## LATEX

### The Hilbert-Einstein action

This is the first post of a series of four posts concerned with the canonical quantization of general relativity.

## Equivalence principle

The equivalence principle states that spacetime must be a manifold, i.e,  locally it must look like flat Minkowski spacetime. Thus, curvature can be cancelled by freely falling observers in the gravitational field associated with the metric and as a consequence the spacetime manifold will be seen as flat by these observes.

Indeed, by choosing the so-called normal coordinates near any point of the spacetime manifold it is seen that the trajectories of freely falling objects looks locally like straight lines. Effectively, the manifold is approximated there by its tangent vector space. The difference between the metric $g_{\mu\nu}$ and the Minkowski metric $\eta_{\mu\nu}$ vanish to first order while the difference at second order is characterized by the so-called Riemann curvature tensor $R_{\mu\alpha\nu\beta}$. Explicitly, we have
\begin{eqnarray}
g_{\mu\nu}=\eta_{\mu\nu}-\frac{1}{3}R_{\mu\alpha\nu\beta}x^{\alpha}x^{\beta}+...
\end{eqnarray}

## Review of General Relativity

We consider a Riemannian (curved) manifold ${\cal M}$ with a metric $g_{\mu\nu}$. A coordinates transformation is given by
\begin{eqnarray}
x^{\mu}\longrightarrow x^{'\mu}=x^{'\mu}(x).
\end{eqnarray}
The vectors and one-forms on the manifold are quantities which are defined to transform under the above coordinates transformation respectively as follows
\begin{eqnarray}
V^{'\mu}=\frac{\partial x^{'\mu}}{\partial x^{\nu}}V^{\nu}.\label{contr}
\end{eqnarray}
\begin{eqnarray}
V^{'}_{\mu}=\frac{\partial x^{\nu}}{\partial x^{'\mu}}V_{\nu}.\label{cov}
\end{eqnarray}
The spaces of vectors and one-forms are the tangent and co-tangent bundles.

A tensor is a quantity with multiple indices (covariant and contravariant) transforming in a similar way, i.e. any contravariant index is transforming as (\ref{contr}) and any covariant index is transforming as (\ref{cov}). For example, the metric $g_{\mu\nu}$ is a second rank symmetric tensor which transforms as
\begin{eqnarray}
g^{'}_{\mu\nu}(x^{'})=\frac{\partial x^{\alpha}}{\partial x^{'\mu}}\frac{\partial x^{\beta}}{\partial x^{'\nu}}g_{\alpha\beta}(x).
\end{eqnarray}
The interval $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$ is therefore invariant. In fact, all scalar quantities are invariant under coordinate transformations. For example, the volume element $d^4x\sqrt{-{\rm det}g}$ is a scalar under coordinate transformation.

The derivative of a tensor does not transform as a tensor. However, the so-called covariant derivative of a tensor will transform as a tensor. The covariant derivatives of vectors and one-forms are given by
\begin{eqnarray}
\nabla_{\mu}V^{\nu}=\partial_{\mu}V^{\nu}+\Gamma_{\alpha\mu}^{\nu}V^{\alpha}.
\end{eqnarray}
\begin{eqnarray}
\nabla_{\mu}V_{\nu}=\partial_{\mu}V_{\nu}-\Gamma_{\mu\nu}^{\alpha}V_{\alpha}.
\end{eqnarray}
These transform indeed as tensors as one can easily check. Generalization to tensors is obvious. The Christoffel symbols $\Gamma_{\mu\nu}^{\alpha}$ are given in terms of the metric $g_{\mu\nu}$ by
\begin{eqnarray}
\Gamma_{\mu\nu}^{\alpha}=\frac{1}{2}g^{\alpha\beta}\big(\partial_{\mu}g_{\nu\beta}+\partial_{\nu}g_{\mu\beta}-\partial_{\beta}g_{\mu\nu}\big).
\end{eqnarray}
There exists a unique covariant derivative, and thus a unique choice of Christoffel symbols,  for which the metric is covariantly constant, viz
\begin{eqnarray}
\nabla_{\mu}g_{\alpha\beta}=0.
\end{eqnarray}
The straightest possible lines on the curved manifolds are given by the geodesics. A geodesic is a curve whose tangent vector is parallel transported along itself.  It is given explicitly by the Newton's second law on the curved manifold
\begin{eqnarray}
\frac{d^2x^{\mu}}{d\lambda}+\Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}=0.
\end{eqnarray}
The $\lambda$ is an affine parameter along the curve. The timelike geodesics define the trajectories of freely falling particles in the gravitational field encoded in the curvature of the Riemannian manifold.

