Page Menu
Home
c4science
Search
Configure Global Search
Log In
Files
F91298161
spectral_statistical_moments.tex
No One
Temporary
Actions
Download File
Edit File
Delete File
View Transforms
Subscribe
Mute Notifications
Award Token
Subscribers
None
File Metadata
Details
File Info
Storage
Attached
Created
Sat, Nov 9, 18:56
Size
10 KB
Mime Type
text/x-tex
Expires
Mon, Nov 11, 18:56 (2 d)
Engine
blob
Format
Raw Data
Handle
22190882
Attached To
rTAMAAS tamaas
spectral_statistical_moments.tex
View Options
\documentclass[10pt]{article}
\usepackage[T1]{fontenc}
\usepackage[latin1]{inputenc}
\usepackage{color,graphicx}
\usepackage{url}
\usepackage{xifthen}
\usepackage{wrapfig}
\usepackage{fullpage}
\newcommand{\phx}{\frac{2\pi k_x}{N}}
\newcommand{\phy}{\frac{2\pi k_y}{N}}
\newcommand{\cox}{\cos\left(\phx x\right)}
\newcommand{\coy}{\cos\left(\phy y\right)}
\newcommand{\six}{\sin\left(\phx x\right)}
\newcommand{\siy}{\sin\left(\phy y\right)}
\newcommand{\ncox}[1]{\cos^{#1}\left(\phx x\right)}
\newcommand{\ncoy}[1]{\cos^{#1}\left(\phy y\right)}
\newcommand{\nsix}[1]{\sin^{#1}\left(\phx x\right)}
\newcommand{\nsiy}[1]{\sin^{#1}\left(\phy y\right)}
\newcommand{\scox}{\ncox{2}}
\newcommand{\scoy}{\ncoy{2}}
\newcommand{\ssix}{\nsix{2}}
\newcommand{\ssiy}{\nsiy{2}}
\begin{document}
Our surface is described with:
\begin{eqnarray*}
h(x,y) = \cox \coy
\end{eqnarray*}
\section{Root mean square of heights}
The RMS is defined as:
\begin{eqnarray*}
<h^2(x,y)> &=& \frac{1}{N^2} \int_0^N \int_0^N \scox \scoy dx dy = \frac{1}{N^2} \int_0^N \scox dx \int_0^N \scoy dy \\
&=& \frac{1}{N^2} \frac{N^2}{4 \pi^2 k_x k_y} \int_0^{2\pi k_x} \cos^2(x) dx \int_0^{2\pi k_y} \cos^2(y) dy
\end{eqnarray*}
Let us focus on the integral:
\begin{eqnarray}
\label{I1}
\int_0^{2\pi k_x} \cos^2(x) dx = \frac{1}{2} \left[ x + \frac{\sin(2x)}{2} \right]_0^{2\pi k_x}
= \pi k_x
\end{eqnarray}
We finally obtain:
\begin{eqnarray*}
<h^2(x,y)> &=& \frac{1}{4 \pi^2 k_x k_y} \pi^2 k_x k_y = \frac{1}{4}
\end{eqnarray*}
\section{Root mean square of slopes}
The gradient is expressed as:
\begin{eqnarray*}
\nabla h(x,y) = \left(
\begin{array}{c}
- \phx \six \coy \\
- \phy \cox \siy \\
\end{array}
\right)
\end{eqnarray*}
The length of the gradient is:
\begin{eqnarray*}
\nabla h(x,y)^2 = \frac{4\pi^2}{N^2} \left\{
k_x^2 \ssix \scoy + k_y^2 \scox \ssiy
\right\}
\end{eqnarray*}
\subsection{General direction $k_x \neq 0$, $k_y \neq 0$ }
Let us define the value:
\begin{eqnarray*}
A &=& \int_0^N \int_0^N \ssix \scoy dxdy \\
&=& \int_0^N \ssix dx \int_0^N \scoy dy
= \frac{N^2}{4\pi^2 k_x k_y} \int_0^{2\pi k_x} \cos^2(x) dx \int_0^{2\pi k_y} \cos^2(y) dy \\
%&=& \int_0^N \ssix dx \frac{N}{2\pi k_y} \left[ \frac{1}{2} \left( y + \frac{\sin(2y)}{2} \right) \right]_0^{2\pi k_y}
