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\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}
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