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  LENNARD-JONES LINES


 
  • 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 $u_{LJ}(r)=4 \,(r^{-12}-r^{-6})$.
 
Let's try:
  • 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.

  • Find the "proxy points" of shortest relative distance $r_{0}$ on the two carrier lines.

  • 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}$.

  • 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} $

  • 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.


  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} } $


Calculate the volume of a quadratic piece from a bivariate Gaussian function:





Help comes from unexpected quarters ...   click to enlarge   ... Money math!

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.



  • 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 $He_{n+1}(x)=x He_{n}(x)-n He_{n-1}(x)$;  $He_{0}(x)=1$; $He_{1}(x)=x$.


  • For $\rho^{2} > 0.5$ use Vasicek's expansion:

    • 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.

    • 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} $


    • 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 $




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.




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


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vesely 2006/08