%= \frac{2\pi k_y N}{4\pi k_y} \int_0^N \ssix dx \\
%&=& \frac{N}{2} \int_0^N \ssix dx = \frac{N}{2} \frac{N}{2\pi k_x} \left[ \frac{1}{2} \left( x + \frac{\sin(2x)}{2} \right) \right]_0^{2\pi k_x} = \frac{N^2}{8\pi k_x} \left[ x + \frac{\sin(2x)}{2} \right]_0^{2\pi k_x} \\
%&=& \frac{N^2 2\pi k_x}{8\pi k_x} = \frac{N^2}{4}
\end{eqnarray*}
With equation (\ref{I1}) we then we obtain:
\begin{eqnarray*}
A &=& \frac{N^2}{4\pi^2 k_x k_y} \int_0^{2\pi k_x} \cos^2(x) dx \int_0^{2\pi k_y} \cos^2(y) dy \\
&=& \frac{N^2}{4\pi^2 k_x k_y} \pi^2 k_x k_y = \frac{N^2}{4}
\end{eqnarray*}
So that the average length of the gradient is becoming:
\begin{eqnarray*}
< \nabla h(x,y)^2 > = \frac{1}{N^2} \int_0^N \int_0^N \nabla h(x,y)^2 dx dy =
\frac{4\pi^2}{N^4} A \left( k_x^2 + k_y^2 \right) =
\frac{\pi^2}{N^2} \left( k_x^2 + k_y^2 \right)
\end{eqnarray*}
\subsection{Single direction case $k_y = 0$}
\begin{eqnarray*}
\nabla h(x,y)^2 = \frac{4\pi^2}{N^2} k_x^2 \ssix
\end{eqnarray*}
et:
\begin{eqnarray*}
A &=& \int_0^N \int_0^N \ssix dxdy \\
&=& N \int_0^N \ssix dx
= N \frac{N}{2\pi k_x} \int_0^N \sin^2(x) dx = \frac{N^2}{2\pi k_x} \pi k_x
= \frac{N^2}{2}
\end{eqnarray*}
and finally:
\begin{eqnarray*}
< \nabla h(x,y)^2 > = \frac{1}{N^2} \int_0^N \int_0^N \nabla h(x,y)^2 dx dy =
\frac{4\pi^2}{N^4} k_x^2 \frac{N^2}{2} =
\frac{2\pi^2}{N^2} k_x^2
\end{eqnarray*}
\section{Root mean square of Gauss Curvature}
The gauss curvature is defined as the product of the principal curvatures.
It is thus extracted from the determinent of the square gradient matrix:
\begin{eqnarray*}
\nabla^2 h(x,y) &= \left(
\begin{array}{cc}
\frac{4\pi^2 k_x^2}{N^2} \cox \coy & \frac{4\pi^2 k_x k_y}{N^2} \six \siy \\
\frac{4\pi^2 k_x k_y}{N^2} \six \siy & \frac{4\pi^2 k_y^2}{N^2} \cox \coy \\
\end{array}
\right)
\\
&= \frac{4\pi^2}{N^2} \left(
\begin{array}{cc}
k_x^2 \cox \coy & k_x k_y \six \siy \\
k_x k_y \six \siy & k_y^2 \cox \coy \\
\end{array}
\right)
\end{eqnarray*}
we obtain:
\begin{eqnarray*}
\kappa = det (\nabla^2 h(x,y))
&=& \frac{16\pi^4 k_x^2 k_y^2}{N^4} \left( \scox \scoy
- \ssix \ssiy \right) \\
\end{eqnarray*}
we want to compute $<\kappa^2>$ :
\begin{eqnarray*}
<\kappa^2> &=&
\frac{1}{N^2} \left( \frac{16\pi^4 k_x^2 k_y^2}{N^4} \right)^2 \int_0^N \int_0^N \left[ \ncox{2} \ncoy{2}
- \nsix{2} \nsiy{2} \right]^2 dxdy \\
&=& \frac{2^8\pi^8 k_x^4 k_y^4}{N^{10}} \int_0^N \int_0^N
\left[ \quad \ncox{4} \ncoy{4}
+ \nsix{4} \nsiy{4} \right. \\
& & \qquad \qquad \qquad \qquad \qquad \left. - 2 \ncox{2} \ncoy{2} \nsix{2} \nsiy{2} \qquad \right]dx dy\\
\end{eqnarray*}
\subsection{General direction $k_x \neq 0$, $k_y \neq 0$ }
Let us focus on the integral:
\begin{eqnarray*}
A &=& \int_0^N \int_0^N \ncox{4} \ncoy{4} dxdy = \int_0^N \ncox{4} dx \int_0^N \ncoy{4} dy \\
