# Deriving the realtion between the connection and metric from the palatini formalism

1. May 25, 2012

### luckyboots

Hey all, making my way through Landau and Lifgarbagez classical theory of fields and i had a specific question on the Einstein equations. Following the palatini approach, we assume that the connection and metric are independent variables and are not related a priori. In the footnote, they say by calculating the variation of the gravitational action and requiring that it vanish,
$\delta S_g=\int\left(R_{jk}-\frac{1}{2}g_{jk}R\right)\delta g^{jk}\sqrt{-g}d^4x$
we are supposedly able to determine the relation between the two. However, i am at a loss as to how they make this leap from what is given above.

2. May 26, 2012

### atyy

3. May 31, 2012

### luckyboots

Re: deriving the realtion between the connection and metric from the palatini formali

I have read that article, and others like it but still do not understand the reasoning. I can recover the Einstein equations but the second equation (#3 in the article that was posted $\nabla_\lambda(\sqrt{-g}f'(R)g^{\mu\nu})=0$) , the one obtained by varying with respect to the connect, still eludes me. To be honest i have only a vague idea of what this variational is and do not understand the difference between varying with respect to the metric and the connection.

4. May 31, 2012

### Bill_K

Re: deriving the realtion between the connection and metric from the palatini formali

Is your question about the concepts or about the algebra? The concept, as you already stated, is that you vary L treating gμν and Γσμν as independent variables. What you get is of the form δL = ∫ (Aμν δgμν + Bσμν δΓσμν) dV = 0. This implies that the coefficients of δgμν and δΓσμν must vanish separately. Aμν = 0 is the Einstein equations and Bσμν = 0 gives Γσμν in terms of gμν and its derivatives.

How do you actually do it? For example, the first term in R√-g is gμν√-g Γσμν,σ. Integrate by parts to isolate Γ, getting -(gμν√-g) Γσμν. This is linear in Γ, so the variation of this wrt Γσμν is just the factor in front, namely -(gμν√-g).

Some of the other terms are quadratic in Γ, so when you take the variation of these you'll get a term from each of the two factors of Γ. Like from gμν√-g Γσμν Γτστ you'll get gμν√-g Γτστ from varying the first Γ.

5. May 31, 2012

### julian

Re: deriving the realtion between the connection and metric from the palatini formali

Connections are general objects and are defined in the absense of a metric (refered to as affine connections). A manifold can be given an affine connection and a metric at the same time with their respective geodesics not coinciding. Under certain conditions however the connection is given in terms of the metric (then called the metric connection), more precisely:

"If $\nabla_a$ denotes the covariant derivative with respect to the affine connection, the necessary and sufficient condition for the covariant derivative of the metric to vanish is that the connection is the metric connection."

The geodesics then coincide. Setting the covariant derivative of the metric to zero has the physical meaning that the length of a parallel propagated vector is preserved.

In Ray D'Inverno's book, pages 152-153 he proves that variation with respect to the connection leads to the condition that the $\nabla_c g_{ab} = 0$, identifying the affine connection with the metric connection, and giving results that are equivalent to the equations derived from the, original, purely metric formulation.

In the Palantini approach, variation with respect to the densitized inverse metric give the vacuum field equations, $R_{ab} = 0$ directly, all be it in terms of the affine connection. Basically, varing the connection in order to minamize the action then gives the metric connection. This condition can be substituted into the the field equations, recovering the same answer you get from varying the original metric action - see D'Inverno on this.

Last edited: May 31, 2012
6. May 31, 2012

### julian

Re: deriving the realtion between the connection and metric from the palatini formali

Just noticed that in Landau and Lifgarbagez they show the Einstein tensor is zero instead of the Ricci tensor. You'll have to modify what is in D'Inverno to get out $\nabla_c g_{ab} = 0$.

7. May 31, 2012

### julian

Re: deriving the realtion between the connection and metric from the palatini formali

The key equation in getting the result from the variation of the connection comes from the Palantini equation:

$\delta R_{bd} = \nabla_a (\delta \Gamma^a_{bd} - \nabla_d (\delta \Gamma^a_{ba})$

(see D'Inverno page 145 for simple derivation). A simple modification of this should help derive the metric connection in the calculation of Landau and Lifgarbagez.

Last edited: May 31, 2012
8. Jun 1, 2012

### julian

Re: deriving the realtion between the connection and metric from the palatini formali

The relationship between the metric action and the Palantini action?...with Palantini do the connection variation first, get the metric connection, hold this 'constant' as you make your further variation. When you consider the variation of the metric action instead you are implicitly keeping the metric connection 'constant' (b/c that's just the convention of the metric formulation) - this is just made explcit, and taken advantage of, in the Palantini approach.

Last edited: Jun 1, 2012
9. Jun 1, 2012

### julian

Re: deriving the realtion between the connection and metric from the palatini formali

The relationship between Landau and D'Inverno is:

$\sqrt{-g} G_{ab} \delta g^{ab} = \delta (\sqrt{-g} g^{ab}) R_{ab}$

The proof that the connection is the metric connection goes through the same regardless of which 'metric' variable you vary.

Last edited: Jun 1, 2012
10. Jan 16, 2013

Re: deriving the realtion between the connection and metric from the palatini formali

what if one wants to get to the variation with respect to the connection?I did the following steps but I cannot go further more.

$\delta$ S= $\int$ d$^{4}$ x g$^{\mu \nu}$
[$\nabla$ $_{\lambda}$($\sqrt{-g}$ F $\delta$ $\Gamma$ $^{\lambda}_{\mu \nu}$ )
-$\delta$ $\Gamma$ $^{\lambda}_{\mu \nu}$ $\nabla$ $_{\lambda}$ ($\sqrt{-g}$ F)
- [$\nabla$ $_{\nu}$($\sqrt{-g}$ F $\delta$ $\Gamma$ $^{\lambda}_{\lambda \mu}$ )
+$\delta$ $\Gamma$ $^{\lambda}_{\mu \lambda}$ $\nabla$ $_{\nu}$ ($\sqrt{-g}$ F)]

but it does not go further!!!
And I have no idea what to do to get it into the famous equation which has been stated in
equation (16) of arXiv:0805.1726v4 [gr-qc] 4 Jun 2010
too.