Ngô Quốc Anh

January 11, 2013

The stress-energy tensor and Einstein constraint equations in the presence of scalar fields

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 6:26

In this notes, we are particularly interested in some special model of general relativity. More precisely, here we consider the scalar fields since, for example, a real scalar field more or less provides one of the simplest sources of stress-energy in GR.

1. Derivation of stress-energy tensor.

To derive the stress-energy tensor associated to a real scalar field, we make use of the Einstein-Hilbert action. Suppose that the full action of the theory is given by the Einstein–Hilbert term plus a term $\mathcal{L}_\mathrm{M}$ describing any matter fields appearing in the theory, then we have

$\displaystyle S = \int ( \text{Scal} + \mathcal{L}_\mathrm{M} ) \sqrt{-g}\, \mathrm{d}^n x .$

The action principle then tells us that the variation of this action with respect to the inverse metric is zero, yielding $\delta S=0$. Calculating this equation gives

$\displaystyle \frac{\delta R}{\delta g^{\mu\nu}} + \frac{R}{\sqrt{-g}} \frac{\delta \sqrt{-g}}{\delta g^{\mu\nu}} =-2 \frac{\delta \mathcal{L}_\mathrm{M}}{\delta g^{\mu\nu}} + g_{\mu\nu} \mathcal{L}_\mathrm{M},$

where the right hand side is nothing but the stress-energy tensor $T_{\mu\nu}$.

In modern cosmology, one can introduce on the spacetime $(V,\mathbf g)$ a real scalar field $\boldsymbol \psi$ with potential $U$ as a smooth function of $\boldsymbol\psi$. A particular Einstein field theory is specified by the choice of an action principle with

$\displaystyle\mathcal{L}_\mathrm{M}=-\frac{1}{2}|\nabla\boldsymbol\psi|^2_{\mathbf g}-U(\boldsymbol\psi).$

To find its associated stress-energy tensor, we first look at $\frac{{\delta {\mathcal{L}_{\text{M}}}}}{{\delta {g^{\mu \nu }}}}$. Since the term $U(\boldsymbol\psi)$ does not depend on the metric, we get

$\displaystyle\frac{{\delta {\mathcal{L}_{\text{M}}}}}{{\delta {\mathbf g^{\mu \nu }}}} = \frac{1}{{\delta {\mathbf g^{\mu \nu }}}}\left( { - \frac{1}{2}(\delta {\mathbf g^{\mu \nu }}){\partial _{\mu }}\boldsymbol\psi {\partial _\nu }\boldsymbol\psi } \right) = - \frac{1}{2}{\partial _{\mu }}\boldsymbol\psi {\partial _\nu }\boldsymbol\psi,$

where we have used the fact that partial derivatives also do not depend on the variation of metric. Therefore,

$\displaystyle\begin{array}{lcl} {T_{\mu \nu }} &=&\displaystyle - 2\frac{{\delta {\mathcal{L}_{\text{M}}}}}{{\delta {\mathbf g^{\mu \nu }}}} + {\mathbf g_{\mu \nu }}{\mathcal{L}_{\text{M}}} \hfill \\ &=&\displaystyle {\partial _{\mu }}\boldsymbol\psi {\partial _\nu }\boldsymbol\psi - \frac{1}{2}{\mathbf g_{\mu \nu }}|\nabla \boldsymbol\psi |_{\mathbf g}^2 - {\mathbf g_{\mu \nu }}U(\boldsymbol\psi ). \hfill \\ \end{array}$

2. Derivation of the momentum constraint equation.

Following the notes where we have derived the momentum constrain equation, we know that the momentum constraint equation is nothing but

$\displaystyle \text{div}_gK - d(\text{trace}_gK)=T(\cdot,n),$

here we simply assume that $\kappa=1$. Using components and assuming $\partial_0=n$ known as the timelike vector sitting in the frame, we write

$\displaystyle {\nabla _\mu }{K^{\mu \nu }} - {g^{\mu \nu }}{\partial _\mu }({\text{trac}}{{\text{e}}_g}K) = T_0^\nu .$

It is important to note that $T_0^\nu$ is a projection on $M$ and on the normal to $M$ of the stress-energy tensor $T$.

Therefore, by raising one index and let the other to be zero, we found that the last two terms in the formula for $T$ vanish, which yields

$\displaystyle T_0^\nu = - {g^{\nu \rho }}{\partial _0}\psi {\partial _\rho }\psi,$

where $\psi$ is the induced of $\boldsymbol\psi$ on $M$ and $g$ is also the induced metric on $M$ of the spacetime metric $\mathbf g$.

