[gmx-users] Forces vector problem

[Erik Lindahl]

On Nov 15, 2006, at 4:37 PM, Alexander Yakovenko wrote:

> Hi All.
> I have rather fundamental question that can be addressed to nearly  
> all force field based energy minimization routines. Why do the  
> minimizer can`t go down if the force vector exists? I can realize  
> this problem due to machine precision, but some systems I have met  
> consists of several tens of atoms only and have derivative vector  
> that is incomparable with machine precision level. If we assume  
> that the partial derivatives of elementary energy function are  
> calculated correctly (with incomparable small machine errors) and  
> their number is relatively small (machine_errors*function_number <<  
> gradient_lenght), what would be the origin of the problem?
> Thank you a lot,

It depends on several things. First, if you are using tabulated  
switch/shift/pme functions the force (=gradient) won't quite be  
machine precision unless you manually change the table density (this  
has improved in CVS, though). Second, some of the minimization  
algorithms are very efficient at the actual minimization, but due to  
the large amount of floating-point operations involved they can have  
numerical problems well before the gradient has reached machine  
precision. Third, if you work out the math around a local minimum in  
energy it turns out that the smallest step resolution (along the  
search gradient) is roughly sqrt(EPS) rather than EPS - with large  
numbers of atoms involved this can easily mean the largest force on  
an individual atom is significantly larger than EPS.

Finally, we've noticed numerical problems in a couple of the  
algorithms when you get very close to the local minimum and are too  
agressive; with L-BFGS this could lead to instabilities, so we prefer  
to play it safe and be a bit more conservative (you can for instance  
get situations where the gradient in the search direction is just  
slightly positive, but if you take the smallest possible step you  
will still go to a slightly lower energy, due to numerical limitations).

Cheers,

Erik