Hi Thomas,
The previous problem I have described was not about the modification I
have done by changing the source code. I just wanted to see the
initial (nstep=0) energy and dv/dl results of eth->meth and meth->eth
transformations. Theoretically, the dv/dl results for both cases should be
equal to each other for the same lambda value (with a sign difference),
but I have showed that this depends on the cutoff we chose. (In my case,
I have chosen 2 cutoffs, cut=10 and cut=30. In cut=30, the expected
result was satisfied, but for cut=10, it was not; which I could not
explain why) But in all these test cases, I used the original sander.
I have also changed the source code such that the system is transforming
from one state to another smoothly. In order to do that, I have used a
different 'mixing function' than (1-x)^k. I am testing this new/modified
sander on my test transformations; meth->eth and eth->meth. I have done
700 ps simulation for 64 different lambda value for meth->eth and
eth->meth perturbation. Theoretically, the integral should give me the
same absolute value (with different signs) for both cases. But this was
not satisfied %100. By that, I mean the following: for eth->meth case, I
have found the free energy difference as 0.239 kcal/mol. For the
meth->eth, I found the free energy difference as -0.103 kcal/mol. As I
said before, this is just a test case. I cannot compare it with anything
found/calculated in the literature.
I am thinking that the reason for this difference lies in the perturbation
I am intended. In one case (meth->eth), I am changing a proton to a
carbon, and in another one (eth->meth), I am changing a carbon to a
proton. This might be the reason why I am getting such a difference.
(Maybe the method I am following is no good for creating/annihilating
heavy atoms) So, I will test my 'new' sander to a case where I am
creating/annihilating just protons.
Which 'mixing function' did u use rather (1-x)^k?
Best regards,
On Tue, 26 Apr 2005, Thomas Steinbrecher wrote:
> Dear Ilyas, Amber List
>
> I have attempted the same modification of the sander mixing rule that you
> tried, to enable the starting state to have dummy atoms.
>
> We probably did very similar code modifications, and I ran into a problem
> that sounds a little like yours.
>
> That is, my modifikation works fine for a nonperiodic system (giving
> exactly symmetrical dvdl curves, only different in the sign) but they give
> strange results when I activate periodic boundary conditions.
>
> I think this is not a TI problem but something in the way sander handles
> dummy atom in the starting state. Unfortunately I could not pinpoint the
> error any better than this, due to lack of programming skills...
>
> Kind Regards,
>
> Thomas
>
> -----------------------------------------------------------------------
> The AMBER Mail Reflector
> To post, send mail to amber.scripps.edu
> To unsubscribe, send "unsubscribe amber" to majordomo.scripps.edu
>
>
--
Ilyas Yildirim
---------------------------------------------------------------
- Department of Chemisty - -
- University of Rochester - -
- Hutchison Hall, # B10 - -
- Rochester, NY 14627-0216 - Ph.:(585) 275 67 66 (Office) -
- http://www.pas.rochester.edu/~yildirim/ -
---------------------------------------------------------------
-----------------------------------------------------------------------
The AMBER Mail Reflector
To post, send mail to amber.scripps.edu
To unsubscribe, send "unsubscribe amber" to majordomo.scripps.edu
Received on Thu Apr 28 2005 - 05:53:01 PDT
