# Ngô Quốc Anh

## January 29, 2013

### Say HI in MuPAD

Filed under: Linh Tinh — Tags: — Ngô Quốc Anh @ 5:37

I found this interesting note regarding to Matlab. In that note, they plotted the word HI using Matlab. Here I try to use MuPAD in order to get a slightly better picture.

As mentioned in the note, the full function we need to use is

$\displaystyle e^{-x^2-\frac{1}{2}y^2} \cos(4x) + e^{-3\big( (x+\frac{1}{2})^2+\frac{1}{2}y^2 \big)}.$

If you plot that full function, what you are going to have is the following picture

## January 22, 2013

### PhD Thesis Defense

Filed under: Linh Tinh, Luận Văn — Ngô Quốc Anh @ 15:46

I just passed my PhD defense on 18 Jan, 2013. Following is the front page of the slides I used during the defense.

The committee of my defense consists of

Since the thesis contains some unpublished results, I cannot provide the slides here but you can email me if interested.

Title page of the slides used in my PhD thesis defense

## January 17, 2013

### Conformal Killing Operator

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 16:14

In this notes, I want to summarize some properties of the so-called Conformal Killing Operator relative to the metric $g$, say $\mathbb L_g$.

1. First, we start with its definition. Roughly speaking, the conformal killing operator is a generalization of the Killing operator relative to the metric $g$. It maps any vector field on $M$ to some tensor of type $(2,0)$. More precisely, in components, we have

$\displaystyle (\mathbb L_g v)^{ij} = \nabla^iv^j + \nabla^jv^i - \frac{2}{n}\nabla_kv^k g^{ij},$

where $v$ is a vector field on $M$.

Immediately, one can check that $\mathbb L_gv$ is traceless as can be seen in the following

$\displaystyle {g_{ij}}{({\mathbb{L}_g}v)^{ij}} = {g_{ij}}{\nabla ^i}{v^j} + {g_{ij}}{\nabla ^j}{v^i} - 2{\nabla _k}{v^k} = 0.$

2. We now define the so-called Conformal Vector Laplacian associated to the metric $g$. Basically, it is given by

$\displaystyle (\Delta_{g,\text{conf}}v)^i= \nabla_j(\mathbb L_g v)^{ij}.$

In components, we have

$\displaystyle\begin{array}{lcl} {({\Delta _{g,{\text{conf}}}}v)^i} &=&\displaystyle {\nabla _j}{\nabla ^i}{v^j} + {\nabla _j}{\nabla ^j}{v^i} - \frac{2}{n}{\nabla ^i}{\nabla _k}{v^k} \hfill \\ &=&\displaystyle {\nabla ^i}{\nabla _j}{v^j} + R_j^i{v^j} + {\nabla _j}{\nabla ^j}{v^i} - \frac{2}{n}{\nabla ^i}{\nabla _k}{v^k} \hfill \\ &=&\displaystyle \frac{{n - 2}}{n}{\nabla ^i}{\nabla _j}{v^j} + R_j^i{v^j} + {\nabla _j}{\nabla ^j}{v^i}. \hfill \\ \end{array}$

## January 14, 2013

### Hyperbolicity of the 3+1 system of the Einstein equations under the harmonic slicing

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 2:20

Let us first recall the evolution equation of $K$ in this note

$\displaystyle \frac{\partial }{{\partial t}}{K_{\alpha \beta }} = - {\nabla _\alpha }{\nabla _\beta }N + N({\text{Ric}_{\alpha \beta }}-\gamma _\alpha ^\mu \gamma _\beta ^\nu {\overline {\text{Ric}} _{\mu \nu }} + K{K_{\alpha \beta }} - 2{K_{\alpha \mu }}K_\beta ^\mu ) + {\mathcal L_{\vec \beta}}{K_{\alpha \beta }},$

