1
$\begingroup$

When looking at a Gaussian log file today, I noticed the following information in the convergence criterion section after an analytical frequency calculation was performed:

Item Value Threshold Converged? Maximum Force 0.000001 0.000450 YES RMS Force 0.000000 0.000300 YES Maximum Displacement 0.000190 0.001800 YES RMS Displacement 0.000030 0.001200 YES

This struck me as odd. How could the forces be so low but the maximum displacement still be non-negligible? Does anyone have any hypothesis about why this phenomenon would occur? This makes me wonder if the convergence condition for the maximum displacement is unreasonably low if essentially no acting forces can result in a maximum displacement that's merely 1 order of magnitude lower than the default threshold. I believe the force units are Hartrees/Bohr and the distance units are angstroms.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ It is pretty common to see this: that is why we have four different convergence criteria in Gaussian $\endgroup$ – Greg Feb 8 '18 at 0:07
  • $\begingroup$ I get that, but I'm wondering why it occurs. $\endgroup$ – Argon Feb 8 '18 at 1:43
  • $\begingroup$ I'm not sure what units those numbers are in, but geometry convergence and its criteria are a bit of a dark art. $\endgroup$ – TAR86 Feb 8 '18 at 6:09
  • $\begingroup$ I assume that's a displacement in atomic units, and that's just not a very large displacement at all. $\endgroup$ – jheindel Feb 9 '18 at 7:00
  • $\begingroup$ @Argon What why? Why do you expect similar force constant associated to the stretch of a strong bond and eg a rotation? There is no mystery in the fact that they can have very different forces/ displacements. $\endgroup$ – Greg Feb 10 '18 at 14:00
3
$\begingroup$

For clarity I will assume Gaussian performs a generic Newton-Raphson minimization (NR), which should suffice to explain the phenomenon. In NR, the linear problem $$ \nabla\nabla^\ast E \Delta + \nabla E = 0 $$ is solved, where $\Delta$ is the displacement. In order to arrive at "large" displacements despite "small" (but non-zero) forces ($-\nabla E$), it suffices for the Hessian ($\nabla\nabla^\ast E$) to have eigenvalues close to 0 itself, because a small number is divided by an even smaller number. This happens when the potential energy surface is very flat.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks for the reply! I realized shortly after posting this that it was an optimization detail and came to the same conclusion as you did. Indeed, the PES is expected to be very flat in my case. $\endgroup$ – Argon Feb 12 '18 at 20:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.