{1}{2}g_{\mu\nu} (\nabla F)^2)[/tex]
     
  8. Jan 17, 2013 #7
    Yes this is the exact equation.
    But I do not know how they reach to this by combining
    [itex]\nabla_{\lambda}[/itex] ( [itex]\sqrt{-g}[/itex] G g[itex]^{\mu\nu}[/itex])=0

    and

    F R[itex]_{\mu\nu}[/itex] - [itex]\frac{1}{2}[/itex] f g[itex]_{\mu\nu}[/itex] = [itex]\kappa[/itex] [itex]^{2}[/itex] T [itex]_{\mu\nu}[/itex]

    !!!
     
  9. Jan 17, 2013 #8
    It is interesting to mention that there are actually two different R in this equation.
    The first one on the left hand side is R(g) and the second one in the right hand side is R(T)
     
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