General Relativity – Why Does Palatini Variation of Einstein-Hilbert Action Yield Levi-Civita Connection?

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There are two ways to do the variation of Einstein-Hilbert action.

First one is Einstein formalism which takes only metric independent. After variation of action, we get the Einstein field equation.
Second one is Palatini formalism which takes metric and connection are independent. After variation, we get two equations, first is field equation and second is that connection is Levi-Civita connection.

So my question is why it is so coincident that Palatini variation of Einstein-Hilbert action will obtain an equation that connection is Levi-Civita connection and Palatini formalism coincides with Einstein formalism? While for $f(R)$ action they are generally different. Are there some deeper mathematical or physical structures of Einstein-Hilbert action which can account for it.

Best Answer

I) In Palatini $f(R)$ gravity, the Lagrangian density is$^1$

$$ {\cal L}(g,\Gamma)~=~ \frac{1}{2\kappa}\sqrt{-g} f(R) + {\cal L}_{\rm m}; \tag{1}$$

with matter Lagrangian density ${\cal L}_{\rm m}$; with scalar curvature

$$R~:=~ g^{\mu\nu} R_{\mu\nu}(\Gamma);\tag{2}$$

with Ricci curvature $R_{\mu\nu}(\Gamma)$; and where

$$\Gamma^{\lambda}_{\mu\nu}~=~\Gamma^{\lambda}_{\nu\mu}\tag{3}$$

is an arbitrary torsionfree$^2$ connection.

II) As OP mentions, the word Palatini refers to that the metric $g_{\mu\nu}$ and the connection $\Gamma^{\lambda}_{\mu\nu}$ are independent variables$^3$. We therefore get two types of EL equations:

The EL equations $$ f^{\prime}(R)R_{\mu\nu} -\frac{1}{2}f(R)g_{\mu\nu}~\stackrel{(1)+(5)}{\approx}~\kappa T_{\mu\nu} \tag{4}$$ for the metric $g_{\mu\nu}$ are the generalization of EFE, where $$T^{\mu\nu}~:=~\frac{2}{\sqrt{-g}}\frac{\delta S_{\rm m}}{\delta g_{\mu\nu}} \tag{5} $$ is the matter Hilbert stress-energy-momentum (SEM) tensor. [In eq. (4) the $\approx$ symbol means equality modulo equations of motion. In this answer, we use $(-,+,\ldots,+)$ Minkowski sign convention in $d$ spacetime dimensions.]
If the matter action $S_{\rm m}$ doesn't depend on the connection $\Gamma^{\lambda}_{\mu\nu}$, then the EL equations $$ \nabla_{\lambda}\hat{\mathfrak{g}}^{\mu\nu} ~\stackrel{(1)}{\approx}~0,\qquad \hat{\mathfrak{g}}^{\mu\nu} ~:=~\sqrt{-g} f^{\prime}(R) g^{\mu\nu} ~\stackrel{(8)}{=}~\sqrt{-\hat{g}} \hat{g}^{\mu\nu}, \tag{6}$$ for the connection $\Gamma^{\lambda}_{\mu\nu}$ turn out to be the metric compatibility condition $$\nabla_{\lambda}\hat{g}_{\mu\nu} ~\stackrel{(6)+(8)}{\approx}~0\tag{7}$$ for a conformally equivalent metric $$ \hat{g}_{\mu\nu}~:=~f^{\prime}(R)^{\frac{2}{d-2}} g_{\mu\nu}, \tag{8}$$ known as the Einstein frame metric. In other words, the classical solution for $\Gamma^{\lambda}_{\mu\nu}$ is the Levi-Civita connection for the Einstein frame metric $\hat{g}_{\mu\nu}$.

III) So Einstein gravity (GR) with a possible cosmological constant

$$ f(R)~=~R-2\Lambda,\tag{9} $$

or equivalently

$$ f^{\prime}(R)~=~1,\tag{10}$$

corresponds to the special case where the two metrics $g_{\mu\nu}$ and $\hat{g}_{\mu\nu}$ coincide, and hence $\Gamma^{\lambda}_{\mu\nu}$ becomes the Levi-Civita connection for $g_{\mu\nu}$.