&=& \frac{N^2}{4\pi^2 k_x k_y} \int_0^{2\pi k_x} \cos^4(x) dx \int_0^{2\pi k_y} \cos^4(y) dy
\end{eqnarray*}
If we focus again only on the simple integral:
\begin{eqnarray*}
\int_0^{2\pi k_x} \cos^4(x) dx = \left[ \frac{\cos^{3}(x) \sin(x)}{4} + \frac{3}{4} \int_0^x \cos^2(x')dx' \right]_0^{2\pi k_x}
= \frac{3}{4} \left[ \int_0^{2\pi k_x} \cos^2(x')dx' \right] = \frac{3 \pi k_x}{4}
\end{eqnarray*}
Then:
\begin{eqnarray*}
A = \frac{N^2}{4\pi^2 k_x k_y} \int_0^{2\pi k_x} \cos^4(x) dx \int_0^{2\pi k_y} \cos^4(y) dy =
\frac{N^2}{4\pi^2 k_x k_y} \frac{9 \pi^2 k_xk_y}{16} = N^2 \frac{9}{2^6}
\end{eqnarray*}
It can be demonstrated that:
\begin{eqnarray*}
\int_0^N \int_0^N \nsix{4} \nsix{4} dxdy = N^2 \frac{9}{2^6}
\end{eqnarray*}
The other integral we need to focus on is:
\begin{eqnarray*}
B &=& \int_0^N \int_0^N \ncox{2} \ncoy{2} \nsix{2} \nsiy{2} dxdy \\
&=& \int_0^N \ncox{2} \nsix{2} dx \int_0^N \ncoy{2} \nsiy{2} dy \\
&=& \frac{N^2}{4\pi^2 k_x k_y} \int_0^{2\pi k_x} \cos^2(x) \sin^2(x) dx \int_0^{2\pi k_y} \cos^2(y) \sin^2(y) dy
\end{eqnarray*}
Again we focus on a sub integral:
\begin{eqnarray*}
\int_0^{2\pi k_x} \cos^2(x) \sin^2(x) dx &=& \int_0^{2\pi k_x} \cos^2(x) - \int_0^{2\pi k_x} \cos^4(x) dx = \frac{2\pi k_x}{2} - \frac{3\pi k_x}{4} \\
&=& \frac{\pi k_x}{4} \\
\end{eqnarray*}
So that:
\begin{eqnarray*}
B &=& \frac{N^2}{4\pi^2 k_x k_y} \frac{\pi^2 k_x k_y}{16} = \frac{N^2}{2^6}
\end{eqnarray*}
Finally:
\begin{eqnarray*}
<\kappa^2> &=&
\frac{2^8\pi^8 k_x^4 k_y^4}{N^{10}} \left( 2 A - 2 B \right)
= \frac{2^9\pi^8 k_x^4 k_y^4}{N^{10}} \left( N^2 \frac{9}{2^6} - N^2 \frac{1}{2^6} \right)
= \frac{2^9\pi^8 k_x^4 k_y^4}{2^6 N^8} \left( 9 - 1 \right) \\
&=& \frac{2^{12} \pi^8 k_x^4 k_y^4}{2^6 N^8} = \frac{2^6 \pi^8 k_x^4 k_y^4}{N^4}
\end{eqnarray*}
\subsection{Single direction case $k_y = 0$}
In that case the double derivative of heights becomes:
\begin{eqnarray*}
\nabla^2 h(x,y) = \left(
\begin{array}{cc}
\frac{4\pi^2 k_x^2}{N^2} \cox & 0 \\
0 & 0 \\
\end{array}
\right)
\end{eqnarray*}
Since the determinent is null, we should define the single direction curvature as:
$$\kappa = \frac{4\pi^2 k_x^2}{N^2} \cox$$
and the root mean square of curvatures we want to compute becomes:
\begin{eqnarray*}
<\kappa^2> &=&
\frac{1}{N^2} \left( \frac{4\pi^2 k_x^2 }{N^2} \right)^2 \int_0^N \int_0^N \ncox{2} dxdy \\
&=& \frac{2^4\pi^4 k_x^4}{N^{6}} N \int_0^N \ncox{2} dx = \frac{2^4\pi^4 k_x^4}{N^{5}} \frac{N}{2\pi k_x} \int_0^{2\pi k_x} \cos^2(x) dx = \frac{2^4\pi^4 k_x^4}{N^{4}} \frac{1}{2\pi k_x} \pi k_x \\
&=& \frac{2^3\pi^4 k_x^4}{N^{4}}
\end{eqnarray*}
\section{Root mean square of Mean Curvature}
\subsection{General direction $k_x \neq 0$, $k_y \neq 0$ }
The mean curvature is half the sum of the principal curvatures.
