I am trying to understand how the friction metric (last collum in the awh .xvg output file) is calculated from the components of the friction tensor in the optional output file when running awh with two reaction cordinates (e.g. z coordinate and tilt angle). From my understanding, the friction tensor is supposed to be a positive semidefinitive symmetric matrix. My friction tensor output file from gmx awh consists of two collums with the reaction cordinates followed by three collums for the friction tensor (see example below). I assume these three values are the lower half of the tensor. However, if I assume a symmetric matrix (here 2x2) and then calculate the square root of the determinant, I sometimes get negative values which makes no sense when considering the relationship to the diffusion constant. I am at a loss what is the relationship between the diffusion constant and the friction tensor for 2D awh.
Example from the friction tensor .xvg file from gmx awh output:
The legend says what the different column represent. The determinant is the product of the third and fifth column, which is always positive for the lines your posted (and is by definition positive).
EDIT: I forgot to subtract the square of the fourth column.
What is the fourth column here then? I was thinking the determinant is column3 * column5 - column4 * column4 (assuming a symmetric matrix). Thanks for the answer.
A follow-up question to this. The gmx awh -more produces a file that contains @ s6 legend “sqrt(friction metric)”.
I assume that this is the value from Eq. 361 of the manual, which is equal to sqrt(determinant).
However, if I use the friction file to calculate sqrt(column3 * column5 - column4 * column 4), I get wildly different values.
For reference, these are the values from the same time and space point:
sqrt(friction metric) from the awh.xvg file = 13.9373