$^1$ It is natural to replace the Lagrangian density (1) with the extended Lagrangian density

$$ \tilde{\cal L}(g,\Gamma,\Phi)~=~ \frac{1}{2\kappa}\sqrt{-g}\{\Phi R-V(\Phi)\} + {\cal L}_{\rm m}; \tag{11}$$

with auxiliary scalar dilaton field $\Phi$; and where the potential

$$V(\Phi)~:=~\sup_r(\Phi r -f(r))\tag{12}$$

is the Legendre transform of the function $f$. If we integrate out the auxiliary scalar field $\Phi$, we then return to the $f(R)$ Lagrangian density (1) which we started from! The EL equations

$$ \nabla_{\lambda}\hat{\mathfrak{g}}^{\mu\nu} ~\stackrel{(1)}{\approx}~0,\qquad \hat{\mathfrak{g}}^{\mu\nu} ~:=~\sqrt{-g} \Phi g^{\mu\nu} ~\stackrel{(14)}{=}~\sqrt{-\hat{g}} \hat{g}^{\mu\nu}, \tag{13}$$

for the connection $\Gamma^{\lambda}_{\mu\nu}$ become the metric compatibility condition (7) for the Einstein frame metric

$$ \hat{g}_{\mu\nu}~:=~\Phi^{\frac{2}{d-2}} g_{\mu\nu}. \tag{14}$$

After the connection $\Gamma^{\lambda}_{\mu\nu}$ has been integrated out, the Lagrangian density (11) becomes

$$ \tilde{\cal L}(g,\Phi)~=~ \frac{1}{2\kappa}\sqrt{-\hat{g}}\{R(\hat{g})-\Phi^{\frac{d}{2-d}}V(\Phi)\} + {\cal L}_{\rm m} , \tag{15}$$

where eq. (14) is implicitly assumed.

However, unfortunately the Legendre transform $V$ does not exist for Einstein gravity (9), so we will not consider the extended Lagrangian density (11) further in this answer.

$^2$ One could allow a non-dynamical torsion piece, but we will not pursue this here for simplicity. For more on torsion, see e.g. also this Phys.SE post.

$^3$ Normally in non-Palatini formulations, we integrate out the connection $\Gamma^{\lambda}_{\mu\nu}$ and keep the metric $g_{\mu\nu}$. Eddington & Schrödinger proposed the opposite! Let us analyze this possibility here. Define for later convenience a double-index notation $M=\mu\mu^{\prime}$ and the following shorthand notation

$$\frac{f(R)}{2f^{\prime}(R)}~=:~\hat{f}(R) ~\equiv~ \hat{f}_0+\hat{f}_1R +\hat{f}_2(R).\tag{16}$$

Let us only consider the vacuum

$$T_{\mu\nu}~=~0.\tag{17}$$

from now on. We then have

$$g^M~\stackrel{(4)+(16)+(17)}{\approx}~\hat{f}(R)R^M,\tag{18}$$

where $R^M$ is the inverse Ricci tensor. Equivalently, we have

$$\left(\delta^M_N - \hat{f}_1 R^M R_N\right) g^N~\stackrel{(16)+(18)}{\approx}~\left(\hat{f}_0+f_2(R) \right)R^M.\tag{19}$$

So we get a fixed-point equation for the inverse metric

$$g^N~\stackrel{(19)}{\approx}~\left(\delta^N_M +\frac{\hat{f}_1}{1-d\hat{f}_1} R^N R_M\right)\left(\hat{f}_0+\hat{f}_2(R) \right)R^M$$ $$~=~ \frac{1}{1-d\hat{f}_1}\left(\hat{f}_0+\hat{f}_2\left(g^MR_M\right) \right)R^N.\tag{20}$$

