If one were to calculate the non-bonded interaction energy between two atoms, this would equate to the sum of the vdW + electrostatic potential energies:

$$ E_{\text{non-bond}} = E_{\text{vdW}} + E_{\text{electrostatic}} $$

Could anyone please explain how I could calculate these potentials providing a formula?

Perhaps given an example with zinc ($\ce{Zn^2+}$) and oxygen ($\ce{O}$), where $\sigma$ and $\varepsilon$ values (defined by Charmm27) are given as:

\begin{array}{ccc} \hline \text{Species} & \sigma & \varepsilon \\ \hline \ce{Zn} & 1.942 & 0.25 \\ \ce{O} & 3.029 & 0.12 \\ \hline \end{array}

Ultimately I just wish to understand how they are calculated.

Simply put, I am trying to calculate non-bonded interaction energy between a ligand and a protein. However, after trying to understand each of these terms I am a little lost.

  • 2
    $\begingroup$ What are sigma and epsilon? $\endgroup$ Sep 11 '15 at 13:28

You've got the Lennard-Jones potential parameters for zinc and oxygen from a certain version of the CHARMM force field. The commonly-used notation is $\varepsilon$ for potential well depth and $\sigma$ is the distance where the pair potential is zero, and $r_{min}$ is the distance which minimizes the L-J potential energy function. The linked Wikipedia article gives the potential as:

$$E_{\rm L-J}(r) = 4\varepsilon\left[\left({\sigma\over r}\right)^{12} - \left({\sigma\over r}\right)^{6}\right] = \varepsilon\left[\left({r_{min}\over r}\right)^{12} - 2\left({r_{min}\over r}\right)^{6}\right] $$

At $r = r_{min}$, $E_{\rm L-J} = -\varepsilon$ and we can map $r_{min} = 2^{1/6}\sigma$ to work with either form of the L-J potential function.

Accurate Calculation of Hydration Free Energies using Pair-Specific Lennard-Jones Parameters in the CHARMM Drude Polarizable Force Field. J. Chem. Theory Comput. 6(4) 1181–1198 (2010) describes (among many other things) how the L-J potential term is calculated for 2 interacting species (as in your case). See equations (1), (2), and (3) in that paper, reproduced below (where I have used $r_{min}$ instead of $R_{min}$ for consistency).

In the paper, the Lennard-Jones potential is given as

$$E_{\rm L-J}(r) = \varepsilon\left[\left({r_{min}\over r}\right)^{12} - 2\left({r_{min}\over r}\right)^{6}\right]$$

For a pair of interacting species $i,j$:

$$r_{min} = {r_{min,i} + r_{min,j}\over 2}$$


$$\varepsilon = (\varepsilon_{i}\cdot\varepsilon_{j})^{1/2}$$

For your system, we compute $r_{min}$ from the given values of $\sigma$ (using the mean of the two values of $\sigma$, as specified in the manuscript):

$$r_{min} = 2^{1/6}\sigma = 2^{1/6}\cdot {1.942 + 3.029\over 2} = 2.789\;\unicode{xc5}$$


$$\varepsilon = (0.25\cdot 0.12)^{1/2} = 0.1732\;\mathrm{kcal\;mol^{-1}}$$

Now, at some distance $r$, you can compute $E_{\rm L-J}(r)$.

In the linked publication, the authors also go through the process of how the L-J parameters are derived/fit, which may be of interest to you.

Finally, see this vintage web page authored by a familiar guy which goes through a procedure for taking parameters for computed hydration free energies of cations and deriving L-J parameters for use in the AMBER force field. The procedure is very similar to that used for CHARMM.

  • $\begingroup$ Thank you for your detailed reply. however, in the case where we have two differing atoms with separate $\sigma$ and $\epsilon$ values am I correct in saying the following: $\sigma_{ij}$ = $\sqrt{(\sigma_i\sigma_j)}$ where in the cited paper we have $\epsilon_{ij}$ = $\sqrt{(\epsilon_i\epsilon_j)}$ ? $\endgroup$
    – dsmoore
    Sep 11 '15 at 14:24
  • $\begingroup$ You want $\sigma = R_{min}$ and $\epsilon = \varepsilon$ in their notation - I edited my answer to include the math with your numbers, so it should hopefully be a little clearer. $\endgroup$
    – Todd Minehardt
    Sep 11 '15 at 14:39
  • $\begingroup$ Thank you so much, You have really helped me understand this. with regard to the -2 in the LJ 6 component, could you kindly explain what this means? It does not appear in other literature which I have read. $\endgroup$
    – dsmoore
    Sep 11 '15 at 15:57
  • $\begingroup$ Also after further reading and for those who are also interested in this question: the $\sigma_{ij}$ value is dependant on the force field used for instance: CHARMM27 uses the Lorentz-Bertelot method for determining $\sigma$ i.e. $\sigma_{ij}=1/2(\sigma_{ij}+\sigma_{ij})$ while another OPLS uses a Geometrical average such that: $\sigma = (\sigma_{i}\sigma_{j})^{1/2}$ $\endgroup$
    – dsmoore
    Sep 11 '15 at 16:04
  • $\begingroup$ The -2 in the L-J expression is a remnant of whether or not $\sigma$ or $R_{min}$ is used in the analytical expression (see the linked Wikipedia page in my answer, above). That said, it's not clear to me that I mapped your $\sigma$ values to $R_{min}$ properly above - $R_{min} = 2^{1/6}\sigma$ - so let me edit that in above. $\endgroup$
    – Todd Minehardt
    Sep 11 '15 at 16:18

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