Dear Ioana,
> Ioana Cozmuta wrote:
>
> Hi amber users,
>
> I have a box with 216 rigid water molecules (using SPC/E model) and I am
> equilibrating it at 300K (using amber 7.0, Irix version).
> The density of the system is fine, the only thing that puzzles me is the
> positive VDW energy. I did check with the rdparm program the distances
> between the oxygen atoms in the model (only for O-O the vdw is defined)
> and the average is 2.5A. For this distance, using the A (629362.17) and C
> (625.27) and a 6-12 potential I get that the value should be negative. So
> in principle the sum over all the atoms within the cutoff should be
> negative as well. But the value in the output file is positive.
> What am I missing or doing wrong?
>
> A second question: According to Berendsen paper (1987 in which the
> parameters for SPC/E were developed) the LJ parameters for the O-O
> interaction are A=0.37 (kJ/mol)^(1/6)*nm and B = 0.3428 (kJ/mol)^(1/12)*nm
> (V_LJ = (B/r)^12-(A/r)^6).
> In amber the potential is V_amber = A/r^12-C/r^6 thus
> A_amber=B_berendsen^12=2.074e-5(kcal/mol)^(1/12)*Angs and C_amber=
> A_berendsen^6=0.0206 (kcal/mol)^(1/6)*Angs
Using V_ber = (B_ber/r)^12-(A_ber/r)^6 and V_amb = A_amb/r^12-C_amb/r^6
and your values for A_ber and B_ber from above, I get:
A_amb = B_ber^12 = 0.3428^12 * kJ/mol * nm^12
= 0.3428^12 * kcal/(mol*4.184) * 10^12 Angstrom^12
= 629358 kcal/mol * Angstrom^12
C_amb = A_ber^6 = 0.37^6 * kJ/mol * nm^6
= 0.37^6 * kcal/(mol*4.184) * 10^6 Angstrom^6
= 613 kcal/mol * Angstrom^6
This is close to what you get from rdparm.
> However the rdparm gives values of 629362.17 and 625.27 respectively.
> Why are these values different?
Using the A_amb and C_amb values and a distance of 2.5A, I get a
positive energy value.
>
> Thank you,
> Ioana
>
To your second email:
> The printTypes command gives
> Type r* eps
> OW 1.7767 0.1553
>
> The vdw energy here corresponding to r = 2.7A is a negative number.
>
> I thought that A = eps*(r*)^12 =153.65 and
> C = 2*eps *(r*)^6 = 9.77
To relate r* and eps to A and C, you have to consider that r*(ij) =
r*(i) + r*(j) in U = eps(r*(ij)/r)^12 - 2eps(r*(ij)/r)^6 (see the "Van
der Waals parameters for cations in Amber format" page on the AMBER
homepage).
In your case, r*(ij) = 2 * r* = 2*1.7767 = 3.5534 Angstrom. This then
gives
A = eps * r*(ij)^12 = 629362.17 kcal/mol * Angstrom^12 and
C = 2eps * r*(ij)^6 = 625.27 kcal/mol * Angstrom^6,
in agreement to what you get from rdparm.
Using r = 2.7 Angstrom, I also get U > 0 here.
Best regards
Holger
--
+++++++++++++++++++++++++++++++++++++++++++++
Dr. Holger Gohlke
Dept. of Molecular Biology, TPC15
The Scripps Research Institute
10550 N. Torrey Pines Rd.
La Jolla CA 92037 USA
phone: +1-858-784-9788
fax: +1-858-784-8896
email: gohlke.scripps.edu
+++++++++++++++++++++++++++++++++++++++++++++
Received on Mon May 05 2003 - 00:53:01 PDT
