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I'm reading about water models and their dispersion coefficients, and going back to S. C. Wang's work according to the citation of an equation (Google books here)

enter image description here

As you can see one citation's to a biographical work, and the other to a German work.

Nowhere in the chapter I'm reading (chapter 5, ‘Resolution’, in Rowlinson’s 2002 Cohesion: A Scientific History of Intermolecular Forces) does it say what $C_6$ stands for, and it doesn't seem to be a series of $C_n$, so I'm a little confused.

There's the equation, apparently a Casimir-Polder integral

$$E_\mathrm{DS} = ‒\sum_{n=6,8,10,...} \left(\frac{C_n}{r_n}\right)$$

in something else I'm reading, but it's biochemistry and so they're skipping over actually defining these things (so I can see DFT would be density functional theory but can't actually figure out what either $E_\mathrm{DS}$ is, nor the meaning of a $C_6$ coefficient. Looking for $E_\mathrm{DS}$ is just bringing up a bunch of articles' editors in search results (Eds.).

Can anyone enlighten me as to the latter - the meaning of a $C_6$ coefficient with respect to [London] dispersion forces?

My own answer to this after more reading

The oscillating electrons in a molecule generate not only instantaneous dipoles but also quadrupoles and higher multipoles. It is to be expected, therefore, that the London dispersion energy is only the first term in a series expansion for the attractive energy

that starts in $$u(r) = −C_6r^{−6}$$

The term comes originally from the wave equation (as Rowlinson's text does eventually describe), the number six arising from a Cartesian coordinate system for each of the two interaction partners

enter image description here

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    $\begingroup$ I have updated your post with chemistry/math markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details. $\endgroup$ – Martin - マーチン Mar 25 '15 at 8:24
  • $\begingroup$ en.wikipedia.org/wiki/Lennard-Jones_potential contains terms r^-12 and r^-6 with respective coefficients, could it help? $\endgroup$ – ssavec Mar 25 '15 at 10:41
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    $\begingroup$ I've looked into it in some detail now, should probably delete this... :-/ Apologies, all looked a bit too quantal but wasn't so bad $\endgroup$ – Louis Maddox Mar 25 '15 at 11:39
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    $\begingroup$ It may be useful for other guys and you can post your solution as an answer $\endgroup$ – Mithoron Mar 25 '15 at 11:50
  • $\begingroup$ This is good, but you should cut out the portion that is an answer and post it an an answer (this is not just better practice here but gets you more points per up vote). $\endgroup$ – matt_black Aug 2 '17 at 20:09
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From Calculation of Coefficients in the Power Series Expansion of the Long‐Range Dispersion Force between Atoms J. Chem. Phys. 56, 2801 (1972)

Highly accurate calculations of $C_6$, the coefficient of the $R^{-6}$ term, resulting from an induced dipole-induced dipole interaction, have been made for many atomic systems, while values of $C_8$ (dipole-quadrupole interaction) and $C_{10}$ (dipoleoctopole plus quadrupole-quadrupole interactions) are...

In other words, in London dispersion interaction is usually represented as induced dipole - induced dipole interaction, the energy of which is proportional to $R^{-6}$, but there are other components of the interaction.

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  • $\begingroup$ Nice, I found my way to that paper just a few minutes ago! My seconday source says their estimation of C6 was only ~14% over current accepted value :-) $\endgroup$ – Louis Maddox Mar 25 '15 at 11:41
  • $\begingroup$ Actually no I was looking at their 1971 paper, improved error bounds for the long-range forces between atoms, which just calls it "the constant C(6)ab" - thanks for this $\endgroup$ – Louis Maddox Mar 25 '15 at 12:56
  • $\begingroup$ @DavePhD is there a way to calculate "c6" with gaussian or gamess? I am trying to parameterize a molecule I finished bond-angle-dihedral-improper values. But I don't know how to calculate "well depth". All articles talking about "c6" value but I don't know how to obtain it. Could you suggest me any source or any way to calculate it? $\endgroup$ – Baris Vvolf Feb 15 '19 at 10:13

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