Derivatives of RMSD

[lizhuzh]

Hi PLUMED developers,

In the definition of RMSD:

there is a M matrix calculated from Kearsley algorithm. When calculating the derivative of RMSD, there should be a term containing

I took a look at the source code OptimalRMSD.cpp and I did not find the code is calculating terms containing this derivative. Am I understanding the source code wrong? Or is this derivative automatically zero after the minimization in Kearsley algorithm?

Thanks,
Lizhu

[HanatoK]

RMSD is probably a special case where the derivatives of rotation matrix with respect to the coordinates vanish (see also Colvars/colvars#570 (comment)).

[GiovanniBussi]

@lizhuzh one easy way of seeing this is this: M is defined as the matrix that minimizes the MSD. Hence, the derivative of the MDS with respect to M is zero.

The term that seems to be missing is: d(MSD) / dM times dM / dX.

dM / dX (derivatives of rotation matrix with respect to the coordinates) is not zero. However, d(MSD) / dM (derivatives of the MSD with respect to the rotational matrix) is zero. So the product is zero as well.

The term is instead present when the groups (or weights) used for alignment and for displacement calculations differ.

I hope this helps

[lizhuzh]

@lizhuzh one easy way of seeing this is this: M is defined as the matrix that minimizes the MSD. Hence, the derivative of the MDS with respect to M is zero.

The term that seems to be missing is: d(MSD) / dM times dM / dX.

dM / dX (derivatives of rotation matrix with respect to the coordinates) is not zero. However, d(MSD) / dM (derivatives of the MSD with respect to the rotational matrix) is zero. So the product is zero as well.

The term is instead present when the groups (or weights) used for alignment and for displacement calculations differ.

I hope this helps

This helps a lot! Thanks for clarifying it!