which can be rewritten as the following

$\displaystyle {\overline {{\text{Ric}}} _{\alpha \beta }} = - \frac{1}{N}{\nabla _\alpha }{\nabla _\beta }N + {\text{Ri}}{{\text{c}}_{\alpha \beta }} + K{K_{\alpha \beta }} - 2{K_{\alpha \mu }}K_\beta ^\mu - \frac{1}{N}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){K_{\alpha \beta }}$

for all $\alpha ,\beta > 0$ since in our setting $\gamma_i^j=\delta_i^j$ for all $i,j>0$. The quantities $\text{Ric}_{0\beta}$ and $\text{Ric}_{00}$ come from the Codazzi and Ricci equations. To be precise, we have, by the Codazzi equation,

$\displaystyle {\overline {{\text{Ric}}} _{0\beta }} = {\nabla _h}K_\beta ^h - {\partial _\beta }K$

and, by the Ricci equation,

$\displaystyle {\overline {{\text{Ric}}} _{00}} = \frac{1}{N}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})K + \frac{1}{N}{\nabla ^i}{\nabla _i}N + {K_{ij}}{K^{ij}}.$

It is clear that the infinitesimal variation of the Ricci curvature $\delta R_{mu\nu}$ corresponding to an infinitesimal $\delta g$ variation of the space metric is

$\displaystyle\delta {R_{\mu \nu }} = \frac{1}{2}({\nabla ^h}{\nabla _{(i}}\delta {g_{j)h}} - {\nabla _h}{\nabla ^h}\delta {g_{ij}} - {\nabla _j}{\partial _i}({g^{hk}}\delta {g_{hk}})),$

where the notation $(i j)$ is nothing but $ij+ji$. This formula can be applied to $\frac{\partial}{\partial t}$ and $\mathcal L_{\vec \beta}$ to get

$\displaystyle (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){R_{\mu \nu }} = \frac{1}{2} \Big({\nabla ^h}{\nabla _{(i}}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){g_{j)h}} - {\nabla _h}{\nabla ^h}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){g_{ij}} - {\nabla _j}{\partial _i}({g^{hk}}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){g_{hk}}) \Big).$

## 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

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

## January 1, 2013

### The harmonic slicing

Filed under: Uncategorized — Ngô Quốc Anh @ 17:46

As we have already seen in this post, the Einstein equations are essentially hyperbolic if one assumes that the harmonic condition holds, i.e. $\square_g x^\alpha=0$. This type of condition was first introduced by De Donder in 1921 and have played an important role in theoretical developments, notably in the Choquet-Bruhat work of the well-posedness of the Cauchy problem for $3+1$ Einstein equations.

The harmonic slicing is defined by requiring that the harmonic condition holds only for the coordinate $x^0=t$, i.e. $\square_g t=0$, leaving freedom to choose any coordinate $(x^\alpha)$, $\alpha>0$ in each hypersurface $M$.

Using the formula for the d’Alembertian operator, which is the Laplace-Beltrami operator in the Minkowski space, we obtain

$\displaystyle\frac{1}{{\sqrt { - \det g} }}\frac{\partial }{{\partial {x^\mu }}} \Big(\sqrt { - \det g} {g^{\mu \nu }}\underbrace {\frac{{\partial t}}{{\partial {x^\nu }}}}_{\delta _\nu ^0}\Big) = 0,$

that is to say

$\displaystyle\frac{\partial }{{\partial {x^\mu }}}(\sqrt { - \det g} {g^{\mu 0}}) = 0.$

Thanks to the identity $\sqrt { - \det g} = N\sqrt {\det \gamma }$, one can see that

$\displaystyle\frac{\partial }{{\partial t}}(N\sqrt {\det \gamma } {g^{00}}) + \frac{\partial }{{\partial {x^i}}}(N\sqrt {\det \gamma } {g^{i0}}) = 0.$

Since

$\displaystyle {g^{00}} = - \frac{1}{{{N^2}}}, \quad {g^{i0}} = \frac{{{\beta ^i}}}{{{N^2}}}$