Let us specialize to Einstein gravity (9). Then

$$\hat{f}_0~=~-\Lambda;\qquad\hat{f}_1~=~\frac{1}{2};\qquad\hat{f}_2(R)~=~0.\tag{21}$$

The inverse metric becomes

$$g^N~\stackrel{(20)+(21)}{\approx}~\frac{2\Lambda}{d-2}R^N.\tag{22}$$

And hence $$R~\stackrel{(22)}{\approx}~\frac{2d}{d-2}\Lambda,\tag{23}$$

and

$$g_{\mu\nu}~\stackrel{(2)+(22)}{\approx}~\frac{d-2}{2\Lambda}R_{\mu\nu}(\Gamma).\tag{24}$$

So the EH Lagrangian density becomes Born-Infeld-like:

$${\cal L}(\Gamma)~\stackrel{(1)+(17)+(23)+(24)}{\approx}~\frac{1}{\kappa}\left(\frac{d-2}{2\Lambda}\right)^{\frac{d}{2}-1}\sqrt{-\det(R_{\mu\nu}(\Gamma))}.\tag{25}$$

Note that the Eddington-Schrödinger action (25) only works for a non-zero cosmological constant $\Lambda\neq 0$.

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A long time ago, when I was in graduate school, a friend and I went to Bryce DeWitt with a similar question. He thought about it for a while, and then said to us:

 "Do you know what your problem is? Too much book learning."

He explained that the answer could almost certainly be found in a book or paper, if we looked hard enough, but we would learn much more by working out the answer ourselves. This may be the best advice I ever received as a graduate student.

Why am I telling you this? Because the answer to your question about the first variation of the action (and the notes I linked to) provides you with everything you need to solve this problem. Give it a try! If you get stuck on a specific step in the derivation, come back here and ask about it -- I'll be happy to work through the details with you. But this is a great chance for you to apply what you learned from your last question.

General Relativity – Variation of Modified Einstein-Hilbert Action Explained

Find the answer to question 1 below. Question 2. is weird since $\nabla_\mu g_{\alpha\beta} = 0$ (when the connection is metric compatible) as mentioned by @Trimok. In any case, the variation of the action can be derived using the method described below.

