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Re: [ESPResSo-users] Re: Time step question

From: Ulf Schiller
Subject: Re: [ESPResSo-users] Re: Time step question
Date: Tue, 22 Mar 2011 17:06:14 -0500
User-agent: Mozilla/5.0 (Windows; U; Windows NT 6.1; en-US; rv: Gecko/20110303 Thunderbird/3.1.9

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 */[0] += (SQR(p1->m.v[0]) + SQR(p1->m.v[1]) +

   /* the rotational part is added to the total kinetic energy;
      at the moment, we assume unit inertia tensor I=(1,1,1)  */[0] += (SQR(p1->[0]) + SQR(p1->[1]) +

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.

How do you thermalize the rotational degrees of freedom?


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