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units.tex
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units.tex
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This document is not UTF8. It was detected as Shift JIS and converted to UTF8 for display.
\documentclass
[10pt]
{
article
}
\usepackage
{
graphicx
}
\usepackage
{
graphics
}
\usepackage
{
amsmath
}
\usepackage
[french]
{
babel
}
\usepackage
[latin1]
{
inputenc
}
\title
{
Unit conversion table for molecular dynamics codes
}
\author
{
Guillaume ANCIAUX
}
\usepackage
{
fancyhdr
}
\setlength
{
\hoffset
}{
-18pt
}
\setlength
{
\oddsidemargin
}{
0pt
}
% Marge gauche sur pages impaires
\setlength
{
\evensidemargin
}{
9pt
}
% Marge gauche sur pages paires
\setlength
{
\marginparwidth
}{
54pt
}
% Largeur de note dans la marge
\setlength
{
\textwidth
}{
481pt
}
% Largeur de la zone de texte (17cm)
\setlength
{
\voffset
}{
-18pt
}
% Bon pour DOS
\setlength
{
\marginparsep
}{
7pt
}
% S駱aration de la marge
\setlength
{
\topmargin
}{
0cm
}
% Pas de marge en haut
\setlength
{
\headheight
}{
0cm
}
% Haut de page
\setlength
{
\headsep
}{
0cm
}
% Entre le haut de page et le texte
\setlength
{
\footskip
}{
1cm
}
% Bas de page + s駱aration
\setlength
{
\textheight
}{
25cm
}
% Hauteur de la zone de texte (25cm)
%\pagestyle{fancy}
\fancyhf
{}
\begin
{
document
}
\maketitle
\begin
{
align*
}
1 calorie
&
= 4.184 Joule
\\
\end
{
align*
}
\begin
{
align*
}
1 Newton
&
= 1
\frac
{
Joule
}{
meter
}
= 10
^{
-10
}
\frac
{
Joule
}{
\AA
}
\\
&
=
\frac
{
10
^{
-13
}}{
4.185
}
\frac
{
Kcalorie
}{
\AA
}
\\
&
=
\frac
{
6.0228
}{
4.184
}
10
^{
10
}
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
\\
&
= 1.439483747609942639
\cdot
10
^{
10
}
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
\\
\end
{
align*
}
Velocity, mass, time and mass are related through the newton expression:
\begin
{
align*
}
m
\dot
{
v
}
&
= f
\\
1 Kg
\cdot
1 meter
\cdot
second
^{
-2
}
&
= 1 Newton
\end
{
align*
}
In lammps, the real units are grams/mol for mass, angstroems for distance, femtosecond for time and
Kcalorie/mol-
\AA
{}
for force (unit style : real). Then we can change the relation with:
\begin
{
align*
}
1 Kg
\cdot
1 meter
\cdot
second
^{
-2
}
&
= 1 Newton
\\
10
^
3 g
\cdot
10
^{
10
}
\AA
\cdot
10
^{
-30
}
fs
^{
-2
}
&
= 1 Newton
\\
6.0228
\cdot
10
^{
26
}
\frac
{
g
}{
mol
}
\cdot
10
^{
10
}
\AA
\cdot
10
^{
-30
}
fs
^{
-2
}
&
= 1 Newton
\\
6.0228
\cdot
10
^{
-4
}
\frac
{
g
}{
mol
}
\cdot
1
\AA
\cdot
fs
^{
-2
}
&
=
\frac
{
6.0228
}{
4.184
}
10
^{
10
}
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
\\
10
^{
-4
}
\frac
{
g
}{
mol
}
\cdot
1
\AA
\cdot
fs
^{
-2
}
&
=
\frac
{
10
^{
10
}}{
4.184
}
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
\\
4.184
\cdot
10
^{
-14
}
\frac
{
g
}{
mol
}
\cdot
1
\AA
\cdot
fs
^{
-2
}
&
= 1
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
\\
\end
{
align*
}
In the context of a velocity update we have:
\begin
{
align*
}
v
&
= f
\cdot
m
\cdot
t
\\
1 meter
\cdot
second
^{
-1
}
&
= 1
\frac
{
Newton
}{
Kg
}
\cdot
1 second
\end
{
align*
}
By the same procedure we have:
\begin
{
align*
}
1 meter
\cdot
second
^{
-1
}
&
= 1
\frac
{
Newton
}{
Kg
}
\cdot
1 second
\\
10
^{
10
}
\AA
\cdot
10
^{
-15
}
fs
^{
-1
}
&
=
\frac
{
6.0228
}{
4.184
}
10
^{
10
}
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
\cdot
\frac
{
1
}{
Kg
}
\cdot
1 second
\\
10
^{
-5
}
\AA
\cdot
fs
^{
-1
}
&
=
\frac
{
6.0228
}{
4.184
}
10
^{
10
}
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
\cdot
\frac
{
1
}{
6.0228
\cdot
10
^{
26
}}
\frac
{
1
}{
g/mol
}
\cdot
10
^{
15
}
fs
\\
10
^{
-5
}
\AA
\cdot
fs
^{
-1
}
&
=
\frac
{
1
}{
4.184
}
10
^{
10
}
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
\cdot
10
^{
-26
}
\frac
{
1
}{
g/mol
}
\cdot
10
^{
15
}
fs
\\
10
^{
-5
}
\AA
\cdot
fs
^{
-1
}
&
=
\frac
{
10
^{
-1
}}{
4.184
}
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
\cdot
\frac
{
1
}{
g/mol
}
\cdot
fs
\\
\underbrace
{
\frac
{
4.184
}{
10
^{
4
}}}_{
fmt2v
}
\AA
\cdot
fs
^{
-1
}
&
=
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
\cdot
\frac
{
1
}{
g/mol
}
\cdot
fs
\\
\end
{
align*
}
Here,
$
fmt
2
v
$
is a coefficient that is used in lammps and the numercal value is
$
1
/
2390
.