We start with the BD action $$ S = \frac{1}{16\pi}\int d^4 x \sqrt{-g} \left[ \phi R - \frac{\omega}{\phi} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi \right] + S_M $$ where $S_M$ is the matter action. To determine Einstein's field equations, we vary the action w.r.t to the metric. We will use the formulae (ref. wikipedia) \begin{equation} \begin{split} \delta R = R_{\mu\nu} \delta g^{\mu\nu} + \nabla_\sigma \left( g^{\mu\nu} \delta \Gamma^\sigma_{\mu\nu} - g^{\mu\sigma} \delta \Gamma^\rho_{\rho\mu} \right) \end{split} \end{equation} The variation of the Christoffel tensor is \begin{equation} \begin{split} \delta \Gamma^\lambda_{\mu\nu} & = \delta g^{\lambda\rho} g_{\rho\alpha} \Gamma^\alpha_{\mu\nu} + \frac{1}{2} g^{\lambda\rho} \left( \partial_\mu \delta g_{\nu\rho} + \partial_\nu \delta g_{\mu\rho} - \partial_\rho \delta g_{\mu\nu} \right) \\ & = \frac{1}{2} g^{\lambda\rho} \left( \nabla_\mu \delta g_{\nu\rho} + \nabla_\nu \delta g_{\mu\rho} - \nabla_\rho \delta g_{\mu\nu} \right) \\ & = - \frac{1}{2} \left( g_{\nu\alpha} \nabla_\mu \delta g^{\alpha\lambda} + g_{\mu\alpha} \nabla_\nu \delta g^{\alpha\lambda} - g_{\mu\alpha} g_{\nu\beta} \nabla^\lambda \delta g^{\alpha\beta} \right) \end{split} \end{equation} where we used $\delta g_{\mu\nu} = - g_{\mu\alpha} g_{\nu\beta} \delta g^{\alpha\beta}$. This implies \begin{equation} \begin{split} g^{\mu\nu} \delta \Gamma^\sigma_{\mu\nu} &= -\nabla_\alpha \delta g^{\alpha\sigma} + \frac{1}{2} g_{\alpha\beta}\nabla^\sigma\delta g^{\alpha\beta} \\ g^{\mu\sigma} \delta \Gamma^\lambda_{\lambda\mu} &= - \frac{1}{2}g_{\alpha\beta} \nabla^\sigma \delta g^{\alpha\beta} \end{split} \end{equation} which implies \begin{equation} \begin{split} \delta R = R_{\mu\nu} \delta g^{\mu\nu} - \nabla_\mu \nabla_\nu \delta g^{\mu\nu} + g_{\mu\nu} \nabla^2 \delta g^{\mu\nu} \end{split} \end{equation} Finally, from 1, we also have $$ \delta \sqrt{-g} = - \frac{1}{2} \sqrt{-g} g_{\mu\nu} \delta g^{\mu\nu} $$ Finally, we are ready to compute the variation of the action. We have \begin{equation} \begin{split} \delta S &= \frac{1}{16\pi}\int d^4 x \delta \sqrt{-g} \left[ \phi R - \frac{\omega}{\phi} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi \right] \\ &~~~~~~~~~~~~ + \frac{1}{16\pi}\int d^4 x \sqrt{-g} \left[ \phi \delta R - \frac{\omega}{\phi} \delta g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi \right] + \delta S_M \\ &= - \frac{1}{32\pi}\int d^4 x \sqrt{-g} g_{\mu\nu} \left[ \phi R - \frac{\omega}{\phi} g^{\alpha\beta} \partial_\alpha \phi \partial_\beta \phi \right] \delta g^{\mu\nu} + \int d^4 x \frac{ \delta S_M}{\delta g^{\mu\nu}} \delta g^{\mu\nu} \\ &~~~~~~~~~~~~ + \frac{1}{16\pi}\int d^4 x \sqrt{-g} \left[ \left( \phi R_{\mu\nu} - \nabla_\mu \nabla_\nu \phi + g_{\mu\nu} \nabla^2 \phi \right) - \frac{\omega}{\phi} \partial_\mu \phi \partial_\nu \phi \right] \delta g^{\mu\nu} \end{split} \end{equation} Requiring that the variation of the action vanish (to leading order in $\delta g^{\mu\nu}$) gives \begin{equation} \begin{split} G_{\mu\nu} &= - \frac{16\pi}{\phi \sqrt{-g}} \frac{ \delta S_M}{\delta g^{\mu\nu}} + \frac{\omega}{\phi^2} \left[ \partial_\mu \phi \partial_\nu \phi - \frac{1}{2} g_{\mu\nu} \partial_\alpha \phi \partial^\alpha \phi \right] +\frac{1}{\phi} \left[ \nabla_\mu \nabla_\nu \phi - g_{\mu\nu} \nabla^2 \phi \right] \end{split} \end{equation} Recall that the stress-tensor is defined as \begin{equation} \begin{split} T_{\mu\nu} = - \frac{2}{\sqrt{-g}} \frac{ \delta S_M}{\delta g^{\mu\nu}} \end{split} \end{equation} Thus \begin{equation} \begin{split} G_{\mu\nu} &= \frac{8\pi}{\phi} T_{\mu\nu} + \frac{\omega}{\phi^2} \left[ \partial_\mu \phi \partial_\nu \phi - \frac{1}{2} g_{\mu\nu} \partial_\alpha \phi \partial^\alpha \phi \right] +\frac{1}{\phi} \left[ \nabla_\mu \nabla_\nu \phi - g_{\mu\nu} \nabla^2 \phi \right] \end{split} \end{equation} which is the Brans-Dicke equation.

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