Lennard-Jones potential for 3 atoms

[barnflakes]

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:$$a_2 = 24(2r_{21}^{-13} - r_{21}^{-7}) + 24(-2r_{32}^{-13} + r_{32}^{-7});$$

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

Is this correct?

 

[DrClaude]

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.