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LENNARD-JONES LINES
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Definition:
Consider two lines of finite length $L$ containing a
homogeneous density
of LJ centers. Let $\vec{e}_{1,2}$ be the direction vectors,
$\vec{r}_{12}$ the vector between the centers of the two line
segments, and $\lambda, \, \mu$ parameters giving the positions of
the interacting points along $1$ and $2$. The squared distance between
any two such points is given by
$r^{2}= \left| r_{12}+\mu \vec{e}_{2}-\lambda \vec{e}_{1} \right|^{2}$.
The total interaction energy between the two lines is then
$
u(\vec{r}_{12},\vec{e}_{1},\vec{e}_{2}) =
\frac{\textstyle 1}{\textstyle L^{2}}
\int \limits_{1} d\lambda \int \limits_{2} d\mu \, \,
u_{LJ} ( r )
$
with $u_{LJ}(r)=4 \,(r^{-12}-r^{-6})$.
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Let's try:
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At fixed $\lambda$ the ($r^{-6}$) integral over $\mu$ is
$
I^{-}(\vec{r}_{12},\vec{e}_{1},\vec{e}_{2},\lambda) =
\int \limits_{-h}^{h}d\mu
\,
\left| r_{12}+\mu \vec{e}_{2}-\lambda \vec{e}_{1} \right|^{-6}
$
where $h$ is the half length of the stick.
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Find the "proxy points" of shortest relative distance
$r_{0}$ on the two carrier lines.
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Using new parameters $\gamma$, $\delta$ originating at the
proxy points we have
$
I^{-}(\gamma) =
\int \limits_{\delta_{a}}^{\delta_{b}}d\delta
\, \left[ \delta^{2}+p \delta + q^{2} \right]^{-3}
$
where $\delta_{b,a}=-\mu_{0} \pm h$, $p=-2 \gamma \rho$, and
$q^{2}=\gamma^{2}+r_{0}^{2}$; the correlation $\rho$ is
just the scalar product of the direction vectors,
$ \rho = e_{1} \cdot e_{2}$.
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There is in fact an analytic solution to this:
$
\begin{eqnarray}
I^{-}(\gamma) &=&
\left[
3 \,
\frac{p+2\delta}{(4q^{2}-p^{2})^{2}[q^{2}+\delta(p+\delta)]}
+\frac{1}{2} \,
\frac{p+2\delta}{(4q^{2}-p^{2})[q^{2}+\delta(p+\delta)]^{2}}
\right.
\\
&&
\left.
+\frac{12}{(4q^{2}-p^{2})^{5/2}}
\arctan \frac{p+2\delta}{(4q^{2}-p^{2})^{1/2}}
\right]_{\delta=\delta_{a}}^{\delta=\delta_{b}}
\end{eqnarray}
$
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However, the second integration in
$
I^{\pm} \equiv
\int \limits_{\gamma_{a}}^{\gamma_{b}} d\gamma \, I^{\pm}(\gamma)
$
with $\gamma_{b,a}=-\lambda_{0} \pm h$
can in general
not
be performed in closed form.
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Lennard-Jones & Gauss:
Let us fit the
Lennard-Jones potential
by a sum of 3 Gaussians:
Demanding that $u(r)$ equals $u_{LJ}(r)$ at the points
$r=0.7; 0.8; 0.9; 1.0; r_{min}; 1.6$, we find
$
u(r) \approx \sum \limits_{i=1}^{3} A_{i} \,
\exp^{\textstyle -B_{i}\, r^{2}}
$
with
$A_{i}=3.9710294335e+05 \, | \, 7.1326116569e+03 \, | \, -5.9236223976e+00$
$B_{i}=1.6202936970e+01 \, | \, 8.3948109130e+00 \, | \, 1.2789514461e+00$
Insert this in the double integral representing the potential,
remembering that
$r^{2}(\gamma,\delta,\rho) = \gamma^{2}+\delta^{2}
-2 \, \gamma \, \delta \, \rho$
with
$\rho \equiv e_{1} \cdot e_{2}$
$
u(r_{12},e_{1},e_{2})= \frac{\textstyle 1}{\textstyle L}
\sum_{i} A_{i} \, \int \limits_{1} \int \limits_{2}
d \gamma \, d \delta \exp^{
\textstyle - B_{i} \left[
\gamma^{2} + \delta^{2} - 2 \gamma \delta \rho
\right]^{2}
}
$
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Calculate the volume of a quadratic piece from
a bivariate Gaussian function:
Help comes from unexpected quarters ...
