[lammps-users] Anomalous energy during simulation of TIP3P H2O

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I am having a lot of trouble simulating a simple water system. I am running the simplest system possible: two water molecules. I have conducted a hand calculation of this simple system using the TIP3P long-range parameters in the "How-to" section and found that the total energy should be -398.5147 kcal/mol or -17.281 eV in the given geometry.

Am I missing something in my input script, or is my hand-calculation very wrong?

Thanks in advance for you help.

INPUT SCRIPT:

dimension 3
boundary p p p
atom_style full
read_data h2o.tip3p.dat
pair_style lj/cut/coul/long 11.0
pair_coeff 1 1*2 0.000 0.000
pair_coeff 2 2 0.102 3.188
bond_style harmonic
bond_coeff 1 450 0.9572
angle_style harmonic
angle_coeff 1 55 104.52
kspace_style pppm 1.0e-6

neighbor 0.3 bin
neigh_modify delay 0 every 1 check yes
thermo 1

minimize 1.0e-8 1.0e-6 20000 20000

ATOM POSITIONS:

1 1 2 -0.830000 0.58588 0.75695 0.0 # OT TIP3
2 1 1 0.415000 0.0 1.5139 0.0 # HT TIP3
3 1 1 0.415000 0.0 0.0 0.0 # HT TIP3
4 2 2 -0.830000 3.77388 0.75695 0.0 # OT TIP3
5 2 1 0.415000 4.35976 0.0 0.0 # HT TIP3
6 2 1 0.415000 4.35976 1.5139 0.0 # HT TIP3

OUTPUT:

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I am having a lot of trouble simulating a simple water system. I am running the simplest system possible: two water molecules. I have conducted a hand calculation of this simple system using the TIP3P long-range parameters in the "How-to" section and found that the total energy should be -398.5147 kcal/mol or -17.281 eV in the given geometry.

Am I missing something in my input script, or is my hand-calculation very wrong?

i don't see a units keyword.
lammps defaults to reduced units

you'll have to use the setting matching
the units of your force field parameters.

cheers,
    axel.