Palatini f(R) gravity and the variation

[shadi_s10]

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_{\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)

I tried but it does not work!

 

[bcrowell]

If you want your math to show up correctly, you need to surround it with itex tags, like this: a^2+b^2=c^2. Click on the QUOTE button in my post to see how I did that.

 

[shadi_s10]

If you want your math to show up correctly, you need to surround it with itex tags, like this: a^2+b^2=c^2. Click on the QUOTE button in my post to see how I did that.

Thanks!

 

[bcrowell]

If you edit your #1, we'll be able to read it.

 

[shadi_s10]

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


I tried but it does not work!

 

[Mentz114]

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)

 

[shadi_s10]

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)

Yes this is the exact equation.
But I do not know how they reach to this by combining
\nabla_{\lambda} ( \sqrt{-g} G g^{\mu\nu})=0

and

F R_{\mu\nu} - \frac{1}{2} f g_{\mu\nu} = \kappa ^{2} T _{\mu\nu}

!

 

[shadi_s10]

Yes this is the exact equation.
But I do not know how they reach to this by combining
\nabla_{\lambda} ( \sqrt{-g} G g^{\mu\nu})=0

and

F R_{\mu\nu} - \frac{1}{2} f g_{\mu\nu} = \kappa ^{2} T _{\mu\nu}

!

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)