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## Re: [ESPResSo-users] Tabulated angle potential in Espresso

 From: Axel Arnold Subject: Re: [ESPResSo-users] Tabulated angle potential in Espresso Date: Tue, 3 Jan 2012 16:49:20 +0100 User-agent: KMail/1.13.6 (Linux/2.6.38-13-generic; KDE/4.6.5; x86_64; ; )

```On Tuesday 03 January 2012, Javier Ramos wrote:
> I understood from the Espresso manual that the force is, as you say,
> U'(r)/r being the r the distance between two beads in the case of
> tabulated non-bonded potentials. However, in the case of tabulated angle
> potentials I am not be able to understand the meaning of r. Which is the
> r that I have to use?. Is the connecting distance between 1-2 beads? or

What you need to give is the magnitude of the force F(r,phi)/r on one of the
it, the force needs to scale with this distance, as the potential U(phi) only
depends on the enclosed angle. Also, the force on the 1st and 3rd particle are
not equal in magnitude, as their distance to the center particle is different,
and therefore their torsion arms.

Axel

> Best regards,
> Javi
>
>
> Hi!
>
> On Thursday 15 December 2011 10:40:15 Javier Ramos wrote:
> >/  Dear Espresso users,/
> >/  /
> >/  I am trying to use a tabulated angle potential (variant 2 in the
> >section/ /  5.3.7 of the user guide). I have two questions./
> >/  /
> >/       i) Does the angle potetial zero corresponds to the polymer is
> >this/ /  position: (p1)-(p2)-(p3)?/
>
> Yes, if you mean the stretched position, where all particles are in a line.
>
> >/      ii) It is not clear for me which is the scaled length that I need
> >to/ /  use for the forces. Following the explanation in the non-bonded/ /
> > tabulated potential I am assuming that the second column has to be
> >built/ /  as F(theta) = - Vprime(theta) / theta where Vprime is the
> >derivate of/ /  V(theta) respect to theta and theta is the angle between
> >p1,p2 and p3/ /  particles. is that right?/
>
> According to the User's guide, no. It says only that the forces are scaled
> with the inverse length of the connecting vectors, which is what is
> required due to the angle. What you therefore give is the magnitude of the
> force on one of the two outer particles, multiplied by the distance of the
> particle to the center particle.
>
> Axel

--
JP Dr. Axel Arnold      Tel: +49 711 685 67609