Force on a Torus.

  • Thread starter llello
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  • #1
llello
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The problem is with calculating the force of a particle by another particle while both particles lie on a torus. By this I mean, if particle i and j are in a box (LxL) and particle i is on the left edge of the box and particle j is on the right edge of the box, the distance used in the force will be [tex] L - \left| x_i - x_j \right| [/tex].

I have 100 particles in a box where each particle obeys the Lennard Jones potential:

[tex] V(x) = 4(\frac{1}{r^{12}}-\frac{1}{r^6} ) [/tex]

So I put in

[tex]r^{2}=x^{2}+y^{2}[/tex] and use [tex]F_{ij}=-\frac{\partial V}{\partial x}\hat x - \frac{\partial V}{\partial y} \hat y [/tex]

and get

[tex] F_{ij} = (24(\frac{2}{r^{14}} - \frac{1}{r^{8}}))*[(x_i - x_j)\hat x + (y_i - y_j)\hat y ][/tex]

for the force of the jth particle on the ith particle. This is all well and good if we are not going to deal with the periodic boundary conditions. I'm not seeing how this would be modified to account for the torus. Any thoughts?
 
Last edited:

Answers and Replies

  • #2
llello
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Yeah, that was the TeX problem. Thanks :)

Also, it might be of interest to know that:

[tex] \left| r_{i,j} \right| = min( \left| x_i - x_j \right|, L - \left| x_i - x_j \right| [/tex] )
 
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