I'm trying to do the dimensional analysis of the Lennard-Jones force to work out what units are being used in my MD simulation.

The lennard Jones force is given as the negative derivative of the potential:

$F = 24e/r*[2*(s/r)^{12}-(s/r)^6]$

This article:


suggests that the units of r and sigma are in angstroms and epsilon is in kj/mol.

Upon performing dimensional analysis of the bracket I get 0 implying that the force is dimension-less? or 0?

This cannot be correct; could someone tell me where I am going wrong?


  • $\begingroup$ The quantity in brackets, $\left[2\left(\frac s r\right)^{12}-\left(\frac s r\right)^6\right]$, is naturally dimensionless, as it is a function of the ratio between two lengths. The dimensional content of the RHS is in the factor $e/r$. $\endgroup$ Aug 10, 2013 at 19:13
  • $\begingroup$ Remember 1 joule is 1 newton times 1 meter, and pull all the prefixes in front as factors of 10. (alternatively, work with abstract units "length", "force", "energy" to avoid any and all numerical work) Also, you should consider the units of the subtraction, not the subtraction of the units. Otherwise every physically meaningful expression containing a subtraction would yield $0$ in dimensional analysis (since it's only physically meaningful to subtract/add two quantities bearing the same units). If you do it right, you should find units of $10^{13}\,\mathrm{N/mol}$. $\endgroup$
    – Wouter
    Aug 11, 2013 at 0:22
  • $\begingroup$ Your edit is actually a computational question, not a physical one, so I'm going to revert it. But you might consider asking a followup question at Computational Science. $\endgroup$
    – David Z
    Sep 10, 2013 at 3:30

2 Answers 2


I guess to obtain a potential "per one molecule" you should divide your potential by the Avogadro number 6x$10^{23}$/mol, you obtain the potential in (kilo)Joules, and after you differentiate it with respect to the coordinate you'll obtain the force in Newtons.

  • $\begingroup$ Thanks for answering, I have added detail above. If the answer is in newtons then this gives crazy results for the acceleration and hence speed of argon atoms $\endgroup$
    – RNs_Ghost
    Aug 11, 2013 at 1:37
  • $\begingroup$ @RRs_Ghost: Did you divide the force by the Avogadro number? I suspect there may be other errors in your calculation, but your calculation is not detailed enough to see them without too much effort. $\endgroup$
    – akhmeteli
    Aug 11, 2013 at 2:22
  • $\begingroup$ Once again, thank you. No I did not, I assumed that the force equation above would give me the force beween two atoms of argon at some distance r apart. Why is it necessary to divide by avagadro's number? $\endgroup$
    – RNs_Ghost
    Aug 11, 2013 at 2:42
  • $\begingroup$ @RRs_Ghost: Because the value of epsilon is given in kJ/mole, and a mole contains an Avogadro number of molecules. $\endgroup$
    – akhmeteli
    Aug 11, 2013 at 2:47
  • $\begingroup$ holy crap. Cheers buddy. SO - the force equation above tells me the force in newtons between two moles of argon then? Am I inputting the other things in the correct units i.e. converting meters to angstroms? Can you see from what I have posted If I am doing anything else wrong. $\endgroup$
    – RNs_Ghost
    Aug 11, 2013 at 2:50

When you see a subtraction in an expression, it does not mean the units of the quantities being subtracted are "cancelled" out. Actually, the two quantities being subtracted must have the same units. Think of it this way; 5 meters minus 5 meters is 0 meters, not simply 0. (without specification you could construe the 0 to mean energy, for example, which is wrong)

But when you consider division of two quantities, the units of the two can "cancel out" if they are the same. For example, the ratio of $10$ meters to $5$ meters is simply $2$.

In the formula you posted, the expression in the bracket is dimensionless, since $\frac{s}{r}$ has no units ($s$ and $r$ have the same units). However, the remaining part does have units, which should work out to the units of Force as @akhmeteli suggested.

  • $\begingroup$ thanks for answering, if this is the case please see the edit above $\endgroup$
    – RNs_Ghost
    Aug 11, 2013 at 1:36
  • $\begingroup$ @RRs_Ghost I don't think that your program should include unit considerations. As far as I remember, MD simulations are done in unit less quantities. This link might be of help : fisica.uniud.it/~ercolessi/md/md/… $\endgroup$ Aug 11, 2013 at 8:36

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