The Riemann curvature tensor $R_{\mu\nu\beta}^{\alpha}$ is defined in terms of the covariant derivative by
\begin{eqnarray}
(\nabla_{\mu}\nabla_{\nu}-\nabla_{\nu}\nabla_{\mu})t^{\alpha}=-R_{\mu\nu\rho}^{\alpha}t^{\rho}.
\end{eqnarray}
It is given explicitly by
\begin{eqnarray}
R_{\mu\nu\rho}^{\alpha}=\partial_{\nu}\Gamma_{\mu\rho}^{\alpha}-\partial_{\rho}\Gamma_{\mu\nu}^{\alpha}+\Gamma_{\sigma\nu}^{\alpha}\Gamma_{\mu\rho}^{\sigma}-\Gamma_{\rho\sigma}^{\alpha}\Gamma_{\mu\nu}^{\sigma}.
\end{eqnarray}
We define the Ricci tensor $R_{\mu\nu}$ and the Ricci scalar $R$ by the equations
\begin{eqnarray}
R=g^{\mu\nu}R_{\mu\nu}.
\end{eqnarray}
\begin{eqnarray}
R_{\mu\nu}=R_{\mu\alpha\nu}^{\alpha}.
\end{eqnarray}
The Einstein's equations for general relativity reads (with $T_{\mu\nu}$ being the energy-momentum tensor and $G$ is the Newton's constant)
\begin{eqnarray}
R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=8\pi GT_{\mu\nu}.
\end{eqnarray}

## The Hilbert-Einstein Action

The dynamical variable is obviously the metric $g_{\mu\nu}$. The goal is to construct an action principle from which the Einstein's equations follow as the Euler-Lagrange equations of motion for the metric. This action principle will read as
\begin{eqnarray}
S=\int d^nx~{\cal L}(g).\label{dig}
\end{eqnarray}
The first problem with this way of writing is that both $d^nx$ and ${\cal L}$ are tensor densities rather than tensors. We digress briefly to explain this important different.

Let us recall the familiar Levi-Civita symbol in $n$ dimensions defined by
\begin{eqnarray}
\tilde{\epsilon}_{\mu_1...\mu_n}&=&+1~{\rm even}~{\rm permutation}\nonumber\\
&=&-1~{\rm odd}~~{\rm permutation}\nonumber\\
&=&~~0~{\rm otherwise}.
\end{eqnarray}
This is a symbol and not a tensor since it does not change under coordinate transformations. The determinant of a matrix $M$ can be given by the formula
\begin{eqnarray}
\tilde{\epsilon}_{\nu_1^{}...\nu_n^{}}{\rm det} M&=&\tilde{\epsilon}_{\mu_1^{}...\mu_n^{}}M^{\mu_1^{}}~_{\nu_1^{}}...M^{\mu_n^{}}~_{\nu_n^{}}.
\end{eqnarray}
By choosing $M^{\mu}~_{\nu^{}}=\partial x^{\mu}/\partial y^{\nu^{}}$ we get the transformation law
\begin{eqnarray}
\tilde{\epsilon}_{\nu_1^{}...\nu_n^{}}&=&{\rm det}\frac{\partial y^{}}{\partial x}~ \tilde{\epsilon}_{\mu_1^{}...\mu_n^{}}\frac{\partial x^{\mu_1}}{\partial y^{\nu_1^{}}}...\frac{\partial x^{\mu_n}}{\partial y^{\nu_n^{}}}.
\end{eqnarray}
In other words $\tilde{\epsilon}_{\mu_1^{}...\mu_n^{}}$ is not a tensor because of the determinant appearing in this equation. This is an example of a tensor density. Another example of a tensor density is ${\rm det} g$. Indeed from the tensor transformation law of the metric $g^{'}_{\alpha\beta}=g_{\mu\nu}(\partial x^{\mu}/\partial y^{\alpha})( \partial x^{\nu}/\partial y^{\beta})$ we can show in a straightforward way that
\begin{eqnarray}
{\rm det} g^{'}&=&({\rm det}\frac{\partial y^{}}{\partial x})^{-2}~ {\rm det} g.
\end{eqnarray}
The actual Levi-Civita tensor can then be defined by
\begin{eqnarray}
\epsilon_{\mu_1...\mu_n}=\sqrt{{\rm det} g^{}}~\tilde{\epsilon}_{\mu_1...\mu_n}.
\end{eqnarray}
Next under a coordinate transformation $x\longrightarrow y$ the volume element transforms as
\begin{eqnarray}
d^nx\longrightarrow d^ny={\rm det}\frac{\partial y}{\partial x}~d^nx.\label{ele}
\end{eqnarray}
In other words the volume element transforms as a tensor density and not as a tensor. We verify this important point in our language as follows. We write
\begin{eqnarray}
d^nx&=&dx^0\wedge dx^1\wedge ...\wedge dx^{n-1}\nonumber\\
&=&\frac{1}{n!}\tilde{\epsilon}_{\mu_1...\mu_n}dx^{\mu_1}\wedge ...\wedge dx^{\mu_n}.\label{sdf}
\end{eqnarray}
Recall that a differential $p-$form is a $(0,p)$ tensor which is completely antisymmetric. For example scalars are $0-$forms and dual cotangent vectors are $1-$forms. The  Levi-Civita tensor $\epsilon_{\mu_1...\mu_n}$ is a $4-$form. The differentials $dx^{\mu}$ appearing in the second line of equation  (\ref{sdf}) are $1-$forms and hence under a coordinate transformation $x\longrightarrow y$ we have $dx^{\mu}\longrightarrow dy^{\mu}=dx^{\nu}\partial y^{\mu}/\partial x^{\nu}$.  By using this  transformation law we can immediately show that $dx^n$ transforms to $d^ny$ exactly as in equation (\ref{ele}).

