# Why are Coulomb interactions modeled with just r, not r squared?

As far as I understand (as a complete beginner to molecular simulation and modelling), Coulomb's law for two atoms i and j is:

$$F=k\frac{q_iq_j}{(r_{ij})^2}$$

(the force is proportional to the product of the two charges and the inverse square of the distance)

I am trying to understand molecular simulation and force fields and many use this formula but without squaring the distance, is there a reason behind this? My best thought is simply to improve computational time complexity as this is often the most computationally expensive step and quadratic terms scale worse than linear ones.

• The force field does not use the force equations. It uses the Coulomb electrostatic potential instead, that's why.
– Camps
Aug 23 at 14:04

The force $$F$$ is proportional to $$\frac{1}{r^2}$$ but the potential $$V$$ is proportional to $$\frac{1}{r}$$.

This is because $$F$$ is proportional to $$\frac{\partial V}{\partial r}$$, the derivative of $$V$$ with respect to $$r$$ (this is by definition).

Using $$r$$ instead of $$r^2$$ is not to save computational time (it is incorrect to think that calculating $$r*r$$ is more computationally expensive than calculating $$q_1*q_2$$). It is simply the correct Coulomb potential corresponding to the Coulomb force that you correctly wrote.

• Just to double check, you're saying that because force on an atom can be written as the negative derivative of the potential energy ($-\nabla{V}({r})$) then the (negative) derivative of 1/r is 1/r^2 and this is used because the force field is a summation of potential energy terms (not forces)? Thanks so much for your help with this
– Cro
Aug 23 at 16:30
• The derivative of 1/r^2 is not 1/r. However, you have the right idea: when you see 1/r in the context of Coulomb interactions, it's the potential, not the force. A "force field" is technically a field of forces (not potentials), but we tend to use "potential energy fields" in MD software, even if people don't tend to use that term. Here you can see that a "force field" is indeed defined as a field of forces in the first sentence, but then for the majority of the article, energies are discussed instead. Aug 23 at 16:37
• sorry I think I meant the other way around that the derivative of 1/r is (negative) 1/r^2 - though does that not matter? I'm still slightly confused how to get to 1/r for the potential. Other than that, yeah it can be a little confusing reading the literature with no prior background, I appreciate your explainer!
– Cro
Aug 23 at 16:51
• If you want to get to 1/r for the potential, starting with 1/r^2 for the force, then you can do the integral of 1/r^2 and add the negative sign. Aug 23 at 16:54
• ah, yes - that makes complete sense, thank you!! Help and patience so very much appreciated
– Cro
Aug 23 at 16:57