3. Derivation of the Hamiltonian constraint equation.

We now turn to the Hamiltonian constraint equation which can be written as

$\displaystyle \text{Scal}_g - |K|_g^2 + (\text{trace}_gK)^2 = 2T(n, n).$

Clearly, in terms of components and again we assume that $\partial_0 = n$, it is nothing but

$\displaystyle \text{Scal}_g - |K|_g^2 + (\text{trace}_gK)^2 = 2T_{00}.$

To calculate $T_{00}$ we simply get

$\displaystyle {T_{00}} = {\partial _0}\psi {\partial _0}\psi - \frac{1}{2}{g_{00}}{g^{ij}}{\nabla _i}\psi {\nabla _j}\psi - {g_{00}}U(\psi ).$

4. The constraint equations with the adapted frame.

First, we take local coordinates adapted to the product structure $V = M \times \mathbb R$ as follows $(x^\alpha)=(x^0=t, x^i)$. For a natural frame on $M$, we choose

$\displaystyle \partial_i = \frac{\partial}{\partial x_i}, \quad i = \overline{1,n}.$

The dual coframe is found to be such that

$\displaystyle \theta^i = dx^i+\beta^i dt, \quad i = \overline{1,n},$

while the $1$-form $\theta^0$ is nothing but $dt$. The last thing we need to find is the vector $\partial_0$. To make it correct, we choose

$\displaystyle \partial_0 =\frac{d}{dt}-\beta^j\partial_j.$

As in this post and thanks to $\partial_0 = Nn$, we find that

$\displaystyle\begin{array} {lcl}{\mathbf g_{00}} &=&\displaystyle \mathbf g({\partial _0},{\partial _0}) \hfill \\&=&\displaystyle \mathbf g(Nn,Nn ) \hfill \\ &=&\displaystyle - {N^2}. \hfill \\ \end{array}$

Similarly, one can also obtain

$\displaystyle\begin{array}{lcl} {\mathbf g_{0i}} &=&\displaystyle \mathbf g({\partial _0},{\partial _i}) \hfill \\&=&\displaystyle \mathbf g(Nn ,{\partial _i}) \hfill \\&=&0. \hfill \\ \end{array}$

Finally, it is easy to see that

$\displaystyle {\mathbf g_{ij}} = \mathbf g({\partial _i},{\partial _j}) = g({\partial _i},{\partial _j}) = g _{ij}$

for all $i,j \geqslant 1$. By collecting the above calculation, we arrive at

$\displaystyle \mathbf g_{\alpha\beta}=\begin{pmatrix} \mathbf g_{00} & \mathbf g_{0j} \\ \mathbf g_{i0} & \mathbf g_{ij} \end{pmatrix}=\begin{pmatrix} -N^2 & 0 \\ 0 & g_{ij} \end{pmatrix}.$

The components of the inverse matrix $\mathbf g^{\alpha\beta}$ can be calculated as given below

$\displaystyle \mathbf g^{\alpha\beta}=\begin{pmatrix} \mathbf g^{00} & \mathbf g^{0j} \\ \mathbf g^{i0} & \mathbf g^{ij} \end{pmatrix}=\begin{pmatrix} -\frac{1}{N^2} & 0 \\ 0 & g^{ij} \end{pmatrix}.$

Therefore, the stress-energy tensor has the following

$\displaystyle {T_{00}} = {N^2} \Big({N^{ - 2}}{\partial _0}\psi {\partial _0}\psi + \frac{1}{2}{g^{ij}}{\nabla _i}\psi {\nabla _j}\psi + U(\psi ) \Big).$

Thanks to the fact that, in the local frame,

$\displaystyle n_\alpha=(N,0,...,0) \quad n^\alpha = (-\frac{1}{N}, 0,...,0),$

we find that

$\displaystyle T(n,n) = {N^{ - 2}}{T_{00}},$

here $n$ is being considered as a vector field, $n = n^\alpha \partial_\alpha$, which yields the following constraint equation

$\displaystyle {\text{Scal}}_g - |K|_g^2 + {({\text{trac}}{{\text{e}}_g}K)^2} = 2{N^{ - 2}}{\partial _0}\psi {\partial _0}\psi + {g^{ij}}{\nabla _i}\psi {\nabla _j}\psi + 2U(\psi ).$

For the momentum constraint equation, we can check that

$\displaystyle T({\partial _\nu },n) = - \frac{1}{N}{T_{0\nu }} = - \frac{1}{N}{\partial _0}\psi {\partial _\nu }\psi ,$

here $n$ is also being considered as a vector field, $n = n^\alpha \partial_\alpha$. Hence, we have

$\displaystyle {\nabla _\mu }{K^{\mu \nu }} - {g^{\mu \nu }}{\partial _\mu }({\text{trac}}{{\text{e}}_g}K)=- \frac{1}{N}{\partial _0}\psi {\partial _\nu }\psi.$

The term${N^{ - 2}}{\partial _0}\psi {\partial _0}\psi$ is sometime written as $|\pi|^2$ with

$\displaystyle\pi = \frac{1}{N}{\partial _0}\psi = \frac{1}{N}\left( {\frac{\partial }{{\partial t}} - {\beta ^i}{\partial _i}} \right)\psi$

known as the normalized time derivative.