Dear Developers and Users,
Recently I'm simulating a porous graphene rectangular ribbon by applying different temperatures to the two ends. I used fix temp/rescale command for the temperature control in the two ends to maintain a temperature difference. Each thermal bath region contains more than 300 atoms.
According to the lammps manual: "The scalar is the cummulative energy change due to this fix", however, I observed quite large difference between the accumulated energy on the two ends (1000 eV vs. 20 eV) after several nanoseconds. I thought this might indicate the system is undergoing relaxation so the energy difference goes into system potential energy, but when I checked the system potential energy, it only changed for less than 60 eV. Also the kinetic energy is about constant. So I'm confused, where else can the energy go? Did I miss something? Please help, thanks!
The input I used:
The fix is not storing the accumulated energy - it's storing
the total energy the fix has added (subtracted) to its atoms.
If you're holding one end at high T and the kinetic energy
is leaking away into the rest of the sample, the fix will
need to continuously add energy - so the fix value for
that end will be larger than the cool end.
Steve
Dear Steve,
Thanks for the reply! However, I'm still a little confused. When the system reaches steady-state, won't the energy added in the hot end and the energy subtracted from the cold end be the same for a time interval delt? As seen from my simulation in my last email, it's not the case. At about 4ns, the system looks to be in steady-state already. However. the energy injection into the hotter end is still at a much faster rate as compared to the colder end, while the system potential and kinetic energy is about constant. I don't understand where did the energy go. Could you please shed some light on this? Thanks!
Bo
You should reach steady-state. You can look at the source
code - it's not doing anything sophisticated. So you'll have
to figure out what the output means in the context of
your model.
Steve