It is thus extracted from the trace of the square gradient matrix.
\begin{eqnarray*}
\kappa = \frac{1}{2} tr(\nabla^2 h(x,y)) &=&
\frac{2\pi^2 k_x^2 }{N^2} \scox \scoy +
\frac{2\pi^2 k_y^2}{N^2} \scox \scoy
\end{eqnarray*}
Then we obtain:
\begin{eqnarray*}
<\kappa^2> &=&
\frac{1}{N^2} \int_0^N \int_0^N \left[ \frac{2\pi^2 k_x^2 }{N^2} \scox \scoy +
\frac{2\pi^2 k_y^2}{N^2} \scox \scoy \right]^2 dxdy \\
&=& \frac{1}{N^2} \int_0^N \int_0^N \left[ \frac{2\pi^2 (k_x^2 + k_y^2)}{N^2} \scox \scoy \right]^2 dxdy \\
&=& \frac{4\pi^4(k_x^2 + k_y^2)^2}{N^6} \int_0^N \int_0^N \ncox{4} \ncoy{4} dxdy \\
&=& \frac{4\pi^4(k_x^2 + k_y^2)^2}{N^6} \frac{N^2}{4\pi^2 k_x k_y} \int_0^{2\pi k_x} \cos^4(x) dx \int_0^{2\pi k_y} \cos^4(y) dy \\
&=& \frac{\pi^2(k_x^2 + k_y^2)^2}{N^4} \frac{1}{k_x k_y} \int_0^{2\pi k_x} \cos^4(x) dx \int_0^{2\pi k_y} \cos^4(y) dy \\
&=& \frac{\pi^2(k_x^2 + k_y^2)^2}{N^4} \frac{1}{k_x k_y} \frac{3\pi k_x}{4} \frac{3\pi k_y}{4} =
\frac{(k_x^2 + k_y^2)^2}{N^4} \frac{9\pi^4}{16}
\end{eqnarray*}
\subsection{Single direction case $k_y = 0$}
In that case the curvature becomes:
\begin{eqnarray*}
\kappa = tr(\nabla^2 h(x,y)) &=&
\frac{2\pi^2 k_x^2 }{N^2} \scox
\end{eqnarray*}
Then we obtain:
\begin{eqnarray*}
<\kappa^2> &=&
\frac{1}{N^2} \int_0^N \int_0^N \left[ \frac{2\pi^2 k_x^2 }{N^2} \scox \right]^2 dxdy \\
&=& \frac{4\pi^4 k_x^4 }{N^6} \int_0^N \int_0^N \ncox{4} dxdy = \frac{4\pi^4 k_x^4}{N^6} N \frac{N}{2\pi k_x} \int_0^{2\pi k_x} \cos^4(x) dx \\
&=& \frac{2\pi^3 k_x^4}{N^4} \frac{1}{k_x} \int_0^{2\pi k_x} \cos^4(x) dx = \frac{2\pi^3 k_x^4}{N^4} \frac{1}{k_x} \frac{3\pi k_x}{4} = \frac{3\pi^4 k_x^4}{2N^4}
\end{eqnarray*}
\section{computation of Alpha}
By definition:
\begin{eqnarray*}
\alpha = \frac{m_0 m_4}{m_2^2} = \frac{M_0 3/8 M_4}{1/4 M_2^2}
\end{eqnarray*}
Dans le cas d'un unique mode $(k_x,k_y)$ we obtain:
\begin{eqnarray*}
\alpha = \frac{1/4 \cdot 3/8 (k_x^2 + k_y^2)^2}{1/4 (k_x^2 + k_y^2)^2} = \frac{3}{8}
\end{eqnarray*}
or with geometric definition (if $k_x \neq 0$ and $k_y \neq 0$):
\begin{eqnarray*}
\alpha = \frac{1/4 \cdot 3/8 \pi^4 9/16 (k_x^2 + k_y^2)^2}{1/4 \pi^4 (k_x^2 + k_y^2)^2} = \frac{3}{8} \frac{9}{16} = \frac{3^3}{2^7}
\end{eqnarray*}
or with geometric definition (if $k_y = 0$):
\begin{eqnarray*}
\alpha = \frac{1/2 \cdot 3/8 \pi^4 3/2 (k_x^2 + k_y^2)^2}{1/4 4 \pi^4 (k_x^2 + k_y^2)^2} = \frac{9}{4} \frac{1}{8} = \frac{3^2}{2^5}
\end{eqnarray*}
It does not work !!! unless I made an(other) arithmetic error.
\end{document}
Event Timeline
Log In to Comment