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... Money math!
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To estimate risks and expected yields of options and bonds,
one uses the standard bivariate normal distribution with density
$
n_{2}(x,y,\rho) \equiv
\frac{\textstyle 1}{\textstyle 2 \pi \sqrt{1-\rho^{2}}}
\exp \left\{ - \left( x^{2}+y^{2}-2 x y \rho \right)
/ \left( 2 (1-\rho^{2})\right) \right\}
$
and cumulative probability
$
N_{2}(a,b,\rho) \equiv
\int \limits_{- \infty}^{a} d x
\int \limits_{- \infty}^{b} dy \; n_{2}(x,y,\rho)
$
Therefore , algorithms and series expansions for $N_{2}$
are available.
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For $\rho^{2} \leq 0.5$ use the
tetrachoric series:
$
N_{2}(x,y,\rho) =
N(x)\, N(y) +n(x)\, n(y)
\sum \limits_{k=0}^{m} \frac{\textstyle 1}{(\textstyle k+1)!}
He_{k}(x) \, He_{k}(y) \, \rho^{k+1}
$
where $n(x)$ is the 1 D normal density with
cumulative distribution $N(x)$, and
$He_{k}$ are the "probabilists'
Hermite polynomials" defined by
$He_{n+1}(x)=x He_{n}(x)-n He_{n-1}(x)$;
$He_{0}(x)=1$; $He_{1}(x)=x$.
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For $\rho^{2} > 0.5$ use
Vasicek's expansion:
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We only need $N_{2}(x,0,\rho)$, since for all $xy \neq 0$
$
\begin{eqnarray}
N_{2}(x,y,\rho) &=&
N_{2}\left( x,0,\mathrm{sign}(x)
( \rho x - y)/\sqrt{x^{2}-2\rho x y +y^{2}} \right)
\\
&&+N_{2}\left( y,0,\mathrm{sign} (y)
( \rho y - x)/\sqrt{x^{2}-2\rho x y +y^{2}} \right)
-s_{0}
\end{eqnarray}
$
where $s_{0}=0$ or $0.5$ for $xy > 0$ or $xy < 0$, respectively.
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Use the recursion
$
A_{k}= - \, \frac{\textstyle 2k-1}{\textstyle 2k(2k+1)} \,
x^{2} A_{k-1} + B_{k}
\;\;\;\;\;
B_{k}= \frac{\textstyle (2k-1)^{2}}{\textstyle 2k(2k+1)} \,
(1-\rho^{2}) \, B_{k-1}
$
starting with
$
B_{0}=\frac{\textstyle \sqrt{1-\rho^{2}}}{\textstyle 2 \pi}
\exp\{ - \, \frac{\textstyle x^{2}}{\textstyle 2(1-\rho^{2})}\} ;
\;\;\;\;\;
A_{0}=- \frac{\textstyle | x | }{\textstyle \sqrt{2 \pi}} N \left(
- | x | / \sqrt{1-\rho^{2}}\right) + B_{0}
$
(where $N(x)=\left(1 / \sqrt{2 \pi}\right)
\int_{- \infty}^{x} du \exp\{ -u^{2}/2\} $,
the error integral in 1 D) and accumulate the sum
$
Q = \sum \limits_{k=0}^{m} \, A_{k}
$
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If $\rho > 0$,
$
N_{2}(x,0,\rho)=\min \left( N(x), 0.5 \right) - Q
$
and if $\rho < 0$
$
N_{2}(x,0,\rho)=\max \left( N(x)-0.5, 0 \right) + Q
$
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Here is a graph of $N_{2}(x,y)$:
The volume of a quadratic piece under $n_{2}(x,y)$,
with corner points 1-4, is
$
V(1..4)=N_{2}(4)-N_{2}(3)-N_{2}(2)+N_{2}(1)
$
Use this to compute $u(r_{12},e_{1},e_{2})$
as given above.
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Should be approached by many-site model:
Comparison with 4LJ (= standard used to gauge Gay-Berne):
Comparison of LJL and 4LJ multisite models for
selected configurations.
Left ... side-by-side;
middle ... tee; right ... end-to-end;
full lines ... LJL; dotted ... 4LJ
vesely 2006/08
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