It is not difficult to see now that an invariant volume element can be given by the $n-$form defined by the equation
\begin{eqnarray}
dV=\sqrt{{\rm det} g}~ d^nx.
\end{eqnarray}
We can show that
\begin{eqnarray}
dV&=&\frac{1}{n!}\sqrt{{\rm det} g}~\tilde{\epsilon}_{\mu_1...\mu_n}dx^{\mu_1}\wedge ...\wedge dx^{\mu_n}\nonumber\\
&=&\frac{1}{n!}{\epsilon}_{\mu_1...\mu_n}dx^{\mu_1}\wedge ...\wedge dx^{\mu_n}\nonumber\\
&=&{\epsilon}_{\mu_1...\mu_n}dx^{\mu_1}\otimes ...\otimes dx^{\mu_n}\nonumber\\
&=&\epsilon.
\end{eqnarray}
In other words the  invariant volume element is precisely the  Levi-Civita tensor. In the case of Lorentzian signature we replace ${\rm det} g$ with  $-{\rm det} g$.

We go back now to equation (\ref{dig}) and rewrite it as
\begin{eqnarray}
S&=&\int d^nx~{\cal L}(g)\nonumber\\
&=&\int d^n x\sqrt{-{\rm det} g}~\hat{\cal L}(g).
\end{eqnarray}
Clearly ${\cal L}=\sqrt{-{\rm det} g}~\hat{\cal L}$. Since the invariant volume element $d^n x\sqrt{-{\rm det} g}$ is a scalar the function $\hat{\cal L}$ must also be a scalar and as such can be identified with the Lagrangian density.

