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rTAMAAS tamaas
spectral_statistical_moments.tex
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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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