On Tuesday 22 March 2011 23:06:14 Ulf Schiller wrote:
> Hi Michael,
>
> On 3/22/2011 11:40 AM, Michael Winokur wrote:
> > One thing to note in the original code of energy.h is that there is an
> > overt difference in the _expression_ so that the variable time_step only
> > appears in the rotational energy term:
> > */
> > MDINLINE void add_kinetic_energy(Particle *p1)
> > {
> >
> > /* kinetic energy */
> > energy.data.e[0] += (SQR(p1->m.v[0]) + SQR(p1->m.v[1]) +
> >
> > SQR(p1->m.v[2]))*PMASS(*p1);
> >
> > #ifdef ROTATION
> >
> > /* the rotational part is added to the total kinetic energy;
> >
> > at the moment, we assume unit inertia tensor I=(1,1,1) */
> >
> > energy.data.e[0] += (SQR(p1->m.omega[0]) + SQR(p1->m.omega[1]) +
> >
> > SQR(p1->m.omega[2]))*time_step*time_step;
> > #endif
> > }
> >
> > If anyone can comment in regards to this I would be happy to hear it.
>
> This depends on the units the velocities and energies are stored. You
> have to make sure that the units for the translational and rotational
> parts are consistent.