We use the result that the only independent scalar quantity which is constructed from the metric and which is at most second order in its derivatives is the Ricci scalar $R$. In other words the simplest choice for the Lagrangian density $\hat{\cal L}$ is
\begin{eqnarray}
\hat{\cal L}(g)=R.
\end{eqnarray}
The corresponding action is called the Hilbert-Einstein action. We compute
\begin{eqnarray}
\delta S
&=&\int d^n x\delta \sqrt{-{\rm det} g}~g^{\mu\nu}R_{\mu\nu}+\int d^n x \sqrt{-{\rm det} g}~\delta g^{\mu\nu}R_{\mu\nu}+\int d^n x \sqrt{-{\rm det} g}~g^{\mu\nu}\delta R_{\mu\nu}.\nonumber\\
\end{eqnarray}
We have
\begin{eqnarray}
\delta R_{\mu\nu}&=&\delta R_{\mu\rho\nu}~^{\rho}\nonumber\\
&=&\partial_{\rho}\delta \Gamma^{\rho}~_{\mu\nu}-\partial_{\mu}\delta\Gamma^{\rho}~_{\rho\nu}+\delta (\Gamma^{\lambda}~_{\mu\nu}\Gamma^{\rho}~_{\rho\lambda}-\Gamma^{\lambda}~_{\rho\nu}\Gamma^{\rho}~_{\mu\lambda})\nonumber\\
&=&(\nabla_{\rho}\delta \Gamma^{\rho}~_{\mu\nu}-\Gamma^{\rho}~_{\rho\lambda}\delta\Gamma^{\lambda}~_{\mu\nu}+\Gamma^{\lambda}~_{\rho\mu}\delta\Gamma^{\rho}~_{\lambda\nu}+\Gamma^{\lambda}~_{\rho\nu}\delta \Gamma^{\rho}~_{\lambda\mu})-(\nabla_{\mu}\delta \Gamma^{\rho}~_{\rho\nu}-\Gamma^{\rho}~_{\mu\lambda}\delta\Gamma^{\lambda}~_{\rho\nu}+\Gamma^{\lambda}~_{\mu\rho}\delta\Gamma^{\rho}~_{\lambda\nu}\nonumber\\
&+&\Gamma^{\lambda}~_{\mu\nu}\delta \Gamma^{\rho}~_{\rho\lambda})+\delta (\Gamma^{\lambda}~_{\mu\nu}\Gamma^{\rho}~_{\rho\lambda}-\Gamma^{\lambda}~_{\rho\nu}\Gamma^{\rho}~_{\mu\lambda})\nonumber\\
&=&\nabla_{\rho}\delta \Gamma^{\rho}~_{\mu\nu}-\nabla_{\mu}\delta \Gamma^{\rho}~_{\rho\nu}.
\end{eqnarray}
In the second line of the above equation we have used the fact that $\delta \Gamma^{\rho}~_{\mu\nu}$ is a tensor since it is the difference of two connections. Thus

