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.

See also:

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