05736137667304
$
. Exactly the
squared root of this value is used:
$
\sqrt
{fmt
2
v}
=
1
/
48
.
88821290839616117
$
.
If now we might need to convert forces between eV and Kcal/mol/
\AA
{}
. We remind that
\begin
{
align*
}
1 eV
&
= 1.60218
\cdot
10
^{
-19
}
Joule
\\
&
= 1.60218
\cdot
10
^{
-19
}
\frac
{
10
^{
-3
}}{
4.184
}
Kcalories
\\
&
=
\frac
{
1.60218
\cdot
10
^{
-22
}}{
4.184
}
Kcalories
\end
{
align*
}
Then we have:
\begin
{
align*
}
1
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
&
=
\frac
{
1
}{
6.0228
\cdot
10
^{
23
}}
\frac
{
Kcalorie
}{
\AA
}
\\
1
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
&
=
\frac
{
1
}{
6.0228
\cdot
10
^{
23
}}
\frac
{
4.184
}{
1.60218
\cdot
10
^{
-22
}}
\frac
{
eV
}{
\AA
}
\\
1
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
&
=
\underbrace
{
\frac
{
4.184
}{
6.0228
\cdot
16.0218
}}_{
ev2Kcal/mol/
\AA
}
\frac
{
eV
}{
\AA
}
\\
\end
{
align*
}
The numerical value for
$
ev
2
Kcal
/
mol
/
\AA
$
is:
$$
ev
2
Kcal
/
mol
/
\AA
=
0
.
04335926662676967479
$$
Then the previous relation can be changed as:
\begin
{
align*
}
\AA
\cdot
fs
^{
-1
}
&
=
\frac
{
10
^{
4
}}{
4.184
}
\frac
{
Kcalorie
}{
mol
\cdot\AA
}
\cdot
\frac
{
1
}{
g/mol
}
\cdot
fs
\\
\AA
\cdot
fs
^{
-1
}
&
=
\frac
{
10
^{
4
}}{
4.184
}
\frac
{
4.184
}{
6.0228
\cdot
16.0218
}
\frac
{
eV
}{
\AA
}
\cdot
\frac
{
1
}{
g/mol
}
\cdot
fs
\\
\AA
\cdot
fs
^{
-1
}
&
=
\frac
{
10
^{
4
}}{
6.0228
\cdot
16.0218
}
\frac
{
eV
}{
\AA
}
\cdot
\frac
{
1
}{
g/mol
}
\cdot
fs
\\
\AA
\cdot
fs
^{
-1
}
&
= 103.6311343852047677
\frac
{
eV
}{
\AA
}
\cdot
\frac
{
1
}{
g/mol
}
\cdot
fs
\\
\end
{
align*
}
Now if we consider Kg,
\AA
{}
, and eV as the units:
\begin
{
align*
}
1 meter
\cdot
second
^{
-1
}
&
= 1
\frac
{
Newton
}{
Kg
}
\cdot
1 second
\\
10
^{
-5
}
\AA
\cdot
fs
^{
-1
}
&
= 1
\frac
{
Joule
}{
Kg
\cdot
metre
}
\cdot
10
^{
15
}
fs
\\
10
^{
-5
}
\AA
\cdot
fs
^{
-1
}
&
= 10
^{
15
}
\cdot
10
^{
-10
}
\frac
{
Joule
}{
Kg
\cdot
\AA
}
\cdot
fs
\\
10
^{
-10
}
\AA
\cdot
fs
^{
-1
}
&
=
\frac
{
1
}{
1.60218
\cdot
10
^{
-19
}}
\frac
{
eV
}{
Kg
\cdot
\AA
}
\cdot
fs
\\
1.60218
\cdot
10
^{
-29
}
\AA
\cdot
fs
^{
-1
}
&
= 1
\frac
{
eV
}{
Kg
\cdot
\AA
}
\cdot
fs
\\
\end
{
align*
}
\end
{
document
}
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