\begin{eqnarray}
\int d^n x \sqrt{-{\rm det} g}~g^{\mu\nu}\delta R_{\mu\nu}&=&\int d^n x \sqrt{-{\rm det} g}~g^{\mu\nu}\bigg(\nabla_{\rho}\delta \Gamma^{\rho}~_{\mu\nu}-\nabla_{\mu}\delta \Gamma^{\rho}~_{\rho\nu}\bigg)\nonumber\\
&=&\int d^n x \sqrt{-{\rm det} g}~\nabla_{\rho}\bigg(g^{\mu\nu}\delta \Gamma^{\rho}~_{\mu\nu}-g^{\rho\nu}\delta \Gamma^{\mu}~_{\mu\nu}\bigg).
\end{eqnarray}
We compute also (with $\delta g_{\mu\nu}=-g_{\mu\alpha}g_{\nu\beta}\delta g^{\alpha \beta}$)
\begin{eqnarray}
\delta\Gamma^{\rho}~_{\mu\nu}&=&\frac{1}{2}g^{\rho\lambda}\bigg(\nabla_{\mu}\delta g_{\nu\lambda}+\nabla_{\nu}\delta g_{\mu\lambda}-\nabla_{\lambda}\delta g_{\mu\nu}\bigg)\nonumber\\
&=&-\frac{1}{2}\bigg(g_{\nu\lambda}\nabla_{\mu}\delta g^{\lambda\rho}+g_{\mu\lambda}\nabla_{\nu}\delta g^{\lambda\rho}-g_{\mu\alpha}g_{\nu\beta}\nabla^{\rho}\delta g^{\alpha\beta}\bigg).
\end{eqnarray}
Thus
\begin{eqnarray}
\int d^n x \sqrt{-{\rm det} g}~g^{\mu\nu}\delta R_{\mu\nu}
&=&\int d^n x \sqrt{-{\rm det} g}~\nabla_{\rho}\bigg(g_{\mu\nu}\nabla^{\rho}\delta g^{\mu\nu}-\nabla_{\mu}\delta g^{\mu\rho}\bigg).
\end{eqnarray}
By Stokes's theorem this integral is equal to the integral over the boundary of spacetime of the expression $g_{\mu\nu}\nabla^{\rho}\delta g^{\mu\nu}-\nabla_{\mu}\delta g^{\mu\rho}$ which is $0$ if we assume that the metric and its first derivatives are held fixed on the boundary. The variation of the action reduces to
\begin{eqnarray}
\delta S
&=&\int d^n x\delta \sqrt{-{\rm det} g}~g^{\mu\nu}R_{\mu\nu}+\int d^n x \sqrt{-{\rm det} g}~\delta g^{\mu\nu}R_{\mu\nu}.
\end{eqnarray}
Next we use the result
\begin{eqnarray}
\delta \sqrt{-{\rm det} g}=-\frac{1}{2}\sqrt{-{\rm det} g}~ g_{\mu\nu}\delta g^{\mu\nu}.
\end{eqnarray}
Hence
\begin{eqnarray}
\delta S
&=&\int d^n x \sqrt{-{\rm det} g}~\delta g^{\mu\nu}(R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R).
\end{eqnarray}
This will obviously lead to Einstein's equations in vacuum  which is partially our goal. We want also to include the effect of matter  which requires considering the more general actions of the form
\begin{eqnarray}
S=\frac{1}{16\pi G}\int d^nx~\sqrt{-{\rm det}g}~R+S_M.\label{HE}
\end{eqnarray}
\begin{eqnarray}
S_M=\int d^nx~\sqrt{-{\rm det}g}~\hat{\cal L}_M.
\end{eqnarray}
The variation of the action becomes
\begin{eqnarray}
\delta S
&=&\frac{1}{16\pi G}\int d^n x \sqrt{-{\rm det} g}~\delta g^{\mu\nu}(R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R)+\delta S_M\nonumber\\
&=&\int d^n x \sqrt{-{\rm det} g}~\delta g^{\mu\nu}\bigg[\frac{1}{16\pi G}(R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R)+\frac{1}{\sqrt{-{\rm det} g}}\frac{\delta S_M}{\delta g^{\mu\nu}} \bigg].
\end{eqnarray}
In other words
\begin{eqnarray}
\frac{1}{\sqrt{-{\rm det} g}}\frac{\delta S}{\delta g^{\mu\nu}} &=&\frac{1}{16\pi G}(R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R)+\frac{1}{\sqrt{-{\rm det} g}}\frac{\delta S_M}{\delta g^{\mu\nu}}.
\end{eqnarray}
Einstein's equations are therefore given by
\begin{eqnarray}
R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=8\pi G T_{\mu\nu}.
\end{eqnarray}
The stress-energy-momentum tensor must therefore be defined by the equation
\begin{eqnarray}
T_{\mu\nu}=-\frac{2}{\sqrt{-{\rm det} g}}\frac{\delta S_M}{\delta g^{\mu\nu}}.
\end{eqnarray}
As a first example we consider the action of a scalar field in curved spacetime given by
\begin{eqnarray}
S_{\phi}=\int d^nx \sqrt{-{\rm det} g}~\bigg[-\frac{1}{2}g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi-V(\phi)\bigg].
\end{eqnarray}
The corresponding stress-energy-momentum tensor is calculated to be given by
\begin{eqnarray}
T_{\mu\nu}^{(\phi)}=\nabla_{\mu}\phi\nabla_{\nu}\phi-\frac{1}{2}g_{\mu\nu}g^{\rho\sigma}\nabla_{\rho}\phi\nabla_{\sigma}\phi-g_{\mu\nu}V(\phi).
\end{eqnarray}
As a second example we consider the action of the  electromagnetic field in curved spacetime given by
\begin{eqnarray}
S_{A}=\int d^nx \sqrt{-{\rm det} g}~\bigg[-\frac{1}{4}g^{\mu\nu}g^{\alpha\beta}F_{\mu\nu}F_{\alpha\beta}\bigg].
\end{eqnarray}
In this case the  stress-energy-momentum tensor is calculated to be given by
\begin{eqnarray}
T_{\mu\nu}^{(A)}=F^{\mu\lambda}F^{\nu}~_{\lambda}-\frac{1}{4}g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}.
\end{eqnarray}
The cosmological constant is one of the simplest matter action that one can add to the Hilbert-Einstein action. It is given by
\begin{eqnarray}
S_{\rm cc}=-\frac{1}{8\pi G}\int d^4x\sqrt{-{\rm det}g}\Lambda.
\end{eqnarray}
In this case the energy-momentum tensor and the Einstein equations read
\begin{eqnarray}
T_{\mu\nu}=-\frac{\Lambda}{8\pi G}g_{\mu\nu}.
\end{eqnarray}
\begin{eqnarray}
R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R+\Lambda g_{\mu\nu}=0.
\end{eqnarray}