Lennard-Jones potential for 3 atoms

  • Thread starter barnflakes
  • Start date
If I have a 3 atom lattice laid out at positions ##x_1 = 1, x_2 = 2, x_3 = 3##

##x_1## and ##x_3## are stationary, but ##x_2## interacts with ##x_1## and ##x_3## via the Lennard-Jones interatomic potential.

I therefore worked out that the total acceleration for ##x_2## for ##m = \sigma = \epsilon = 1## is:


[tex]a_2 = 24(2r_{21}^{-13} - r_{21}^{-7}) + 24(-2r_{32}^{-13} + r_{32}^{-7});[/tex]

where ##r_{21} = |x_2 - x_1| , r_{32} = |x_3 - x_2|##

Is this correct?
 
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DrClaude

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First, I assume that the in this particular case the Lennard-Jones potential is written as
$$
V_\mathrm{LJ}(r) = 4 \epsilon \left[ \left( \frac{\sigma}{r} \right)^{12} - \left( \frac{\sigma}{r} \right)^{6}
\right]
$$

Is this correct?
Yes, since
$$
a = \frac{F}{m} = \frac{1}{m} (- \nabla V)
$$

It is always good to make a sanity check. If ##r_{21} \rightarrow 0##, then atoms 1 and 2 are getting very close and the repulsive force dominates, so that should lead to an positive acceleration (i.e., to the right), and conversely for ##r_{32} \rightarrow 0##. The sign does appear to be correct.
 

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