# Palatini f(R) gravity and the variation

In summary, the equation states that there is a gravity force between two masses, where the first mass is G heavier than the second.
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!

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bcrowell said:
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Thanks!

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!

$$R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R =\frac{\kappa^2 T_{\mu\nu}}{F} - \frac{FR-f}{2F}g_{\mu\nu} + \frac{1}{F}(\nabla_\mu \nabla_\nu F - g_{\mu\nu} \Box F)- \frac{3}{2F^2}(\partial_\mu F\partial_\nu F - \frac{1}{2}g_{\mu\nu} (\nabla F)^2)$$

Mentz114 said:
$$R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R =\frac{\kappa^2 T_{\mu\nu}}{F} - \frac{FR-f}{2F}g_{\mu\nu} + \frac{1}{F}(\nabla_\mu \nabla_\nu F - g_{\mu\nu} \Box F)- \frac{3}{2F^2}(\partial_\mu F\partial_\nu F - \frac{1}{2}g_{\mu\nu} (\nabla F)^2)$$

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 $_{\mu\nu}$

!

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)

## What is Palatini f(R) gravity and how does it differ from standard general relativity?

Palatini f(R) gravity is a modified version of general relativity that includes an additional term in the Einstein-Hilbert action. In standard general relativity, the metric and the connection are treated as independent variables, while in Palatini f(R) gravity, the connection is treated as an independent variable and the metric is derived from it through the use of the Palatini variational principle.

## What is the significance of the "f(R)" term in Palatini f(R) gravity?

The "f(R)" term is a function that represents the extra curvature term added to the Einstein-Hilbert action in Palatini f(R) gravity. This term allows for modifications to the standard general relativity equations and can lead to different predictions for the behavior of gravity on large scales.

## How does Palatini f(R) gravity explain the observed acceleration of the expansion of the universe?

One possible explanation for the observed acceleration of the expansion of the universe is the existence of dark energy, which is represented by the "f(R)" term in Palatini f(R) gravity. This extra curvature term can lead to a repulsive force that counteracts the attractive force of gravity, causing the expansion of the universe to accelerate.

## What are some of the potential implications and consequences of Palatini f(R) gravity?

Some potential implications and consequences of Palatini f(R) gravity include the possibility of explaining the observed acceleration of the expansion of the universe without the need for dark energy, the potential to resolve the singularity problem in the center of black holes, and the ability to make predictions for the behavior of gravity on large scales that differ from those of standard general relativity.

## How does the variation of the "f(R)" function affect the behavior of Palatini f(R) gravity?

Varying the "f(R)" function can lead to different predictions for the behavior of Palatini f(R) gravity. For example, different choices of the "f(R)" function can lead to different cosmological solutions, such as accelerated expansion or the avoidance of a big bang singularity. The specific behavior of Palatini f(R) gravity will depend on the specific form of the "f(R)" function used.

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