# lj unit, epsilon & temperature

Dear LAMMPS users,

I am running my simulation in lj unit and I’m confused by the relationship between epsilon in Lennard-Jones potential and temperature in ‘fix langevin’ in my case. Here are the codes below:

``````pair_style lj/cut 1.12
pair_modify shift yes
pair_coeff * * 0  0.0  0.0
pair_coeff 1 1 1  1.0  1.12  # excluded volume
pair_coeff 2 2 1  0.6  0.67  # repulsive
pair_coeff 3 3 1  0.6  0.67  # repulsive
``````
``````pair_coeff 2 3 4  0.2  0.50  # attractive
``````

``````fix 4 all nve
``````

fix 5 all langevin 0.2 0.2 10 123456

As can be seen above, I set the epsilon energy for 2-3 pair as 4, and others as 0. The temperature in langevin is chosen as 0.2. As explained in the doc, the real temperature T* = T k_B/ epsilon, then I will assume that in this case T=0.2, but then epsilon = ? 1 or 4?

The simulation relies on the system temperature quite a lot - at least the 2-3 pair interaction depends on the temperature. When I set the temperature to 0.4 in fix langevin, there is almost no chance for 2-3 pair to be attractive. If I would like to calculate the temperature-dependence of the connected fraction of 2-3 pair, is it better to change the temperature in ‘fix langevin’ or change the epsilon in ‘pair_coeff 2 3’?

Best wishes,
Sunnia

Dear LAMMPS users,

I am running my simulation in lj unit and I'm confused by the relationship
between epsilon in Lennard-Jones potential and temperature in 'fix
langevin' in my case. Here are the codes below:

pair_style lj/cut 1.12
pair_modify shift yes
pair_coeff * * 0 0.0 0.0
pair_coeff 1 1 1 1.0 1.12 # excluded volume
pair_coeff 2 2 1 0.6 0.67 # repulsive
pair_coeff 3 3 1 0.6 0.67 # repulsive

pair_coeff 2 3 4 0.2 0.50 # attractive

......

fix 4 all nve

fix 5 all langevin 0.2 0.2 10 123456

As can be seen above, I set the epsilon energy for 2-3 pair as 4, and
others as 0. The temperature in langevin is chosen as 0.2. As explained in
the doc, the real temperature T* = T k_B/ epsilon, then I will assume that
in this case T=0.2, but then epsilon = ? 1 or 4?

neither. remember your lennard-jones epsilon parameters are reduced by the
"real epsilon" (e.g. some number in regular units) and in reduced units,
you only are specifying that the epsilon between atom types 2 and 3 is 4x
stronger than that between atom types 1 and 1 or 2 and 2 or 3 and 3, which
is the same as your "reference epsilon".

axel.

Dear Axel,

Can I reinterpret what you mean as below to check if I understand it correctly?

The epsilon in Lennard-Jones potential is the not epsilon in the unitless system: the former one is a prefactor of LJ potential, which is in the energy unit; the latter one is the factor for reducing the unit?

So for 2-3 pair LJ potential, the minimum energy E* = 4, which corresponds to E = E* times epsilon= 4 epsilon; for the temperature in fix langevin, T* = 0.2, which corresponds to T = T* times epsilon over kB. However, kB = 1 in lj unit, so T= 0.2 epsilon. To adapt the T & E to the real system, I only need to choose the right number and unit for epsilon - is that right?

My next question will be that in thermodynamics, the temperature is often associated with energy, but there is no relationship between them at all?

Many thanks,
Sunnia

sorry, but this is drifting into a classroom discussion about MD and statistical thermodynamics fundamentals and is no longer about using LAMMPS. i suggest you catch up on those topics by studying relevant text books and discussing with local folks that teach on these topics.

That’s all right - I understand it. Thank you all the same!

Sunnia