Difference in computed quantities evaluated from flux file

Dear Users,

I calculated the heat flux using the Green Kubo method as well as the autocorrelation heatflux autocorrelation using lammps.

I used the following line

fix JJ all ave/correlate 1 1 1 c_flux[1] c_flux[2] c_flux[3] type auto file profile.heatflux ave running

I have chosen specific parameter purposely.

I believe the profile.heatflux contains the Cij(t) at each time i.e.

Cij(0)=vi(0)*vj(0)

Cij(1)=vi(0)*vj(1)

Cij(2)=vi(0)*vj(2)


and so on.

Heat flux as a function of time is given as

0 27.8875382066987 81.1519133148716 -28.7945144229034
1 27.515787175563 80.5363732774365 -31.5297000480196
2 27.067407835748 79.809128430772 -34.1896916707998
3 26.5444845352641 78.9789306880053 -36.7708188432657
4 25.9575235252814 78.0636869983545 -39.2592049503619
5 25.3136009964808 77.0831769527867 -41.64164149417
6 24.6161305396799 76.0505070369271 -43.9141063897921
7 23.8705060712651 74.9748025444434 -46.081784162989
8 23.0856364157348 73.8624313814461 -48.1586081359605
9 22.2737051366499 72.7184262964691 -50.1603514280137
10 21.4509823972621 71.5531859621367 -52.1010444750821

the lammps calculated correlation values are

0 1
1 0 1 777.715 6585.63 829.124
1 1
1 0 2 767.417 6535.87 911.623
2 1
1 0 3 755.826 6480.41 997.394
3 1
1 0 4 743.022 6419.73 1086.07
4 1
1 0 5 729.176 6354.57 1177.11
5 1
1 0 6 714.443 6285.78 1269.93
6 1
1 0 7 698.945 6214.05 1364
7 1
1 0 8 682.802 6139.95 1458.95
8 1
1 0 9 666.151 6063.91 1554.54
9 1
1 0 10 649.148 5986.32 1650.69
10 1

I am getting different values when I computed these values manually using the heat flux values at t>0.
am I missing anything?

Thanks,
Mayank

Dear Users,

I calculated the heat flux using the Green Kubo method as well as the autocorrelation heatflux autocorrelation using lammps.

I used the following line

fix JJ all ave/correlate 1 1 1 c_flux[1] c_flux[2] c_flux[3] type auto file profile.heatflux ave running

I have chosen specific parameter purposely.

I believe the profile.heatflux contains the Cij(t) at each time i.e.

Cij(0)=vi(0)*vj(0)

Cij(1)=vi(0)*vj(1)

Cij(2)=vi(0)*vj(2)


and so on.

no it doesn’t.

with Nevery = 1, Nrepeat = 1 and Nfreq = 1 as well as ave running what you get is the running average of Cij(0). for every step.
the documentation of the fix gives a detailed example of what is computed by it and how.

axel.