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

HI 0


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Since the thesis contains some unpublished results, I cannot provide the slides here but you can email me if interested.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

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.


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


Blog at