myprojects-hg-reborn

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An attempt to calculate analytically the projected area of a monomer.
The projected area is calculated in the literature by considering an
aggregate in 3D, randomly oriented, and projecting it on a plane (chosen to be the $xy$ plane
here).
The area of the projection is evaluated and the procedure is repeated
for many random orientations and the averaged (on many orientations)
area is called projected area.





\begin{figure}[htbp]
\begin{center}
 
\input{test.pdf_t}  
       
\caption{Projection of a dimer in the $xy$ plane.}
\label{figure:example}
\end{center}
\end{figure}

In the case of a dimer, the projection always consists of two
partially overlapping circles.
The orientation of the two circles in the $xy$ plane is totally
irrelevant, the area being determined only by the distances $d$
between the centres of the two circles.
Here I claim that this distance depends only on the angle $\theta$
between the longitudinal symmetry axis of the dimer and the $z$ axis.
The distance $d$ is then given by
\be
d=2r|\sin(\theta)|,
\ee
where $r$ is the circle radius.

According to the link you can find   \href{http://bit.ly/T1t9ZU}{here}
the area of the overlap between the two circles is given by

\be
A_{\cap}=2r^{2}\arccos\(\f{d}{2r}\)-\f{1}{2}d\sqrt{4r^{2}-d^{2}}
\ee
which for $d=2r|\sin(\theta)|$ leads to

\be
A_{\cap}=2r^{2}\[ \arccos(|\sin(\theta)|) -|\sin(\theta)|\sqrt{1-|\sin(\theta)|^{2}}  \],
\ee
i.e.


\be
A_{\cap}=2r^{2}\[ \arccos(|\sin(\theta)|) -|\sin(\theta)\cos(\theta)|  \],
\ee


for a random orientation of the dimer, $\theta$ should be  $\theta \in
U[0, 2\pi] $ i.e. uniformly distributed between $0$ and $2\pi$.

At this point, numerically I find $\langle A_{\cap} \rangle\simeq
0.95r^{2}$, meaning that the projected area
$A_{pro}=2\pi r^{2}-A_{\cap}$ is about $5.33r^{2}$.
Unfortunately, with an entirely numerical procedure, I do not get this
value (I have not tested it thoroughly though).
Right now: are you convinced by the argument above?
I hope I understood what is meant by projected area.... 

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